id,problem_uid,problem_source,year,type,statement,solution,stmt_len,sol_len,type_label,problem_num
219,jbmo_2010_p1,jbmo,2010,a,"The real numbers $a, b, c, d$ satisfy simultaneously the equations
\[
abc - d = 1,\quad bcd - a = 2,\quad cda - b = 3,\quad dab - c = -6.
\]
Prove that $a + b + c + d 
eq 0$.","Suppose that $a + b + c + d = 0$. Then
\[
abc + bcd + cda + dab = 0. \tag{1}
\]
If $abcd = 0$, then one of the numbers, say $d$, must be $0$. In this case $abc = 0$, and so at least two of the numbers $a, b, c, d$ will be equal to $0$, making one of the given equations impossible. Hence $abcd 
eq 0$ and, from (1),
\[
\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} = 0,
\]
implying
\[
\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{a+b+c}.
\]
It follows that $(a+b)(b+c)(c+a) = 0$, which is impossible (for instance, if $a+b = 0$, then adding the second and third given equations would lead to $0 = 2 + 3$, a contradiction). Thus $a + b + c + d 
eq 0$.",174,664,Algebra,1
220,jbmo_2010_p2,jbmo,2010,n,"Find all integers $n$, $n \geqslant 1$, such that $n \cdot 2^{n+1} + 1$ is a perfect square.","Answer: $n = 0$ and $n = 3$.

Clearly $n \cdot 2^{n+1} + 1$ is odd, so, if this number is a perfect square, then $n \cdot 2^{n+1} + 1 = (2x+1)^2$, $x \in \mathbb{N}$, whence $n \cdot 2^{n-1} = x(x+1)$.

The integers $x$ and $x+1$ are coprime, so one of them must be divisible by $2^{n-1}$, which means that the other must be at most $n$. This shows that $2^{n-1} \leqslant n + 1$.

An easy induction shows that the above inequality is false for all $n \geqslant 4$, and a direct inspection confirms that the only convenient values in the case $n \leqslant 3$ are $0$ and $3$.",92,575,Number Theory,2
221,jbmo_2010_p3,jbmo,2010,g,"Let $AL$ and $BK$ be angle bisectors in the non-isosceles triangle $ABC$ ($L$ lies on the side $BC$, $K$ lies on the side $AC$). The perpendicular bisector of $BK$ intersects the line $AL$ at point $M$. Point $N$ lies on the line $BK$ such that $LN$ is parallel to $MK$. Prove that $LN = NA$.","The point $M$ lies on the circumcircle of $\triangle ABK$ (since both $AL$ and the perpendicular bisector of $BK$ bisect the arc $BK$ of this circle). Then $\angle CBK = \angle ABK = \angle AMK = \angle NLA$. Thus $ABLN$ is cyclic, whence $\angle NAL = \angle NBL = \angle CBK = \angle NLA$. Now it follows that $LN = NA$.",292,322,Geometry,3
222,jbmo_2010_p4,jbmo,2010,c,"A $9 \times 7$ rectangle is tiled with tiles of the two types shown in the picture below (the tiles are composed by three, respectively four unit squares and the L-shaped tiles can be rotated repeatedly with $90^\circ$).

Let $n \geqslant 0$ be the number of the $2 \times 2$ tiles which can be used in such a tiling. Find all the values of $n$.","Answer: $0$ or $3$.

Denote by $x$ the number of pieces of the L-shaped type and by $y$ the number of pieces of the type $2 \times 2$. Mark $20$ squares of the rectangle as in the figure.

Obviously, each piece covers at most one marked square. Thus,
\[
x + y \geq 20 \tag{1}
\]
and consequently $3x + 3y \geq 60$ (2). On the other hand $3x + 4y = 63$ (3). From (2) and (3) it follows $y \leq 3$ and from (3), $3 \mid y$.

The proof is finished if we produce tilings with $3$, respectively $0$, $2 \times 2$ tiles.",345,514,Combinatorics,4
223,jbmo_2012_p1,jbmo,2012,a,"Let $a$, $b$ and $c$ be positive real numbers such that $a + b + c = 1$. Prove that
\[
\frac{a}{b} + \frac{b}{a} + \frac{b}{c} + \frac{c}{b} + \frac{c}{a} + \frac{a}{c} + 6 \geq 2\sqrt{2}\!\left(\frac{1-a}{a} + \frac{1-b}{b} + \frac{1-c}{c}
ight).
\]
When does equality hold?","Replacing $1-a, 1-b, 1-c$ with $b+c, c+a, a+b$ respectively on the right hand side, the given inequality becomes
\[
\frac{b+c}{a} + \frac{c+a}{b} + \frac{a+b}{c} + 6 \geq 2\sqrt{2}\!\left(\frac{b+c}{a} + \frac{c+a}{b} + \frac{a+b}{c}
ight)
\]
and equivalently
\[
\left(\frac{b+c}{a} - 2\sqrt{2}\cdot\frac{b+c}{a} + 2
ight) + \left(\frac{c+a}{b} - 2\sqrt{2}\cdot\frac{c+a}{b} + 2
ight) + \left(\frac{a+b}{c} - 2\sqrt{2}\cdot\frac{a+b}{c} + 2
ight) \geq 0,
\]
which can be written as
\[
\left(\sqrt{\frac{b+c}{a}} - \sqrt{2}
ight)^2 + \left(\sqrt{\frac{c+a}{b}} - \sqrt{2}
ight)^2 + \left(\sqrt{\frac{a+b}{c}} - \sqrt{2}
ight)^2 \geq 0,
\]
which is true.

The equality holds if and only if $\dfrac{b+c}{a} = \dfrac{c+a}{b} = \dfrac{a+b}{c}$, which together with the given condition $a+b+c=1$ gives $a = b = c = \tfrac{1}{3}$.",275,823,Algebra,1
224,jbmo_2012_p2,jbmo,2012,g,"Let the circles $k_1$ and $k_2$ intersect at two distinct points $A$ and $B$, and let $t$ be a common tangent of $k_1$ and $k_2$, that touches $k_1$ and $k_2$ at $M$ and $N$, respectively. If $t \perp AM$ and $MN = 2AM$, evaluate $\angle NMB$.","\textit{Solution 1.} Let $P$ be the symmetric of $A$ with respect to $M$. Then $AM = MP$ and $t \perp AP$, hence the triangle $APN$ is isosceles with $AP$ as its base, so $\angle NAP = \angle NPA$. We have $\angle BAP = \angle BAM = \angle BMN$ and $\angle BAN = \angle BNM$.

Thus we have
\[
180^\circ - \angle NBM = \angle BNM + \angle BMN = \angle BAN + \angle BAP = \angle NAP = \angle NPA,
\]
so the quadrangle $MBNP$ is cyclic (since the points $B$ and $P$ lie on different sides of $MN$). Hence $\angle APB = \angle MPB = \angle MNB$ and the triangles $APB$ and $MNB$ are congruent ($MN = 2AM = AM + MP = AP$). From that we get $AB = MB$, i.e.\ the triangle $AMB$ is isosceles, and since $t$ is tangent to $k_1$ and perpendicular to $AM$, the centre of $k_1$ is on $AM$, hence $AMB$ is a right-angled triangle. From the last two statements we infer $\angle AMB = 45^\circ$, and so $\angle NMB = 90^\circ - \angle AMB = 45^\circ$.

\medskip

\textit{Solution 2.} Let $C$ be the common point of $MN$ and $AB$. Then $CN^2 = CB \cdot CA$ and $CM^2 = CB \cdot CA$. So $CM = CN$. But $MN = 2AM$, so $CM = CN = AM$, thus the right triangle $ACM$ is isosceles, hence $\angle NMB = \angle CMB = \angle BCM = 45^\circ$.",243,1216,Geometry,2
225,jbmo_2012_p3,jbmo,2012,c,"On a board there are $n$ nails each two connected by a string. Each string is colored in one of $n$ given distinct colors. For each three distinct colors, there exist three nails connected with strings in these three colors. Can $n$ be
\begin{enumerate}[label=\alph*)]
\item $6$?
\item $7$?
\end{enumerate}","\textit{Solution.} (a) The answer is \textbf{no}.

Suppose it is possible. Consider some color, say blue. Each blue string is the side of $4$ triangles formed with vertices on the given points. As there exist $\binom{5}{2} = \tfrac{5 \cdot 4}{2} = 10$ pairs of colors other than blue, and for any such pair of colors together with the blue color there exists a triangle with strings in these colors, we conclude that there exist at least $3$ blue strings (otherwise the number of triangles with a blue string as a side would be at most $2 \cdot 4 = 8$, a contradiction). The same is true for any color, so altogether there exist at least $6 \cdot 3 = 18$ strings, while we have just $\binom{6}{2} = \tfrac{6 \cdot 5}{2} = 15$ of them.

(b) The answer is \textbf{yes}.

Put the nails at the vertices of a regular $7$-gon and color each one of its sides in a different color. Now color each diagonal in the color of the unique side parallel to it. It can be checked directly that each triple of colors appears in some triangle (because of symmetry, it is enough to check only the triples containing the first color).

\medskip

\textit{Remark.} The argument in (a) can be applied to any even $n$. The argument in (b) can be applied to any odd $n = 2k+1$ as follows: first number the nails as $0, 1, 2, \ldots, 2k$ and similarly number the colors as $0, 1, 2, \ldots, 2k$. Then connect nail $x$ with nail $y$ by a string of color $x + y \pmod{n}$. For each triple of colors $(p, q, r)$ there are vertices $x, y, z$ connected by these three colors. Indeed, we need to solve (mod $n$) the system
\[
x + y \equiv p,\quad x + z \equiv q,\quad y + z \equiv r.
\]
Adding all three, we get $2(x + y + z) \equiv p + q + r$ and multiplying by $k+1$ we get $x + y + z \equiv (k+1)(p+q+r)$. We can now find $x, y, z$ from these identities.",306,1825,Combinatorics,3
226,jbmo_2012_p4,jbmo,2012,n,"Find all positive integers $x, y, z$ and $t$ such that
\[
2^x \cdot 3^y + 5^z = 7^t.
\]","\textit{Solution.} Reducing modulo $3$ we get $5^z \equiv 1$, therefore $z$ is even, $z = 2c$, $c \in \mathbb{N}$.

Next we prove that $t$ is even. Obviously $t \geq 2$. Let us suppose that $t$ is odd, say $t = 2d+1$, $d \in \mathbb{N}$. The equation becomes $2^x \cdot 3^y + 25^c = 7 \cdot 49^d$. If $x \geq 2$, reducing modulo $4$ we get $1 \equiv 3$, a contradiction. And if $x = 1$, we have $2 \cdot 3^y + 25^c = 7 \cdot 49^d$ and reducing modulo $24$ we obtain
\[
2 \cdot 3^y + 1 \equiv 7 \Rightarrow 24 \mid 2(3^y - 3),\quad \text{i.e. } 4 \mid 3^{y-1} - 1,
\]
which means that $y - 1$ is even. Then $y = 2b+1$, $b \in \mathbb{N}$. We obtain $6 \cdot 9^b + 25^c = 7 \cdot 49^d$, and reducing modulo $5$ we get $(-1)^b \equiv 2 \cdot (-1)^d$, which is false for all $b, d \in \mathbb{N}$. Hence $t$ is even, $t = 2d$, $d \in \mathbb{N}$, as claimed.

Now the equation can be written as
\[
2^x \cdot 3^y + 25^c = 49^d \iff 2^x \cdot 3^y = (7^d - 5^c)(7^d + 5^c).
\]
As $\gcd(7^d - 5^c, 7^d + 5^c) = 2$ and $7^d + 5^c > 2$, there exist exactly three possibilities:
\[
(1)\begin{cases}7^d - 5^c = 2^{x-1}\\ 7^d + 5^c = 2 \cdot 3^y\end{cases},\quad
(2)\begin{cases}7^d - 5^c = 2 \cdot 3^y\\ 7^d + 5^c = 2^{x-1}\end{cases},\quad
(3)\begin{cases}7^d - 5^c = 2\\ 7^d + 5^c = 2^{x-1} \cdot 3^y\end{cases}.
\]

\textit{Case (1):} We have $7^d = 2^{x-2} + 3^y$ and reducing modulo $3$, we get $2^{x-2} \equiv 1 \pmod{3}$, hence $x - 2$ is even, i.e.\ $x = 2a+2$, $a \in \mathbb{N}$, where $a > 0$, since $a = 0$ would mean $3^y + 1 = 7^d$, which is impossible (even $=$ odd). We obtain
\[
7^d - 5^c = 2 \cdot 4^a \Rightarrow 7^d \equiv 1 \pmod{4} \Rightarrow d = 2e,\ e \in \mathbb{N}.
\]
Then $49^e - 5^c = 2 \cdot 4^a \Rightarrow 5^c \equiv 1 \pmod{8} \Rightarrow c = 2f$, $f \in \mathbb{N}$.
We obtain $49^e - 25^f = 2 \cdot 4^a \Rightarrow 0 \equiv 2 \pmod{3}$, false. In conclusion, in this case there are no solutions.

\textit{Case (2):} From $2^{x-1} = 7^d + 5^c \geq 12$ we obtain $x \geq 5$. Then $7^d + 5^c \equiv 0 \pmod{4}$, i.e.\ $3 + 1 \equiv 0 \pmod{4}$, hence $d$ is odd. As $7^d \geq 11$, we get $d \geq 2$, hence $d = 2e+1$, $e \in \mathbb{N}$. As in the previous case, from $7^d = 2^{x-2} + 3^y$ reducing modulo $3$ we obtain $x = 2a+2$ with $a \geq 2$. We get $7 \cdot 49^e = 4^a + 3^y$, i.e., reducing modulo $8$ we obtain $7 \equiv 3^y$ which is false (since $3^y$ is congruent to $1$ or $3$ modulo $8$). In conclusion, there are no solutions.

\textit{Case (3):} From $7^d = 5^c + 2$ it follows that the last digit of $7^d$ is $7$, hence $d = 4k+1$, $k \in \mathbb{N}$. If $c \geq 2$, from $7^{4k+1} = 5^c + 2$ reducing modulo $25$ we obtain $7 \equiv 2 \pmod{25}$ which is false. For $c = 1$ we get $d = 1$ and the solution $x = 3,\ y = 1,\ z = t = 2$.",87,2772,Number Theory,4
197,shl_jbmo_2012_a1,shl_jbmo,2012,a,"\textit{Let $a$, $b$ and $c$ be positive real numbers such that $a + b + c = 1$. Prove the inequality}
\[
\frac{a}{b} + \frac{b}{a} + \frac{b}{c} + \frac{c}{b} + \frac{c}{a} + \frac{a}{c} + 6 \geq 2\sqrt{2}\left(\sqrt{\frac{1-a}{a}} + \sqrt{\frac{1-b}{b}} + \sqrt{\frac{1-c}{c}}
ight)
\]
\textit{When does equality hold?}","\textbf{Solution.} Replacing $1-a, 1-b, 1-c$ with $b+c, a+c, a+b$ respectively on the right hand side, the given inequality becomes
\[
\frac{a+c}{b} + \frac{b+c}{a} + \frac{a+b}{c} + 6 \geq 2\sqrt{2}\left(\sqrt{\frac{b+c}{a}} + \sqrt{\frac{a+c}{b}} + \sqrt{\frac{a+b}{c}}
ight)
\]
and equivalently
\[
\left(\frac{a+c}{b} - 2\sqrt{2}\sqrt{\frac{a+c}{b}} + 2
ight) + \left(\frac{b+c}{a} - 2\sqrt{2}\sqrt{\frac{b+c}{a}} + 2
ight) + \left(\frac{a+b}{c} - 2\sqrt{2}\sqrt{\frac{a+b}{c}} + 2
ight) \geq 0,
\]
which can be written as
\[
\left(\sqrt{\frac{a+c}{b}} - \sqrt{2}
ight)^2 + \left(\sqrt{\frac{b+c}{a}} - \sqrt{2}
ight)^2 + \left(\sqrt{\frac{a+b}{c}} - \sqrt{2}
ight)^2 \geq 0,
\]
which is true. The equality holds if and only if $(a+c)/b = (b+c)/a = (a+b)/c = 2$, which together with the given condition $a + b + c = 1$ immediately give $a = b = c = 1/3$. \hfill$\blacksquare$",321,878,Algebra,1
198,shl_jbmo_2012_a2,shl_jbmo,2012,a,"\textit{Let $a$, $b$, $c$ be positive real numbers such that $abc = 1$. Show that}
\[
\frac{1}{a^3 + bc} + \frac{1}{b^3 + ca} + \frac{1}{c^3 + ab} \leq \frac{(ab + bc + ca)^2}{6}.
\]","\textbf{Solution.} By the AM-GM inequality we have $a^3 + bc \geq 2\sqrt{a^3 bc} = 2\sqrt{a^2(abc)} = 2a$ and so
\[
\frac{1}{a^3 + bc} \leq \frac{1}{2a}.
\]
Similarly, $\dfrac{1}{b^3 + ca} \leq \dfrac{1}{2b}$, $\dfrac{1}{c^3 + ab} \leq \dfrac{1}{2c}$ and then
\[
\frac{1}{a^3 + bc} + \frac{1}{b^3 + ca} + \frac{1}{c^3 + ab} \leq \frac{1}{2a} + \frac{1}{2b} + \frac{1}{2c} = \frac{1}{2}\frac{ab + bc + ca}{abc} \leq \frac{(ab + bc + ca)^2}{6}.
\]
Therefore, it is enough to prove $\dfrac{(ab+bc+ca)^2}{6} \leq \dfrac{(ab+bc+ca)^2}{6}$. This inequality is trivially shown to be equivalent to $3 \leq ab + bc + ca$ which is true because of the AM-GM inequality, $3 = \sqrt[3]{(abc)^2} \leq ab + bc + ca$. \hfill$\blacksquare$",182,722,Algebra,2
199,shl_jbmo_2012_a3,shl_jbmo,2012,a,"\textit{Let $a$, $b$, $c$ be positive real numbers such that $a + b + c = a^2 + b^2 + c^2$. Show that}
\[
\frac{a^2}{a^2 + ab} + \frac{b^2}{b^2 + bc} + \frac{c^2}{c^2 + ca} \geq \frac{a+b+c}{2}.
\]","\textbf{Solution.} By the Cauchy-Schwarz inequality it is
\[
\left(\frac{a^2}{a^2+ab} + \frac{b^2}{b^2+bc} + \frac{c^2}{c^2+ca}
ight)\left((a^2+ab)+(b^2+bc)+(c^2+ca)
ight) \geq (a+b+c)^2
\]
\[
\Rightarrow \quad \frac{a^2}{a^2+ab} + \frac{b^2}{b^2+bc} + \frac{c^2}{c^2+ca} \geq \frac{(a+b+c)^2}{a^2+b^2+c^2+ab+bc+ca}
\]
So it is enough to prove $\dfrac{(a+b+c)^2}{a^2+b^2+c^2+ab+bc+ca} \geq \dfrac{a+b+c}{2}$, that is to prove
\[
2(a+b+c) \geq a^2 + b^2 + c^2 + ab + bc + ca.
\]
Substituting $a^2 + b^2 + c^2$ for $a + b + c$ into the left hand side we wish equivalently to prove
\[
a^2 + b^2 + c^2 \geq ab + bc + ca.
\]
But the $a^2 + b^2 \geq 2ab$, $b^2 + c^2 \geq 2bc$, $c^2 + a^2 \geq 2ca$ which by addition imply the desired inequality. \hfill$\blacksquare$",197,761,Algebra,3
200,shl_jbmo_2012_a4,shl_jbmo,2012,a,"\textit{Solve the following equation for $x, y, z \in \mathbb{N}$}
\[
\left(1 + \frac{x}{y+z}
ight)^2 + \left(1 + \frac{y}{z+x}
ight)^2 + \left(1 + \frac{z}{x+y}
ight)^2 = \frac{27}{4}
\]","\textbf{Solution 1.} Call $a = 1 + \dfrac{x}{y+z}$, $b = 1 + \dfrac{y}{z+x}$, $c = 1 + \dfrac{z}{x+y}$ to get
\[
a^2 + b^2 + c^2 = \frac{27}{4}.
\]
Since it is also true that
\[
\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 2,
\]
the quadratic-harmonic means inequality implies
\[
\frac{3}{2} = \sqrt{\frac{a^2+b^2+c^2}{3}} \geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}} = \frac{3}{2}
\]

So the inequality in the middle holds as an equality, and this happens whenever $a = b = c$, from which $1 + \dfrac{x}{y+z} = 1 + \dfrac{y}{z+x} = 1 + \dfrac{z}{x+y}$.

But $1 + \dfrac{x}{y+z} = 1 + \dfrac{y}{z+x} \Leftrightarrow x^2 + xz = y^2 + yz \Leftrightarrow (x-y)(x+y) = z(y-x)$ and the two sides of this equality will be of different sign, unless $x = y$ in which case both sides become 0. So $x = y$, and similarly $y = z$, thus $x = y = z$.

Indeed, any triad of equal natural numbers $x = y = z$ is a solution for the given equation, and so these are all its solutions. \hfill$\blacksquare$

\textbf{Solution 2.} The given equation is equivalent to
\[
\frac{27}{4} = (x+y+z)^2\left(\frac{1}{(y+z)^2} + \frac{1}{(z+x)^2} + \frac{1}{(x+y)^2}
ight).
\]
Now observe that by the well known inequality $a^2 + b^2 + c^2 \geq ab + bc + ca$, with $\dfrac{1}{y+z}$, $\dfrac{1}{z+x}$, $\dfrac{1}{x+y}$ in place of $a, b, c$, we get:
\begin{align*}
\frac{27}{4} &= (x+y+z)^2\left(\frac{1}{(y+z)^2} + \frac{1}{(z+x)^2} + \frac{1}{(x+y)^2}
ight) \\
&\geq (x+y+z)^2\left(\frac{1}{(y+z)(z+x)} + \frac{1}{(z+x)(x+y)} + \frac{1}{(x+y)(y+z)}
ight) = \frac{2(x+y+z)^3}{(x+y)(y+z)(z+x)} \\
&= \frac{(2(x+y+z))^3}{4(x+y)(y+z)(z+x)} = \frac{((x+y)+(y+z)+(z+x))^3}{4(x+y)(y+z)(z+x)} \overset{\text{AM-GM}}{\geq} \frac{\left(3\sqrt[3]{(x+y)(y+z)(z+x)}
ight)^3}{4(x+y)(y+z)(z+x)} \\
&= \frac{27}{4}.
\end{align*}
This means all inequalities in the above calculations are equalities, and this holds exactly whenever $x + y = y + z = z + x$, that is $x = y = z$. By the statement's demand we need to have $a, b, c$ integers. And conversely, any triad of equal natural numbers $x = y = z$ is indeed a solution for the given equation, and so these are all its solutions. \hfill$\blacksquare$",187,2162,Algebra,4
201,shl_jbmo_2012_a5,shl_jbmo,2012,a,"\textit{Find the largest positive integer $n$ for which the inequality}
\begin{equation}
\frac{a+b+c}{abc+1} + \sqrt[n]{abc} \leq \frac{5}{2} \tag{1}
\end{equation}
\textit{holds for all $a, b, c \in [0,1]$. Here $\sqrt[n]{abc} = abc$.}","\textbf{Solution.} Let $n_{max}$ be the sought largest value of $n$, and let $E_{a,b,c}(n) = \dfrac{a+b+c}{abc+1} + \sqrt[n]{abc}$. Then $E_{a,b,c}(m) - E_{a,b,c}(n) = \sqrt[m]{abc} - \sqrt[n]{abc}$ and since $abc \leq 1$ we clearly have $E_{a,b,c}(m) \geq E_{a,b,c}(n)$ for $m \geq n$. So if $E_{a,b,c}(n) \geq \frac{5}{2}$ for some choice of $a, b, c \in [0,1]$, it must be $n_{max} \leq n$.

We use this remark to determine the upper bound $n_{max} \leq 3$ by plugging some particular values of $a, b, c$ into the given inequality as follows:

For $(a,b,c) = (1,1,c)$, $c \in [0,1]$, inequality (1) implies $\dfrac{c+2}{c+1} + \sqrt[n]{c} \leq \dfrac{5}{2} \Leftrightarrow \dfrac{1}{c+1} + \sqrt[n]{c} \leq$

$\dfrac{3}{2}$. Obviously, every $x \in [0;1]$ is written as $\sqrt[n]{c}$ for some $c \in [0;1]$. So the last inequality is equivalent to:
\begin{align*}
\frac{1}{x^n+1} + x &\leq \frac{3}{2} \Leftrightarrow 2 + 2x^{n+1} + 2x \leq 3x^n + 3 \Leftrightarrow 3x^n + 1 \geq 2x^{n+1} + 2x \\
&\Leftrightarrow 2x^n(1-x) + (1-x) + (x-1)(x^{n-1} + \cdots + x) \geq 0 \\
&\Leftrightarrow (1-x)[2x^n + 1 - (x^{n-1} + x^{n-2} + \ldots + x)] \geq 0, \quad \forall x \in [0,1].
\end{align*}

For $n = 4$, the left hand side of the above becomes $(1-x)(2x^4 + 1 - x^3 - x^2 - x) = (1-x)(x-1)(2x^3 + x^2 - 1) = -(1-x)^2(2x^3 + x^2 - 1)$ which for $x = 0.9$ is negative. Thus $n_{max} \leq 3$ as claimed.

Now, we shall prove that for $n = 3$ inequality (1) holds for all $a, b, c \in [0,1]$, and this would mean $n_{max} = 3$. We shall use the following Lemma:

\underline{Lemma.} For all $a, b, c \in [0;1]$: $a + b + c \leq abc + 2$.

Proof of the Lemma: The required result comes by adding the following two inequalities side by side
\begin{align*}
0 &\leq (a-1)(b-1) \Leftrightarrow a + b \leq ab + 1 \Leftrightarrow a + b - ab \leq 1 \\
0 &\leq (ab-1)(c-1) \Leftrightarrow ab + c \leq abc + 1.
\end{align*}

Because of the Lemma, our inequality (1) for $n = 3$ will be proved if the following weaker inequality is proved for all $a, b, c \in [0,1]$:
\[
\frac{abc+2}{abc+1} + \sqrt[3]{abc} \leq \frac{5}{2} \Leftrightarrow \frac{1}{abc+1} + \sqrt[3]{abc} \leq \frac{3}{2}.
\]
Denoting $\sqrt[3]{abc} =: y \in [0;1]$, this inequality becomes:
\begin{align*}
\frac{1}{y^3+1} + y &\leq \frac{3}{2} \Leftrightarrow 2 + 2y^4 + 2y \leq 3y^3 + 3 \Leftrightarrow -2y^4 + 3y^3 - 2y + 1 \geq 0 \\
&\Leftrightarrow 2y^3(1-y) + (y-1)y(y+1) + (1-y) \geq 0 \Leftrightarrow (1-y)(2y^3 + 1 - y^2 - y) \geq 0.
\end{align*}
The last inequality is obvious because $1 - y \geq 0$ and $2y^3 + 1 - y^2 - y = y^3 + (y-1)^2(y+1) \geq 0$. \hfill$\blacksquare$",236,2636,Algebra,5
202,shl_jbmo_2012_g1,shl_jbmo,2012,g,"\textit{Let $ABC$ be an equilateral triangle, and $P$ a point on the circumcircle of the triangle $ABC$ and distinct from $A$, $B$ and $C$. If the lines through $P$ and parallel to $BC$, $CA$, $AB$ intersect the lines $CA$, $AB$, $BC$ at $M$, $N$ and $Q$ respectively, prove that $M$, $N$ and $Q$ are collinear.}","\textbf{Solution.} Without any loss of generality, let $P$ be in the minor arc of the chord $AC$ as in Figure 1. Since $\angle PNA = \angle NPM = 60^\circ$ and $\angle NAM = \angle PMA = 120^\circ$, it follows that the points $A$, $M$, $P$ and $N$ are concyclic. This yields
\begin{equation}
\angle NMP = \angle NAP. \tag{2}
\end{equation}

\begin{center}
[Figure 1: Exercise G1.]
\end{center}

Similarly, since $\angle PMC = \angle MCQ = 60^\circ$ and $\angle CQP = 60^\circ$, it follows that the points $P$, $M$, $Q$ and $C$ are concyclic. Thus
\[
\angle PMQ = 180^\circ - \angle PCQ = 180^\circ - \angle NAP \overset{(2)}{=} 180^\circ - \angle NMP.
\]
This implies $\angle PMQ + \angle NMP = 180^\circ$, which shows that $M$, $N$ and $Q$ belong to the same line. \hfill$\blacksquare$",312,786,Geometry,6
203,shl_jbmo_2012_g2,shl_jbmo,2012,g,"\textit{Let $ABC$ be an isosceles triangle with $AB = AC$. Let also $c(K, KC)$ be a circle tangent to the line $AC$ at point $C$ which it intersects the segment $BC$ again at an interior point $H$. Prove that $HK \perp AB$.}","\textbf{Solution 1.} Let lines $KH$, $AB$ intersect at $M$ (Figure 5a). From the quadrilateral $KMAC$ we have

$\angle KMA = 360^\circ - \angle A - \angle ACK - \angle CKM = 360^\circ - \angle A - 90^\circ - (180^\circ - 2\angle KCH) = 90 - \angle A + 2\angle KCH = 90 - \angle A + 2(90^\circ - \angle ACB) = 270^\circ - \angle A - 2\angle ACB = 270 - \angle A - \angle ACB - \angle ABC = 270^\circ - 180^\circ = 90^\circ,$

so $KH \perp AB$ as wanted. \hfill$\blacksquare$

\textbf{Solution 2.} Let $D$ be a point on $c$ such that $AD < AC$, and let $E$, $Z$ be the second points of intersection of lines $AD$ and $BD$ with $c$ respectively. Let also $N$ be the second point of intersection of line $BE$ with the circle $c$. Figure 5b shows $Z$ between $B$, $D$. The argument below can be trivially modified to apply in case $D$ is in the segment $B$, $Z$ as well.

It is
\[
AB^2 = AC^2 = AD \cdot AE \Rightarrow \frac{AB}{AE} = \frac{AD}{AB}.
\]
This relation and the fact that $\angle BAE = \angle BAD$ implies that the triangles $ABE$, $ADB$ are similar. Thus $\angle ABE = \angle ADB$. Also from the cyclic quadrilateral we get $\angle ADB = \angle ZNE$. Therefore $\angle ABE = \angle ZNE$, so $AB \| NZ$.

Call $P$ the intersection point of $BC$, $NZ$. Since $AC$ is tangent to $c$ it is
\begin{equation}
\angle CEH = \angle BCA \tag{3}
\end{equation}
and then
\begin{align*}
\angle ZNH + \angle CNZ &= \angle HNC = \angle CEH \overset{(3)}{=} \angle BCA = \angle ABC = \angle ZPC = \angle BCN + \angle CNZ \\
\Rightarrow \quad \angle ZNH &= \angle BCN \\
\Rightarrow \quad \angle ZNH &= \angle HZN
\end{align*}
Therefore $H$ is the midpoint of the arc $NZ$, so $KH \perp NZ$ and as $AB \| NZ$ we finally get $KH \perp AB$ as wanted. \hfill$\blacksquare$",224,1761,Geometry,7
204,shl_jbmo_2012_g3,shl_jbmo,2012,g,"\textit{Let $AB$ and $CD$ be chords in a circle of center $O$ with $A$, $B$, $C$, $D$ distinct, and let the lines $AB$ and $CD$ meet at a right angle at point $E$. Let also $M$ and $N$ be the midpoints of $AC$ and $BD$ respectively. If $MN \perp OE$, prove that $AD \| BC$.}","\textbf{Solution.} $E$ can be inside, or outside the circle (Figure 3) but the proof below holds in both cases; notice that $E$ cannot be on the circle as $A$, $B$, $C$, $D$ are distinct. Let lines $AC$ and $NE$ meet at point $P$. Then $EN = DN = BN$ (median in a right triangle), so $\angle PEC = \angle NED = \angle NDE = \angle BDC = \angle BAC = \angle EAP$. Now $AB \perp CD$ so $EN \perp AC$. But $OM \perp AC$ so $OM \| EN$. Similarly $ON \| EM$ so $NEMO$ is a parallelogram (possibly degenerated). As $MN \perp OE$, this parallelogram is a rhombus. Then the chords $AC$ and $BD$, being equidistant from $O$, are equal. Hence their minor arcs are equal, which means that either $AD \| BC$ or $AB \| CD$; the latter contradicts the fact that $AB$ and $CD$ meet at $E$. \hfill$\blacksquare$",274,795,Geometry,8
205,shl_jbmo_2012_g4,shl_jbmo,2012,g,"\textit{Let $ABC$ be an acute-angled triangle with circumcircle $\Gamma$, and let $O$, $H$ be the triangle's circumcenter and orthocenter respectively. Let also $A'$ be the point where the angle bisector of angle $BAC$ meets $\Gamma$. If $A'H = AH$, find the measure of angle $BAC$.}","\textbf{Solution.} The segment $AA'$ bisects $\angle OAH$: if $\angle BCA = y$ (Figure 4), then $\angle BOA = 2y$, and since $OA = OB$, it is $\angle OAB = \angle OBA = 90^\circ - y$. Also since $AH \perp BC$, it is $\angle HAC = 90^\circ - y = \angle OAB$ and the claim follows.

Since $AA'$ bisects $\angle OAH$ and $A'H = AH$, $OA' = OA$, we have that the isosceles triangles $OAA'$, $HAA'$ are equal. Thus
\begin{equation}
AH = OA = R \tag{4}
\end{equation}
where $R$ is the circumradius of triangle $ABC$.

Call $\angle ACH = a$ and recall by the law of sines that $AH = 2R' \sin a$, where $R'$ is the circumradius of triangle $AHC$. Then (4) implies
\begin{equation}
R = 2R' \sin a \tag{5}
\end{equation}
But notice that $R = R'$ because $\dfrac{AC}{\sin(AHC)} = 2R'$, $\dfrac{AC}{\sin(ABC)} = 2R$ and $\sin(AHC) = \sin(180^\circ - ABC) = \sin(ABC)$. So (5) gives $1 = 2\sin a$, and $a$ as an acute angle can only be $30^\circ$. Finally, $\angle BAC = 90^\circ - a = 60^\circ$.

\underline{Remark.} The steps in the above proof can be traced backwards making the converse also true, that is: If $\angle BAC = 60^\circ$ then $A'H = AH$. \hfill$\blacksquare$",283,1162,Geometry,9
206,shl_jbmo_2012_g5,shl_jbmo,2012,g,"\textit{Let the circles $k_1$ and $k_2$ intersect at two distinct points $A$ and $B$, and let $t$ be a common tangent of $k_1$ and $k_2$ that touches them at $M$ and $N$ respectively. If $t \perp AM$ and $MN = 2AM$, evaluate $\angle NMB$.}","\textbf{Solution.} Let $P$ be the symmetric of $A$ with respect to $M$ (Figure 5). Then $AM = MP$ and $t \perp AP$, hence the triangle $APN$ is isosceles with $AP$ as its base, so $\angle NAP = \angle NPA$. We have $\angle BAP = \angle BAM = \angle BMN$ and $\angle BAN = \angle BNM$. Thus

\[
180^\circ - \angle NBM = \angle BNM + BMN = \angle BAN + \angle BAP = \angle NAP = \angle NPA,
\]
so the quadrangle $MBNP$ is cyclic (since the points $B$ and $P$ lie on different sides of $MN$). Hence $\angle APB = \angle MPB = \angle MNB$ and the triangles $APB$ and $MNB$ are congruent $(MN = 2AM = AM + MP = AP)$. From that we get $AB = MB$, i.e.\ the triangle $AMB$ is

isosceles, and since $t$ is tangent to $k_1$ and perpendicular to $AM$, the center of $k_1$ is on $AM$, hence $AMB$ is a right-angled triangle. From the last two statements we infer $\angle AMB = 45^\circ$, and so $\angle NMB = 90^\circ - \angle AMB = 45^\circ$. \hfill$\blacksquare$",239,952,Geometry,10
207,shl_jbmo_2012_g6,shl_jbmo,2012,g,"\textit{Let $O_1$ be a point in the exterior of the circle $c(O, R)$ and let $O_1N$, $O_1D$ be the tangent segments from $O_1$ to the circle. On the segment $O_1N$ consider the point $B$ such that $BN = R$. Let the line from $B$ parallel to $ON$ intersect the segment $O_1D$ at $C$. If $A$ is a point on the segment $O_1D$ other than $C$ so that $BC = BA = a$, and if $c'(K, r)$ is the incircle of the triangle $O_1AB$; find the area of $ABC$ in terms of $a$, $R$, $r$.}","\textbf{Solution.} Obviously, the segment $BC$ is tangent to the circle $c$. Let $M$ be the point of tangency (Figure 6). Call $Q$, $M$ the tangency points of $BA$, $BC$ with $c'$ and $c$ respectively, and call $H$ the midpoint of segment $AC$. It is well known that
\[
AQ = \frac{1}{2}(AO_1 + AB - BO_1) \quad \text{and} \quad CM = \frac{1}{2}(BO_1 + BC - CO_1).
\]
and so
\[
AQ + CM = \frac{1}{2}(2BC - AC) = a - \frac{1}{2}AC = a - HC.
\]

The triangles $KAQ$ and $OCM$ are similar and this implies
\begin{equation}
\frac{KQ}{OM} = \frac{AQ}{CM} \Leftrightarrow \frac{KQ}{AQ} = \frac{OM}{CM} \Leftrightarrow \frac{r}{AQ} = \frac{R}{CM} = \frac{R+r}{AQ+CM} = \frac{R+r}{a - HC} \Leftrightarrow \frac{r}{AQ} = \frac{R+r}{a - HA}. \tag{6}
\end{equation}

If $AZ$ is the bisector segment of triangle $BAH$ it holds
\[
\angle AZH = 90^\circ = \frac{1}{2}\angle BAC \quad \text{and} \quad \angle KAQ = \frac{1}{2}(180^\circ - \angle BAC) = 90^\circ - \frac{1}{2}\angle BAC.
\]
Therefore, from the similar triangles $KQA$ and $AHZ$ we get
\begin{equation}
\frac{AH}{ZH} = \frac{r}{AQ} \overset{(6)}{=} \frac{R+r}{a - HA}. \tag{7}
\end{equation}

Also, from the bisector-theorem in triangle $ABH$ it holds
\[
ZH = \frac{AH \cdot BH}{a + AH}
\]
and from (7) it follows
\[
\frac{R+r}{a - HA} = \frac{AH}{\frac{AH \cdot BH}{a + AH}} \Rightarrow R + r = \frac{a^2 - AH^2}{BH} = \frac{BH^2}{BH} = BH.
\]
So
\[
HA^2 = a^2 - (R+r)^2 \Leftrightarrow HA = \sqrt{a^2 - (R+r)^2}
\]
and finally the area of triangle $ABC$ in terms of $a$, $R$, $r$ is:
\[
(ABC) = AH \cdot BH = (R+r)\sqrt{a^2 - (R+r)^2}
\]",470,1588,Geometry,11
208,shl_jbmo_2012_g7,shl_jbmo,2012,g,"\textit{Let $MNPQ$ be a square of side length 1, and $A$, $B$, $C$, $D$ points on the sides $MN$, $NP$, $PQ$, and $QM$ respectively such that $AC \cdot BD = \dfrac{5}{4}$. Can the set $\{AB, BC, CD, DA\}$ be partitioned into two subsets $S_1$ and $S_2$ of two elements each, so that each one of the sums of the elements of $S_1$ and $S_2$ are positive integers?}","\textbf{Solution.} The answer is negative.

Suppose such a partitioning was possible (Figure 7). Then $AB + BC + CD + DA \in \mathbb{N}$. But $(AB + BC) + (CD + DA) > AC + AC \geq 2$, hence $AB + BC + CD + DA > 2$. On the other hand, $AB + BC + CD + DA < (AN + NB) + (BP + PC) + (CQ + QD) + (DM + MA) = 4$, hence $AB + BC + CD + DA < 4$.

Obviously one of the sums of the elements of $S_1$ and $S_2$ must be 1 and the other 2. Without any loss of generality, we may assume that the sum of the elements of $S_1$ is 1 and the sum of the elements of $S_2$ is 2. As $AB + BC > AC \geq 1$ we find that $S_1 
eq \{AB, BC\}$. Similarly, $S_1$ cannot contain two adjacent sides of the quadrilateral $ABCD$. Therefore, without any loss of generality, we may assume that $S_1 = \{AD, BC\}$ and $S_2 = \{AB, CD\}$. Then $AD + BC = 1$ and $AB + CD = 2$.

We have $AD \cdot BC \leq \dfrac{1}{4} \cdot (AD + BC)^2 = \dfrac{1}{4}$ and $AB \cdot CD \leq \dfrac{1}{4} \cdot (AB + CD)^2 = 1$.

According to Ptolemy's inequality, we have
\[
\frac{5}{4} = AC \cdot BD \leq AB \cdot CD + AD \cdot BC = \frac{1}{4} + 1 = \frac{5}{4},
\]
hence we have equality all around, which means the quadrilateral $ABCD$ is cyclic, $AD = BC = \dfrac{1}{2}$ and $AB = CD = 1$, hence $ABCD$ is a rectangle of dimensions 1 and $\dfrac{1}{2}$.

There are many different ways of proving that this configuration is not possible. For example:

Suppose $ABCD$ is a rectangle with $AD = \dfrac{1}{2}$, $AB = 1$. Then we have $AC = BD = \dfrac{\sqrt{5}}{2}$ and $\triangle ANB \equiv \triangle CQD$ (Angle-Side-Angle). Denoting $AM = x$, $MD = y$ we have $AN =$

$1 - x$, $BN = 1 - y$ and the following conditions need to be fulfilled for some $x, y \in [0;1]$ (Pythagorean Theorem in triangles $AMD$, $ANB$, $BB'C$, where $B'$ is the projection of $B$ on $MQ$):
\begin{equation}
x^2 + y^2 = \frac{1}{4}, \quad (1-x)^2 + (1-y)^2 = 1 \quad \text{and} \quad 1 + (2y-1)^2 = \frac{5}{4}. \tag{8}
\end{equation}
But $1 + (2y-1)^2 = \dfrac{5}{4}$ implies $y \in \left\{\dfrac{1}{4}, \dfrac{3}{4}
ight\}$. If $y = \dfrac{3}{4}$, then $x^2 + y^2 = \dfrac{1}{4}$ cannot hold. If on the other hand $y = \dfrac{1}{4}$, then $(1-x)^2 + (1-y)^2 = 1$ implies $x = 0$, but then $(1-x)^2 + (1-y)^2 = 1$ cannot hold. Therefore such a configuration is not possible. \hfill$\blacksquare$",362,2324,Geometry,12
209,shl_jbmo_2012_c1,shl_jbmo,2012,c,\textit{Along a round table are arranged 11 cards with the names (all distinct) of the 11 members of the $16^{th}$ JBMO Problem Selection Committee. The distances between each two consecutive cards are equal. Assume that in the first meeting of the Committee none of its 11 members sits in front of the card with his name. Is it possible to rotate the table by some angle so that at the end at least two members of sit in front of the card with their names?},"\textbf{Solution.} Yes it is: Rotating the table by the angles $\dfrac{360^\circ}{11}$, $2 \cdot \dfrac{360^\circ}{11}$, $3 \cdot \dfrac{360^\circ}{11}$, \ldots, $10 \cdot \dfrac{360^\circ}{11}$, we obtain 10 new positions of the table. By the assumption, it is obvious that every one of the 11 members of the Committee will be seated in front of the card with his name in exactly one of these 10 positions. Then by the Pigeonhole Principle there should exist one among these 10 positions in which at least two of the $11 (> 10)$ members of the Committee will be placed in their positions, as claimed. \hfill$\blacksquare$",458,622,Combinatorics,13
210,shl_jbmo_2012_c2,shl_jbmo,2012,c,"\textit{$n$ nails nailed on a board are connected by two via a string. Each string is colored in one of $n$ given colors. For any three colors there exist three nails connected by two with strings in these three colors. Can $n$ be: (a) 6, (b) 7?}","\textbf{Solution.} (a) The answer is no:

Suppose it is possible. Consider some color, say blue. Each blue string is the side of 4 triangles formed with vertices on the given points. As there exist $\binom{5}{2} = \frac{5 \cdot 4}{2} = 10$ pairs of colors other than blue, and for any such pair of colors together with the blue color there exists a triangle with strings in these colors, we conclude that there exist at least 3 blue strings (otherwise the number of triangles with a blue string as a side would be at most $2 \cdot 4 = 8$, a contradiction). The same is true for any color, so altogether there exist at least $6 \cdot 3 = 18$ strings, while we have just $\binom{6}{2} = \frac{6 \cdot 5}{2} = 15$ of them.

(b) The answer is yes (Figure 8):

Put the nails at the vertices of a regular 7-gon (Figure 8) and color each one of its sides in a different color. Now color each diagonal in the color of the unique side parallel to it. It can be checked directly that each triple of colors appears in some triangle (because of symmetry, it is enough to check only the triples containing the first color).

\underline{Remark.} The argument in (a) can be applied to any even $n$. The argument in (b) can be applied to any odd $n = 2k + 1$ as follows: first number the nails as $0, 1, 2, \ldots, 2k$ and similarly number the colors as $0, 1, 2, \ldots, 2k$. Then connect nail $x$ with nail $y$ by a string of color $x + y \pmod{n}$. For each triple of colors $(p, q, r)$ there are vertices $x, y, z$ connected by these three colors. Indeed, we need to solve $\pmod{n}$ the system
\[
(*) \quad (x + y \equiv p, \ x + z \equiv q, \ y + z \equiv r)
\]
Adding all three, we get $2(x+y+z) \equiv p+q+r$ and multiplying by $k+1$ we get $x+y+z \equiv (k+1)(p+q+r)$. We can now find $x, y, z$ from the identities $(*)$.",246,1814,Combinatorics,14
211,shl_jbmo_2012_c3,shl_jbmo,2012,c,\textit{In a circle of diameter 1 consider 65 points no three of which are collinear. Prove that there exist 3 among these points which form a triangle with area less than or equal to $\dfrac{1}{72}$.},"\textbf{Solution.} \underline{Lemma}: If a triangle $ABC$ lies in a rectangle $KLMN$ with sides $KL = a$ and $LM = b$, then the area of the triangle is less than or equal to $\dfrac{ab}{2}$.

Proof of the lemma: Without any loss of generality assume that among the distance of $A$, $B$, $C$ from $KL$, that of $A$ is between the other two. Let $\ell$ be the line through $A$ and parallel to $KL$. Let $D$ be the intersection of $\ell$, $BC$ and $x$, $y$ the distances of $B$, $C$ from $\ell$ respectively. Then the area of $ABC$ equals $\dfrac{AD(x+y)}{2} \leq \dfrac{ab}{2}$, since $AD \leq a$ and $x + y \leq b$ and we are done.

Now back to our problem, let us cover the circle with 24 squares of side $\dfrac{1}{6}$ and 8 other irregular and equal figures as shown in Figure 9, with boundary consisting of an arc on the circle and three line segments. Call $S = ADNM$ one of these figures. One of the line segments in the boundary of $S$ is of length $AD = AB - DB = \sqrt{AC^2 - BC^2} - \dfrac{2}{6} =$

\[
\sqrt{\left(\frac{1}{2}
ight)^2 - \left(\frac{1}{6}
ight)^2} - \frac{1}{3} = \frac{\sqrt{2}-1}{3}.
\]
The boundary segment $MN$ goes through the center $C$ of the circle, forming with the horizontal lines an angle of $45^\circ$. The point in $S$ with maximum distance from the boundary segment $AB$ is the endpoint $M$ of the arc on the boundary of $S$. This distance equals $ME = MF - EF \overset{CMF = \text{isosceles}}{=} \dfrac{\sqrt{2}}{2}CM - \dfrac{1}{6} = \dfrac{\sqrt{2}}{4} - \dfrac{1}{6} = \dfrac{3\sqrt{2}-2}{12}$.

So $S$ can be put inside a rectangle $R$ with sides parallel to $AD$, $ND$ of lengths $\dfrac{\sqrt{2}-1}{3}$ and $\dfrac{3\sqrt{2}-2}{12}$. So the triangle formed by any three points inside this figure, has an area less or equal to
\[
\frac{1}{2} \cdot \frac{\sqrt{2}-1}{3} \cdot \frac{3\sqrt{2}-2}{12} = \frac{8 - 5\sqrt{2}}{72} < \frac{1}{72}.
\]
Also, the triangle formed by any three points inside any square of side $\dfrac{1}{6}$, has an area less or equal to $\dfrac{1}{2} \cdot \dfrac{1}{6} \cdot \dfrac{1}{6} = \dfrac{1}{72}$.

By the Pigeonhole Principle, we know that among the 65 given points there exist 3 inside the same one of the 32 squares and irregular figures of the picture covering the given circle. Then according to the above, the triangle formed by these 3 points has an area not exceeding $\dfrac{1}{72}$ as wanted. \hfill$\blacksquare$",201,2402,Combinatorics,15
212,shl_jbmo_2012_n1,shl_jbmo,2012,n,"\textit{If $a$, $b$ are integers and $s = a^3 + b^3 - 60ab(a+b) \geq 2012$, find the least possible value of $s$.}","\textbf{Solution.} It is $s = (a+b)^3 - 63ab(a+b)$ which gives the same residue modulo 7 as $(a+b)^3$. But the residues modulo 7 of perfect cubes can only be 0, 1 or 6. So the residue of $s$ modulo 7 is 0, 1 or 6. Now for $a = 6$, $b = -1$ we get $s = 2015 \geq 2012$ and this is the least possible value of $s$ because the numbers 2012, 2013, 2014 give 3, 4 and 5 as residues mod 7 which are distinct from 0, 1, 6, and so 2012, 2013, 2014 cannot be $s$ for any choice of $a$, $b$. \hfill$\blacksquare$",114,502,Number Theory,16
213,shl_jbmo_2012_n2,shl_jbmo,2012,n,\textit{Do there exist prime numbers $p$ and $q$ such that $p^2(p^3 - 1) = q(q+1)$?},"\textbf{Solution.} Write the given equation in the form
\begin{equation}
p^2(p-1)(p^2+p+1) = q(q+1). \tag{9}
\end{equation}
First observe that it must not be $p = q$, since in this case the left hand side of (9) is greater than its right hand side. Hence, since $p$ and $q$ are distinct primes, (9) immediately yields $p^2 \mid q+1$, that is
\begin{equation}
q = ap^2 - 1 \tag{10}
\end{equation}
for some $a \in \mathbb{N}$. Since $p$ and $q$ are both primes, by (9) we get the following cases:

\textit{Case 1}: $q \mid p - 1$, that is
\begin{equation}
p = bq + 1 \tag{11}
\end{equation}
for some $b \in \mathbb{N}$. Substituting (11) into (10), and using the fact that $a \geq 1$ and $b \geq 1$, we obtain
\[
q = a(bq+1)^2 - 1 \geq (q+1)^2 - 1 = q^2 + 2q,
\]
a contradiction.

\textit{Case 2}: $q \mid p^2 + p + 1$, that is
\begin{equation}
p^2 + p + 1 = bq \tag{12}
\end{equation}
for some $b \in \mathbb{N}$. Substituting (10) into (12), we get
\begin{equation}
p^2 + p + 1 = b(ap^2 - 1) \tag{13}
\end{equation}
If $a \geq 2$, then from (13) it follows that
\[
p^2 + p + 1 \geq 2p^2 - 1,
\]
or equivalently, $p + 1 \geq (p-1)(p+1)$, that is, $(p+1)(2-p) \geq 0$. This implies that $p = 2$, and so $q \mid 2^2 + 2 + 1 = 7$. Hence, $q = 7$, but the pair $p = 2$ and $q = 7$ does not satisfy the equation (9).

Hence, it must be $a = 1$. Then if $b \geq 3$, (13) implies
\[
p^2 + p + 1 \geq 3(p^2 - 1),
\]
or equivalently, $4 \geq p(2p-1)$, which is obviously impossible.

Thus, it must be $a = 1$ and $b \in \{1, 2\}$. For $a = b = 1$, (13) implies that $p = 2$, which by (12) again yields $q = 7$, which is impossible. Finally, for $a = 1$ and $b = 2$, (13) gives $p(p-1) = 3$, which is clearly not satisfied for any prime $p$.

Hence, there do not exist prime numbers $p$ and $q$ which satisfy given equation. \hfill$\blacksquare$",84,1834,Number Theory,17
214,shl_jbmo_2012_n3,shl_jbmo,2012,n,"\textit{Decipher the equality}
\[
(\overline{VER} - \overline{IA}) : (\overline{GRE} + \overline{ECE}) = G^{R^E},
\]
\textit{assuming that the number $\overline{GREECE}$ has a maximum value. It is supposed that each letter corresponds to a unique digit from 0 to 9 and different letters correspond to different digits, and also that all letters $G$, $E$, $V$ and $I$ are different from 0. Also, the notation $\overline{a_n \ldots a_1 a_0}$ stands for the number $a_n \cdot 10^n + \cdots + 10^1 \cdot a_1 + a_0$.}","\textbf{Solution.} Denote
\[
x = \overline{VER} - \overline{IA}, \quad y = \overline{GRE} + \overline{ECE}, \quad z = G^{R^E}.
\]
Then obviously, we have
\begin{align*}
(201 + 131 \ \text{or} \ 231 + 101) &\leq y \leq (879 + 969 \ \text{or} \ 869 + 979 \ \text{or} \ 769 + 989) \\
\Rightarrow \quad 332 &\leq y \leq 1848 \Rightarrow 102 - 98 \leq x \leq 987 - 10 \Rightarrow 4 \leq x \leq 977,
\end{align*}
hence it follows that
\[
\frac{4}{1848} \leq \frac{x}{y} = z \leq \frac{977}{332} \Rightarrow 1 \leq z \leq 2.
\]
This shows that $z = G^{R^E} \in \{1, 2\}$. Hence, if $R \geq 1$, then $R^E \geq 1$, which implies that $2 \geq G^{R^E} \geq G$. Thus, if $R \geq 1$, then it must be $G \leq 2$. In view of this and the assumption of the problem that the number $\overline{GREECE}$ has a maximum value, we will consider the case when $R = 0$ hoping to get a solution with $G > 2$. Then $G^{R^E} = G^0 = 1$ for all digits $G$ and $E$ with $1 \leq G, E \leq 9$, and therefore, the above equality becomes
\[
\overline{VER} - \overline{IA} = \overline{GRE} + \overline{ECE},
\]
which substituting $R = 0$, can be written as
\begin{equation}
\overline{VE0} = \overline{G0E} + \overline{ECE} + \overline{IA}. \tag{14}
\end{equation}
Now we consider the following cases:

\textit{Case 1}: $G = 9$. Then $V \leq 8$, so $\overline{VE0} \leq 900$, while the right hand side of (1) is greater than 900. This is impossible, and no solution exists in this case.

\textit{Case 2}: $G = 8$. Then (14) becomes
\begin{equation}
\overline{VE0} = \overline{80E} + \overline{ECE} + \overline{IA}, \tag{15}
\end{equation}
hence it immediately follows that $V = 9$. For $V = 9$, (15) becomes
\begin{equation}
\overline{9E0} = \overline{80E} + \overline{ECE} + \overline{IA}, \tag{16}
\end{equation}
Notice that for $E \geq 2$, the right hand side of (16) is greater than 1000, while the left hand side of (16) is less than 1000. Therefore, it must be $E \leq 1$, that is, $E = 1$ in view of the fact that $R = 0$. Substituting $E = 1$ into (16), we get
\begin{equation}
910 = 801 + \overline{1C1} + \overline{IA}, \tag{17}
\end{equation}
hence it follows that
\begin{equation}
109 = \overline{1C1} + \overline{IA}. \tag{18}
\end{equation}
But the right hand side of (18) is greater than 121. This shows that $G = 8$ does not lead to any solution.

\textit{Case 3}: $G = 7$. Then (14) becomes
\begin{equation}
\overline{VE0} = \overline{70E} + \overline{ECE} + \overline{IA}. \tag{19}
\end{equation}
Thus it must be $V \geq 8$.

\textit{Subcase 3(a)}: $V = 8$. Then (19) gives
\begin{equation}
\overline{8E0} = \overline{70E} + \overline{ECE} + \overline{IA}, \tag{20}
\end{equation}
hence we immediately obtain $E = 1$ (since the right hand side of (6) must be less than 900). For $E = 1$, (20) reduces to
\begin{equation}
109 = \overline{1C1} + \overline{IA}, \tag{21}
\end{equation}
which is impossible since $\overline{1C1} \geq 121$.

\textit{Subcase 3(b)}: $V = 9$. Then (19) gives
\begin{equation}
\overline{9E0} = \overline{70E} + \overline{ECE} + \overline{IA}, \tag{22}
\end{equation}
hence we immediately obtain $E \leq 2$ (since the right hand side of (7) must be less than 1000). For $E = 2$, (22) reduces to
\begin{equation}
218 = \overline{2C2} + \overline{IA}, \tag{23}
\end{equation}
which is impossible since $\overline{2C2} \geq 232$. Finally, for $E = 1$, (22) reduces to
\begin{equation}
209 = \overline{1C1} + \overline{IA}. \tag{24}
\end{equation}
Since it is required that the number $\overline{GREECE}$ has a maximum value, taking $C = 8$ into (24) we find that
\begin{equation}
28 = \overline{IA}, \tag{25}
\end{equation}
which yields $8 = A = C$. This is impossible since must be $A 
eq C$. Since $C 
eq G = 7$, then taking $C = 6$ into (24) we obtain
\begin{equation}
48 = \overline{IA}, \tag{26}
\end{equation}
hence we have $I = 4$ and $A = 8$. Previously, we have obtained $G = 7$, $R = 0$, $V = 9$, $E = 1$ and $C = 6$. For these values, we obtain that $\overline{GREECE} = 701161$ is the desired maximum value. \hfill$\blacksquare$",512,4044,Number Theory,18
215,shl_jbmo_2012_n4,shl_jbmo,2012,n,"\textit{Determine all triples $(m, n, p)$ satisfying}
\begin{equation}
n^{2p} = m^2 + n^2 + p + 1 \tag{27}
\end{equation}
\textit{where $m$ and $n$ are integers and $p$ is a prime number.}","\textbf{Solution.} By Fermat's theorem $n^{2p} \equiv n^2 \pmod{p}$, therefore $m^2 + n^2 + p + 1 \equiv n^2 \pmod{p} \Rightarrow m^2 \equiv -1 \pmod{p}$.

\textit{Case 1}: $p = 4k + 3$. We have $(m^2)^{2k+1} \equiv (-1)^{2k+1} \pmod{p}$. Therefore,
\begin{equation}
m^{p-1} \equiv -1 \pmod{p} \tag{28}
\end{equation}
and $p$ does not divide $m$. On the other hand, by Fermat's theorem
\begin{equation}
m^{p-1} \equiv 1 \pmod{p} \tag{29}
\end{equation}
(28) and (29) yield $p = 2$. Thus, $p 
eq 4k + 3$.

\textit{Case 2}: $p = 4k + 1$. Let us consider (27) in mod 4. $n^2 = 0$ or 1 in mod 4. In both cases $n^{2p} \equiv n^2 \pmod{4}$. From (27) we get $n^2 \equiv m^2 + n^2 + 1 + 1 \pmod{4}$. Therefore, $m^2 \equiv -2 \pmod{4}$, and again there is no solution.

\textit{Case 3}: $p = 2$. The given equation is written as
\[
n^4 - n^2 - 3 = m^2.
\]
Let $l = n^2$. Readily, we do not get any solution for $l = 0, 1$. If $l = 4$, then there are four solutions: $(3, 2, 2)$, $(-3, 2, 2)$, $(3, -2, 2)$, $(-3, -2, 2)$. There is no solution for $l > 4$, since in this case
\[
(l-1)^2 = l^2 - 2l + 1 < m^2 = l^2 - l - 3 < l^2
\]
Thus, (27) has four solutions: $(3, 2, 2)$, $(-3, 2, 2)$, $(3, -2, 2)$, $(-3, -2, 2)$ and we are done. \hfill$\blacksquare$",188,1247,Number Theory,19
216,shl_jbmo_2012_n5,shl_jbmo,2012,n,"\textit{Find all the positive integers $x$, $y$, $z$, $t$ such that $2^x \cdot 3^y + 5^z = 7^t$.}","\textbf{Solution.} Reducing modulo 3 we get $5^z \equiv 1$, therefore $z$ is even, $z = 2c$, $c \in \mathbb{N}$.

Next we prove that $t$ is even. Obviously, $t \geq 2$. Let us suppose that $t$ is odd, $t = 2d + 1$, $d \in \mathbb{N}$. The equation becomes $2^x \cdot 3^y + 25^c = 7 \cdot 49^d$.

If $x \geq 2$, reducing modulo 4, we get $1 \equiv 3$, contradiction.

For $x = 1$, we have $2 \cdot 3^y + 25^c = 7 \cdot 49^d$, and, reducing modulo 24, we obtain $2 \cdot 3^y + 1 \equiv 7 \Rightarrow 24 \mid 2(3^y - 3)$, i.e.\ $4 \mid 3^{y-1} - 1$, which means that $y - 1$ is even. Then, $y = 2b + 1$, $b \in \mathbb{N}$.

We obtain $6 \cdot 9^b + 25^c = 7 \cdot 49^d$, and, reducing modulo 5, we get $(-1)^b \equiv 2 \cdot (-1)^d$, which is false, for all $b, d \in \mathbb{N}$. Hence $t$ is even, $t = 2d$, $d \in \mathbb{N}$.

The equation can be written as $2^x \cdot 3^y + 25^c = 49^d \Leftrightarrow 2^x \cdot 3^y = (7^d - 5^c)(7^d + 5^c)$.

As $\gcd(7^d - 5^c, 7^d + 5^c) = 2$ and $7^c + 5^c > 2$, there exist exactly three possibilities:
\[
(1) \begin{cases} 7^d - 5^c = 2^{x-1} \\ 7^d + 5^c = 2 \cdot 3^y \end{cases}; \qquad (2) \begin{cases} 7^d - 5^c = 2 \cdot 3^y \\ 7^d + 5^c = 2^{x-1} \end{cases}; \qquad (3) \begin{cases} 7^d - 5^c = 2 \\ 7^d + 5^c = 2^{x-1} \cdot 3^y \end{cases}
\]

\textit{Case (1).} We have $7^d = 2^{x-2} + 3^y$ and, reducing modulo 3, we get $2^{x-2} \equiv 1 \pmod{3}$, hence $x - 2$ is even, i.e.\ $x = 2a + 2$, where $a \in \mathbb{N}$, since $a = 0$ would mean $3^y + 1 = 7^d$, which is impossible (even $=$ odd).

We obtain $7^d - 5^c = 2 \cdot 4^a \overset{\bmod 4}{=} 7^d \equiv 1 \pmod{4} \Rightarrow d = 2e$, $e \in \mathbb{N}$. Then $49^e - 5^c = 2 \cdot 4^a \overset{\bmod 8}{=} 5^c \equiv 1 \pmod{8} \Rightarrow c = 2f$, $f \in \mathbb{N}$. We obtain $49^e - 25^f = 2 \cdot 4^a \overset{\bmod 3}{=} 0 \equiv 2 \pmod{3}$, false. In conclusion, in this case there are no solutions to the equation.

\textit{Case (2).} From $2^{x-1} = 7^d + 5^c \geq 12$, we obtain $x \geq 5$. Then $7^d + 5^c \equiv 0 \pmod{4}$, i.e.\ $3^d + 1 \equiv 0 \pmod{4}$, hence $d$ is odd. As $7^d = 5^c + 2 \cdot 3^y \geq 11$, we get $d \geq 2$, hence $d = 2e + 1$, $e \in \mathbb{N}$.

As in the previous case, from $7^d = 2^{x-2} + 3^y$, reducing modulo 3, we obtain $x = 2a + 2$, with $a \geq 2$ (because $x \geq 5$). We get $7^d = 4^a + 3^y$, i.e.\ $7 \cdot 49^e = 4^a + 3^y$, hence, reducing modulo 8, we obtain $7 \equiv 3^y$, which is false, because $3^y$ is congruent mod 8 either to 1 (if $y$ is even) or to 3 (if $y$ is odd). In conclusion, in this case there are no solutions to the equation.

\textit{Case (3).} From $7^d = 5^c + 2$, it follows that the last digit of $7^d$ is 7, hence $d = 4k + 1$, $k \in \mathbb{N}$.

If $c \geq 2$, from $7^{4k+1} = 5^c + 2$, reducing modulo 25, we obtain $7 \equiv 2 \pmod{25}$, which is false.

For $c = 1$ we get $d = 1$, and the solution $x = 3$, $y = 1$, $z = t = 2$. \hfill$\blacksquare$",97,2965,Number Theory,20
217,shl_jbmo_2012_n6,shl_jbmo,2012,n,"\textit{If $a$, $b$, $c$, $d$ are integers and $A = 2(a - 2b + c)^4 + 2(b - 2c + a)^4 + 2(c - 2a + b)^4$, $B = d(d+1)(d+2)(d+3) + 1$, prove that $\left(\sqrt{A} + 1
ight)^2 + B$ cannot be a perfect square.}","\textbf{Solution.} First we prove the following Lemma

\underline{Lemma}: \textit{If $x$, $y$, $z$ real numbers such that $x + y + z = 0$, then $2(x^4 + y^4 + z^4) = (x^2 + y^2 + z^2)^2$.}

Proof of the Lemma:
\begin{align*}
x^4 + y^4 + z^4 &= x^2 x^2 + y^2 y^2 + z^2 z^2 = x^2(y+z)^2 + y^2(z+x)^2 + z^2(x+y)^2 \\
&= 2(x^2y^2 + y^2z^2 + z^2x^2) + 2xyz(x + y + z) \\
&= (x^2 + y^2 + z^2)^2 - x^4 - y^4 - z^4
\end{align*}
and the claim follows.

Now back to our problem notice that $(a - 2b + c) + (b - 2c + a) + (c - 2a + b) = 0$, thus according to the lemma it holds
\begin{align*}
A &= 2(a - 2b + c)^4 + 2(b - 2c + a)^4 + 2(c - 2a + b)^4 \\
&= \left[(a - 2b + c)^2 + (b - 2c + a)^2 + (c - 2a + b)^2
ight]^2 = \left[6(a^2 + b^2 + c^2 - ab - bc - ca)
ight]^2.
\end{align*}
Since $a^2 + b^2 + c^2 \geq ab + bc + ca$ we have that
\[
\sqrt{A} + 1 = 6(a^2 + b^2 + c^2 - ab - bc - ca) + 1
\]
In addition, it is easy to check that
\[
B = d(d+1)(d+2)(d+3) = (d^2 + 3d + 1)^2
\]
Let us set $6(a^2 + b^2 + c^2 - ab - bc - ca) + 1 = m$, $d^2 + 3d + 1 = n$. We need to prove that the number $\left(\sqrt{A} + 1
ight)^2 + B = m^2 + n^2$ is not a perfect square.

Since both $m$, $n$ are odd integers, both $m^2$, $n^2$ are integers of the form $4k + 1$, so the number $m^2 + n^2$ is an integer of the form $4k + 2$. But it is well known that all perfect squares are of the form $4k$ or $4k + 1$, and we are done. \hfill$\blacksquare$",206,1420,Number Theory,21
218,shl_jbmo_2012_n7,shl_jbmo,2012,n,"\textit{Find all natural numbers $a$, $b$, $c$ for which $1997^a + 15^b = 2012^c$.}","\textbf{Solution.} $1997^a + 15^b = 2012^c \Rightarrow 1 + (-1)^b \equiv 0 \pmod{4}$, so $b$ is an odd number.

$1997^a + 15^b = 2012^c \Rightarrow 1 + 0 \equiv 2^c \pmod{3}$, so $c$ is even, say $c = 2c_1$.

We intend to consider the given equation modulo 8 and for this reason we discern two cases:

(1): $c = 1$. Clearly then $a = b = 1$ and $a = b = c = 1$ is a solution. This is actually the only solution of the given equation since in the remaining case where $c > 1$ it will be shown that there exist no solution.

(2): $c > 1$. Then $2012^c = (4 \cdot 503)^c$ is a multiple of 8 and $1997^a + 15^b = 2012^c \Rightarrow 5^a + (-1)^b = 5^a + (-1) \equiv 0 \pmod{8}$, so $a$ is even, say $a = 2a_1$. Hence
\[
3^{4b} \cdot 5^{4b} = 15^b = 2012^c - 1997^a = (2012^{c_1} - 1997^{a_1}) \cdot (2012^{c_1} + 1997^{a_1}).
\]
Observe that $2012^{c_1} - 1997^{a_1}$, $2012^{c_1} + 1997^{a_1}$ are both greater than 1 and prime to each other as $\gcd(2012^{c_1} - 1997^{a_1}, 2012^{c_1} + 1997^{a_1}) = \gcd(2012^{c_1} - 1997^{a_1}, 2 \cdot 1997^{a_1}) = 1$. So there exist two cases:
\[
\text{Case 1:} \quad \frac{2012^{c_1} - 1997^{a_1} = 5^b}{2012^{c_1} + 1997^{a_1} = 3^b}, \qquad \text{Case 2:} \quad \frac{2012^{c_1} - 1997^{a_1} = 3^b}{2012^{c_1} + 1997^{a_1} = 5^b}
\]

\textit{Case 1}:
\begin{align*}
2012^{c_1} - 1997^{a_1} &= 5^b \Rightarrow 2^{c_1} - 2^{a_1} \equiv 0 \pmod{5} \Rightarrow c_1 \equiv a_1 \pmod{5} \\
2012^{c_1} + 1997^{a_1} &= 3^b \Rightarrow 2^{c_1} + 2^{a_1} \equiv 0 \pmod{5} \Rightarrow c_1 \equiv a_1 + 1 \pmod{5}
\end{align*}
a contradiction.

\textit{Case 2}:
\begin{align*}
2012^{c_1} - 1997^{a_1} &= 3^b \\
2012^{c_1} + 1997^{a_1} &= 5^b
\end{align*}
Since $b$ is an odd number we get $2012^{c_1} + 1997^{a_1} \equiv 5^b \pmod{3} \Rightarrow 2^{c_1} + 2^{a_1} \equiv 5^b \equiv 2 \pmod{3}$ so $a_1$, $c_1$ are even numbers, say $a_1 = 2a_2$, $c_1 = 2c_2$. Then
\[
(2012^{c_2} - 1997^{a_2}) \cdot (2012^{c_2} + 1997^{a_2}) = 3^b
\]
But $\gcd(2012^{c_2} - 1997^{a_2}, 2012^{c_2} + 1997^{a_2}) = 1$ and the above implies $2012^{c_2} - 1997^{a_2} = 1$. But then mod 4, we get $0 - 1 \equiv \pmod{4}$, a contradiction.

Therefore there exists no solution for $c > 1$.

Hence $a = b = c = 1$ is the only solution. \hfill$\blacksquare$",83,2261,Number Theory,22
227,jbmo_2013_p1,jbmo,2013,n,"Find all ordered pairs $(a, b)$ of positive integers for which the numbers $\dfrac{a^3b - 1}{a+1}$ and $\dfrac{b^3a + 1}{b - 1}$ are both positive integers.","\textbf{Solution.} As $a^3b - 1 = b(a^3 + 1) - (b + 1)$ and $a + 1 \mid a^3 + 1$, we have $a + 1 \mid b + 1$.

As $b^3a + 1 = a(b^3 - 1) + (a + 1)$ and $b - 1 \mid b^3 - 1$, we have $b - 1 \mid a + 1$.

So $b - 1 \mid b + 1$ and hence $b - 1 \mid 2$.

\begin{itemize}
  \item If $b = 2$, then $a + 1 \mid b + 1 = 3$ gives $a = 2$. Hence $(a, b) = (2, 2)$ is the only solution in this case.
  \item If $b = 3$, then $a + 1 \mid b + 1 = 4$ gives $a = 1$ or $a = 3$. Hence $(a, b) = (1, 3)$ and $(3, 3)$ are the only solutions in this case.
\end{itemize}

To summarize, $(a, b) = (1, 3)$, $(2, 2)$ and $(3, 3)$ are the only solutions.",156,631,Number Theory,1
228,jbmo_2013_p2,jbmo,2013,g,"Let $ABC$ be an acute triangle with $AB < AC$ and let $O$ be the center of its circumcircle $\omega$. Let $D$ be a point on the line segment $BC$ such that $\angle BAD = \angle CAO$. Let $E$ be the second point of intersection of $\omega$ and the line $AD$. If $M$, $N$ and $P$ are the midpoints of the line segments $BE$, $OD$ and $AC$, respectively, show that the points $M$, $N$ and $P$ are collinear.","\textbf{Solution.} We will show that $MOPD$ is a parallelogram. From this it follows that $M$, $N$, $P$ are collinear.

Since $\angle BAD = \angle CAO = 90^\circ - \angle ABC$, $D$ is the foot of the perpendicular from $A$ to side $BC$. Since $M$ is the midpoint of the line segment $BE$, we have $BM = ME = MD$ and hence $\angle MDE = \angle MED = \angle ACB$.

Let the line $MD$ intersect the line $AC$ at $D_1$. Since $\angle ADD_1 = \angle MDE = \angle ACD$, $MD$ is perpendicular to $AC$. On the other hand, since $O$ is the center of the circumcircle of triangle $ABC$ and $P$ is the midpoint of the side $AC$, $OP$ is perpendicular to $AC$. Therefore $MD$ and $OP$ are parallel.

Similarly, since $P$ is the midpoint of the side $AC$, we have $AP = PC = DP$ and hence $\angle PDC = \angle ACB$. Let the line $PD$ intersect the line $BE$ at $D_2$. Since $\angle BDD_2 = \angle PDC = \angle ACB = \angle BED$, we conclude that $PD$ is perpendicular to $BE$. Since $M$ is the midpoint of the line segment $BE$, $OM$ is perpendicular to $BE$ and hence $OM$ and $PD$ are parallel.",404,1082,Geometry,2
229,jbmo_2013_p3,jbmo,2013,a,"Show that
\[
\left(a + 2b + \frac{2}{a+1}
ight)\!\left(b + 2a + \frac{2}{b+1}
ight) \geq 16
\]
for all positive real numbers $a$ and $b$ such that $ab \geq 1$.","\textbf{Solution 1.} By the AM-GM Inequality we have:
\[
  \frac{a+1}{2} + \frac{2}{a+1} \ge 2
\]
Therefore
\[
  a + 2b + \frac{2}{a+1} \ge \frac{a+3}{2} + 2b,
\]
and, similarly,
\[
  b + 2a + \frac{2}{b+1} \ge 2a + \frac{b+3}{2}.
\]
On the other hand,
\[
  (a + 4b + 3)(b + 4a + 3) \ge \left(\sqrt{ab} + 4\sqrt{ab} + 3
ight)^2 \ge 64
\]
by the Cauchy-Schwarz Inequality as $ab \ge 1$, and we are done.

\textbf{Solution 2.} Since $ab \ge 1$, we have $a + b \ge a + 1/a \ge 2\sqrt{a \cdot (1/a)} = 2$.

Then
\[
  a + 2b + \frac{2}{a+1}
    = b + (a+b) + \frac{2}{a+1}
    \ge b + 2 + \frac{2}{a+1}
    = \frac{b+1}{2} + \frac{b+1}{2} + 1 + \frac{2}{a+1}
    \ge 4\sqrt[4]{\frac{(b+1)^2}{2(a+1)}}
\]
by the AM-GM Inequality. Similarly,
\[
  b + 2a + \frac{2}{b+1} \ge 4\sqrt[4]{\frac{(a+1)^2}{2(b+1)}}.
\]

Now using these and applying the AM-GM Inequality another time we obtain:
\[
  \left(a + 2b + \frac{2}{a+1}
ight)\!\left(b + 2a + \frac{2}{b+1}
ight)
    \ge 16\sqrt[4]{\frac{(a+1)(b+1)}{4}}
    \ge 16\sqrt[4]{\frac{(2\sqrt{a})(2\sqrt{b})}{4}}
    = 16\sqrt[8]{ab}
    \ge 16
\]

\textbf{Solution 3.} We have
\begin{align*}
  \left(a + 2b + \frac{2}{a+1}
ight)\!\left(b + 2a + \frac{2}{b+1}
ight)
  &= \left((a+b) + b + \frac{2}{a+1}
ight)\!\left((a+b) + a + \frac{2}{b+1}
ight) \\
  &\ge \left(a + b + \sqrt{ab} + \frac{2}{\sqrt{(a+1)(b+1)}}
ight)^2
\end{align*}
by the Cauchy-Schwarz Inequality.

On the other hand,
\[
  \frac{2}{\sqrt{(a+1)(b+1)}} \ge \frac{4}{a+b+2}
\]
by the AM-GM Inequality and
\[
  a + b + \sqrt{ab} + \frac{2}{\sqrt{(a+1)(b+1)}}
    \ge a + b + 1 + \frac{4}{a+b+2}
    = \frac{(a+b+1)(a+b-2)}{a+b+2} + 4
    \ge 4
\]
as $a + b \ge 2\sqrt{ab} \ge 2$, finishing the proof.",159,1702,Algebra,3
230,jbmo_2013_p4,jbmo,2013,c,"Let $n$ be a positive integer. Two players, Alice and Bob, are playing the following game:
\begin{itemize}
\item Alice chooses $n$ real numbers, not necessarily distinct.
\item Alice writes all pairwise sums on a sheet of paper and gives it to Bob (there are $\tfrac{n(n-1)}{2}$ such sums, not necessarily distinct).
\item Bob wins if he finds correctly the initial $n$ numbers chosen by Alice with only one guess.
\end{itemize}
Can Bob be sure to win for the following cases?
\begin{enumerate}[label=\alph*.]
\item $n = 5$
\item $n = 6$
\item $n = 8$
\end{enumerate}
Justify your answer(s).


oindent[For example, when $n = 4$, Alice may choose the numbers $1, 5, 7, 9$, which have the same pairwise sums as the numbers $2, 4, 6, 10$, and hence Bob cannot be sure to win.]","\textbf{Solution.}

\textbf{a.} Yes. Let $a \le b \le c \le d \le e$ be the numbers chosen by Alice. As each number appears in a pairwise sum $4$ times, by adding all $10$ pairwise sums and dividing the result by $4$, Bob obtains $a + b + c + d + e$. Subtracting the smallest and the largest pairwise sums $a + b$ and $d + e$ from this he obtains $c$. Subtracting $c$ from the second largest pairwise sum $c + e$ he obtains $e$. Subtracting $e$ from the largest pairwise sum $d + e$ he obtains $d$. He can similarly determine $a$ and $b$.

\textbf{b.} Yes. Let $a \le b \le c \le d \le e \le f$ be the numbers chosen by Alice. As each number appears in a pairwise sum $5$ times, by adding all $15$ pairwise sums and dividing the result by $5$, Bob obtains $a + b + c + d + e + f$. Subtracting the smallest and the largest pairwise sums $a + b$ and $e + f$ from this he obtains $c + d$.

Subtracting the smallest and the second largest pairwise sums $a + b$ and $d + f$ from $a + b + c + d + e + f$ he obtains $c + e$. Similarly he can obtain $b + d$. He uses these to obtain $a + f$ and $b + e$.

Now $a + d$, $a + e$, $b + c$ are the three smallest among the remaining six pairwise sums. If Bob adds these up, subtracts the known sums $c + d$ and $b + e$ from the result and divides the difference by $2$, he obtains $a$. Then he can determine the remaining numbers.

\textbf{c.} No. If Alice chooses the eight numbers $1, 5, 7, 9, 12, 14, 16, 20$, then Bob cannot be sure to guess these numbers correctly as the eight numbers $2, 4, 6, 10, 11, 15, 17, 19$ also give exactly the same $28$ pairwise sums as these numbers.",773,1621,Combinatorics,4
179,shl_jbmo_2013_a1,shl_jbmo,2013,a,"Find all ordered triples $(x, y, z)$ of real numbers satisfying the following system of equations:
\[
x^3 = \frac{z}{y} - 2\frac{y}{z}
\]
\[
y^3 = \frac{x}{z} - 2\frac{z}{x}
\]
\[
z^3 = \frac{y}{x} - 2\frac{x}{y}
\]","\textbf{Solution.} We have
\[
x^3yz = z^2 - 2y^2
\]
\[
y^3zx = x^2 - 2z^2
\]
\[
z^3xy = y^2 - 2x^2
\]
with $xyz 
eq 0$.

Adding these up we obtain $(x^2 + y^2 + z^2)(xyz + 1) = 0$. Hence $xyz = -1$. Now the system of equations becomes:
\[
x^2 = 2y^2 - z^2
\]
\[
y^2 = 2z^2 - x^2
\]
\[
z^2 = 2x^2 - y^2
\]

Then the first two equations give $x^2 = y^2 = z^2$. As $xyz = -1$, we conclude that $(x, y, z) = (1, 1, -1)$, $(1, -1, 1)$, $(-1, 1, 1)$ and $(-1, -1, -1)$ are the only solutions.",215,486,Algebra,1
180,shl_jbmo_2013_a2,shl_jbmo,2013,a,"Find the largest possible value of the expression $\left|\sqrt{x^2 + 4x + 8} - \sqrt{x^2 + 8x + 17}
ight|$ where $x$ is a real number.","\textbf{Solution.} We observe that
\[
\left|\sqrt{x^2 + 4x + 8} - \sqrt{x^2 + 8x + 17}
ight| = \left|\sqrt{(x-(-2))^2 + (0-2)^2} - \sqrt{(x-(-4))^2 + (0-1)^2}
ight|
\]
is the absolute difference of the distances from the point $P(x, 0)$ in the $xy$-plane to the points $A(-2, 2)$ and $B(-4, 1)$.

By the Triangle Inequality, $|PA - PB| \leq |AB|$ and the equality occurs exactly when $P$ lies on the line passing through $A$ and $B$, but not between them.

If $P$, $A$, $B$ are collinear, then $(x - (-2))/(0 - 2) = ((-4) - (-2))/(1 - 2)$. This gives $x = -6$, and as $-6 < -4 < -2$,
\[
\left|\sqrt{(-6)^2 + 4(-6) + 8} - \sqrt{(-6)^2 + 8(-6) + 17}
ight| = \left|\sqrt{20} - \sqrt{5}
ight| = \sqrt{5}
\]
is the largest possible value of the expression.",134,751,Algebra,2
181,shl_jbmo_2013_a3,shl_jbmo,2013,a,"Show that
\[
\left(a + 2b + \frac{2}{a+1}
ight)\left(b + 2a + \frac{2}{b+1}
ight) \geq 16
\]
for all positive real numbers $a$, $b$ satisfying $ab \geq 1$.","\textbf{Solution 1.} By the AM-GM Inequality we have:
\[
\frac{a+1}{2} + \frac{2}{a+1} \geq 2
\]
Therefore
\[
a + 2b + \frac{2}{a+1} \geq \frac{a+3}{2} + 2b\,.
\]
and, similarly,
\[
b + 2a + \frac{2}{b+1} \geq 2a + \frac{b+3}{2}\,.
\]
On the other hand,
\[
(a + 4b + 3)(b + 4a + 3) \geq (\sqrt{ab} + 4\sqrt{ab} + 3)^2 \geq 64
\]
by the Cauchy-Schwarz Inequality as $ab \geq 1$, and we are done.

\textbf{Solution 2.} Since $ab \geq 1$, we have $a + b \geq a + 1/a \geq 2\sqrt{a \cdot (1/a)} = 2$.

Then
\begin{align*}
a + 2b + \frac{2}{a+1} &= b + (a+b) + \frac{2}{a+1}\\
&\geq b + 2 + \frac{2}{a+1}\\
&= \frac{b+1}{2} + \frac{b+1}{2} + 1 + \frac{2}{a+1}\\
&\geq 4\sqrt[4]{\frac{(b+1)^2}{2(a+1)}}
\end{align*}
by the AM-GM Inequality. Similarly,
\[
b + 2a + \frac{2}{b+1} \geq 4\sqrt[4]{\frac{(a+1)^2}{2(b+1)}}\,.
\]
Now using these and applying the AM-GM Inequality another time we obtain:
\begin{align*}
\left(a + 2b + \frac{2}{a+1}
ight)\left(b + 2a + \frac{2}{b+1}
ight) &\geq 16\sqrt[4]{\frac{(a+1)(b+1)}{4}}\\
&\geq 16\sqrt[4]{\frac{(2\sqrt{a})(2\sqrt{b})}{4}}\\
&= 16\sqrt[8]{ab}\\
&\geq 16
\end{align*}

\textbf{Solution 3.} We have
\begin{align*}
\left(a + 2b + \frac{2}{a+1}
ight)\left(b + 2a + \frac{2}{b+1}
ight) &= \left((a+b) + b + \frac{2}{a+1}
ight)\left((a+b) + a + \frac{2}{b+1}
ight)\\
&\geq \left(a + b + \sqrt{ab} + \frac{2}{\sqrt{(a+1)(b+1)}}
ight)^2
\end{align*}
by the Cauchy-Schwarz Inequality.

On the other hand,
\[
\frac{2}{\sqrt{(a+1)(b+1)}} \geq \frac{4}{a+b+2}
\]
by the AM-GM Inequality and
\[
a + b + \sqrt{ab} + \frac{2}{\sqrt{(a+1)(b+1)}} \geq a + b + 1 + \frac{4}{a+b+2} = \frac{(a+b+1)(a+b-2)}{a+b+2} + 4 \geq 4
\]
as $a + b \geq 2\sqrt{ab} \geq 2$, finishing the proof.",155,1707,Algebra,3
182,shl_jbmo_2013_c1,shl_jbmo,2013,c,"Find the largest number of distinct integers that can be chosen from the set $\{1, 2, \ldots, 2013\}$ so that the difference of no two of them is equal to 17.","\textbf{Solution.} Consider the sets $A_{mn} = \{34m + n - 34, 34m + n - 17\}$ for $1 \leq m \leq 59$ and $1 \leq n \leq 17$, and $B_k = \{2006 + k\}$ for $1 \leq k \leq 7$. As we cannot choose more than one number from each of these sets, we can choose at most $59 \cdot 17 + 7 = 1010$ numbers. On the other hand, choosing the smaller element of each of these sets gives exactly 1010 numbers satisfying the condition.

\textbf{Comment.} The original problem proposal asks the question with the numbers 55 and 5.",158,512,Combinatorics,4
183,shl_jbmo_2013_c2,shl_jbmo,2013,c,"On a billiards table in the shape of a rectangle $ABCD$ with $AB = 2013$ and $AD = 1000$, a billiard ball is shot along the bisector of the angle $\angle BAD$. Assuming that the ball is reflected from the sides at the same angle it comes in, determine whether it will ever go to the corner $B$.","\textbf{Solution 1.} The ball travels a horizontal distance of 1000 units between two bounces from the sides $AB$ and $CD$ as it always moves on a line making a $45^\circ$ angle with the sides. Hence it is always at a distance of even number of units to the line $AD$ when it hits $AB$ or $CD$. Hence it can never hit $AB$ at $B$.

\textbf{Solution 2.} Consider a rectangle $A'B'C'D'$ which is wider $1/2$ units on all sides, divide it into unit squares, and color them black and white alternatingly with the vertex $A$ being the center of a black unit square. Then the ball always moves along the diagonals of the black unit squares. As $B$ lies at the center of a white unit square, the ball never reaches $B$.

\textbf{Solution 3.} The vertical lines $x = 2013m$ and the horizontal lines $y = 1000n$, where $m$ and $n$ are integers, divide the $xy$-plane into rectangles congruent to the rectangle $ABCD$. Let $A(0,0)$, $B(2013, 0)$, $C(2013, 1000)$, $D(0, 1000)$, and identify the other rectangles with $ABCD$ via reflections across these lines. Under this identification, the ball moves along the line $y = x$ and the coordinates of the points identified with $B$ have the form $(2013k, 1000l)$ where $k$ is an odd integer and $l$ is an even one. Hence the ball never goes to $B$.",294,1285,Combinatorics,5
184,shl_jbmo_2013_c3,shl_jbmo,2013,c,"All possible pairs of $n$ apples are weighed and the results are given to us in an arbitrary order. Can we determine the weights of the apples if \textbf{a.} $n = 4$, \textbf{b.} $n = 5$, \textbf{c.} $n = 6$?","\textbf{Solution. a.} No. Four apples with weights $1, 5, 7, 9$ and with weights $2, 4, 6, 10$ both give the results $6, 8, 10, 12, 14, 16$ when weighed in pairs.

\textbf{b.} Yes. Let $a \leq b \leq c \leq d \leq e$ be the weights of the apples. As each apple is weighed 4 times, by adding all 10 pairwise weights and dividing the sum by 4, we obtain $a+b+c+d+e$. Subtracting the smallest and the largest pairwise weights $a+b$ and $d+e$ from this we obtain $c$. Subtracting $c$ from the second largest pairwise weight $c+e$ we obtain $e$. Subtracting $e$ from the largest pairwise weight $d+e$ we obtain $d$. $a$ and $b$ are similarly determined.

\textbf{c.} Yes. Let $a \leq b \leq c \leq d \leq e \leq f$ be the weights of the apples. As each apple is weighed 5 times, by adding all 15 pairwise weights and dividing the sum by 5, we obtain $a+b+c+d+e+f$. Subtracting the smallest and the largest pairwise weights $a+b$ and $e+f$ from this we obtain $c+d$.

Subtracting the smallest and the second largest pairwise weights $a+b$ and $d+f$ from $a+b+c+d+e+f$ we obtain $c+e$. Similarly we obtain $b+d$. We use these to obtain $a+f$ and $b+e$.

Now $a+d$, $a+e$, $b+c$ are the three smallest among the remaining six pairwise weights. If we add these up, subtract the known weights $c+d$ and $b+e$ form the sum and divide the difference by 2, we obtain $a$. Then the rest follows.",208,1381,Combinatorics,6
185,shl_jbmo_2013_g1,shl_jbmo,2013,g,"Let $AB$ be a diameter of a circle $\omega$ with center $O$ and $OC$ be a radius of $\omega$ which is perpendicular to $AB$. Let $M$ be a point on the line segment $OC$. Let $N$ be the second point of intersection of the line $AM$ with $\omega$, and let $P$ be the point of intersection of the lines tangent to $\omega$ at $N$ and at $B$. Show that the points $M$, $O$, $P$, $N$ are concyclic.","\textbf{Solution.} Since the lines $PN$ and $BP$ are tangent to $\omega$, $NP = PB$ and $OP$ is the bisector of $\angle NOB$. Therefore the lines $OP$ and $NB$ are perpendicular. Since $\angle ANB = 90^\circ$, it follows that the lines $AN$ and $OP$ are parallel. As $MO$ and $PB$ are also parallel and $AO = OB$, the triangles $AMO$ and $OPB$ are congruent and $MO = PB$. Hence $MO = NP$. Therefore $MOPN$ is an isosceles trapezoid and therefore cyclic. Hence the points $M$, $O$, $P$, $N$ are concyclic.",393,505,Geometry,7
186,shl_jbmo_2013_g2,shl_jbmo,2013,g,"$\omega_1$ and $\omega_2$ are two circles that are externally tangent to each other at the point $M$ and internally tangent to a circle $\omega_3$ at the points $K$ and $L$, respectively. Let $A$ and $B$ be the two points where the common tangent line at $M$ to $\omega_1$ and $\omega_2$ intersects $\omega_3$. Show that if $\angle KAB = \angle LAB$ then the line segment $AB$ is a diameter of $\omega_3$.","\textbf{Solution.} Let $C$ be the intersection point of the tangent lines to the circles $\omega_1$ at $K$ and $\omega_2$ at $L$. Point $C$ lies on the radical axis of circles $\omega_1$ and $\omega_3$, and also on the radical axis of the circles $\omega_2$ and $\omega_3$. Therefore $C$ lies on the radical axis of the circles $\omega_1$ and $\omega_2$ too. Therefore the points $A$, $B$, $C$ are collinear.

Since $\angle KAB = \angle LAB$, the chords $KB$ and $BL$ have the same length. As we also have $CK = CL$, the triangles $KBC$ and $LBC$ are congruent. In particular, $\angle KBA = \angle LBA$. Therefore, $\angle BKA = 180^\circ - (\angle ABK + \angle BAK) = 180^\circ - (\angle LBK + \angle LAK)/2 = 180^\circ - 90^\circ = 90^\circ$, and $AB$ is a diameter.

\textbf{Comment.} The original problem proposal gives $\angle KAB = \angle LAB = 15^\circ$ and asks the measures of the angles of the quadrilateral $AKBL$.",405,925,Geometry,8
187,shl_jbmo_2013_g3,shl_jbmo,2013,g,"\textbf{G3.} Let $D$ be a point on the side $BC$ of an acute triangle $ABC$ such that $\angle BAD = \angle CAO$ where $O$ is the center of the circumcircle $\omega$ of the triangle $ABC$. Let $E$ be the second point of intersection of $\omega$ and the line $AD$. Let $M$, $N$, $P$ be the midpoints of the line segments $BE$, $OD$, $AC$, respectively. Show that $M$, $N$, $P$ are collinear.","\textbf{Solution.} We will show that $MOPD$ is a parallelogram. From this it follows that $M$, $N$, $P$ are collinear.

Since $\angle BAD = \angle CAO = 90^\circ - \angle ABC$, $D$ is the foot of the perpendicular from $A$ to side $BC$. Since $M$ is the midpoint of the line segment $BE$, we have $BM = ME = MD$ and hence $\angle MDE = \angle MED = \angle ACB$.

Let the line $MD$ intersect the line $AC$ at $D_1$. Since $\angle ADD_1 = \angle MDE = \angle ACD$, $MD$ is perpendicular to $AC$. On the other hand, since $O$ is the center of the circumcircle of triangle $ABC$ and $P$ is the midpoint of the side $AC$, $OP$ is perpendicular to $AC$. Therefore $MD$ and $OP$ are parallel.

Similarly, since $P$ is the midpoint of the side $AC$, we have $AP = PC = DP$ and hence $\angle PDC = \angle ACB$. Let the line $PD$ intersect the line $BE$ at $D_2$. Since $\angle BDD_2 = \angle PDC = \angle ACB = \angle BED$, we conclude that $PD$ is perpendicular to $BE$. Since $M$ is the midpoint of the line segment $BE$, $OM$ is perpendicular to $BE$ and hence $OM$ and $PD$ are parallel.",389,1082,Geometry,9
188,shl_jbmo_2013_g4,shl_jbmo,2013,g,"\textbf{G4.} Let $I$ be the incenter and $AB$ the shortest side of a triangle $ABC$. The circle with center $I$ and passing through $C$ intersects the ray $AB$ at the point $P$ and the ray $BA$ at the point $Q$. Let $D$ be the point where the excircle of the triangle $ABC$ belonging to angle $A$ touches the side $BC$, and let $E$ be the symmetric of the point $C$ with respect to $D$. Show that the lines $PE$ and $CQ$ are perpendicular.","\textbf{Solution.} First we will show that points $P$ and $Q$ are not on the line segment $AB$.

Assume that $Q$ is on the line segment $AB$. Since $CI = QI$ and $\angle IBQ = \angle IBC$, either the triangles $CBI$ and $QBI$ are congruent or $\angle ICB + \angle IQB = 180^\circ$. In the first case, we have $BC = BQ$ which contradicts $AB$ being the shortest side.

In the second case, we have $\angle IQA = \angle ICB = \angle ICA$ and the triangles $IAC$ and $IAQ$ are congruent. Hence this time we have $AC = AQ$, contradicting $AB$ being the shortest side.

Now we will show that the lines $PE$ and $CQ$ are perpendicular.

Since $\angle IQB \leq \angle IAB = (\angle CAB)/2 < 90^\circ$ and $\angle ICB = (\angle ACB)/2 < 90^\circ$, the triangles $CBI$ and $QBI$ are congruent. Hence $BC = BQ$ and $\angle CQP = \angle CQB = 90^\circ - (\angle ABC)/2$. Similarly, we have $AC = AP$ and hence $BP = AC - AB$.

On the other hand, as $DE = CD$ and $CD + AC = u$, where $u$ denotes the semiperimeter of the triangle $ABC$, we have $BE = BC - 2(u - AC) = AC - AB$. Therefore $BP = BE$ and $\angle QPE = (\angle ABC)/2$.

Hence, $\angle CQP + \angle QPE = 90^\circ$.",439,1166,Geometry,10
189,shl_jbmo_2013_g5,shl_jbmo,2013,g,"\textbf{G5.} A circle passing through the midpoint $M$ of the side $BC$ and the vertex $A$ of a triangle $ABC$ intersects the sides $AB$ and $AC$ for the second time at the points $P$ and $Q$, respectively. Show that if $\angle BAC = 60^\circ$ then
\[
AP + AQ + PQ < AB + AC + \tfrac{1}{2}\,BC\,.
\]","\textbf{Solution.} Since the quadrilateral $APMQ$ is cyclic, we have $\angle PMQ = 180^\circ - \angle PAQ = 180^\circ - \angle BAC = 120^\circ$. Therefore $\angle PMB + \angle QMC = 180^\circ - \angle PMQ = 60^\circ$.

Let the point $B'$ be the symmetric of the point $B$ with respect to the line $PM$ and the point $C'$ be the symmetric of the point $C$ with respect to the line $QM$. The triangles $B'MP$ and $BMP$ are congruent and the triangles $C'MQ$ and $CMQ$ are congruent. Hence $\angle B'MC' = \angle PMQ - \angle B'MP - \angle C'MQ = 120^\circ - \angle BMP - \angle CMQ = 120^\circ - 60^\circ = 60^\circ$. As we also have $B'M = BM = CM = C'M$, we conclude that the triangle $B'MC'$ is equilateral and $B'C' = BC/2$.

On the other hand, we have $PB' + B'C' + C'Q \geq PQ$ by the Triangle Inequality, and hence $PB + BC/2 + QC \geq PQ$. This gives the inequality $AB + BC/2 + AC \geq AP + PQ + AQ$.

We get an equality only when the points $B'$ and $C'$ lie on the line segment $PQ$. If this is the case, then $\angle PQC + \angle QPB = 2(\angle PQM + \angle QPM) = 120^\circ$ and therefore $\angle APQ + \angle AQP = 240^\circ 
eq 120^\circ$, a contradiction.",299,1169,Geometry,11
190,shl_jbmo_2013_g6,shl_jbmo,2013,g,"Let $P$ and $Q$ be the midpoints of the sides $BC$ and $CD$, respectively, of a rectangle $ABCD$. Let $K$ and $M$ be the points of intersection of the line $PD$ with $QB$ and $QA$, respectively, and let $N$ be the point of intersection of the lines $PA$ and $QB$.

Let $X$, $Y$, $Z$ be the midpoints of the line segments $AN$, $KN$, $AM$, respectively. Let $\ell_1$ be the line passing through $X$ and perpendicular to $MK$, $\ell_2$ be the line passing through $Y$ and perpendicular to $AM$, $\ell_3$ be the line passing through $Z$ and perpendicular to $KN$. Show that $\ell_1$, $\ell_2$, $\ell_3$ are concurrent.","\textbf{Solution.} Let $R$ be the midpoint of the side $AD$. Then the lines $BR$ and $PD$ are parallel. Since $\angle MAN = \angle QAP = \angle QBR = \angle QKM$, the points $A$, $N$, $K$, $M$ are concyclic.

Let $\ell_4$ be the line passing through the midpoint $W$ of the line segment $MK$ and perpendicular to the line $AN$. Let $E_1$ be the point of intersection of $\ell_1$ and $\ell_4$, and $E_2$ be the point of intersection of $\ell_2$ and $\ell_3$. We will show that the points $E_1$ and $E_2$ coincide.

Let $O$ be the circumcenter of the cyclic quadrilateral $ANKM$. $OW$ is perpendicular to the side $MK$ and $OX$ is perpendicular to the side $AN$. Hence $OW$ is parallel to $\ell_1$, $OX$ is parallel to $\ell_4$, and $XOWE_1$ is a parallelogram. Therefore the midpoints of the line segments $OE_1$ and $WX$ coincide. Similarly, the midpoints of the line segments $OE_2$ and $YZ$ coincide.

On the other hand, as $X$, $Y$, $Z$, $W$ are midpoints of the sides of the quadrilateral $ANKM$, $XYWZ$ is a parallelogram and therefore the midpoints of the line segments $WX$ and $YZ$ coincide. Hence the midpoints of the line segments $OE_1$ and $OE_2$ coincide. In other words, $E_1$ and $E_2$ are the same point, and the lines $\ell_1$, $\ell_2$, $\ell_3$ are concurrent.

\textbf{Comment.} The problem can be asked in the following form:

Let $ANKM$ be a cyclic quadrilateral and let $X$, $Y$, $Z$ be the midpoints of the sides $AN$, $KN$, $AM$, respectively. Let $\ell_1$ be the line passing through $X$ and perpendicular to $MK$, $\ell_2$ be the line passing through $Y$ and perpendicular to $AM$, $\ell_3$ be the line passing through $Z$ and perpendicular to $KN$. Show that $\ell_1$, $\ell_2$, $\ell_3$ are concurrent.",615,1731,Geometry,12
191,shl_jbmo_2013_n1,shl_jbmo,2013,n,Find all positive integers $n$ for which $1^3 + 2^3 + \cdots + 16^3 + 17^n$ is a perfect square.,"\textbf{Solution.} We have $1^3 + 2^3 + \cdots + 16^3 = (1 + 2 + \cdots + 16)^2 = 8^2 \cdot 17^2$. Hence, if $1^3 + 2^3 + \cdots + 16^3 + 17^n = m^2$ for a positive integer $m$, then $17 \mid m$. If $m = 17k$ for some positive integer $k$, then $17^{n-2} = (k+8)(k-8)$. As $(k+8) - (k-8) = 16$, this can only happen when $k + 8 = 17^{n-2}$ and $k - 8 = 1$. Hence $k = 9$, and $n = 3$ is the only solution.",96,405,Number Theory,13
192,shl_jbmo_2013_n2,shl_jbmo,2013,n,"Find all ordered triples $(x, y, z)$ of integers satisfying $20^x + 13^y = 2013^z$.","\textbf{Solution.} We have $1^3 + 2^3 + \cdots + 16^3 = (1 + 2 + \cdots + 16)^2 = 8^2 \cdot 17^2$. Hence, if $1^3 + 2^3 + \cdots + 16^3 + 17^n = m^2$ for a positive integer $m$, then $17 \mid m$. If $m = 17k$ for some positive integer $k$, then $17^{n-2} = (k+8)(k-8)$. As $(k+8) - (k-8) = 16$, this can only happen when $k + 8 = 17^{n-2}$ and $k - 8 = 1$. Hence $k = 9$, and $n = 3$ is the only solution.


ewpage

\textbf{N2.} Find all ordered triples $(x, y, z)$ of integers satisfying $20^x + 13^y = 2013^z$.

\textbf{Solution.} As $20 \cdot 13 = 2^2 \cdot 5 \cdot 13$ and $2013 = 3 \cdot 11 \cdot 61$ are relatively prime, $x$, $y$ and $z$ must be nonnegative.

Considering the equation modulo 3, we observe that $x$ must be odd. Now considering the equation modulo 7, we obtain $(-1) + (-1)^y \equiv 4^z \pmod{7}$, which is impossible as the right hand side can only be 1, 2 and 4 modulo 7.

There are no solutions.",83,921,Number Theory,14
193,shl_jbmo_2013_n3,shl_jbmo,2013,n,"Find all ordered pairs $(a, b)$ of positive integers for which the numbers $\dfrac{a^3b - 1}{a + 1}$ and $\dfrac{b^3a + 1}{b - 1}$ are positive integers.","\textbf{Solution.} As $a^3b - 1 = b(a^3 + 1) - (b + 1)$ and $a + 1 \mid a^3 + 1$, we have $a + 1 \mid b + 1$.

As $b^3a + 1 = a(b^3 - 1) + (a + 1)$ and $b - 1 \mid b^3 - 1$, we have $b - 1 \mid a + 1$.

So $b - 1 \mid b + 1$ and hence $b - 1 \mid 2$.

\begin{itemize}
    \item If $b = 2$, then $a + 1 \mid b + 1 = 3$ gives $a = 2$. Hence $(a, b) = (2, 2)$ is the only solution in this case.
    \item If $b = 3$, then $a + 1 \mid b + 1 = 4$ gives $a = 1$ or $a = 3$. Hence $(a, b) = (1, 3)$ and $(3, 3)$ are the only solutions in this case.
\end{itemize}

To summarize, $(a, b) = (1, 3)$, $(2, 2)$ and $(3, 3)$ are the only solutions.",153,635,Number Theory,15
194,shl_jbmo_2013_n4,shl_jbmo,2013,n,"A rectangle in the $xy$-plane is called \textit{latticed} if all its vertices have integer coordinates.

\textbf{a.} Find a latticed rectangle with area 2013 whose sides are not parallel to the axes.

\textbf{b.} Show that if a latticed rectangle has area 2011, then its sides are parallel to the axes.","\textbf{Solution. a.} The rectangle $PQRS$ with $P(0, 0)$, $Q(165, 198)$, $R(159, 203)$, $S(-6, 5)$ has area 2013.

\textbf{b.} Suppose that the latticed rectangle $PQRS$ has area 2011 and its sides are not parallel to the axes. Without loss of generality we may assume that its vertices are $P(0, 0)$, $Q(a, b)$, $S(c, d)$, $R(a + c, b + d)$ where $a$, $b$, $c$, $d$ are integers and $abcd 
eq 0$.

Then $(a^2 + b^2)(c^2 + d^2) = 2011^2$. As 2011 is a prime, $2011 \mid a^2 + b^2$ or $2011 \mid c^2 + d^2$. Assume that the first one is the case. Since $2011 \equiv 3 \pmod{4}$, this can happen only if $a = 2011a_0$ and $b = 2011b_0$ for some integers $a_0$ and $b_0$. Now we have $(a_0^2 + b_0^2)(c^2 + d^2) = 1$. This means $c^2 + d^2 = 1$ and hence $c = 0$ or $d = 0$, contradicting $cd 
eq 0$.

\textbf{Comment.} The original problem proposal has the numbers 13 and 11 instead.",302,882,Number Theory,16
195,shl_jbmo_2013_n5,shl_jbmo,2013,n,"Find all ordered triples $(x, y, z)$ of positive integers satisfying the equation
\[
\frac{1}{x^2} + \frac{y}{xz} + \frac{1}{z^2} = \frac{1}{2013}\,.
\]","\textbf{Solution.} We have $x^2z^2 = 2013(x^2 + xyz + z^2)$. Let $d = \gcd(x, z)$ and $x = da$, $z = db$. Then $a^2b^2d^2 = 2013(a^2 + aby + b^2)$.

As $\gcd(a, b) = 1$, we also have $\gcd(a^2, a^2 + aby + b^2) = 1$ and $\gcd(b^2, a^2 + aby + b^2) = 1$. Therefore $a^2 \mid 2013$ and $b^2 \mid 2013$. But $2013 = 3 \cdot 11 \cdot 61$ is squarefree and therefore $a = 1 = b$.

Now we have $x = z = d$ and $d^2 = 2013(y + 2)$. Once again as 2013 is squarefree, we must have $y + 2 = 2013n^2$ where $n$ is a positive integer.

Hence $(x, y, z) = (2013n, 2013n^2 - 2, 2013n)$ where $n$ is a positive integer.",152,604,Number Theory,17
196,shl_jbmo_2013_n6,shl_jbmo,2013,n,"Find all ordered triples $(x, y, z)$ of integers satisfying the following system of equations:
\[
x^2 - y^2 = z
\]
\[
3xy + (x - y)z = z^2
\]","\textbf{Solution.} If $z = 0$, then $x = 0$ and $y = 0$, and $(x, y, z) = (0, 0, 0)$.

Let us assume that $z 
eq 0$, and $x + y = a$ and $x - y = b$ where $a$ and $b$ are nonzero integers such that $z = ab$. Then $x = (a + b)/2$ and $y = (a - b)/2$, and the second equation gives $3a^2 - 3b^2 + 4ab^2 = 4a^2b^2$.

Hence
\[
b^2 = \frac{3a^2}{4a^2 - 4a + 3}
\]
and
\[
3a^2 \geq 4a^2 - 4a + 3
\]
which is satisfied only if $a = 1$, 2 or 3.

\begin{itemize}
    \item If $a = 1$, then $b^2 = 1$. $(x, y, z) = (1, 0, 1)$ and $(0, 1, -1)$ are the only solutions in this case.
    \item If $a = 2$, then $b^2 = 12/11$. There are no solutions in this case.
    \item If $a = 3$, then $b^2 = 1$. $(x, y, z) = (1, 2, -3)$ and $(2, 1, 3)$ are the only solutions in this case.
\end{itemize}

To summarize, $(x, y, z) = (0, 0, 0)$, $(1, 0, 1)$, $(0, 1, -1)$, $(1, 2, -3)$ and $(2, 1, 3)$ are the only solutions.

\textbf{Comments. 1.} The original problem proposal asks for the solutions when $z = p$ is a prime number.

\textbf{2.} The problem can be asked with a single equation in the form:
\[
3xy + (x-y)^2(x+y) = (x^2 - y^2)^2
\]",141,1121,Number Theory,18
231,jbmo_2014_p1,jbmo,2014,n,"Find all distinct prime numbers $p$, $q$ and $r$ such that
\[
3p^4 - 5q^4 - 4r^2 = 26.
\]","\subsection*{Solution}

First notice that if both primes $q$ and $r$ differ from $3$, then
$q^2 \equiv r^2 \equiv 1 \pmod{3}$,
hence the left hand side of the given equation is congruent to zero modulo $3$, which is
impossible since $26$ is not divisible by $3$. Thus, $q = 3$ or $r = 3$. We consider two cases.

\textbf{Case 1.} $q = 3$.

The equation reduces to $(p^4 - r^2)(3^4 + 3r^2 + 1) = p^4 - r^4 \cdot 431 = 1$.

If $p 
eq 5$, by Fermat's little theorem, $p^4 \equiv 1 \pmod{5}$, which yields
$r^{2} \cdot 3 - 4 \equiv 1 \pmod{5}$, or equivalently, $r^2 + 2 \equiv 0 \pmod{5}$.
The last congruence is impossible in view of the fact that a residue of a square of a
positive integer belongs to the set $\{0, 1, 4\}$. Therefore $p = 5$ and $r = 19$.

\textbf{Case 2.} $r = 3$.

The equation becomes $(p^4 - 3^5)(q^2 \cdot 6 - 2 \cdot 2) = p^4 \cdot q$.

Obviously $p 
eq 5$. Hence, Fermat's little theorem gives $p^4 \equiv 1 \pmod{5}$. But then
$q^4 \cdot 5 \equiv 1 \pmod{5}$, which is impossible.

Hence, the only solution of the given equation is $p = 5$, $q = 3$, $r = 19$.",89,1084,Number Theory,1
232,jbmo_2014_p2,jbmo,2014,g,"Consider an acute triangle $ABC$ with area $S$. Let $CD \perp AB$ ($D \in AB$), $DM \perp AC$ ($M \in AC$) and $DN \perp BC$ ($N \in BC$). Denote by $H_1$ and $H_2$ the orthocenters of the triangles $MNC$ and $MND$ respectively. Find the area of the quadrilateral $AH_1BH_2$ in terms of $S$.","\subsection*{Solution 1}

Let $O$, $P$, $K$, $R$ and $T$ be the mid-points of the segments $CD$, $MN$, $CN$, $CH_1$
and $MH_1$, respectively. From $\triangle MNC$ we have that $PK = \tfrac{1}{2}MC$ and
$PK \parallel MC$. Analogously, from $\triangle MH_1C$ we have that $TR = \tfrac{1}{2}MC$ and
$TR \parallel MC$. Consequently, $PK = TR$ and $PK \parallel TR$. Also $OK \parallel DN$
(from $\triangle CDN$) and since $DN \perp BC$ and $MH_1 \perp BC$, it follows that
$TH_1 \parallel OK$. Since $O$ is the circumcenter of $\triangle CMN$, $OP \perp MN$.
Thus, $CH_1 \perp MN$ implies $OP \parallel CH_1$. We conclude
$\triangle TRH_1 \cong \triangle KPO$ (they have parallel sides and $TR = PK$), hence
$RH_1 = PO$, i.e.\ $CH_1 = PO \cdot 2$ and $CH_1 \parallel PO$.

Analogously, $DH_2 = PO \cdot 2$ and $DH_2 \parallel PO$. From $CH_1 = PO \cdot 2 = DH_2$
and $CH_1 \parallel PO \parallel DH_2$ the quadrilateral $CH_1H_2D$ is a parallelogram,
thus $H_1H_2 = CD$ and $H_1H_2 \parallel CD$. Therefore the area of the quadrilateral
$AH_1BH_2$ is
\[
\frac{AB \cdot H_1H_2}{2} = \frac{AB \cdot CD}{2} = S.
\]

\subsection*{Solution 2}

Since $MH_1 \parallel DN$ and $NH_1 \parallel DM$, $MDN H_1$ is a parallelogram. Similarly,
$NH_2 \parallel CM$ and $MH_2 \parallel CN$ imply $MCN H_2$ is a parallelogram. Let $P$ be
the midpoint of the segment $MN$. Then $P$ is the image of $D$ under the symmetry
$\sigma(H_1)$ and $P$ is the image of $C$ under the symmetry $\sigma(H_2)$, thus
$CD \parallel H_1H_2$ and $CD = H_1H_2$. From $CD \perp AB$ we deduce
\[
\frac{1}{2} \cdot AB \cdot CD = S \implies [AH_1BH_2] = S.
\]",291,1615,Geometry,2
233,jbmo_2014_p3,jbmo,2014,a,"Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that
\[
\left(a + \frac{1}{b}
ight)^2 + \left(b + \frac{1}{c}
ight)^2 + \left(c + \frac{1}{a}
ight)^2 \geq 3(a + b + c + 1).
\]
When does equality hold?","\subsection*{Solution 1}

By using AM-GM ($x^2 + y^2 + z^2 \geq xy + yz + zx$) we have
\[
\left(a+\frac{1}{b}
ight)^2+\left(b+\frac{1}{c}
ight)^2+\left(c+\frac{1}{a}
ight)^2
\geq
\left(a+\frac{1}{b}
ight)\!\left(b+\frac{1}{c}
ight)
+\left(b+\frac{1}{c}
ight)\!\left(c+\frac{1}{a}
ight)
+\left(c+\frac{1}{a}
ight)\!\left(a+\frac{1}{b}
ight)
\]
\[
= \left(ab + \frac{a}{c} + 1 + \frac{1}{bc}
ight)
+ \left(bc + \frac{b}{a} + 1 + \frac{1}{ca}
ight)
+ \left(ca + \frac{c}{b} + 1 + \frac{1}{ab}
ight).
\]

Notice that by AM-GM we have $\frac{b}{c} + a \geq 2$,\; $\frac{a}{b} + b \geq 2$,\;
and $\frac{c}{b} + c \geq 2$,\; and $\frac{a}{c} + \frac{b}{a} \geq 2$. Thus,
\[
\left(a+\frac{1}{b}
ight)^2+\left(b+\frac{1}{c}
ight)^2+\left(c+\frac{1}{a}
ight)^2
\geq ab + bc + ca + \frac{a}{c}+\frac{b}{a}+\frac{c}{b}
+ \frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab} + 3
\geq 3(1 + a + b + c).
\]

The equality holds if and only if $a = b = c = 1$.

\subsection*{Solution 2}

From QM-AM we obtain
\[
\frac{\left(a+\frac{1}{b}
ight)^2+\left(b+\frac{1}{c}
ight)^2+\left(c+\frac{1}{a}
ight)^2}{3}
\geq
\left(\frac{\left(a+\frac{1}{b}
ight)+\left(b+\frac{1}{c}
ight)+\left(c+\frac{1}{a}
ight)}{3}
ight)^2 \tag{1}
\]
\[
\implies
\left(a+\frac{1}{b}
ight)^2+\left(b+\frac{1}{c}
ight)^2+\left(c+\frac{1}{a}
ight)^2
\geq \frac{\left[\left(a+\frac{1}{b}
ight)+\left(b+\frac{1}{c}
ight)+\left(c+\frac{1}{a}
ight)
ight]^2}{3}.
\]

From AM-GM we have $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \geq 3\sqrt[3]{\dfrac{1}{abc}} = 3$,
and substituting in $(1)$ we get
\[
\left(a+\frac{1}{b}
ight)^2+\left(b+\frac{1}{c}
ight)^2+\left(c+\frac{1}{a}
ight)^2
\geq
\frac{(a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2}{3}
\geq
\frac{(a+b+c+3)^2}{3}
\]
\[
\geq
\frac{(a+b+c)^2 + 6(a+b+c) + 9}{3}
\geq
\frac{9 + 6(a+b+c) + 9}{3}
= \frac{18 + 6(a+b+c)}{3}
= 3(1 + a + b + c).
\]

The equality holds if and only if $a = b = c = 1$.

\subsection*{Solution 3}

By using $x^2 + y^2 + z^2 \geq xy + yz + zx$:
\begin{align*}
\left(a+\frac{1}{b}
ight)^2+\left(b+\frac{1}{c}
ight)^2+\left(c+\frac{1}{a}
ight)^2
&= a^2+b^2+c^2+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}
+ 2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}
ight) \\
&\geq ab+bc+ca+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}
+2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}
ight).
\end{align*}

Clearly
\[
\frac{a}{b}+\frac{b}{c}+\frac{c}{a} = abc \cdot \frac{a}{b}+\frac{b}{c}+\frac{c}{a} = a+b+c,
\]
\[
\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq a+b+c \quad (\text{by AM-GM: } ab+bc+ca \geq a+b+c),
\]
\[
\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq 3\sqrt[3]{\frac{abc}{bca}} = 3.
\]

Hence
\[
\left(a+\frac{1}{b}
ight)^2+\left(b+\frac{1}{c}
ight)^2+\left(c+\frac{1}{a}
ight)^2
\geq 2(a+b+c) + 2 \cdot abc \cdot \frac{a}{b}+\frac{b}{c}+\frac{c}{a}
+ 3 \geq 3(1+a+b+c).
\]

The equality holds if and only if $a = b = c = 1$.

\subsection*{Solution 4}

Let $x = a/y$, $y = bz$, $z = c/x$:
\[
\left(\frac{x}{z}+\frac{y}{x}+\frac{z}{y}
ight)^2 \geq 3\left(1+\frac{x}{y}+\frac{y}{z}+\frac{z}{x}
ight).
\]
It suffices to show
\[
(x^2z + x^2z + y^2x + y^2x + z^2y + z^2y)^2
\geq (xz + yx + zy + xyz)(x^2z + y^2x + z^2y + xyz)^2.
\]

This follows from four applications of AM-GM:
\begin{align*}
1)&\quad xy^2z + yz^2x + zx^2y \geq 3xyz\sqrt[3]{xy^2z \cdot yz^2x \cdot zx^2y} = 3x^2y^2z^2 \\
2)&\quad xz^4 + zx^4 + xy^3 \geq 3xz \cdot xy^3\\
3)&\quad xy^4 + yx^4 + yz^3 \geq 3xy \cdot yz^3\\
4)&\quad zy^4 + yz^4 + zx^3 \geq 3zy \cdot zx^3
\end{align*}

Equality holds when $x = y = z$, i.e., $a = b = c = 1$.

\subsection*{Solution 5}

\[
\sum_{\mathrm{cyc}} \left(a + \frac{1}{b}
ight)^2 \geq 2\sum_{\mathrm{cyc}}\left(a + \frac{1}{b}
ight) + 3
\iff
\sum_{\mathrm{cyc}}\left[\left(a+\frac{1}{b}
ight)^2 - 2\left(a+\frac{1}{b}
ight) - 1 - 0
ight] \geq 0.
\]
\[
\sum_{\mathrm{cyc}} \frac{a}{b} \geq 3 \cdot \frac{abc}{bca} = 3 \tag{1}
\]

\[
\left(a - \frac{3}{4}
ight)^2 \geq 0 \implies a^2 - \frac{3}{2}a + \frac{9}{16} \geq 0 \implies a^2 \geq \frac{3}{2}a - \frac{9}{16} \geq \frac{3}{4}a - \frac{9}{16}
\]
\[
\implies \frac{1}{a} - 1 \geq 0 \implies \frac{1}{a} \geq 1
\]
\[
\sum_{\mathrm{cyc}}\left[\left(a+\frac{1}{b}
ight)^2 - 2 - \frac{1}{1}
ight]
\geq 3\cdot\frac{1}{abc} - 9 - 15 - 6 = \frac{1}{1} - 9 = -6 \cdot \frac{1}{abc} \tag{2}
\]

Using $(1)$ and $(2)$ we obtain
\[
\sum_{\mathrm{cyc}}\left[\left(a+\frac{1}{b}
ight)^2 - 2\left(a+\frac{1}{b}
ight) - 1
ight]
\geq -6 + 6 = 0.
\]

Equality holds when $a = b = c = 1$.",218,4413,Algebra,3
234,jbmo_2014_p4,jbmo,2014,c,"For a positive integer $n$, two players $A$ and $B$ play the following game: Given a pile of $s$ stones, the players take turns alternatively with $A$ going first. On each turn the player is allowed to take either one stone, or a prime number of stones, or a positive multiple of $n$ stones. The winner is the one who takes the last stone. Assuming both $A$ and $B$ play perfectly, for how many values of $s$ the player $A$ cannot win?","Denote by $k$ the sought number and let $\{s_1, s_2, \ldots, s_k\}$ be the corresponding values
for $s$. We call each $s_i$ a \emph{losing number} and every other nonnegative integer a
\emph{winning number}.

Clearly every multiple of $n$ is a winning number.

Suppose there are two different losing numbers $s_i > s_j$, which are congruent modulo $n$.
Then, on his first turn of play, player $A$ may remove $s_i - s_j$ stones (since $n \mid s_i - s_j$),
leaving a pile with $s_j$ stones for $B$. This is in contradiction with both $s_i$ and $s_j$ being
losing numbers.

Hence, there are at most $n - 1$ losing numbers, i.e.\ $k \leq n - 1$.

Suppose there exists an integer $r \in \{1, 2, \ldots, n-1\}$, such that $mn + r$ is a winning number
for every $m \in \mathbb{N}_0$. Let us denote by $u$ the greatest losing number (if $k > 0$) or $0$
(if $k = 0$), and let $s = \mathrm{LCM}(u + 2, 3, \ldots, n+1)$. Note that all the numbers
$s + 2$, $s + 3$, \ldots, $s + u + n + 1$ are composite. Let $m' \in \mathbb{N}_0$ be such that
\[
s + u \leq m'n + r \leq s + u + n + 1.
\]
In order for $m'n + r$ to be a winning number, there
must exist an integer $p$, which is either one, or prime, or a positive multiple of $n$, such
that $m'n + r - p$ is a losing number or $0$, and hence less than or equal to $u$. Since
\[
s + 2 \leq m'n + r - p \leq s + u + n + 1,
\]
$p$ must be composite, hence $p$ is a
multiple of $n$ (say $p = qn$). But then $m'n + r - p = (m' - q)n + r$ must be a winning
number, according to our assumption. This contradicts our assumption that all numbers
$mn + r$, $m \in \mathbb{N}_0$ are winning.

Hence, each nonzero residue class modulo $n$ contains a losing number.

\textbf{There are exactly $n - 1$ losing numbers.}

\bigskip

\textbf{Lemma:} No pair $(u, n)$ of positive integers satisfies the following property:

$(*)$ There exists an arithmetic progression $(a_t)_{t=1}^{\infty}$ with difference $n$ such that each
segment $[a_i - u, a_i + u]$ contains a prime.

\textit{Proof of the lemma:} Suppose such a pair $(u, n)$ and a corresponding arithmetic
progression $(a_t)_{t=1}^{\infty}$ exist. There exist arbitrarily long patches of consecutive composites.
Take such a patch $P$ of length $3un$. Then, at least one segment $[a_i - u, a_i + u]$ is fully
contained in $P$, a contradiction.

Suppose such a nonzero residue class modulo $n$ exists (hence $n > 1$). Let $u \in \mathbb{N}$ be greater
than every losing number. Consider the members of the supposed residue class which are
greater than $u$. They form an arithmetic progression with the property $(*)$, a
contradiction (by the lemma).",435,2624,Combinatorics,4
154,shl_jbmo_2014_a1,shl_jbmo,2014,a,"\textit{For any real number $a$, let $\lfloor a 
floor$ denote the greatest integer not exceeding $a$. In positive real numbers solve the following equation}
\[
n + \left\lfloor \sqrt{n} 
ight
floor + \left\lfloor \sqrt[3]{n} 
ight
floor = 2014.
\]","
oindent\textbf{Solution 1.} Obviously $n$ must be a positive integer. Now note that $44^2 = 1936 < 2014 < 2025 = 45^2$ and $12^3 < 1900 < 2014 < 13^3$.

If $n < 1950$ then $2014 = n + \lfloor\sqrt{n}
floor + \lfloor\sqrt[3]{n}
floor < 1950 + 44 + 12 = 2006$, a contradiction!

So $n \geq 1950$. Also if $n > 2000$ then $2014 = n + \lfloor\sqrt{n}
floor + \lfloor\sqrt[3]{n}
floor > 2000 + 44 + 12 = 2056$, a contradiction!

So $1950 \leq n \leq 2000$, therefore $\lfloor\sqrt{n}
floor = 44$ and $\lfloor\sqrt[3]{n}
floor = 12$. Plugging that into the original equation we get:
\[
n + \lfloor\sqrt{n}
floor + \lfloor\sqrt[3]{n}
floor = n + 44 + 12 = 2014
\]
From which we get $n = 1958$, which is the only solution.

\medskip


oindent\textbf{Solution 2.} Obviously $n$ must be a positive integer. Since $n \leq 2014$, $\sqrt{n} < 45$ and $\sqrt[3]{n} < 13$.

Form $n = 2014 - \lfloor\sqrt{n}
floor - \lfloor\sqrt[3]{n}
floor > 2014 - 45 - 13 = 1956$, $\sqrt{n} > 44$ and $\sqrt[3]{n} > 12$, thus $\lfloor\sqrt{n}
floor = 44$ and $\lfloor\sqrt[3]{n}
floor = 12$ and $n = 2014 - \lfloor\sqrt{n}
floor - \lfloor\sqrt[3]{n}
floor = 2014 - 44 - 12 = 1958$.",248,1152,Algebra,1
155,shl_jbmo_2014_a2,shl_jbmo,2014,a,"\textit{Let $a$, $b$ and $c$ be positive real numbers such that $abc = \dfrac{1}{8}$. Prove the inequality}
\[
a^2 + b^2 + c^2 + a^2b^2 + b^2c^2 + c^2a^2 \geq \frac{15}{16}.
\]
\textit{When does equality hold?}","
oindent\textbf{Solution 1.} By using the Arithmetic-Geometric Mean Inequality for 15 positive numbers, we find that
\begin{align*}
&a^2 + b^2 + c^2 + a^2b^2 + b^2c^2 + c^2a^2 \\
&= \frac{a^2}{4}+\frac{a^2}{4}+\frac{a^2}{4}+\frac{a^2}{4}+\frac{b^2}{4}+\frac{b^2}{4}+\frac{b^2}{4}+\frac{b^2}{4}+\frac{c^2}{4}+\frac{c^2}{4}+\frac{c^2}{4}+\frac{c^2}{4}+a^2b^2+b^2c^2+c^2a^2 \\
&\geq 15\sqrt[15]{\frac{a^{12}b^{12}c^{12}}{4^{12}}} = 15\sqrt[15]{\left(\frac{abc}{4}
ight)^{4}} = 15\sqrt[15]{\left(\frac{1}{32}
ight)^4} \cdot \frac{1}{1} = \frac{15}{16}
\end{align*}
as desired. Equality holds if and only if $a = b = c = \dfrac{1}{2}$.

\medskip


oindent\textbf{Solution 2.} By using AM-GM we obtain
\begin{align*}
\left(a^2+b^2+c^2
ight)+\left(a^2b^2+b^2c^2+c^2a^2
ight) &\geq 3\sqrt[3]{a^2b^2c^2}+3\sqrt[3]{a^4b^4c^4}\\
&= 3\sqrt[3]{\left(\frac{1}{8}
ight)^2}+3\sqrt[3]{\left(\frac{1}{8}
ight)^4} = \frac{3}{4}+\frac{3}{16} = \frac{15}{16}.
\end{align*}
The equality holds when $a^2=b^2=c^2$, i.e.\ $a=b=c=\dfrac{1}{2}$.",210,1018,Algebra,2
156,shl_jbmo_2014_a3,shl_jbmo,2014,a,"\textit{Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that:}
\[
\left(a+\frac{1}{b}
ight)^2+\left(b+\frac{1}{c}
ight)^2+\left(c+\frac{1}{a}
ight)^2 \geq 3(a+b+c+1).
\]
\textit{When does equality hold?}","
oindent\textbf{Solution 1.} By using AM-GM ($x^2+y^2+z^2 \geq xy+yz+zx$) we have
\begin{align*}
\left(a+\frac{1}{b}
ight)^2+\left(b+\frac{1}{c}
ight)^2+\left(c+\frac{1}{a}
ight)^2 &\geq \left(a+\frac{1}{b}
ight)\left(b+\frac{1}{c}
ight)+\left(b+\frac{1}{c}
ight)\left(c+\frac{1}{a}
ight)+\left(c+\frac{1}{a}
ight)\left(a+\frac{1}{b}
ight)\\
&= \left(ab+1+\frac{a}{c}+\frac{1}{b}
ight) \cdot 1 + \left(bc+1+\frac{b}{a}+\frac{1}{c}
ight)\cdot 1\\
&\quad + \left(ca+1+\frac{c}{b}+\frac{1}{a}
ight)\cdot 1\\
&= ab+bc+ca+\frac{a}{c}+\frac{c}{b}+\frac{b}{a}+3+a+b+c.
\end{align*}
Notice that by AM-GM we have $ab+\dfrac{b}{a} \geq 2b$, $bc+\dfrac{c}{b} \geq 2c$, and $ca+\dfrac{a}{c} \geq 2a$.

Thus,
\[
\left(a+\frac{1}{b}
ight)^2+\left(b+\frac{1}{c}
ight)^2+\left(c+\frac{1}{a}
ight)^2 \geq \left(ab+\frac{b}{a}
ight)+\left(bc+\frac{c}{b}
ight)+\left(ca+\frac{a}{c}
ight)+3+a+b+c \geq 3(a+b+c+1).
\]
The equality holds if and only if $a=b=c=1$.

\medskip


oindent\textbf{Solution 2.} From QM-AM we obtain
\[
\sqrt{\frac{\left(a+\frac{1}{b}
ight)^2+\left(b+\frac{1}{c}
ight)^2+\left(c+\frac{1}{a}
ight)^2}{3}} \geq \frac{a+\frac{1}{b}+b+\frac{1}{c}+c+\frac{1}{a}}{3}
\]
\[
\Longleftrightarrow \left(a+\frac{1}{b}
ight)^2+\left(b+\frac{1}{c}
ight)^2+\left(c+\frac{1}{a}
ight)^2 \geq \frac{\left(a+\frac{1}{b}+b+\frac{1}{c}+c+\frac{1}{a}
ight)^2}{3} \tag{1}
\]
From AM-GM we have $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \geq 3\sqrt[3]{\dfrac{1}{abc}}=3$, and substituting in (1) we get
\begin{align*}
\left(a+\frac{1}{b}
ight)^2+\left(b+\frac{1}{c}
ight)^2+\left(c+\frac{1}{a}
ight)^2 &\geq \frac{\left(a+\frac{1}{b}+b+\frac{1}{c}+c+\frac{1}{a}
ight)^2}{3} \geq \frac{(a+b+c+3)^2}{3}\\
&= \frac{(a+b+c)(a+b+c)+6(a+b+c)+9}{3}\\
&\geq \frac{(a+b+c)\cdot 3\sqrt[3]{abc}+6(a+b+c)+9}{3}\\
&= \frac{9(a+b+c)+9}{3} = 3(a+b+c+1)
\end{align*}
The equality holds if and only if $a=b=c=1$.",217,1873,Algebra,3
157,shl_jbmo_2014_a4,shl_jbmo,2014,a,"\textit{Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that}
\[
\frac{7+2b}{1+a}+\frac{7+2c}{1+b}+\frac{7+2a}{1+c} \geq \frac{69}{4}.
\]
\textit{When does equality hold?}","
oindent\textbf{Solution 1.} The inequality can be written as:
\[
\frac{5+2(1+b)}{1+a}+\frac{5+2(1+c)}{1+b}+\frac{5+2(1+a)}{1+c} \geq \frac{69}{4}.
\]
We substitute $1+a=x, 1+b=y, 1+c=z$.

So, we have to prove the inequality
\[
\frac{5+2y}{x}+\frac{5+2z}{y}+\frac{5+2x}{z} \geq \frac{69}{4} \Longleftrightarrow 5\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}
ight)+2\left(\frac{y}{x}+\frac{z}{y}+\frac{x}{z}
ight) \geq \frac{69}{4}
\]
where $x,y,z>1$ are real numbers and $x+y+z=4$.

We have
\[
\frac{x+y+z}{3} \geq \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}} \Longleftrightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geq \frac{9}{x+y+z} \Longleftrightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geq \frac{9}{4}
\]
\[
\frac{y}{x}+\frac{z}{y}+\frac{x}{z} \geq 3\cdot\sqrt[3]{\frac{y}{x}\cdot\frac{z}{y}\cdot\frac{x}{z}} = 3
\]
Thus,
\[
\frac{5+2y}{x}+\frac{5+2z}{y}+\frac{5+2x}{z} = 5\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}
ight)+2\left(\frac{y}{x}+\frac{z}{y}+\frac{x}{z}
ight) \geq 5\cdot\frac{9}{4}+2\cdot 3 = \frac{69}{4}.
\]
The equality holds, when $\left(x=y=z,\, \frac{y}{x}=\frac{z}{y}=\frac{x}{z},\, x+y+z=4
ight)$, thus $x=y=z=\dfrac{4}{3}$, i.e.\ $a=b=c=\dfrac{1}{3}$.",187,1176,Algebra,4
158,shl_jbmo_2014_a5,shl_jbmo,2014,a,"\textit{Let $x,y,z$ be non-negative real numbers satisfying $x+y+z=xyz$. Prove that}
\[
2\left(x^2+y^2+z^2
ight) \geq 3(x+y+z),
\]
\textit{and determine when equality occurs.}","
oindent\textbf{Solution.} Equality holds when $x=y=z=0$.

Apply AM-GM to $x+y+z=xyz$,
\begin{align*}
xyz = x+y+z &\geq 3\sqrt[3]{xyz} \Rightarrow (xyz)^3 \geq \left(3\sqrt[3]{xyz}
ight)^3\\
&\Rightarrow x^3y^3z^3 \geq 27xyz\\
&\Rightarrow x^2y^2z^2 \geq 27\\
&\Rightarrow \sqrt[4]{x^2y^2z^2} \geq 3
\end{align*}
Also by AM-GM we have, $x^2+y^2+z^2 \geq 3\sqrt[3]{x^2y^2z^2} \geq 9$.

Therefore we get $x^2+y^2+z^2 \geq 9$.

Now,
\begin{align*}
2(x^2+y^2+z^2) \geq 3(x+y+z) &\Longleftrightarrow \frac{2(x^2+y^2+z^2)}{3} \geq (x+y+z)\\
&\Longleftrightarrow 2\cdot\frac{(x^2+y^2+z^2)}{3} \geq 2\cdot(x+y+z)\\
&\Longleftrightarrow \frac{4(x^2+y^2+z^2)}{3} \geq 2\cdot(x+y+z)\\
&\Longleftrightarrow x^2+y^2+z^2+\frac{(x^2+y^2+z^2)}{3} \geq 2\cdot(x+y+z)\\
&\Longleftrightarrow 3+\frac{x^2}{3}+3+\frac{y^2}{3}+3+\frac{z^2}{3} \geq 2(x+y+z)\\
&\Longleftrightarrow 2\sqrt{3\cdot\frac{x^2}{3}}+2\sqrt{3\cdot\frac{y^2}{3}}+2\sqrt{3\cdot\frac{z^2}{3}} \geq 2(x+y+z)
\end{align*}

Equality holds if $3 = \dfrac{x^2}{3} = \dfrac{y^2}{3} = \dfrac{z^2}{3}$, i.e.\ $x=y=z=3$, for which $x+y+z 
eq xyz$.

\medskip


oindent\textbf{Remark.} The inequality can be improved: $x^2+y^2+z^2 \geq \sqrt{3}(x+y+z)$.

\medskip


oindent\textbf{Solution (improved).} If one of the numbers is zero, then from $x+y+z=xyz$ all three numbers are zero and the equality trivially holds.

From AM-GM $x^2+y^2+z^2 \geq 3\sqrt[3]{x^2y^2z^2} = 3\dfrac{x+y+z}{\sqrt[3]{xyz}} \geq 3\dfrac{x+y+z}{\frac{x+y+z}{3}} = 9 \tag{1}$.

From QM-AM $\dfrac{x^2+y^2+z^2}{3} \geq \left(\dfrac{x+y+z}{3}
ight)^2 \tag{2}$.

Multiplying (1) and (2) we get $\dfrac{(x^2+y^2+z^2)^2}{3} \geq 9\cdot\dfrac{(x+y+z)^2}{9} = (x+y+z)^2$. By taking square root on both sides we deduce the stated inequality.

Equality holds only when $x=y=z=\sqrt{3}$ or $x=y=z=0$.",175,1802,Algebra,5
159,shl_jbmo_2014_a6,shl_jbmo,2014,a,"\textit{Let $a,b,c$ be positive real numbers. Prove that}
\[
\left((3a^2+1)^2+2\left(1+\frac{3}{b}
ight)^2
ight)\left((3b^2+1)^2+2\left(1+\frac{3}{c}
ight)^2
ight)\left((3c^2+1)^2+2\left(1+\frac{3}{a}
ight)^2
ight) \geq 48^3.
\]
\textit{When does equality hold?}","
oindent\textbf{Solution.} Let $x$ be a positive real number. By AM-GM we have $\dfrac{1+x+x+x}{4} \geq x^{3/4}$, or equivalently $1+3x \geq 4x^{3/4}$. Using this inequality we obtain:
\[
(3a^2+1)^2 \geq 16a^3 \quad \text{and} \quad 2\left(1+\frac{3}{b}
ight)^2 \geq 32b^{-3/2}.
\]
Moreover, by the inequality of arithmetic and geometric means we have
\[
f(a,b) = (3a^2+1)^2+2\left(1+\frac{3}{b}
ight)^2 \geq 16a^3+32b^{-3/2} = 16\left(a^3+b^{-3/2}+b^{-3/2}
ight) \geq 48\frac{a}{b}.
\]
Therefore, we obtain
\[
f(a,b)f(b,c)f(c,a) \geq 48\cdot\frac{a}{b}\cdot 48\cdot\frac{b}{c}\cdot 48\cdot\frac{c}{a} = 48^3.
\]
Equality holds only when $a=b=c=1$.",262,648,Algebra,6
160,shl_jbmo_2014_a7,shl_jbmo,2014,a,"\textit{Let $a,b,c$ be positive real numbers such that $a^2+b^2+c^2=48$. Prove}
\[
a^2\sqrt{2b^3+16}+b^2\sqrt{2c^3+16}+c^2\sqrt{2a^3+16} \leq 24^2.
\]
\textit{When does equality hold?}","
oindent\textbf{Solution.} Observe that $2x^3+16 = 2(x^3+8) = 2(x+2)(x^2-2x+4)$. From AM-GM:
\[
\sqrt{2x^3+16} = \sqrt{(2x+4)(x^2-2x+4)} \leq \frac{2x+4+x^2-2x+4}{2} = \frac{x^2+8}{2} \tag{1}.
\]
By adding the inequality (1) obtained for $x=a$, $x=b$ and $x=c$ it suffices to prove:
\[
a^2b^2+8a^2+b^2c^2+8b^2+c^2a^2+8c^2 \leq 2\cdot 24^2.
\]
Since $a^2b^2+b^2c^2+c^2a^2 \leq \dfrac{(a^2+b^2+c^2)^2}{3}$, using $a^2+b^2+c^2=48$, we get the stated inequality.

Equality holds only when $a=b=c=4$.",184,495,Algebra,7
161,shl_jbmo_2014_a8,shl_jbmo,2014,a,"\textit{Let $x$, $y$ and $z$ be positive real numbers such that $xyz=1$. Prove the inequality}
\[
\frac{1}{x(ay+b)}+\frac{1}{y(az+b)}+\frac{1}{z(ax+b)} \geq 3, \quad \text{if:}
\]
\begin{enumerate}[label=\alph*)]
  \item $a=0$ and $b=1$;
  \item $a=1$ and $b=0$;
  \item $a+b=1$ for $a,b>0$
\end{enumerate}
\textit{When does the equality hold true?}

\medskip


oindent\textbf{Remark.} The problem can be reformulated:

\textit{Let $a$, $b$, $x$, $y$ and $z$ be nonnegative real numbers such that $xyz=1$ and $a+b=1$. Prove the inequality}
\[
\frac{1}{x(ay+b)}+\frac{1}{y(az+b)}+\frac{1}{z(ax+b)} \geq 3.
\]
\textit{When does the equality hold true?}","
oindent\textbf{Solution.}


oindent a) The inequality reduces to $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} \geq 3$, which follows directly from the AM-GM inequality.

Equality holds only when $x=y=z=1$.


oindent b) Here the inequality reduces to $\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx} \geq 3$, i.e.\ $x+y+z \geq 3$, which also follows from the AM-GM inequality.

Equality holds only when $x=y=z=1$.


oindent c) Let $m$, $n$ and $p$ be such that $x=\dfrac{m}{n}$, $y=\dfrac{n}{p}$ and $z=\dfrac{p}{m}$. The inequality reduces to
\[
\frac{np}{amn+bmp}+\frac{pm}{anp+bnm}+\frac{mn}{apm+bpn} \geq 3 \tag{1}.
\]
By substituting $u=np$, $v=pm$ and $w=mn$, (1) becomes
\[
\frac{u}{aw+bv}+\frac{v}{au+bw}+\frac{w}{av+bu} \geq 3.
\]
The last inequality is equivalent to
\[
\frac{u^2}{auw+buv}+\frac{v^2}{auv+bvw}+\frac{w^2}{avw+buw} \geq 3.
\]
Cauchy-Schwarz Inequality implies
\[
\frac{u^2}{auw+buv}+\frac{v^2}{auv+bvw}+\frac{w^2}{avw+buw} \geq \frac{(u+v+w)^2}{auw+buv+auv+bvw+avw+buw} = \frac{(u+v+w)^2}{uw+vu+wv}.
\]
Thus, the problem simplifies to $(u+v+w)^2 \geq 3(uw+vu+wv)$, which is equivalent to $(u-v)^2+(v-w)^2+(w-u)^2 \geq 0$.

Equality holds only when $u=v=w$, that is only for $x=y=z=1$.",650,1198,Algebra,8
162,shl_jbmo_2014_a9,shl_jbmo,2014,a,"\textit{Let $n$ be a positive integer, and let $x_1,\ldots,x_n,y_1,\ldots,y_n$ be positive real numbers such that $x_1+\ldots+x_n = y_1+\ldots+y_n = 1$. Show that}
\[
|x_1-y_1|+\ldots+|x_n-y_n| \leq 2 - \min_{1\leq i\leq n}\frac{x_i}{y_i} - \min_{1\leq i\leq n}\frac{y_i}{x_i}.
\]","
oindent\textbf{Solution.} Up to reordering the real numbers $x_i$ and $y_i$, we may assume that $\dfrac{x_1}{y_1} \leq \ldots \leq \dfrac{x_n}{y_n}$. Let $A = \dfrac{x_1}{y_1}$ and $B = \dfrac{x_n}{y_n}$, and $S = |x_1-y_1|+\ldots+|x_n-y_n|$. Our aim is to prove that $S \leq 2-A-\dfrac{1}{B}$.

First, note that we cannot have $A>1$, since that would imply $x_i > y_i$ for all $i \leq n$, hence $x_1+\ldots+x_n > y_1+\ldots+y_n$. Similarly, we cannot have $B<1$, since that would imply $x_i < y_i$ for all $i \leq n$, hence $x_1+\ldots+x_n < y_1+\ldots+y_n$.

If $n=1$, then $x_1=y_1=A=B=1$ and $S=0$, hence $S \leq 2-A-\dfrac{1}{B}$. For $n \geq 2$ let $1 \leq k < n$ be some integer such that $\dfrac{x_k}{y_k} \leq 1 \leq \dfrac{x_{k+1}}{y_{k+1}}$. We define the positive real numbers $X_1 = x_1+\ldots+x_k$, $X_2 = x_{k+1}+\ldots+x_n$, $Y_1 = y_1+\ldots+y_k$, $Y_2 = y_{k+1}+\ldots+y_n$. Note that $Y_1 \geq X_1 \geq AY_1$ and $Y_2 \leq X_2 \leq BY_2$.

Thus, $A \leq \dfrac{X_1}{Y_1} \leq 1 \leq \dfrac{X_2}{Y_2} \leq B$. In addition, $S = Y_1-X_1+X_2-Y_2$.

From $0 < X_2, Y_1 \leq 1$, $0 \leq Y_1-X_1$ and $0 \leq X_2-Y_2$, follows
\[
S = Y_1-X_1+X_2-Y_2 = \frac{Y_1-X_1}{Y_1}+\frac{X_2-Y_2}{X_2} = 2-\frac{X_1}{Y_1}-\frac{Y_2}{X_2} \leq 2-A-\frac{1}{B}.
\]",280,1266,Algebra,9
163,shl_jbmo_2014_c1,shl_jbmo,2014,c,"\textit{Several (at least two) segments are drawn on a board. Select two of them, and let $a$ and $b$ be their lengths. Delete the selected segments and draw a segment of length $\dfrac{ab}{a+b}$. Continue this procedure until only one segment remains on the board. Prove:}

\begin{enumerate}[label=\alph*)]
  \item \textit{the length of the last remaining segment does not depend on the order of the deletions.}
  \item \textit{for every positive integer $n$, the initial segments on the board can be chosen with distinct integer lengths, such that the last remaining segment has length $n$.}
\end{enumerate}","
oindent\textbf{Solution.} a) Observe that $\dfrac{1}{\frac{ab}{a+b}} = \dfrac{1}{a}+\dfrac{1}{b}$. Thus, if the lengths of the initial segments on the board were $a_1, a_2, \ldots, a_n$, and $c$ is the length of the last remaining segment, then
\[
\frac{1}{c} = \frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_n},
\]
proving a).

b) From a) and the equation $\dfrac{1}{n} = \dfrac{1}{2n}+\dfrac{1}{3n}+\dfrac{1}{6n}$ it follows that if the lengths of the starting segments are $2n$, $3n$ and $6n$, then the length of the last remaining segment is $n$.",609,550,Combinatorics,10
164,shl_jbmo_2014_c2,shl_jbmo,2014,c,"\textit{In a country with $n$ cities, all direct airlines are two-way. There are $r>2014$ routes between pairs of different cities that include no more than one intermediate stop (the direction of each route matters). Find the least possible $n$ and the least possible $r$ for that value of $n$.}","
oindent\textbf{Solution.} Denote by $X_1, X_2, \ldots, X_n$ the cities in the country and let $X_i$ be connected to exactly $m_i$ other cities by direct two-way airline. Then $X_i$ is a final destination of $m_i$ direct routes and an intermediate stop of $m_i(m_i-1)$ non-direct routes. Thus $r = m_1^2+\ldots+m_n^2$. As each $m_i$ is at most $n-1$ and $13\cdot12^2 < 2014$, we deduce $n \geq 14$.

Consider $n=14$. As each route appears in two opposite directions, $r$ is even, so $r \geq 2016$. We can achieve $r=2016$ by arranging the 14 cities uniformly on a circle and connect (by direct two-way airlines) all of them, except the diametrically opposite pairs. This way, there are exactly $14\cdot12^2 = 2016$ routes.",296,722,Combinatorics,11
165,shl_jbmo_2014_c3,shl_jbmo,2014,c,"\textit{For a given positive integer $n$, two players A and B play the following game: Given is a pile of $a$ stones. The players take turn alternatively with A going first. On each turn the player is allowed to take one stone, a prime number of stones, or a multiple of $n$ stones. The winner is the one who takes the last stone. Assuming perfect play, find the number of values for $a$, for which A cannot win.}","
oindent\textbf{Solution.} Denote by $k$ the sought number and let $\{a_1, a_2, \ldots, a_k\}$ be the corresponding values for $a$. We will call each $a_i$ a losing number and every other positive integer a winning number. Clearly every multiple of $n$ is a winning number.

Suppose there are two different losing numbers $a_i > a_j$, which are congruent modulo $n$. Then, on his first turn of play, the player A may remove $a_i-a_j$ stones (since $n \mid a_i-a_j$), leaving a pile with $a_j$ stones for B. This is in contradiction with both $a_i$ and $a_j$ being losing numbers. Therefore there are at most $n-1$ losing numbers, i.e.\ $k \leq n-1$.

Suppose there exists an integer $r \in \{1,2,\ldots,n-1\}$, such that $mn+r$ is a winning number for every $m \in \mathbb{N}_0$. Let us denote by $u$ the greatest losing number (if $k>0$) or $0$ (if $k=0$), and let $s = \mathrm{lcm}(2,3,\ldots,u+n+1)$. Note that all the numbers $s+2$, $s+3$, \ldots, $s+u+n+1$ are composite. Let $m' \in \mathbb{N}_0$, be such that $s+u+2 \leq m'n+r \leq s+u+n+1$. In order for $m'n+r$ to be a winning number, there must exist an integer $p$, which is either one, or prime, or a positive multiple of $n$, such that $m'n+r-p$ is a losing number or $0$, and hence lesser than or equal to $u$. Since $s+2 \leq m'n+r-u \leq p \leq m'n+r \leq s+u+n+1$, $p$ must be a composite, hence $p$ is a multiple of $n$ (say $p=qn$). But then $m'n+r-p = (m'-q)n+r$ must be a winning number, according to our assumption. This contradicts our assumption that all numbers $mn+r$, $m \in \mathbb{N}_0$ are winning.

Hence there are exactly $n-1$ losing numbers (one for each residue $r \in \{1,2,\ldots,n-1\}$).",413,1676,Combinatorics,12
166,shl_jbmo_2014_c4,shl_jbmo,2014,c,\textit{Let $A = 1\cdot4\cdot7\cdots2014$ be the product of the numbers less or equal to $2014$ that give remainder $1$ when divided by $3$. Find the last non-zero digit of $A$.},"
oindent\textbf{Solution.} Grouping the elements of the product by ten we get:
\begin{align*}
&(30k+1)(30k+4)(30k+7)(30k+10)(30k+13)(30k+16)\\
&(30k+19)(30k+22)(30k+25)(30k+28) =\\
&= (30k+1)(15k+2)(30k+7)(120k+40)(30k+13)(15k+8)\\
&\quad (30k+19)(15k+11)(120k+100)(15k+14)
\end{align*}
(We divide all even numbers not divisible by five, by two and multiply all numbers divisible by five with four.)

We denote $P_k = (30k+1)(15k+2)(30k+7)(30k+13)(15k+8)(30k+19)(15k+11)(15k+14)$. For all the numbers not divisible by five, only the last digit affects the solution, since the power of two in the numbers divisible by five is greater than the power of five. Considering this, for even $k$, $P_k$ ends with the same digit as $1\cdot2\cdot7\cdot3\cdot8\cdot9\cdot1\cdot4$, i.e.\ six and for odd $k$, $P_k$ ends with the same digit as $1\cdot7\cdot7\cdot3\cdot3\cdot9\cdot6\cdot9$, i.e.\ six. Thus $P_0P_1\ldots P_{66}$ ends with six. If we remove one zero from the end of all numbers divisible with five, we get that the last nonzero digit of the given product is the same as the one from $6\cdot2011\cdot2014\cdot4\cdot10\cdot16\cdots796\cdot802$. Considering that $4\cdot6\cdot2\cdot8$ ends with four and removing one zero from every fifth number we get that the last nonzero digit is the same as in $4\cdot4^{26}\cdot784\cdot796\cdot802\cdot1\cdot4\cdots76\cdot79$. Repeating the process we did for the starting sequence we conclude that the last nonzero number will be the same as in $2\cdot6\cdot6\cdot40\cdot100\cdot160\cdot220\cdot280\cdot61\cdot32\cdot67\cdot73\cdot38\cdot79$, which is two.",178,1596,Combinatorics,13
167,shl_jbmo_2014_g1,shl_jbmo,2014,g,\textit{Let $ABC$ be a triangle with $\angle B = \angle C = 40^\circ$. The bisector of $\angle B$ meets $AC$ at the point $D$. Prove that $\overline{BD}+\overline{DA}=\overline{BC}$.},"
oindent\textbf{Solution.} Since $\angle BAC = 100^\circ$ and $\angle BDC = 120^\circ$ we have $\overline{BD} < \overline{BC}$. Let $E$ be the point on $\overline{BC}$ such that $\overline{BD}=\overline{BE}$. Then $\angle DEC = 100^\circ$ and $\angle EDC = 40^\circ$, hence $\overline{DE}=\overline{EC}$, and $\angle BAC + \angle DEB = 180^\circ$. So $A, B, E$ and $D$ are concyclic, implying $\overline{AD}=\overline{DE}$ (since $\angle ABD = \angle DBC = 20^\circ$), which completes the proof.",183,495,Geometry,14
168,shl_jbmo_2014_g2,shl_jbmo,2014,g,"\textit{Let $ABC$ be an acute triangle with $\overline{AB} < \overline{AC} < \overline{BC}$ and $c(O,R)$ be its circumcircle. Denote with $D$ and $E$ be the points diametrically opposite to the points $B$ and $C$, respectively. The circle $c_1\!\left(A,\overline{AE}
ight)$ intersects $\overline{AC}$ at point $K$, the circle $c_2\!\left(A,\overline{AD}
ight)$ intersects $BA$ at point $L$ ($A$ lies between $B$ and $L$). Prove that the lines $EK$ and $DL$ meet on the circle $c$.}","
oindent\textbf{Solution.} Let $M$ be the point of intersection of the line $DL$ with the circle $c(O,R)$ (we choose $M \equiv D$ if $LD$ is tangent to $c$ and $M$ to be the second intersecting point otherwise). It is sufficient to prove that the points $E$, $K$ and $M$ are collinear.

We have that $\angle EAC = 90^\circ$ (since $EC$ is diameter of the circle $c$). The triangle $AEK$ is right-angled and isosceles ($\overline{AE}$ and $\overline{AK}$ are radii of the circle $c_1$). Therefore
\[
\angle AEK = \angle AKE = 45^\circ.
\]
Similarly, we obtain that $\angle BAD = 90^\circ = \angle DAL$. Since $\overline{AD}=\overline{AL}$ the triangle $ADL$ is right-angled and isosceles, we have
\[
\angle ADL = \angle ALD = 45^\circ.
\]
If $M$ is between $D$ and $L$, then $\angle ADM = \angle AEM$, because they are inscribed in the circle $c(O,R)$ and they correspond to the same arc $\widehat{AM}$. Hence $\angle AEK = \angle AEM = 45^\circ$ i.e.\ the points $E,K,M$ are collinear.

If $D$ is between $M$ and $L$, then $\angle ADM + \angle AEM = 180^\circ$ as opposite angles in cyclic quadrilateral. Hence $\angle AEK = \angle AEM = 45^\circ$ i.e.\ the points $E,K,M$ are collinear.",481,1187,Geometry,15
169,shl_jbmo_2014_g3,shl_jbmo,2014,g,"\textit{Let $CD \perp AB$ ($D \in AB$), $DM \perp AC$ ($M \in AC$) and $DN \perp BC$ ($N \in BC$) for an acute triangle $ABC$ with area $S$. If $H_1$ and $H_2$ are the orthocentres of the triangles $MNC$ and $MND$ respectively. Evaluate the area of the quadrilateral $AH_1BH_2$.}","
oindent\textbf{Solution 1.} Let $O$, $P$, $K$, $R$ and $T$ be the midpoints of the segments $CD$, $MN$, $CN$, $CH_1$ and $MH_1$, respectively. From $\triangle MNC$ we have that $\overline{PK} = \frac{1}{2}\overline{MC}$ and $PK \parallel MC$. Analogously, from $\triangle MH_1C$ we have that $\overline{TR} = \frac{1}{2}\overline{MC}$ and $TR \parallel MC$. Consequently, $\overline{PK} = \overline{TR}$ and $PK \parallel TR$. Also $OK \parallel DN$ (from $\triangle CDN$) and since $DN \perp BC$ and $MH_1 \perp BC$, it follows that $TH_1 \parallel OK$. Since $O$ is the circumcenter of $\triangle CMN$, $OP \perp MN$. Thus, $CH_1 \perp MN$ implies $OP \parallel CH_1$. We conclude $\triangle TRH_1 \cong \triangle KPO$ (they have parallel sides and $\overline{TR} = \overline{PK}$), hence $\overline{RH_1} = \overline{PO}$, i.e.\ $\overline{CH_1} = 2\overline{PO}$ and $CH_1 \parallel PO$.

Analogously, $\overline{DH_2} = 2\overline{PO}$ and $DH_2 \parallel PO$. From $\overline{CH_1} = 2\overline{PO} = \overline{DH_2}$ and $CH_1 \parallel PO \parallel DH_2$ the quadrilateral $CH_1H_2D$ is a parallelogram, thus $\overline{H_1H_2} = \overline{CD}$ and $H_1H_2 \parallel CD$. Therefore the area of the quadrilateral $AH_1BH_2$ is $\dfrac{\overline{AB}\cdot\overline{H_1H_2}}{2} = \dfrac{\overline{AB}\cdot\overline{CD}}{2} = S$.

\medskip


oindent\textbf{Solution 2.} Since $MH_1 \parallel DN$ and $NH_1 \parallel DM$, $MDNH_1$ is a parallelogram. Similarly, $NH_2 \parallel CM$ and $MH_2 \parallel CN$ imply $MCNH_2$ is a parallelogram. Let $P$ be the midpoint of the segment $\overline{MN}$. Then $\sigma_P(D) = H_1$ and $\sigma_P(C) = H_2$, thus $CD \parallel H_1H_2$ and $\overline{CD} = \overline{H_1H_2}$. From $CD \perp AB$ we deduce $A_{AH_1BH_2} = \dfrac{1}{2}\overline{AB}\cdot\overline{CD} = S$.",279,1812,Geometry,16
170,shl_jbmo_2014_g4,shl_jbmo,2014,g,"\textit{Let $ABC$ be a triangle such that $\overline{AB} 
eq \overline{AC}$. Let $M$ be a midpoint of $\overline{BC}$, $H$ the orthocenter of $ABC$, $O_1$ the midpoint of $\overline{AH}$ and $O_2$ the circumcenter of $BCH$. Prove that $O_1AMO_2$ is a parallelogram.}","
oindent\textbf{Solution 1.} Let $O_2'$ be the point such that $O_1AMO_2'$ is a parallelogram. Note that $\overrightarrow{MO_2'} = \overrightarrow{AO_1} = \overrightarrow{O_1H}$. Therefore, $O_1HO_2'M$ is a parallelogram and $\overrightarrow{MO_1} = \overrightarrow{O_2'H}$.

Since $M$ is the midpoint of $\overline{BC}$ and $O_1$ is the midpoint of $\overline{AH}$, it follows that $4\overrightarrow{MO_1} = \overrightarrow{BA}+\overrightarrow{BH}+\overrightarrow{CA}+\overrightarrow{CH} = 2(\overrightarrow{CA}+\overrightarrow{BH})$. Moreover, let $B'$ be the midpoint of $\overline{BH}$. Then,
\begin{align*}
2\overrightarrow{O_2'B'}\cdot\overrightarrow{BH} &= (\overrightarrow{O_2'H}+\overrightarrow{O_2'B})\cdot\overrightarrow{BH} = (2\overrightarrow{O_2'H}+\overrightarrow{HB})\cdot\overrightarrow{BH}\\
&= (2\overrightarrow{MO_1}+\overrightarrow{HB})\cdot\overrightarrow{BH} = (\overrightarrow{CA}+\overrightarrow{BH}+\overrightarrow{HB})\cdot\overrightarrow{BH} = \overrightarrow{CA}\cdot\overrightarrow{BH} = 0.
\end{align*}
By $\vec{a}\cdot\vec{b}$ we denote the inner product of the vectors $\vec{a}$ and $\vec{b}$.

Therefore, $O_2'$ lies on the perpendicular bisector of $\overline{BH}$. Since $B$ and $C$ play symmetric roles, $O_2'$ also lies on the perpendicular bisector of $\overline{CH}$, hence $O_2'$ is the circumcenter of $\triangle BCH$ and $O_2 = O_2'$.

\medskip


oindent\textbf{Note:} The condition $\overline{AB} 
eq \overline{AC}$ just aims at ensuring that the parallelogram $O_1ANO_2$ is not degenerate, hence at helping students to focus on the ``general'' case.

\medskip


oindent\textbf{Solution 2.} We use the following two well-known facts:
\begin{align}
\sigma_{BC}(H) &\text{ lies on the circumcircle of } \triangle ABC. \tag{1}\\
\overrightarrow{AH} &= -2\overrightarrow{MO}, \text{ where } O \text{ is the circumcenter of } \triangle ABC. \tag{2}
\end{align}
The statement ``$O_1AMO_2$ is parallelogram'' is equivalent to ``$\sigma_{BC}(O_2)=O$''. The latter is true because the circumcircles of $\triangle ABC$ and $\triangle BCH$ are symmetrical with respect to $BC$, from (1).",266,2120,Geometry,17
171,shl_jbmo_2014_g5,shl_jbmo,2014,g,"\textit{Let $ABC$ be a triangle with $\overline{AB} 
eq \overline{BC}$, and let $BD$ be the internal bisector of $\angle ABC$ ($D \in AC$). Denote the midpoint of the arc $AC$ which contains point $B$ by $M$. The circumcircle of the triangle $BDM$ intersects the segment $AB$ at point $K 
eq B$, and let $J$ be the reflection of $A$ with respect to $K$. If $DJ \cap AM = \{O\}$, prove that the points $J$, $B$, $M$, $O$ belong to the same circle.}","
oindent\textbf{Solution 1.}

Let the circumcircle of the triangle $BDM$ intersect the line segment $BC$ at point $L 
eq B$. From $\angle CBD = \angle DBA$ we have $\overline{DL} = \overline{DK}$. Since $\angle LCM = \angle BCM = \angle BAM = \angle KAM$, $\overline{MC} = \overline{MA}$ and
\begin{align*}
\angle LMC &= \angle LMK - \angle CMK = \angle LBK - \angle CMK = \angle CBA - \angle CMK\\
&= \angle CMA - \angle CMK = \angle KMA,
\end{align*}
it follows that triangles $MLC$ and $MKA$ are congruent, which implies $\overline{CL} = \overline{AK} = \overline{KJ}$.

Furthermore, $\angle CLD = 180^\circ - \angle BLD = \angle DKB = \angle DKJ$ and $\overline{DL} = \overline{DK}$, it follows that triangles $DCL$ and $DJK$ are congruent. Hence, $\angle DCL = \angle DJK = \angle BJO$. Then
\[
\angle BJO + \angle BMO = \angle DCL + \angle BMA = \angle BCA + 180^\circ - \angle BCA = 180^\circ
\]
so the points $J, B, M, O$ belong to the same circle, q.e.d.

\medskip


oindent\textbf{Solution 2.}

Since $\overline{MC} = \overline{MA}$ and $\angle CMA = \angle CBA$, we have $\angle ACM = \angle CAM = 90^\circ - \dfrac{\angle CBA}{2}$. It follows that
\[
\angle MBD = \angle MBA + \angle ABD = \angle ACM + \angle ABD = 90^\circ - \frac{\angle CBA}{2} + \frac{\angle CBA}{2} = 90^\circ.
\]
Denote the midpoint of $\overline{AC}$ by $N$. Since $\angle DNM = \angle CNM = 90^\circ$, $N$ belongs to the circumcircle of the triangle $BDM$. Since $NK$ is the midline of the triangle $ACJ$ and $NK \parallel CJ$, we have
\[
\angle BJC = \angle BKN = 180^\circ - \angle NDB = \angle CDB.
\]
Hence, the quadrilateral $CDJB$ is cyclic (this can also be obtained from the power of a point theorem, because $\overline{AN}\cdot\overline{AD} = \overline{AK}\cdot\overline{AB}$ implies $\overline{AC}\cdot\overline{AD} = \overline{AJ}\cdot\overline{AB}$), and
\[
\angle BJO = \angle 180^\circ - \angle BJD = \angle BCD = \angle BCA = 180^\circ - \angle BMA = 180^\circ - \angle BMO,
\]
so the points $J, B, M, O$ belong to the same circle, q.e.d.

\medskip


oindent\textbf{Remark.} If $J$ is between $A$ and $K$ the solution can be easily adapted.",447,2142,Geometry,18
172,shl_jbmo_2014_g6,shl_jbmo,2014,g,"\textit{Let $ABCD$ be a quadrilateral whose sides $AB$ and $CD$ are not parallel, and let $O$ be the intersection of its diagonals. Denote with $H_1$ and $H_2$ the orthocenters of the triangles $OAB$ and $OCD$, respectively. If $M$ and $N$ are the midpoints of the segments $\overline{AB}$ and $\overline{CD}$, respectively, prove that the lines $MN$ and $H_1H_2$ are parallel if and only if $\overline{AC} = \overline{BD}$.}","
oindent\textbf{Solution.}

Let $A'$ and $B'$ be the feet of the altitudes drawn from $A$ and $B$ respectively in the triangle $AOB$, and $C'$ and $D'$ are the feet of the altitudes drawn from $C$ and $D$ in the triangle $COD$. Obviously, $A'$ and $D'$ belong to the circle $c_1$ of diameter $\overline{AD}$, while $B'$ and $C'$ belong to the circle $c_2$ of diameter $\overline{BC}$.

It is easy to see that triangles $H_1AB$ and $H_1B'A'$ are similar. It follows that $\overline{H_1A}\cdot\overline{H_1A'} = \overline{H_1B}\cdot\overline{H_1B'}$. (Alternatively, one could notice that the quadrilateral $ABA'B'$ is cyclic and obtain the previous relation by writing the power of $H_1$ with respect to its circumcircle.) It follows that $H_1$ has the same power with respect to circles $c_1$ and $c_2$. Thus, $H_1$ (and similarly, $H_2$) is on the radical axis of the two circles.

The radical axis being perpendicular to the line joining the centers of the two circles, one concludes that $H_1H_2$ is perpendicular to $PQ$, where $P$ and $Q$ are the midpoints of the sides $\overline{AD}$ and $\overline{BC}$, respectively. ($P$ and $Q$ are the centers of circles $c_1$ and $c_2$.)

The condition $H_1H_2 \parallel MN$ is equivalent to $MN \perp PQ$. As $MPNQ$ is a parallelogram, we conclude that $H_1H_2 \parallel MN \Longleftrightarrow MN \perp PQ \Longleftrightarrow MPNQ$ a rhombus $\Longleftrightarrow \overline{MP} = \overline{MQ} \Longleftrightarrow \overline{AC} = \overline{BD}$.",425,1491,Geometry,19
173,shl_jbmo_2014_n1,shl_jbmo,2014,n,"\textit{Each letter of the word OHRID corresponds to a different digit belonging to the set $\{1,2,3,4,5\}$. Decipher the equality $(O+H+R+I+D)^2:(O-H-R+I+D)=O^{H^{R^{I^D}}}$.}","
oindent\textbf{Solution.} Since $O$, $H$, $R$, $I$ and $D$ are distinct numbers from $\{1,2,3,4,5\}$, we have $O+H+R+I+D=15$ and $O-H-R+I+D = O+H+R+I+D-2(H+R)<15$. From this
\[
O^{H^{R^{I^D}}} = \frac{(O+H+R+I+D)^2}{O-H-R+I+D} = \frac{225}{15-2(H+R)},
\]
hence $O^{H^{R^{I^D}}} > 15$ and divides $225$, which is only possible for $O^{H^{R^{I^D}}} = 25$ (must be a power of three or five). This implies that $O=5$, $H=2$ and $R=1$. It's easy to check that both $I=3$, $D=4$ and $I=4$, $D=3$ satisfy the stated equation.",176,519,Number Theory,20
174,shl_jbmo_2014_n2,shl_jbmo,2014,n,"\textit{Find all triples $(p,q,r)$ of distinct primes $p$, $q$ and $r$ such that}
\[
3p^4 - 5q^4 - 4r^2 = 26.
\]","
oindent\textbf{Solution.} First notice that if both primes $q$ and $r$ differ from 3, then $q^2 \equiv r^2 \equiv 1\pmod{3}$, hence the left hand side of the given equation is congruent to zero modulo 3, which is impossible since 26 is not divisible by 3. Thus, $q=3$ or $r=3$. We consider two cases.

\medskip


oindent\textbf{Case 1.} $q=3$.

The equation reduces to $3p^4-4r^2 = 431 \tag{1}$.

If $p 
eq 5$, by Fermat's little theorem, $p^4 \equiv 1\pmod{5}$, which yields $3-4r^2 \equiv 1\pmod{5}$, or equivalently, $r^2+2 \equiv 0\pmod{5}$. The last congruence is impossible in view of the fact that a residue of a square of a positive integer belongs to the set $\{0,1,4\}$. Therefore $p=5$ and $r=19$.

\medskip


oindent\textbf{Case 2.} $r=3$.

The equation becomes $3p^4-5q^4 = 62 \tag{2}$.

Obviously $p 
eq 5$. Hence, Fermat's little theorem gives $p^4 \equiv 1\pmod{5}$. But then $5q^4 \equiv 1\pmod{5}$, which is impossible.

Hence, the only solution of the given equation is $p=5$, $q=3$, $r=19$.",112,1011,Number Theory,21
175,shl_jbmo_2014_n3,shl_jbmo,2014,n,"\textit{Find the integer solutions of the equation}
\[
x^2 = y^2(x+y^4+2y^2).
\]","
oindent\textbf{Solution.} If $x=0$, then $y=0$ and conversely, if $y=0$, then $x=0$. It follows that $(x,y)=(0,0)$ is a solution of the problem. Assume $x 
eq 0$ and $y 
eq 0$ satisfy the equation. The equation can be transformed in the form $x^2-xy^2 = y^6+2y^4$. Then $4x^2-4xy^2+y^4 = 4y^6+9y^4$ and consequently
\[
\left(\frac{2x}{y^2}-1
ight)^2 = 4y^2+9 \tag{1}.
\]
Obviously $\dfrac{2x}{y^2}-1$ is integer. From (1), we get that the numbers $\dfrac{2x}{y^2}-1$, $2y$ and $3$ are Pythagorean triplets. It follows that $\dfrac{2x}{y^2}-1 = \pm 5$ and $2y = \pm 4$. Therefore, $x = 3y^2$ or $x=-2y^2$ and $y=\pm 2$. Hence $(x,y)=(12,-2)$, $(x,y)=(12,2)$, $(x,y)=(-8,-2)$ and $(x,y)=(-8,2)$ are the possible solutions. By substituting them in the initial equation we verify that all the 4 pairs are solutions. Thus, together with the couple $(x,y)=(0,0)$ the problem has 5 solutions.",80,886,Number Theory,22
176,shl_jbmo_2014_n4,shl_jbmo,2014,n,"\textit{Prove there are no integers $a$ and $b$ satisfying the following conditions:}
\begin{enumerate}[label=
oman*)]
  \item $16a-9b$ is a prime number
  \item $ab$ is a perfect square
  \item $a+b$ is a perfect square
\end{enumerate}","
oindent\textbf{Solution.} Suppose $a$ and $b$ be integers satisfying the given conditions. Let $p$ be a prime number, $n$ and $m$ be integers. Then we can write the conditions as follows:
\begin{align}
16a-9b &= p \tag{1}\\
ab &= n^2 \tag{2}\\
a+b &= m^2 \tag{3}
\end{align}
Moreover, let $d = \gcd(a,b)$ and $a=dx$, $b=dy$ for some relatively prime integers $x$ and $y$. Obviously $a 
eq 0$ and $b 
eq 0$, $a$ and $b$ are positive (by (2) and (3)).

From (2) follows that $x$ and $y$ are perfect squares, say $x=l^2$ and $y=s^2$.

From (1), $d \mid p$ and hence $d=p$ or $d=1$. If $d=p$, then $16x-9y=1$, and we obtain $x=9k+4$, $y=16k+7$ for some nonnegative integer $k$. But then $s^2 = y \equiv 3\pmod{4}$, which is a contradiction.

If $d=1$ then $16l^2-9s^2 = p \Rightarrow (4l-3s)(4l+3s) = p \Rightarrow (4l+3s=p \wedge 4l-3s=1)$.

By adding the last two equations we get $8l=p+1$ and by subtracting them we get $6s=p-1$. Therefore $p=24t+7$ for some integer $t$ and $a=(3t+1)^2$ and $b=(4t+1)^2$ satisfy the conditions (1) and (2). By (3) we have $m^2=(3t+1)^2+(4t+1)^2=25t^2+14t+2$, or equivalently $25m^2=(25t+7)^2+1$.

Since the difference between two nonzero perfect squares cannot be 1, we have a contradiction. As a result there is no solution.",236,1259,Number Theory,23
177,shl_jbmo_2014_n5,shl_jbmo,2014,n,"\textit{Find all nonnegative integers $x$, $y$, $z$ such that}
\[
2013^x + 2014^y = 2015^z.
\]","
oindent\textbf{Solution.} Clearly, $y>0$, and $z>0$. If $x=0$ and $y=1$, then $z=1$ and $(x,y,z)=(0,1,1)$ is a solution. If $x=0$ and $y \geq 2$, then modulo 4 we have $1+0 \equiv (-1)^z$, hence $z$ is even ($z=2z_1$ for some integer $z_1$). Then $2^y\cdot1007^y = (2015^{z_1}-1)(2015^{z_1}+1)$, and since $\gcd(1007,2015^{z_1}+1)=1$ we obtain $2\cdot1007^y \mid 2015^{z_1}-1$ and $2015^{z_1}+1 \mid 2^{y-1}$. From this we get $2015^{z_1}+1 \leq 2^{y-1} < 2\cdot1007^y \leq 2015^{z_1}-1$, which is impossible.

Now for $x>0$, modulo 3 we get $0+1 \equiv (-1)^z$, hence $z$ must be even ($z=2z_1$ for some integer $z_1$). Modulo 2014 we get $(-1)^x+0 \equiv 1$, thus $x$ must be even ($x=2x_1$ for some integer $x_1$). We transform the equation to $2^y\cdot1007^y = (2015^{z_1}-2013^{x_1})(2015^{z_1}+2013^{x_1})$ and since $\gcd(2015^{z_1}-2013^{x_1}, 2015^{z_1}+2013^{x_1})=2$, $1007^y$ divides $2015^{z_1}-2013^{x_1}$ or $2015^{z_1}+2013^{x_1}$ but not both. If $1007^y \mid 2015^{z_1}-2013^{x_1}$, then $2015^{z_1}+2013^{x_1} \leq 2^y < 1007^y \leq 2015^{z_1}-2013^{x_1}$, which is impossible. Hence $1007 \mid 2015^{z_1}+2013^{x_1}$, and from $2015^{z_1}+2013^{x_1} \equiv 1+(-1)^{x_1}\pmod{1007}$, $x_1$ is odd ($x = 2x_1 = 4x_2+2$ for some integer $x_2$).

Now modulo 5 we get $-1+(-1)^y \equiv (-2)^{4x_2+2}+(-1)^y \equiv 0$, hence $y$ must be even ($y=2y_1$ for some integer $y_1$). Finally modulo 31, we have $(-2)^{4x_2+2}+(-1)^{2y_1} \equiv 0$ or $4^{2x_2+1} \equiv -1$. This is impossible since the remainders of the powers of 4 modulo 31 are 1, 2, 4, 8 and 16.",94,1574,Number Theory,24
178,shl_jbmo_2014_n6,shl_jbmo,2014,n,"\textit{Vukasin, Dimitrije, Dusan, Stefan and Filip asked their professor to guess a three consecutive positive integer numbers after they had told him these (true) sentences:}

\medskip

\textit{Vukasin: ``Sum of the digits of one of them is a prime number. Sum of the digits of some of the other two is an even perfect number ($n$ is perfect if $\sigma(n)=2n$). Sum of the digits of the remaining number is equal to the number of its positive divisors.''}

\medskip

\textit{Dimitrije: ``Each of these three numbers has no more than two digits $1$ in its decimal representation.''}

\medskip

\textit{Dusan: ``If we add $11$ to one of them, we obtain a square of an integer.''}

\medskip

\textit{Stefan: ``Each of them has exactly one prime divisor less than $10$.''}

\medskip

\textit{Filip: ``The $3$ numbers are square-free.''}

\medskip

\textit{Their professor gave the correct answer. Which numbers did he say?}","
oindent\textbf{Solution.} Let the middle number be $n$, so the numbers are $n-1$, $n$ and $n+1$. Since 4 does not divide any of them, $n \equiv 2\pmod{4}$. Furthermore, neither 3, 5 nor 7 divides $n$. Also $n+1+11 \equiv 2\pmod{4}$ cannot be a square. Then 3 must divide $n-1$ or $n+1$. If $n-1+11$ is a square, then $3\mid n+1$ which implies $3\mid n+10$ (a square), so $9\mid n+10$ hence $9\mid n+1$, which is impossible. Thus must be $n+11=m^2$.

Further, 7 does not divide $n-1$, nor $n+1$, because $1+11 \equiv 5\pmod{7}$ and $-1+11 \equiv 3\pmod{7}$ are quadratic nonresidues modulo 7. This implies $5\mid n-1$ or $5\mid n+1$. Again, since $n+11$ is a square, it is impossible $5\mid n-1$, hence $5\mid n+1$ which implies $3\mid n-1$. This yields $n \equiv 4\pmod{10}$ hence $S(n+1)=S(n)+1=S(n-1)+2$ ($S(n)$ is sum of the digits of $n$). Since the three numbers are square-free, their numbers of positive divisors are powers of 2. Thus, we have two even sums of digits -- they must be $S(n-1)$ and $S(n+1)$, so $S(n)$ is prime. From $3\mid n-1$, follows $S(n-1)$ is an even perfect number, and $S(n+1) = 2^p$. Consequently $S(n) = 2^p-1$ is a prime, so $p$ is a prime number. One easily verifies $p 
eq 2$, so $p$ is odd implying $3\mid 2^p-2$. Then
\[
\sigma(2^p-2) \geq (2^p-2)\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{6}
ight) = 2(2^p-2).
\]
Since this number is perfect, $\dfrac{2^p-2}{6}$ must be one, i.e.\ $p=3$ and $S(n-1)=6$, $S(n)=7$ and $S(n+1)=8$.

Since 4 does not divide $n$, the 2-digit ending of $n$ must be 14 or 34. But $n=34$ is impossible, since $n+11=45$ is not a square. Hence, $n=10^a+10^b+14$ with $a \geq b \geq 2$. If $a 
eq b$, then $n$ has three digits 1 in its decimal representation, which is impossible. Therefore $a=b$, and $n=2\cdot10^a+14$. Now, $2\cdot10^a+25=m^2$, hence $5\mid m$, say $m=5t$, and $(t-1)(t+1) = 2^{a+1}5^{a-2}$.

Because $\gcd(t-1,t+1)=2$ there are three possibilities:
\begin{enumerate}
  \item $t-1=2$, $t+1=2^a5^{a-2}$, which implies $a=2$, $t=3$;
  \item $t-1=2^a$, $t+1=2\cdot5^{a-2}$, so $2^a+2=2\cdot5^{a-2}$, which implies $a=3$, $t=9$;
  \item $t-1=2\cdot5^{a-2}$, $t+1=2^a$, so $2\cdot5^{a-2}+2=2^a$, which implies $a=2$, $t=3$, same as case 1).
\end{enumerate}
From the only two possibilities $(n-1,n,n+1)=(213,214,215)$ and $(n-1,n,n+1)=(2013,2014,2015)$ the first one is not possible, because $S(215)=8$ and $\tau(215)=4$. By checking the conditions, we conclude that the latter is a solution, so the professor said the numbers: \textbf{2013, 2014, 2015}.",921,2527,Number Theory,25
235,jbmo_2015_p1,jbmo,2015,n,"Find all prime numbers $a$, $b$, $c$ and positive integers $k$ satisfying the equation
\[
a^2 + b^2 + 16c^2 + 9k^2 = 1.
\]","
oindent\textbf{Solution:}

The relation $9 \cdot k^2 + 1 \equiv 1 \pmod{3}$ implies
$$a^2 + b^2 + 16 \cdot c^2 \equiv 1 \pmod{3} \iff a^2 + b^2 + c^2 \equiv 1 \pmod{3}.$$
Since $a^2 \equiv 0, 1 \pmod{3}$, $b^2 \equiv 0, 1 \pmod{3}$, $c^2 \equiv 0, 1 \pmod{3}$, we have:

\bigskip

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline
$a^2$ & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\
\hline
$b^2$ & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 \\
\hline
$c^2$ & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\
\hline
$a^2+b^2+c^2$ & 0 & 1 & 1 & 2 & 1 & 2 & 2 & 0 \\
\hline
\end{tabular}
\end{center}

\bigskip

From the previous table it follows that two of three prime numbers $a$, $b$, $c$ are equal to 3.


oindent\textbf{Case 1.} $a = b = 3$

We have
\begin{align*}
a^2 + b^2 + 16 \cdot c^2 = 9 \cdot k^2 + 1 &\iff 9 \cdot k^2 - 16 \cdot c^2 = 17 \iff (3k - 4c)(3k + 4c) = 17, \\
&\iff \begin{cases} 3k - 4c = 1, \\ 3k + 4c = 17, \end{cases} \iff \begin{cases} c = 2, \\ k = 3, \end{cases}
\end{align*}
and $(a, b, c, k) = (3, 3, 2, 3)$.


oindent\textbf{Case 2.} $c = 3$

If $(3, b_0, c, k)$ is a solution of the given equation, then $(b_0, 3, c, k)$ is a solution too.

Let $a = 3$. We have
$$a^2 + b^2 + 16 \cdot c^2 = 9 \cdot k^2 + 1 \iff 9 \cdot k^2 - b^2 = 152 \iff (3k - b)(3k + b) = 152.$$
Both factors shall have the same parity and we obtain only 2 cases:
\begin{itemize}
    \item $\begin{cases} 3k - b = 2, \\ 3k + b = 76, \end{cases} \iff \begin{cases} b = 37, \\ k = 13, \end{cases}$ and $(a, b, c, k) = (3, 37, 3, 13)$;
    \item $\begin{cases} 3k - b = 4, \\ 3k + b = 38, \end{cases} \iff \begin{cases} b = 17, \\ k = 7, \end{cases}$ and $(a, b, c, k) = (3, 17, 3, 7)$.
\end{itemize}

So, the given equation has 5 solutions:
$$\{(37,3,3,13),\ (17,3,3,7),\ (3,37,3,13),\ (3,17,3,7),\ (3,3,2,3)\}.$$",122,1785,Number Theory,1
236,jbmo_2015_p2,jbmo,2015,a,"Let $a$, $b$, $c$ be positive real numbers such that $a + b + c = 3$. Find the minimum value of the expression
\[
A = \frac{2 + a^3}{2 + b^3} + \frac{2 + b^3}{2 + c^3} + \frac{2 + c^3}{2 + a^3}.
\]","
oindent\textbf{Solution:}

We can rewrite $A$ as follows:
\begin{align*}
A &= \frac{2 - a^3}{a} + \frac{2 - b^3}{b} + \frac{2 - c^3}{c} = 2\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}
ight) - a^2 - b^2 - c^2 \\
&= 2\left(\frac{ab + bc + ca}{abc}
ight) - (a^2 + b^2 + c^2) = 2\left(\frac{ab + bc + ca}{abc}
ight) - ((a+b+c)^2 - 2(ab+bc+ca)) \\
&= 2\left(\frac{ab + bc + ca}{abc}
ight) - (9 - 2(ab+bc+ca)) = 2\left(\frac{ab+bc+ca}{abc}
ight) + 2(ab+bc+ca) - 9 \\
&= 2(ab + bc + ca)\left(\frac{1}{abc} + 1
ight) - 9.
\end{align*}

Recall now the well-known inequality $(x + y + z)^2 \geq 3(xy + yz + zx)$ and set $x = ab$, $y = bc$, $z = ca$, to obtain
$$(ab + bc + ca)^2 \geq 3abc(a + b + c) = 9abc,$$
where we have used $a + b + c = 3$. By taking the square roots on both sides of the last one we obtain:
$$ab + bc + ca \geq 3\sqrt{abc}. \tag{1}$$
Also by using AM-GM inequality we get that
$$\frac{1}{abc} + 1 \geq 2\sqrt{\frac{1}{abc}}. \tag{2}$$
Multiplication of (1) and (2) gives:
$$(ab + bc + ca)\left(\frac{1}{abc} + 1
ight) \geq 3\sqrt{abc} \cdot 2\sqrt{\frac{1}{abc}} = 6.$$
So $A \geq 2 \cdot 6 - 9 = 3$ and the equality holds if and only if $a = b = c = 1$, so the minimum value is 3.",197,1196,Algebra,2
237,jbmo_2015_p3,jbmo,2015,g,"Let $ABC$ be an acute triangle. The lines $\ell_1$ and $\ell_2$ are perpendicular to $AB$ at the points $A$ and $B$, respectively. The perpendicular lines from the midpoint $M$ of $AB$ to the lines $AC$ and $BC$ intersect $\ell_1$ and $\ell_2$ at the points $E$ and $F$, respectively. If $D$ is the intersection point of the lines $EF$ and $MC$, prove that
\[
\angle ADB = \angle EMF.
\]","
oindent\textbf{Solution:}

Let $H$, $G$ be the points of intersection of $ME$, $MF$ with $AC$, $BC$ respectively. From the similarity of triangles $\triangle MHA$ and $\triangle MAE$ we get $\dfrac{MH}{MA} = \dfrac{MA}{ME}$, thus
$$MA^2 = MH \cdot ME. \tag{1}$$
Similarly, from the similarity of triangles $\triangle MBG$ and $\triangle MFB$ we get $\dfrac{MB}{MF} = \dfrac{MG}{MB}$, thus
$$MB^2 = MF \cdot MG. \tag{2}$$
Since $MA = MB$, from (1), (2), we conclude that the points $E$, $H$, $G$, $F$ are concyclic.

Therefore, we get that $\angle FEH = \angle FEM = \angle HGM$. Also, the quadrilateral $CHMG$ is cyclic, so $\angle CMH = \angle HGC$. We have
$$\angle FEH + \angle CMH = \angle HGM + \angle HGC = 90°,$$
thus $CM \perp EF$. Now, from the cyclic quadrilaterals $FDMB$ and $EAMD$, we get that $\angle DFM = \angle DBM$ and $\angle DEM = \angle DAM$. Therefore, the triangles $\triangle EMF$ and $\triangle ADB$ are similar, so $\angle ADB = \angle EMF$.",387,968,Geometry,3
238,jbmo_2015_p4,jbmo,2015,c,"An L-shape is one of the following four pieces, each consisting of three unit squares. A $5 \times 5$ board, consisting of $25$ unit squares, a positive integer $k \leq 25$ and an unlimited supply of L-shapes are given. Two players, $A$ and $B$, play the following game: starting with $A$ they alternatively mark a previously unmarked unit square until they mark a total of $k$ unit squares. We say that a placement of L-shapes on unmarked unit squares is called \emph{good} if the L-shapes do not overlap and each of them covers exactly three unmarked unit squares of the board. $B$ wins if every good placement of L-shapes leaves uncovered at least three unmarked unit squares. Determine the minimum value of $k$ for which $B$ has a winning strategy.","
oindent\textbf{Solution:}

We will show that player $\boldsymbol{A}$ wins if $k = 1, 2, 3$, but player $\boldsymbol{B}$ wins if $k = 4$. Thus the smallest $k$ for which $\boldsymbol{B}$ has a winning strategy exists and is equal to 4.

If $k = 1$, player $\boldsymbol{A}$ marks the upper left corner of the square and then fills it as follows.

If $k = 2$, player $\boldsymbol{A}$ marks the upper left corner of the square. Whatever square player $\boldsymbol{B}$ marks, then player $\boldsymbol{A}$ can fill in the square in exactly the same pattern as above except that he doesn't put the $L$-figure which covers the marked square of $\boldsymbol{B}$. Player $\boldsymbol{A}$ wins because he has left only two unmarked squares uncovered.

For $k = 3$, player $\boldsymbol{A}$ wins by following the same strategy. When he has to mark a square for the second time, he marks any yet unmarked square of the $L$-figure that covers the marked square of $\boldsymbol{B}$.

Let us now show that for $k = 4$ player $\boldsymbol{B}$ has a winning strategy. Since there will be 21 unmarked squares, player $\boldsymbol{A}$ will need to cover all of them with seven $L$-figures. We can assume that in his first move, player $\boldsymbol{A}$ does not mark any square in the bottom two rows of the chessboard (otherwise just rotate the chessboard). In his first move player $\boldsymbol{B}$ marks the square labeled 1 in the following figure.

If player $\boldsymbol{A}$ in his next move does not mark any of the squares labeled 2, 3 and 4 then player $\boldsymbol{B}$ marks the square labeled 3. Player $\boldsymbol{B}$ wins as the square labeled 2 is left unmarked but cannot be covered with an $L$-figure.

If player $\boldsymbol{A}$ in his next move marks the square labeled 2, then player $\boldsymbol{B}$ marks the square labeled 5. Player $\boldsymbol{B}$ wins as the square labeled 3 is left unmarked but cannot be covered with an $L$-figure.

Finally, if player $\boldsymbol{A}$ in his next move marks one of the squares labeled 3 or 4, player $\boldsymbol{B}$ marks the other of these two squares. Player $\boldsymbol{B}$ wins as the square labeled 2 is left unmarked but cannot be covered with an $L$-figure.

Since we have covered all possible cases, player $\boldsymbol{B}$ wins when $k = 4$.",752,2294,Combinatorics,4
134,shl_jbmo_2015_a1,shl_jbmo,2015,a,"Let $x$, $y$, $z$ be real numbers, satisfying the relations
\[
\begin{cases}
x \geq 20,\\
y \geq 40,\\
z \geq 1675,\\
x + y + z = 2015.
\end{cases}
\]
Find the greatest value of the product $P = x \cdot y \cdot z$.","\subsubsection*{Solution 1:}
By virtue of $z \geq 1675$ we have
\[
y + z < 2015 \iff y < 2015 - z \leq 2015 - 1675 < 1675.
\]
It follows that $(1675 - y)\cdot(1675 - z) \leq 0 \iff y \cdot z \leq 1675 \cdot (y + z - 1675)$.

By using the inequality $u \cdot v \leq \left(\dfrac{u+v}{2}
ight)^2$ for all real numbers $u$, $v$ we obtain
\[
P = x \cdot y \cdot z \leq 1675 \cdot x \cdot (y + z - 1675) \leq 1675 \cdot \left(\frac{x + y + z - 1675}{2}
ight)^2 =
\]
\[
1675 \cdot \left(\frac{2015 - 1675}{2}
ight)^2 = 1675 \cdot 170^2 = 48407500.
\]

We have $P = x \cdot y \cdot z = 48407500 \iff
\begin{cases} x + y + z = 2015,\\ z = 1675,\\ x = y + z - 1675. \end{cases}
\iff
\begin{cases} x = 170,\\ y = 170\\ z = 1675. \end{cases}$

So, the greatest value of the product is $P = x \cdot y \cdot z = 48407500$.

\subsubsection*{Solution 2:}
Let $S = \{(x,y,z)\mid x\geq 20, y\geq 40, z\geq 1675, x+y+z=2015\}$ and $\Pi = \{x\cdot y\cdot z \mid (x,y,z)\in S\}$.
We have to find the biggest element of $\Pi$. By using the given inequalities we obtain:
\[
\begin{cases}
20 \leq x \leq 300,\\
40 \leq y \leq 320,\\
1675 \leq z \leq 1955,\\
y < 1000 < z
\end{cases}
\]
Let $z = 1675 + d$. Since $x \leq 300$ so $(1675+d)\cdot x = 1675x + dx \leq 1675x + 1675d = 1675\cdot(x+d)$.
That means that if $(x, y, 1675+d) \in S$ then $(x+d, y, 1675)\in S$, and
$x \cdot y \cdot (1675+d) \leq (x+d)\cdot y\cdot 1675$. Therefore $z=1675$ must be for the greatest product.

Furthermore, $x\cdot y \leq \left(\dfrac{x+y}{2}
ight)^2 = \left(\dfrac{2015-1675}{2}
ight)^2 = \left(\dfrac{340}{2}
ight)^2 = 170^2$. Since $(170,170,1675)\in S$

that means that the biggest element of $\Pi$ is $170\cdot 170\cdot 1675 = 48407500$.",214,1705,Algebra,1
135,shl_jbmo_2015_a2,shl_jbmo,2015,a,"3) If $x^3 - 3\sqrt{3}\,x^2 + 9x - 3\sqrt{3} - 64 = 0$, find the value of $x^6 - 8x^5 + 13x^4 - 5x^3 + 49x^2 - 137x + 2015$.","\subsubsection*{Solution}
$x^3 - 3\sqrt{3}\,x^2 + 9x - 3\sqrt{3} - 64 = 0 \iff (x-\sqrt{3})^3 = 64 \iff (x-\sqrt{3}) = 4 \iff x - 4 = \sqrt{3} \iff x^2 - 8x + 16 = 3 \iff$
$x^2 - 8x + 13 = 0$
$x^6 - 8x^5 + 13x^4 - 5x^3 + 49x^2 - 137x + 2015 = (x^2-8x+13)(x^4-5x+9)+1898 = 0 + 1898 = 1898$",124,288,Algebra,2
136,shl_jbmo_2015_a3,shl_jbmo,2015,a,"Let $a, b, c$ be positive real numbers. Prove that
\[
\frac{a}{b} + \sqrt{\frac{b}{c}} + \sqrt[3]{\frac{c}{a}} > 2.
\]","\subsubsection*{Solution:}
Starting from the double expression on the left-hand side of given inequality, and applying twice the Arithmetic-Geometric mean inequality, we find that
\begin{align*}
2\frac{a}{b} + 2\sqrt{\frac{b}{c}} + 2\sqrt[3]{\frac{c}{a}}
&= \frac{a}{b} + \left(\frac{a}{b} + \sqrt{\frac{b}{c}} + \sqrt{\frac{b}{c}}
ight) + 2\sqrt[3]{\frac{c}{a}}\\
&\geq \frac{a}{b} + 3\sqrt[3]{\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{b}{c}} + 2\sqrt[3]{\frac{c}{a}}\\
&= \frac{a}{b} + 3\sqrt[3]{\frac{a}{c}} + 2\sqrt[3]{\frac{c}{a}}\\
&= \frac{a}{b} + \sqrt[3]{\frac{a}{c}} + 2\left(\sqrt[3]{\frac{a}{c}} + \sqrt[3]{\frac{c}{a}}
ight)\\
&\geq \frac{a}{b} + \sqrt[3]{\frac{a}{c}} + 2 \cdot 2\sqrt{\sqrt[3]{\frac{a}{c}}\cdot\sqrt[3]{\frac{c}{a}}}\\
&= \frac{a}{b} + \sqrt[3]{\frac{a}{c}} + 4\\
&> 4,
\end{align*}
which yields the given inequality.",118,844,Algebra,3
137,shl_jbmo_2015_a4,shl_jbmo,2015,a,"Let $a, b, c$ be positive real numbers such that $a + b + c = 3$. Find the minimum value of
\[
A = \frac{2 - a^3}{a} + \frac{2 - b^3}{b} + \frac{2 - c^3}{c}.
\]","\subsubsection*{Solution:}
We rewrite $A$ as follows:
\begin{align*}
A &= \frac{2-a^3}{a} + \frac{2-b^3}{b} + \frac{2-c^3}{c}
= 2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}
ight) - a^2 - b^2 - c^2 =\\
&2\left(\frac{ab+bc+ca}{abc}
ight) - (a^2+b^2+c^2)
= 2\left(\frac{ab+bc+ca}{abc}
ight) - ((a+b+c)^2 - 2(ab+bc+ca)) =\\
&2\left(\frac{ab+bc+ca}{abc}
ight) - (9 - 2(ab+bc+ca))
= 2\left(\frac{ab+bc+ca}{abc}
ight) + 2(ab+bc+ca) - 9 =\\
&2(ab+bc+ca)\left(\frac{1}{abc}+1
ight)-9
\end{align*}

Recall now the well-known inequality $(x+y+z)^2 \geq 3(xy+yz+zx)$ and set
$x=ab, y=bc, z=ca$, to obtain $(ab+bc+ca)^2 \geq 3abc(a+b+c) = 9abc$ where we have used
$a+b+c=3$. By taking the square roots on both sides of the last one we obtain:
\[
ab+bc+ca \geq 3\sqrt{abc}. \quad (1)
\]
Also by using AM-GM inequality we get that
\[
\frac{1}{abc}+1 \geq 2\sqrt{\frac{1}{abc}}. \quad (2)
\]
Multiplication of (1) and (2) gives:
\[
(ab+bc+ca)\left(\frac{1}{abc}+1
ight) \geq 3\sqrt{abc}\cdot 2\sqrt{\frac{1}{abc}} = 6.
\]
So $A \geq 2\cdot 6 - 9 = 3$ and the equality holds if and only if $a=b=c=1$, so the minimum value is 3.

\textbf{Remark:} Note that if $f(x) = \dfrac{2-x^3}{x}$, $x\in(0,3)$ then $f''(x) = \dfrac{4}{x^3}-2$, so the function is
convex on $x\in\left(0,\sqrt[3]{2}
ight)$ and concave on $x\in\left(\sqrt[3]{2},3
ight)$. This means that we cannot apply Jensen's
inequality.",160,1374,Algebra,4
138,shl_jbmo_2015_a5,shl_jbmo,2015,a,"Let $x, y, z$ be positive real numbers that satisfy the equality $x^2+y^2+z^2=3$. Prove that
\[
\frac{x^2+yz}{x^2+yz+1} + \frac{y^2+zx}{y^2+zx+1} + \frac{z^2+xy}{z^2+xy+1} \leq 2.
\]","\subsubsection*{Solution:}
We have
\begin{align*}
&\frac{x^2+yz}{x^2+yz+1} + \frac{y^2+zx}{y^2+zx+1} + \frac{z^2+xy}{z^2+xy+1} \leq 2 \iff\\[6pt]
&\frac{x^2+yz+1}{x^2+yz+1} + \frac{y^2+zx+1}{y^2+zx+1} + \frac{z^2+xy+1}{z^2+xy+1} \leq 2 + \frac{1}{x^2+yz+1}+\frac{1}{y^2+zx+1}+\frac{1}{z^2+xy+1} \iff\\[6pt]
&3 \leq 2 + \frac{1}{x^2+yz+1}+\frac{1}{y^2+zx+1}+\frac{1}{z^2+xy+1} \iff\\[6pt]
&1 \leq \frac{1}{x^2+yz+1}+\frac{1}{y^2+zx+1}+\frac{1}{z^2+xy+1}
\end{align*}
\[
\frac{1}{x^2+yz+1}+\frac{1}{y^2+zx+1}+\frac{1}{z^2+xy+1} \geq \frac{9}{x^2+yz+1+y^2+zx+1+z^2+xy+1} =
\]
\[
\frac{9}{x^2+y^2+z^2+xy+yz+zx+3} \geq \frac{9}{2\,x^2+y^2+z^2+3} = 1
\]
The first inequality: AM-GM inequality (also can be achieved with Cauchy-Bunjakowski-Schwarz inequality). The second inequality: $xy+yz+zx \leq x^2+y^2+z^2$ (there are more ways to prove it, AM-GM, full squares etc.)",182,864,Algebra,5
139,shl_jbmo_2015_c1,shl_jbmo,2015,c,"A board $n\times n$ ($n\geq 3$) is divided into $n^2$ unit squares. Integers from 0 to $n$ included are written down: one integer in each unit square, in such a way that the sums of integers in each $2\times2$ square of the board are different. Find all $n$ for which such boards exist.","\subsubsection*{Solution:}
The number of the $2\times2$ squares in a board $n\times n$ is equal to $(n-1)^2$. All possible sums of the numbers in such squares are $0, 1, \ldots, 4n$. A necessary condition for the existence of a board with the required property is $4n+1\geq(n-1)^2$ and consequently $n(n-6)\leq 0$. Thus $n\leq 6$. The examples show the existence of boards $n\times n$ for all $3\leq n\leq 6$.

\[
\begin{array}{|c|c|c|}
\hline
1 & 1 & 1\\
\hline
1 & 0 & 0\\
\hline
0 & 0 & 0\\
\hline
\end{array}
\qquad
\begin{array}{|c|c|c|c|}
\hline
0 & 0 & 2 & 0\\
\hline
0 & 0 & 0 & 2\\
\hline
0 & 1 & 2 & 2\\
\hline
2 & 2 & 2 & 2\\
\hline
\end{array}
\qquad
\begin{array}{|c|c|c|c|c|}
\hline
0 & 0 & 2 & 0 & 4\\
\hline
0 & 0 & 0 & 2 & 1\\
\hline
0 & 1 & 2 & 2 & 4\\
\hline
2 & 2 & 3 & 3 & 4\\
\hline
4 & 3 & 4 & 4 & 4\\
\hline
\end{array}
\qquad
\begin{array}{|c|c|c|c|c|c|}
\hline
6 & 6 & 6 & 6 & 5 & 5\\
\hline
6 & 6 & 5 & 5 & 5 & 5\\
\hline
1 & 2 & 3 & 4 & 4 & 5\\
\hline
3 & 5 & 0 & 5 & 0 & 5\\
\hline
1 & 0 & 2 & 1 & 0 & 0\\
\hline
1 & 0 & 1 & 0 & 0 & 0\\
\hline
\end{array}
\]",286,1087,Combinatorics,6
140,shl_jbmo_2015_c2,shl_jbmo,2015,c,2015 points are given in a plane such that from any five points we can choose two points with distance less than 1 unit. Prove that 504 of the given points lie on a unit disc.,"\subsubsection*{Solution:}
Start from an arbitrary point $A$ and draw a unit disc with center $A$. If all other points belong to this disc then we are done. Otherwise, take any point $B$ outside of the disc. Draw a unit disc with center $B$. If two drawn discs cover all 2015 points, by PHP, one of the discs contains at least 1008 points.

Suppose that there is a point $C$ outside of the two drawn discs. Draw a unit disc with center $C$. If three drawn discs cover all 2015 points, by PHP, one of the discs contains at least 672 points.

Finally, if there is a point $D$ outside of the three drawn discs, draw a unit disc with center $D$. By the given condition, any other point belongs to one of the four drawn discs. By PHP, one of the discs contains at least 504 points, concluding the solution.",175,801,Combinatorics,7
141,shl_jbmo_2015_c3,shl_jbmo,2015,c,"Positive integers are put into the following table:
\[
\begin{array}{ccccccccc}
1 & 3 & 6 & 10 & 15 & 21 & 28 & 36 & \cdots\\
2 & 5 & 9 & 14 & 20 & 27 & 35 & 44 & \cdots\\
4 & 8 & 13 & 19 & 26 & 34 & 43 & 53 & \cdots\\
7 & 12 & 18 & 25 & 33 & 42 & & &\\
11 & 17 & 24 & 32 & 41 & & & &\\
16 & 23 & & & & & & &\\
\cdots & & & & & & & &\\
\cdots & & & & & & & &\\
\end{array}
\]
Find the number of the line and column where the number 2015 stays.","\subsubsection*{Solution 1:}
We shall observe straight lines as on the next picture. We can call these lines diagonals.

On the first diagonal is number 1.\\
On the second diagonal are two numbers: 2 and 3.\\
On the 3rd diagonal are three numbers: 4, 5 and 6.\\
$\ldots$

On the $n$-th diagonal are $n$ numbers. These numbers are greater than $\dfrac{(n-1)n}{2}$ and not greater than $\dfrac{n(n+1)}{2}$ (see the next sentence!).

On the first $n$ diagonals are $1+2+3+\ldots+n = \dfrac{n(n+1)}{2}$ numbers.

If $m$ is in the $k$-th row $l$-th column and on the $n$-th diagonal, then it is $m = \dfrac{(n-1)n}{2}+l$ and $n+1 = k+l$. So, $m = \dfrac{(k+l-2)(k+l-1)}{2}+l$.

We have to find such numbers $n$, $k$ and $l$ for which:
\begin{align}
\frac{(n-1)n}{2} &< 2015 \leq \frac{n(n+1)}{2} \tag{1}\\
n+1 &= k+l \tag{2}\\
2015 &= \frac{(k+l-2)(k+l-1)}{2}+l \tag{3}
\end{align}

$(1),(2),(3) \Rightarrow n^2-n < 4030 \leq n^2+n \Rightarrow n = 63$, $k+l = 64$, $2015 = \dfrac{(64-2)(64-1)}{2}+l \Rightarrow$

$l = 2015 - 31\cdot 63 = 62$, $k = 64-62 = 2$.

Therefore 2015 is located in the second row and 62nd column.

\subsubsection*{Solution 2:}
Firstly, we can see that the first elements of the columns are triangular numbers. If $a_i$ is the first element of the line $i$, we have $a_i = \dfrac{i(i-1)}{2}$.

The second elements of the first row obtained by adding the first element 2.\\
The second elements of the second row is obtained by adding the first element 3.\\
And so on, then the second element on the $n$-th row is obtained by adding the first element $n+1$.

Then the third element of the $n$-th row is obtained by adding $n+2$, and the $k$-th element of it is obtained by adding $k$.

Since the first element of the $n$-th row is $\dfrac{(n-1)n}{2}+1$, the second one is
\[
\frac{(n-1)n}{2}+1+(n+1) = \frac{n(n+1)}{2}+2.
\]
The third one $\dfrac{n(n+1)}{2}+1+(n+2) = \dfrac{(n+1)(n+2)}{2}+3$ so the $k$-th one should be
\[
\frac{(n+k-2)(n+k-1)}{2}+k.
\]

\[
\frac{(n+k-2)(n+k-1)}{2}+k = 2015 \iff n^2+n(2k-3)+k^2-k-4028 = 0
\]
To have a positive integer solution $(2k-3)^2-4(k^2-k-4028) = 16121-8k$ must be a perfect square.

From $16121-8k = x^2$, it is noticed that the maximum of $x$ is 126 (since $k>0$). Simultaneously can be seen that $x$ is odd, so $x\leq 125$.
$16121-8k = x^2 \iff 125^2+496-8k = x^2$\\
So $496-8k=0$, from that $k = 62$.

From that we can find $n=2$.\\
So 2015 is located on the second row and 62nd column.",443,2450,Combinatorics,8
142,shl_jbmo_2015_c4,shl_jbmo,2015,c,"Let $n\geq1$ be a positive integer. A square of side length $n$ is divided by lines parallel to each side into $n^2$ squares of side length 1. Find the number of parallelograms which have vertices among the vertices of the $n^2$ squares of side length 1, with both sides smaller or equal to 2, and which have the area equal to 2.","\subsubsection*{Solution:}
We can divide all these parallelograms into 7 classes (types I -- VII), according to Figure.

\begin{center}
[Figure: Grid showing 7 types of parallelograms labeled I through VII]
\end{center}

Type I: There are $n$ ways to choose the strip for the horizontal (shorter) side of the parallelogram, and $(n-1)$ ways to choose the strip (of the width 2) for the vertical (longer) side. So there are $n(n-1)$ parallelograms of the type I.

Type II: There are $(n-1)$ ways to choose the strip (of the width 2) for the horizontal (longer) side, and $n$ ways to choose the strip for the vertical (shorter) side. So the number of the parallelogram of this type is also $n(n-1)$.

Type III: Each parallelogram of this type is a square inscribed in a unique square $2\times2$ of our grid. The number of such squares is $(n-1)^2$. So there are $(n-1)^2$ parallelograms of type III.

For each of the types IV, V, VI, VII, the strip of the width 1 in which the parallelogram is located can be chosen in $n$ ways and for each such choice there are $n-2$ parallelograms located in the chosen strip.

Summing we obtain that the total number of parallelograms is:
\[
2n(n-1)+(n-1)^2+4n(n-2) = 7n^2-12n+1
\]",329,1216,Combinatorics,9
143,shl_jbmo_2015_c5,shl_jbmo,2015,c,"We have a $5\times5$ chessboard and a supply of L-shaped triominoes, i.e. $2\times2$ squares with one corner missing. Two players $A$ and $B$ play the following game: A positive integer $k\leq 25$ is chosen. Starting with $A$, the players take alternating turns marking squares of the chessboard until they mark a total of $k$ squares. (In each turn a player has to mark exactly one new square.)

At the end of the process, player $A$ wins if he can cover without overlapping all but at most 2 unmarked squares with L-shaped triominoes, otherwise player $B$ wins. It is not permitted any marked squares to be covered.

Find the smallest $k$, if it exists, such that player $B$ has a winning strategy.","\subsubsection*{Solution:}
We will show that player $A$ wins if $k=1, 2$ or 3, but player $B$ wins if $k=4$. Thus the smallest $k$ for which $B$ has a winning strategy exists and is equal to 4.

If $k=1$, player $A$ marks the upper left corner of the square and then fills it as follows.

\begin{center}
[Figure: $5\times5$ chessboard with X in upper left corner and L-shaped triomino tiling]
\end{center}

If $k=2$, player $A$ marks the upper left corner of the square. Whatever square player $B$ marks, then player $A$ can fill in the square in exactly the same pattern as above except that he doesn't put the triomino which covers the marked square of $B$. Player $A$ wins because he has left only two unmarked squares uncovered.

For $k=3$, player $A$ wins by following the same strategy. When he has to mark a square for the second time, he marks any yet unmarked square of the triomino that covers the marked square of $B$.

Let us now show that for $k=4$ player $B$ has a winning strategy. Since there will be 21 unmarked squares, player $A$ will need to cover all of them with seven L-shaped triominoes. We can assume that in his first move, player $A$ does not mark any square in the bottom two rows of the chessboard (otherwise just rotate the chessboard). In his first move player $B$ marks the square labeled 1 in the following figure.

\begin{center}
\[
\begin{array}{|c|c|c|c|c|}
\hline
 &  &  &  & \\
\hline
 &  &  &  & \\
\hline
 &  &  &  & \\
\hline
 &  &  & 1 & 4\\
\hline
 &  & 5 & 3 & 2\\
\hline
\end{array}
\]
\end{center}

If player $A$ in his next move does not mark any of the squares labeled 2, 3 and 4 then player $B$ marks the square labeled 3. Player $B$ wins as the square labeled 2 is left unmarked but cannot be covered with an L-shaped triomino.

If player $A$ in his next move marks the square labeled 2, then player $B$ marks the square labeled 5. Player $B$ wins as the square labeled 3 is left unmarked but cannot be covered with an L-shaped triomino.

Finally, if player $A$ in his next move marks one of the squares labeled 3 or 4, player $B$ marks the other of these two squares. Player $B$ wins as the square labeled 2 is left unmarked but cannot be covered with an L-shaped triomino.

Since we have covered all possible cases, player $B$ wins when $k=4$.",700,2295,Combinatorics,10
144,shl_jbmo_2015_g1,shl_jbmo,2015,g,"Around the triangle $ABC$ the circle is circumscribed, and at the vertex $C$ tangent $t$ to this circle is drawn. The line $p$ which is parallel to this tangent intersects the lines $BC$ and $AC$ at the points $D$ and $E$, respectively. Prove that the points $A, B, D, E$ belong to the same circle.","\subsubsection*{Solution:}
Let $O$ be the center of a circumscribed circle $k$ of the triangle $ABC$, and let $F$ and $G$ be the points of intersection of the line $CO$ with the line $p$ and the circle $k$, respectively (see Figure). From $p\|t$ it follows that $p\perp CO$. Furthermore, $\angle ABC = \angle AGC$, because these angles are peripheral over the same chord. The quadrilateral $AGFE$ has two right angles at the vertices $A$ and $F$, and hence, $\angle AED + \angle ABD = \angle AEF + \angle AGF = 180°$. Hence, the quadrilateral $ABDE$ is cyclic, as asserted.

\begin{center}
[Figure: Triangle $ABC$ with circumscribed circle, tangent $t$ at $C$, line $p \parallel t$ intersecting $BC$ at $D$ and $AC$ at $E$, with center $O$ and point $G$ on circle.]
\end{center}",298,778,Geometry,11
145,shl_jbmo_2015_g2,shl_jbmo,2015,g,"The point $P$ is outside of the circle $\Omega$. Two tangent lines, passing from the point $P$, touch the circle $\Omega$ at the points $A$ and $B$. The median $AM$, $M\in(BP)$, intersects the circle $\Omega$ at the point $C$ and the line $PC$ intersects again the circle $\Omega$ at the point $D$. Prove that the lines $AD$ and $BP$ are parallel.","\subsubsection*{Solution:}
Since $\angle BAC = \angle BAM = \angle MBC$, we have $\triangle MAB \cong \triangle MBC$.

We obtain $\dfrac{MA}{MB} = \dfrac{MB}{MC} = \dfrac{AB}{BC}$. The equality $MB = MP$ implies $\dfrac{MA}{MP} = \dfrac{MP}{MC}$ and $\angle PMC = \angle PMA$ gives the relation $\triangle PMA \cong \triangle CMP$. It follows that
$\angle BPD \equiv \angle MPC = \angle MAP \equiv \angle CAP \equiv \angle CDA \equiv \angle PDA$. So, the lines $AD$ and $BP$ are parallel.",347,488,Geometry,12
146,shl_jbmo_2015_g3,shl_jbmo,2015,g,"Let $c \equiv c(O,K)$ be a circle with center $O$ and radius $R$ and $A, B$ be two points on it, not belonging to the same diameter. The bisector of the angle $\widehat{ABO}$ intersects the circle $c$ at point $C$, the circumcircle of the triangle $AOB$, say $c_1$ at point $K$ and the circumcircle of the triangle $AOC$, say $c_2$, at point $L$. Prove that the point $K$ is the circumcenter of the triangle $AOC$ and the point $L$ is the incenter of the triangle $AOB$.","\subsubsection*{Solution:}
The segments $OB, OC$ are equal, as radii of the circle $c$. Hence $OBC$ is an isosceles triangle and
\[
\hat{B}_1 = \hat{C}_1 = \hat{x}. \tag{1}
\]

\begin{center}
[Figure: Circle $c$ with points $A$, $B$, $C$, center $O$, circles $c_1$, $c_2$, and points $K$, $L$.]
\end{center}

The chord $BC$ is the bisector of the angle $\widehat{OBA}$, and hence
\[
\hat{B}_1 = \hat{B}_2 = \hat{x}. \tag{2}
\]
The angles $\hat{B}_2$ and $\hat{O}_1$ are inscribed to the same arc $OK$ of the circle $c_1$ and hence
\[
\hat{B}_2 = \hat{O}_1 = \hat{x}. \tag{3}
\]
The segments $KO, KC$ are equal, as radii of the circle $c_2$. Hence the triangle $KOC$ is isosceles and so
\[
\hat{O}_2 = \hat{C}_1 = \hat{x}. \tag{4}
\]
From equalities (1), (2), (3) we conclude that
\[
\hat{O}_1 = \hat{O}_2 = \hat{x},
\]
and so $OK$ is the bisector, and hence perpendicular bisector of the isosceles triangle $OAC$.
The point $K$ is the middle of the arc $OK$ (since $BK$ bisects the angle $\widehat{OBA}$). Hence the perpendicular bisector of the chord $AO$ of the circle $c_1$ is passing through point $K$. It means that $K$ is the circumcenter of the triangle $OAC$.

From equalities (1),(2),(3) we conclude that $\hat{B}_2 = \hat{C}_1 = \hat{x}$ and so $AB\,/\!/\,OC \Rightarrow \widehat{OAB} = \widehat{AOC}$,
that is $\hat{A}_1 + \hat{A}_2 = \hat{O}_1 + \hat{O}_2$ and since $\hat{O}_1 = \hat{O}_2 = \hat{x}$, we conclude that
\[
\hat{A}_1 + \hat{A}_2 = 2\hat{O}_1 = 2\hat{x}. \tag{5}
\]

The angles $\hat{A}_1$ and $\hat{C}_1$ are inscribed into the circle $c_2$ and correspond to the same arc $OL$. Hence
\[
\hat{A}_1 = \hat{C}_1 = \hat{x}. \tag{6}
\]
From (5) and (6) we have $\hat{A}_1 = \hat{A}_2$, i.e. $AL$ is the bisector of the angle $\widehat{BAO}$.",470,1763,Geometry,13
147,shl_jbmo_2015_g4,shl_jbmo,2015,g,"Let $\triangle ABC$ be an acute triangle. The lines $(\varepsilon_1)$, $(\varepsilon_2)$ are perpendicular to $AB$ at the points $A$, $B$, respectively. The perpendicular lines from the midpoint $M$ of $AB$ to the sides of the triangle $AC$, $BC$ intersect the lines $(\varepsilon_1)$, $(\varepsilon_2)$ at the points $E$, $F$, respectively. If $I$ is the intersection point of $EF$, $MC$, prove that
\[
\angle AIB = \angle EMF = \angle CAB + \angle CBA
\]","\subsubsection*{Solution:}
Let $H$, $G$ be the points of intersection of $ME$, $MF$, with $AC$, $BC$ respectively. From the similarity of triangles $\triangle MHA$ and $\triangle MAE$ we get
\[
\frac{MH}{MA} = \frac{MA}{ME}
\]
thus, $MA^2 = MH \cdot ME$ \quad (1).

Similarly, from the similarity of triangles $\triangle MBG$ and $\triangle MFB$ we get
\[
\frac{MB}{MF} = \frac{MG}{MB}
\]
thus, $MB^2 = MF \cdot MG$ \quad (2).

Since $MA = MB$, from (1), (2), we have that the points $E, H, G, F$ are concyclic.

\begin{center}
[Figure: Acute triangle $ABC$ with midpoint $M$ of $AB$, lines $(\varepsilon_1)$ and $(\varepsilon_2)$, and points $E$, $F$, $H$, $G$, $I$, $C$.]
\end{center}

Therefore, we get that $\angle FEH = \angle FEM = \angle HGM$. Also, the quadrilateral $CHMG$ is cyclic, so $\angle CMH = \angle HGC$. We have
\[
\angle FEH + \angle CMH = \angle HGM + \angle HGC = 90°
\]
Thus $CM \perp EF$. Now, from the cyclic quadrilaterals $FIMB$ and $EIMA$, we get that $\angle IFM = \angle IBM$ and $\angle IEM = \angle IAM$. Therefore, the triangles $\triangle EMF$ and $\triangle AIB$ are similar, so $\angle AIB = \angle EMF$. Finally,
\[
\angle AIB = \angle AIM + \angle MIB = \angle AEM + \angle MFB = \angle CAB + \angle CBA.
\]",456,1245,Geometry,14
148,shl_jbmo_2015_g5,shl_jbmo,2015,g,"Let $ABC$ be an acute triangle with $AB 
eq AC$. The incircle $\omega$ of the triangle touches the sides $BC$, $CA$ and $AB$ at $D$, $E$ and $F$, respectively. The perpendicular line erected at $C$ onto $BC$ meets $EF$ at $M$, and similarly, the perpendicular line erected at $B$ onto $BC$ meets $EF$ at $N$. The line $DM$ meets $\omega$ again in $P$, and the line $DN$ meets $\omega$ again at $Q$. Prove that $DP = DQ$.","\begin{center}
[Figure: Triangle $ABC$ with incircle $\omega$, tangent points $D$, $E$, $F$, points $M$, $N$, $S$, $P$, $Q$.]
\end{center}

\subsubsection*{Proof 1.1.}
Let $\{T\} = EF \cap BC$. Applying Menelaus' theorem to the triangle $ABC$ and the transversal line $E-F-T$ we obtain $\dfrac{TB}{TC}\cdot\dfrac{EC}{EA}\cdot\dfrac{FA}{FB} = 1$, i.e. $\dfrac{TB}{TC}\cdot\dfrac{s-c}{s-a}\cdot\dfrac{s-a}{s-b} = 1$, or $\dfrac{TB}{TC} = \dfrac{s-b}{s-c}$, where the notations are the usual ones.

This means that triangles $TBN$ and $TCM$ are similar, therefore $\dfrac{TB}{TC} = \dfrac{BN}{CM}$. From the above it follows $\dfrac{BN}{CM} = \dfrac{s-b}{s-c}$, $\dfrac{BD}{CD} = \dfrac{s-b}{s-c}$, and $\angle DBN = \angle DCM = 90°$, which means that triangles $\overline{BDN}$ and $\overline{CDM}$ are similar, hence angles $BDN$ and $CDM$ are equal. This leads to the arcs $DQ$ and $DP$ being equal, and finally to $DP = DQ$.

\subsubsection*{Proof 1.2.}
Let $S$ be the meeting point of the altitude from $A$ with the line $EF$. Lines $BN$, $AS$, $CM$ are parallel, therefore triangles $BNF$ and $ASF$ are similar, as are triangles $ASE$ and $CME$. We obtain $\dfrac{BN}{AS} = \dfrac{BF}{FA}$ and $\dfrac{AS}{CM} = \dfrac{AE}{EC}$.

Multiplying the two relations, we obtain $\dfrac{BN}{CM} = \dfrac{BF}{FA}\cdot\dfrac{AE}{EC} = \dfrac{BF}{EC} = \dfrac{BD}{DC}$ (we have used that $AE=AF$, $BF=BD$ and $CE=CD$).

It follows that the right triangles $BDN$ and $CDM$ are similar (SAS), which leads to the same ending as in the first proof.",420,1537,Geometry,15
149,shl_jbmo_2015_n1,shl_jbmo,2015,n,What is the greatest number of integers that can be selected from a set of 2015 consecutive numbers so that no sum of any two selected numbers is divisible by their difference?,"\subsubsection*{Solution:}
We take any two chosen numbers. If their difference is 1, it is clear that their sum is divisible by their difference. If their difference is 2, they will be of the same parity, and their sum is divisible by their difference. Therefore, the difference between any two chosen numbers will be at least 3. In other words, we can choose at most one number of any three consecutive numbers. This implies that we can choose at most 672 numbers.

Now, we will show that we can choose 672 numbers from any 2015 consecutive numbers. Suppose that these numbers are $a,\, a+1,\, \ldots,\, a+2014$. If $a$ is divisible by 3, we can choose $a+1,\, a+4,\, \ldots,\, a+2014$. If $a$ is not divisible by 3, we can choose $a,\, a+3,\ldots,\, a+2013$.",176,760,Number Theory,16
150,shl_jbmo_2015_n2,shl_jbmo,2015,n,"A positive integer is called \textit{a repunit}, if it is written only by ones. The repunit with $n$ digits will be denoted by $\underbrace{11\ldots1}_{n}$. Prove that:

a) the repunit $\underbrace{11\ldots1}_{n}$ is divisible by 37 if and only if $n$ is divisible by 3;

b) there exists a positive integer $k$ such that the repunit $\underbrace{11\ldots1}_{n}$ is divisible by 41 if and only if $n$ is divisible by $k$.","\subsubsection*{Solution:}
a) Let $n = 3m + r$, where $m$ and $r$ are non-negative integers and $r < 3$.
Denote by $\underbrace{00\ldots0}_{p}$ a recording with $p$ zeroes and $\underbrace{abcabc\ldots abc}_{p\times abc}$ recording with $p$ times $abc$. We
have:
\[
\underbrace{11\ldots1}_{n} = \underbrace{11\ldots1}_{3m+r} = \underbrace{11\ldots1}_{3m}\cdot\underbrace{00\ldots0}_{r} + \underbrace{11\ldots1}_{r} = 111\cdot\underbrace{100100\ldots100}_{(m-1)\times 100}\underbrace{00\ldots0}_{r} + \underbrace{11\ldots1}_{r}.
\]
Since $111 = 37\cdot 3$, the numbers $\underbrace{11\ldots1}_{n}$ and $\underbrace{11\ldots1}_{r}$ are equal modulo 37. On the other hand the numbers 1 and 11 are not divisible by 37. We conclude that $\underbrace{11\ldots1}_{n}$ is divisible by 37 if only if $r = 0$, i.e. if and only if $n$ is divisible by 3.

b) Using the idea from a), we look for a repunit, which is divisible by 41. Obviously, 1 and 11 are not divisible by 41, while the residues of 111 and 1111 are 29 and 4, respectively. We have $11111 = 41\cdot 271$. Since 11111 is a repunit with 5 digits, it follows in the same way as in a) that $\underbrace{11\ldots1}_{n}$ is divisible by 41 if and only if $n$ is divisible by 5.",420,1225,Number Theory,17
151,shl_jbmo_2015_n3,shl_jbmo,2015,n,"\textbf{a)} Show that the product of all differences of possible couples of six given positive integers is divisible by 960 (original from Albania).

\textbf{b)} Show that the product of all differences of possible couples of six given positive integers is divisible by 34560 (modified by problem selecting committee).","\subsubsection*{Solution:}
\textbf{a)} Since we have six numbers then at least two of them have a same residue when divided by 3, so at least one of the differences in our product is divisible by 3.
Since we have six numbers then at least two of them have a same residue when divided by 5, so at least one of the differences in our product is divisible by 5.
We may have:
\begin{enumerate}[label=\alph*)]
\item six numbers with the same parity
\item five numbers with the same parity
\item four numbers with the same parity
\item three numbers with the same parity
\end{enumerate}
There are $C_6^2 = 15$ different pairs, so there are 15 different differences in this product.

a) The six numbers have the same parity; then each difference is divisible by 2, therefore our product is divisible by $2^{15}$.

b) If we have five numbers with the same parity, then the couples that have their difference odd are formed by taking one number from these five numbers, and the second will be the sixth one. Then $C_5^1 = 15 = 5$ differences are odd. So $15 - 5 = 10$ differences are even, so product is divisible by $2^{10}$.

c) If we have four numbers with the same parity, then the couples that have their difference odd are formed by taking one number from these four numbers, and the second will be from the two others numbers. Then $2\cdot C_4^1 = 8$ differences are odd. So $15 - 8 = 7$ differences are even, so our product is divisible by $2^7$.

d) If we have three numbers with the same parity, then the couples that have their difference odd are formed by taking one from each triple. Then $C_3^1 \cdot C_3^1 = 9$ differences are odd, therefore $15 - 9 = 6$ differences are even, so our product is divisible by $2^6$.

Thus, our production is divisible by $2^6\cdot 3\cdot 5 = 960$.

\textbf{b)} Let $a_1, a_2, a_3, a_4, a_5, a_6$ be these numbers. Since we have six numbers then at least two of them when divided by 5 have the same residue, so at least one of these differences in our product is divisible by 5.
Since we have six numbers, and we have three possible residues at the division by 3, then at least three of them replies the residue of previous numbers, so at least three of these differences in our product are divisible by 3.
Since we have six numbers, and we have two possible residues at the division by 2, then at least four of them replies the residue of previous numbers, and two of them replies replied residues, so at least six of these differences in our product are divisible by 2.
Since we have six numbers, and we have four possible residues at the division by 4, then at least two of them replies the residue of previous numbers, so at least two of these differences in our product are divisible by 4. That means that two of these differences are divisible by 4 and moreover four of them are divisible by 2.
Thus, our production is divisible by $2^4\cdot 4^2\cdot 3^3\cdot 5 = 34560$.",318,2914,Number Theory,18
152,shl_jbmo_2015_n4,shl_jbmo,2015,n,"Find all prime numbers $a$, $b$, $c$ and positive integers $k$ which satisfy the equation
\[
a^2 + b^2 + 16\cdot c^2 = 9\cdot k^2 + 1.
\]","\subsubsection*{Solution:}
The relation $9\cdot k^2 + 1 \equiv 1 \pmod{3}$ implies
\[
a^2 + b^2 + 16\cdot c^2 \equiv 1 \pmod{3} \iff a^2 + b^2 + c^2 \equiv 1 \pmod{3}.
\]
Since $a^2 \equiv 0, 1\pmod{3}$, $b^2 \equiv 0, 1\pmod{3}$, $c^2 \equiv 0, 1\pmod{3}$, we have:

\[
\begin{array}{c|cccccccc}
a^2 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1\\
\hline
b^2 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1\\
\hline
c^2 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1\\
\hline
a^2+b^2+c^2 & 0 & 1 & 1 & 2 & 1 & 2 & 2 & 0\\
\end{array}
\]

From the previous table it follows that two of three prime numbers $a$, $b$, $c$ are equal to 3.

\textbf{Case 1.} $a = b = 3$. We have
\[
a^2+b^2+16\cdot c^2 = 9\cdot k^2+1 \iff 9\cdot k^2 - 16\cdot c^2 = 17 \iff (3k-4c)\cdot(3k+4c) = 17.
\]

If $\begin{cases}3k-4c=1,\\3k+4c=17,\end{cases}$ then $\begin{cases}c=2,\\k=3,\end{cases}$ and $(a,b,c,k) = (3,3,2,3)$.

If $\begin{cases}3k-4c=-17,\\3k+4c=-1,\end{cases}$ then $\begin{cases}c=2,\\k=-3,\end{cases}$ and $(a,b,c,k) = (3,3,2,-3)$.

\textbf{Case 2.} $c = 3$. If $(3, b_0, c, k)$ is a solution of the given equation, then $(b_0, 3, c, k)$ is a solution, too.

Let $a = 3$. We have
\[
a^2+b^2+16\cdot c^2 = 9\cdot k^2+1 \iff 9\cdot k^2 - b^2 = 152 \iff (3k-b)\cdot(3k+b) = 152.
\]
Both factors shall have the same parity and we obtain only 4 cases:

If $\begin{cases}3k-b=2,\\3k+b=76,\end{cases}$ then $\begin{cases}b=37,\\k=13,\end{cases}$ and $(a,b,c,k) = (3,37,3,13)$.

If $\begin{cases}3k-b=4,\\3k+b=38,\end{cases}$ then $\begin{cases}b=17,\\k=7,\end{cases}$ and $(a,b,c,k) = (3,17,3,7)$.

If $\begin{cases}3k-b=-76,\\3k+b=-2,\end{cases}$ then $\begin{cases}b=37,\\k=-13,\end{cases}$ and $(a,b,c,k) = (3,37,3,-13)$.

If $\begin{cases}3k-b=-38,\\3k+b=-4,\end{cases}$ then $\begin{cases}b=17,\\k=-7,\end{cases}$ and $(a,b,c,k) = (3,17,3,-7)$.

In addition, $(a,b,c,k)\in\{(37,3,3,13),\ (17,3,3,7),\ (37,3,3,-13),\ (17,3,3,-7)\}$.

So, the given equation has 10 solutions:
\[
S = \left\{
\begin{array}{l}
(37,3,3,13),\ (17,3,3,7),\ (37,3,3,-13),\ (17,3,3,-7),\ (3,37,3,13),\ (3,17,3,7),\ (3,37,3,-13),\\
(3,17,3,-7),\ (3,3,2,3),\ (3,3,2,-3).
\end{array}

ight\}
\]",137,2114,Number Theory,19
153,shl_jbmo_2015_n5,shl_jbmo,2015,n,"Does there exist positive integers $a$, $b$ and a prime $p$ such that
\[
a^3 - b^3 = 4p^2?
\]","\subsubsection*{Solution:}
The given equality may be written as
\begin{equation}
(a-b)(a^2+ab+b^2) = 4p^2. \tag{1}
\end{equation}
Since $a-b < a^2+ab+b^2$, it follows from (1) that
\begin{equation}
a - b < 2p. \tag{2}
\end{equation}
Now consider two cases: 1. $p=2$, and 2. $p$ is an odd prime.

\textbf{Case 1:} $p=2$. Then (1) becomes
\begin{equation}
(a-b)(a^2+ab+b^2) = 16. \tag{3}
\end{equation}
In view of (2) and (3), it must be $a-b=1$ or $a-b=2$. If $a-b=1$, then substituting $a=b+1$ in (3) we obtain
\[
b(b+1) = 5,
\]
which is impossible since $b(b+1)$ is an even integer.
If $a-b=2$, then substituting $a=b+2$ in (3) we get
\[
3b(b+2) = 4,
\]
which is obviously impossible.

\textbf{Case 2:} $p$ is an odd prime. Then (1) yields $a-b\mid 4p^2$. This together with the facts that $p$ is a prime and that by (2) $a-b<2p$, yields $a-b\in\{1,2,4,p\}$.

If $a-b=1$, then substituting $a=b+1$ in (1) we obtain
\[
3b(b+1)+1 = 4p^2,
\]
which is impossible since $3b(b+1)+1$ is an odd integer.

If $a-b=2$, then substituting $a=b+2$ in (1) we obtain
\[
3b^2+6b+4 = 2p^2,
\]
whence it follows that
\[
2(p^2-2) = 3(b^2+2b) \equiv 0\pmod{3},
\]
and hence
\begin{equation}
p^2 \equiv 2 \pmod{3}. \tag{5}
\end{equation}
Since $p^2\equiv 1\pmod{3}$ for each odd prime $p>3$ and $3^2\equiv 0\pmod{3}$, it follows that the congruence (5) is not satisfied for any odd prime $p$.

If $a-b=4$, then substituting $a=b+4$ in (1) we obtain
\[
3b^2+12b+16 = p^2,
\]
whence it follows that $b$ is an odd integer such that
\[
3b^2 \equiv p^2\pmod{4},
\]
whence since $p^2\equiv 1\pmod{4}$ for each odd prime $p$, we have
\begin{equation}
3b^2 \equiv 1\pmod{4}. \tag{6}
\end{equation}
However, since $b^2\equiv 1\pmod{4}$ for each odd integer $b$, it follows that the congruence (6) is not satisfied for any odd integer $b$.

If $a-b=p$, then substituting $a=b+p$ in (1) we obtain
\[
p(3b^2+3bp+p^2-4p) = 0,
\]
i.e.,
\begin{equation}
3b^2+3bp+p^2-4p = 0. \tag{7}
\end{equation}
If $p\geq 5$, then $p^2-4p>0$, and thus (7) cannot be satisfied for any positive integer $b$. If $p=3$, then (7) becomes
\[
3(b^2+3b-1)=0,
\]
which is obviously not satisfied for any positive integer $b$.

Hence, there does not exist positive integers $a$, $b$ and a prime $p$ such that $a^3-b^3=4p^2$.",93,2265,Number Theory,20
239,jbmo_2016_p1,jbmo,2016,g,"A trapezoid $ABCD$ ($AB \parallel CD$, $AB > CD$) is circumscribed. The incircle of the
triangle $ABC$ touches the lines $AB$ and $AC$ at the points $M$ and $N$, respectively. Prove
that the incenter of the trapezoid $ABCD$ lies on the line $MN$.","\subsection*{Solution}

\textbf{Version 1.} Let $I$ be the incenter of triangle $ABC$ and $R$ be the common point of
the lines $BI$ and $MN$. Since
\[
  m(\widehat{ANM}) = 90^\circ - \tfrac{1}{2}\,m(\widehat{MAN})
  \quad\text{and}\quad
  m(\widehat{BIC}) = 90^\circ + \tfrac{1}{2}\,m(\widehat{MAN}),
\]
the quadrilateral $IRNC$ is cyclic. \hfill(1)

It follows that $m(\widehat{BRC}) = 90^\circ$ and therefore
\[
  m(\widehat{BCR}) = 90^\circ - m(\widehat{CBR})
    = 90^\circ - \tfrac{1}{2}\!\left(180^\circ - m(\widehat{BCD})
ight)
    = \tfrac{1}{2}\,m(\widehat{BCD}). \tag{2}
\]

So, $CR$ is the angle bisector of $\widehat{DCB}$ and $R$ is the incenter of the trapezoid. \hfill(3)

\bigskip
\textbf{Version 2.} If $R$ is the incentre of the trapezoid $ABCD$, then $B$, $I$ and $R$ are collinear,
\hfill(1$'$)

and $m(\widehat{BRC}) = 90^\circ$. \hfill(2$'$)

The quadrilateral $IRNC$ is cyclic. \hfill(3$'$)

Then
\[
  m(\widehat{MNC}) = 90^\circ + \tfrac{1}{2}\cdot m(\widehat{BAC}) \tag{4$'$}
\]
and
\[
  m(\widehat{RNC}) = m(\widehat{BIC}) = 90^\circ + \tfrac{1}{2}\cdot m(\widehat{BAC}), \tag{5$'$}
\]
so that $m(\widehat{MNC}) = m(\widehat{RNC})$ and the points $M$, $R$ and $N$ are collinear. \hfill(6$'$)

\bigskip
\textbf{Version 3.} If $R$ is the incentre of the trapezoid $ABCD$, let $M' \in (AB)$ and
$N' \in (AC)$ be the unique points such that $R \in M'N'$ and $(AM') \equiv (AN')$. \hfill(1$''$)

Let $S$ be the intersection point of $CR$ and $AB$. Then $CR = RS$. \hfill(2$''$)

Consider $K \in AC$ such that $SK \parallel M'N'$. Then $N'$ is the midpoint of $(CK)$. \hfill(3$''$)

We deduce
\[
  AN' = \frac{AK + AC}{2} = \frac{AS + AC}{2}
       = \frac{AB - BS + AC}{2} = \frac{AB + AC - BC}{2} = AN. \tag{4$''$}
\]

We conclude that $N = N'$, hence $M = M'$, and $R$, $M$, $N$ are collinear. \hfill(5$''$)",246,1830,Geometry,1
240,jbmo_2016_p2,jbmo,2016,a,"Let $a$, $b$ and $c$ be positive real numbers. Prove that
\[
  \frac{8}{(a+b)^2+4abc}
  + \frac{8}{(b+c)^2+4abc}
  + \frac{8}{(c+a)^2+4abc}
  + a^2 + b^2 + c^2
  \;\geq\;
  \frac{8}{a+3} + \frac{8}{b+3} + \frac{8}{c+3}.
\]","\subsection*{Solution}

Since $2ab \le a^2 + b^2$, it follows that
\[
  (a+b)^2 \le 2(a^2+b^2) \tag{1}
\]
and
\[
  4abc \le 2c(a^2+b^2), \quad\text{for any positive reals } a,b,c. \tag{2}
\]
Adding these inequalities, we find
\[
  (a+b)^2 + 4abc \le 2(a^2+b^2)(c+1), \tag{3}
\]
so that
\[
  \frac{8}{(a+b)^2+4abc} \ge \frac{4}{(a^2+b^2)(c+1)}. \tag{4}
\]

Using the AM-GM inequality, we have
\[
  \frac{4}{(a^2+b^2)(c+1)} + \frac{a^2+b^2}{2}
  \ge 2\sqrt{\frac{2}{c+1}} = \frac{4}{\sqrt{2(c+1)}}, \tag{5}
\]
respectively
\[
  \frac{c+3}{8} = \frac{(c+1)+2}{8} \ge \frac{\sqrt{2(c+1)}}{4}. \tag{6}
\]

We conclude that
\[
  \frac{4}{(a^2+b^2)(c+1)} + \frac{a^2+b^2}{2} \ge \frac{8}{c+3}, \tag{7}
\]
and finally
\[
  \frac{8}{(a+b)^2+4abc}
  + \frac{8}{(a+c)^2+4abc}
  + \frac{8}{(b+c)^2+4abc}
  + a^2+b^2+c^2
  \;\ge\;
  \frac{8}{a+3} + \frac{8}{b+3} + \frac{8}{c+3}. \tag{8}
\]",222,877,Algebra,2
241,jbmo_2016_p3,jbmo,2016,n,"Find all the triples of integers $(a, b, c)$ such that the number
\[
  N = \frac{(a-b)(b-c)(c-a)}{2} + 2
\]
is a power of $2016$.


oindent(A \emph{power of $2016$} is an integer of the form $2016^n$, where $n$ is a non-negative integer.)","\subsection*{Solution}

Let $a, b, c$ be integers and $n$ be a positive integer such that
\[
  (a-b)(b-c)(c-a) + 4 = 2 \cdot 2016^n.
\]
We set $a - b = -x$, $b - c = -y$ and rewrite the equation as
\[
  xy(x+y) + 4 = 2 \cdot 2016^n. \tag{1}
\]

If $n > 0$, then the right-hand side is divisible by $7$, so
\[
  xy(x+y) + 4 \equiv 0 \pmod{7}, \tag{2}
\]
or equivalently
\[
  3xy(x+y) \equiv 2 \pmod{7}, \tag{3}
\]
or
\[
  (x+y)^3 - x^3 - y^3 \equiv 2 \pmod{7}. \tag{4}
\]

By Fermat's Little Theorem, for any integer $k$ the cubic residues satisfy
$k^3 \equiv -1, 0, 1 \pmod{7}$. \hfill(5)

It follows that in (4) some of $(x+y)^3$, $x^3$ and $y^3$ must be divisible by $7$.
But then $xy(x+y)$ is divisible by $7$, which is a contradiction. \hfill(6)

So the only possibility is $n = 0$, giving $xy(x+y) + 4 = 2$, i.e.
\[
  xy(x+y) = -2. \tag{7}
\]
The solutions are $(x,y) \in \{(-1,-1),\,(2,-1),\,(-1,2)\}$, \hfill(8)

so the required triples are $(a,b,c) = (k+2,\,k+1,\,k)$, $k \in \mathbb{Z}$, and all their
cyclic permutations. \hfill(9)

\bigskip
\textbf{Alternative version.} If $n > 0$ then $9$ divides $(a-b)(b-c)(c-a)+4$, i.e.\ the equation
\[
  xy(x+y) + 4 \equiv 0 \pmod{9} \tag{1$'$}
\]
has the solution $x = b-a$, $y = c-b$.
But then $x$ and $y$ must be $1$ modulo $3$, implying $xy(x+y) \equiv 2 \pmod{9}$, a contradiction. \hfill(2$'$)

One can continue as in the first version.",238,1393,Number Theory,3
242,jbmo_2016_p4,jbmo,2016,c,"A $5\times 5$ table is called \emph{regular} if each of its cells contains one of four pairwise
distinct real numbers, such that each of them occurs exactly once in every $2\times 2$ subtable.
The sum of all numbers of a regular table is called the \emph{total sum} of the table. With any
four numbers, one constructs all possible regular tables, computes their total sums and counts the
distinct outcomes. Determine the maximum possible count.","\subsection*{Solution}

We will prove that the maximum number of total sums is $\mathbf{60}$.

The proof is based on the following claim.

\begin{quote}
\textbf{Claim.} In a regular table either each row contains exactly two of the numbers, or each
column contains exactly two of the numbers.
\end{quote}

\textit{Proof of the Claim.} Let $R$ be a row containing at least three of the numbers. Then we can
find three consecutive entries $x, y, z$ in row $R$ (with $\{x,y,z,t\} = \{a,b,c,d\}$). By the
regularity condition the row directly above $R$ (if it exists) must contain $z, t, x$ above $x, y, z$,
and the row above that must again contain $x, y, z$. The same holds for rows below $R$:
\[
\begin{pmatrix}
\cdot & x & y & z & \cdot \\
\cdot & z & t & x & \cdot \\
\cdot & x & y & z & \cdot \\
\cdot & z & t & x & \cdot \\
\cdot & x & y & z & \cdot
\end{pmatrix}
\]
Completing the whole array, each column contains exactly two of the numbers, proving the claim. \hfill(1)

\medskip
Rotating the matrix if necessary, we may assume that each \emph{row} contains exactly two of
the numbers. Removing the first row and column leaves a $4\times 4$ array that splits into four
$2\times 2$ subarrays, each containing every number exactly once, with total sum $4(a+b+c+d)$.

It suffices to count the different ways to fill the first row $R_1$ and the first column $C_1$. \hfill(2)

Denoting by $a_1, b_1, c_1, d_1$ the number of appearances of $a,b,c,d$ in $R_1$ and $C_1$, the
total sum is
\[
  S = 4(a+b+c+d) + a_1 a + b_1 b + c_1 c + d_1 d. \tag{3}
\]

If the first, third, and fifth rows contain the numbers $x, y$ (with $x$ at position $(1,1)$) and
the second and fourth rows contain $z, t$ (with $z$ at position $(2,1)$), then
$x_1 + y_1 = 7$ with $x_1 > 3$, $y_1 \ge 2$, and $z_1 + t_1 = 2$ with $z_1 \ge t_1$.
Thus $\{x_1, y_1\} = \{5,2\}$ or $\{4,3\}$, and $\{z_1, t_1\} = \{2,0\}$ or $\{1,1\}$. \hfill(4)

Therefore $(a_1, b_1, c_1, d_1)$ is a permutation of one of:
\[
  (5,2,2,0),\quad (5,2,1,1),\quad (4,3,2,0),\quad (4,3,1,1). \tag{5}
\]

The number of permutations of each quadruple is:
$\frac{4!}{2!} = 12$ for $(5,2,2,0)$; $12$ for $(5,2,1,1)$; $4! = 24$ for $(4,3,2,0)$; $12$ for
$(4,3,1,1)$. Hence there are at most $12+12+24+12 = 60$ different total sums. \hfill(6)

All $60$ combinations are achievable: use three rows $ababa$ alternating with two rows $cdcdc$
to get $(5,2,2,0)$; three rows $ababa$ with one row $cdcdc$ and one row $dcdcd$ to get
$(5,2,1,1)$; three rows $abcba$ alternating with two rows $cdadc$ to get $(4,3,2,0)$; three rows
$abcda$ alternating with two rows $cdabc$ to get $(4,3,1,1)$. \hfill(7)

Choosing, for example, $a = 10^3$, $b = 10^2$, $c = 10$, $d = 1$ makes all $60$ sums distinct.
\hfill(8)

Hence $60$ is indeed the maximum possible number of different sums. \hfill(9)

\bigskip
\textbf{Alternative Version.} Consider a regular table containing the four distinct numbers
$a, b, c, d$. The four $2\times 2$ corner subtables each contain all four numbers, so that if
$a_1, b_1, c_1, d_1$ are the numbers of appearances of $a, b, c, d$ in the middle row and column,
then
\[
  S = 4(a+b+c+d) + a_1 a + b_1 b + c_1 c + d_1 d. \tag{10}
\]

Let $x$ be at position $(3,3)$, $y$ at $(3,2)$, $y'$ at $(3,4)$, $z$ at $(2,3)$, and $z'$ at $(4,3)$.
If $z 
eq z' = t$, then $y = y'$, and positions $(3,1)$ and $(3,5)$ both contain $x$. \hfill(2$'$)

The second and fourth rows then contain only $z$ and $t$; the first and fifth rows contain only
$x$ and $y$. \hfill(3$'$)

Then $x_1+y_1 = 7$ with $x_1 > 3$, $y_1 \ge 2$, and $z_1+t_1 = 2$ with $z_1 \ge t_1$, giving
$\{x_1,y_1\} = \{5,2\}$ or $\{4,3\}$, and $\{z_1,t_1\} = \{2,0\}$ or $\{1,1\}$. \hfill(4$'$)

One can continue as in the first version.",444,3737,Combinatorics,4
113,shl_jbmo_2016_a1,shl_jbmo,2016,a,"Let $a, b, c$ be positive real numbers such that $abc = 8$. Prove that
\[
\frac{ab + 4}{a + 2} + \frac{bc + 4}{b + 2} + \frac{ca + 4}{c + 2} \geq 6.
\]","
oindent\textbf{Solution.} We have $ab+4 = \frac{8}{c}+4 = \frac{4(c+2)}{c}$ and similarly $bc+4 = \frac{4(a+2)}{a}$ and $ca+4 = \frac{4(b+2)}{b}$.
It follows that
\[
(ab + 4)(bc + 4)(ca + 4) = \frac{64}{abc}(a + 2)(b + 2)(c + 2) = 8(a + 2)(b + 2)(c + 2),
\]
so that
\[
\frac{(ab + 4)(bc + 4)(ca + 4)}{(a + 2)(b + 2)(c + 2)} = 8.
\]
Applying AM-GM, we conclude:
\[
\frac{ab + 4}{a + 2} + \frac{bc + 4}{b + 2} + \frac{ca + 4}{c + 2} \geq 3 \cdot \sqrt[3]{\frac{(ab + 4)(bc + 4)(ca + 4)}{(a + 2)(b + 2)(c + 2)}} = 6.
\]
Alternatively, we can write LHS as
\[
\frac{bc(ab + 4)}{2(bc + 4)} + \frac{ac(bc + 4)}{2(ac + 4)} + \frac{ab(ca + 4)}{2(ab + 4)}
\]
and then apply AM-GM.",151,671,Algebra,1
114,shl_jbmo_2016_a2,shl_jbmo,2016,a,"Given positive real numbers $a, b, c$, prove that
\[
\frac{8}{(a + b)^2 + 4abc} + \frac{8}{(a + c)^2 + 4abc} + \frac{8}{(b + c)^2 + 4abc} + a^2 + b^2 + c^2 \geq \frac{8}{a + 3} + \frac{8}{b + 3} + \frac{8}{c + 3}.
\]","
oindent\textbf{Solution.} Since $2ab \leq a^2 + b^2$, it follows that $(a+b)^2 \leq 2(a^2+b^2)$ and $4abc \leq 2c(a^2+b^2)$,
for any positive reals $a, b, c$. Adding these inequalities, we find
\[
(a + b)^2 + 4abc \leq 2(a^2 + b^2)(c + 1),
\]
so that $\dfrac{8}{(a + b)^2 + 4abc} \geq \dfrac{4}{(a^2 + b^2)(c + 1)}$.

Using the AM-GM inequality, we have
\[
\frac{4}{(a^2 + b^2)(c + 1)} + \frac{a^2 + b^2}{2} \geq 2\sqrt{\frac{2}{c + 1}} = \frac{4}{\sqrt{2(c + 1)}},
\]
respectively
\[
\frac{c + 3}{8} = \frac{(c + 1) + 2}{8} \geq \frac{\sqrt{2(c + 1)}}{4}.
\]
We conclude that
\[
\frac{4}{(a^2 + b^2)(c + 1)} + \frac{a^2 + b^2}{2} \geq \frac{8}{c + 3},
\]
and finally
\[
\frac{8}{(a + b)^2 + 4abc} + \frac{8}{(a + c)^2 + 4abc} + \frac{8}{(b + c)^2 + 4abc} + a^2 + b^2 + c^2 \geq \frac{8}{a + 3} + \frac{8}{b + 3} + \frac{8}{c + 3}.
\]",216,835,Algebra,2
115,shl_jbmo_2016_a3,shl_jbmo,2016,a,"Determine the number of pairs of integers $(m, n)$ such that
\[
\sqrt{n + \sqrt{2016}} + \sqrt{m - \sqrt{2016}} \in \mathbb{Q}.
\]","
oindent\textbf{Solution.} Let $r = \sqrt{n + \sqrt{2016}} + \sqrt{m - \sqrt{2016}}$. Then
\[
n + m + 2\sqrt{n + \sqrt{2016}} \cdot \sqrt{m - \sqrt{2016}} = r^2,
\]
and
\[
(m - n)\sqrt{2016} = \frac{1}{4}(r^2 - m - n)^2 - mn + 2016 \in \mathbb{Q}.
\]
Since $\sqrt{2016} 
otin \mathbb{Q}$, it follows that $m = n$. Then
\[
\sqrt{n^2 - 2016} = \frac{1}{2}(r^2 - 2n) \in \mathbb{Q}.
\]
Hence, there is some nonnegative integer $p$ such that $n^2 - 2016 = p^2$ and (1) becomes
\[
2n + 2p = r^2.
\]
It follows that $2(n + p) = r^2$ is the square of a rational and also an integer, hence a perfect
square. On the other hand, $2016 = (n-p)(n+p)$ and $n+p$ is a divisor of $2016$, larger than
$\sqrt{2016}$. Since $n+p$ is even, so is also $n-p$, and $r^2 = 2(n+p)$ is a divisor of $2016 = 2^5 \cdot 3^2 \cdot 7$,
larger than $2\sqrt{2016} > 88$. The only possibility is $r^2 = 2^4 \cdot 3^2 = 12^2$. Hence, $n + p = 72$ and
$n - p = 28$, and we conclude that $n = m = 50$. Thus, there is only one such pair.",130,1000,Algebra,3
116,shl_jbmo_2016_a4,shl_jbmo,2016,a,"If $x, y, z$ are non-negative real numbers such that $x^2 + y^2 + z^2 = x + y + z$, then show
that:
\[
\frac{x + 1}{\sqrt{x^5 + x + 1}} + \frac{y + 1}{\sqrt{y^5 + y + 1}} + \frac{z + 1}{\sqrt{z^5 + z + 1}} \geq 3.
\]
When does the equality hold?","
oindent\textbf{Solution.} First we factor $x^5 + x + 1$ as follows:
\[
x^5 + x + 1 = x^5 - x^2 + x^2 + x + 1 = x^2(x^3 - 1) + x^2 + x + 1 = x^2(x - 1)(x^2 + x + 1) + x^2 + x + 1
\]
\[
= (x^2 + x + 1)(x^2(x - 1) + 1) = (x^2 + x + 1)(x^3 - x^2 + 1)
\]
Using the AM$-$GM inequality, we have
\[
\sqrt{x^5 + x + 1} = \sqrt{(x^2 + x + 1)(x^3 - x^2 + 1)} \leq \frac{x^2 + x + 1 + x^3 - x^2 + 1}{2} = \frac{x^3 + x + 2}{2}
\]
and since
\[
x^3 + x + 2 = x^3 + 1 + x + 1 = (x+1)(x^2 - x + 1) + x + 1 = (x+1)(x^2 - x + 1 + 1) = (x+1)(x^2 - x + 2),
\]
then
\[
\sqrt{x^5 + x + 1} \leq \frac{(x + 1)(x^2 - x + 2)}{2}
\]
Using $x^2 - x + 2 = \left(x - \frac{1}{2}
ight)^2 + \frac{7}{4} > 0$, we obtain $\dfrac{x+1}{\sqrt{x^5+x+1}} \geq \dfrac{2}{x^2-x+2}$. Applying the Cauchy--Schwarz inequality and the given condition, we get
\[
\sum_{\text{cyc}} \frac{x + 1}{\sqrt{x^5 + x + 1}} \geq \sum_{\text{cyc}} \frac{2}{x^2 - x + 2} \geq \frac{18}{\sum_{\text{cyc}}(x^2 - x + 2)} = \frac{18}{6} = 3
\]
which is the desired result.

For the equality both conditions: $x^2 - x + 2 = y^2 - y + 2 = z^2 - z + 2$ (equality in CBS)
and $x^3 - x^2 + 1 = x^2 + x + 1$ (equality in AM-GM) have to be satisfied.

By using the given condition it follows that $x^2 - x + 2 + y^2 - y + 2 + z^2 - z + 2 = 6$, hence
$3(x^2 - x + 2) = 6$, implying $x = 0$ or $x = 1$. Of these, only $x = 0$ satisfies the second
condition. We conclude that equality can only hold for $x = y = z = 0$.
It is an immediate check that indeed for these values equality holds.

\medskip

oindent\textit{Alternative solution}

Let us present an heuristic argument to reach the key inequality $\dfrac{x+1}{\sqrt{x^5+x+1}} \geq \dfrac{2}{x^2-x+2}$.

In order to exploit the condition $x^2 + y^2 + z^2 = x + y + z$ when applying CBS in Engel
form, we are looking for $\alpha, \beta, \gamma > 0$ such that
\[
\frac{x + 1}{\sqrt{x^5 + x + 1}} \geq \frac{\gamma}{\alpha(x^2 - x) + \beta}.
\]
After squaring and cancelling the denominators, we get
\[
(x + 1)^2(\alpha(x^2 - x) + \beta)^2 \geq \gamma^2(x^5 + x + 1)
\]
for all $x \geq 0$, and, after some manipulations, we reach to $f(x) \geq 0$ for all $x \geq 0$, where
\[
f(x) = \alpha^2 x^6 - \gamma^2 x^5 + (2\alpha\beta - 2\alpha^2)x^4 + 2\alpha\beta x^3 + (\alpha - \beta)^2 x^2 + (2\beta^2 - 2\alpha\beta - \gamma^2)x + \beta^2 - \gamma^2.
\]
As we are expecting the equality to hold for $x = 0$, we naturally impose the condition that
$f$ has $0$ as a double root. This implies $\beta^2 - \gamma^2 = 0$ and $2\beta^2 - 2\alpha\beta - \gamma^2 = 0$, that is, $\beta = \gamma$
and $\gamma = 2\alpha$.

Thus the inequality $f(x) \geq 0$ becomes
\[
\alpha^2 x^6 - 4\alpha^2 x^5 + 2\alpha^2 x^4 + 4\alpha^2 x^3 + \alpha^2 x^2 \geq 0, \quad \forall x \geq 0,
\]
that is,
\[
\alpha^2 x^2(x^2 - 2x - 1)^2 \geq 0 \quad \forall x \geq 0,
\]
which is obviously true.

Therefore, the inequality $\dfrac{x+1}{\sqrt{x^5+x+1}} \geq \dfrac{2}{x^2-x+2}$ holds for all $x \geq 0$ and now we can continue
as in the first solution.",245,3003,Algebra,4
117,shl_jbmo_2016_a5,shl_jbmo,2016,a,"Let $x, y, z$ be positive real numbers such that $x + y + z = \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}$.

\begin{enumerate}[label=\alph*)]
\item Prove the inequality
\[
x + y + z \geq \sqrt{\frac{xy + 1}{2}} + \sqrt{\frac{yz + 1}{2}} + \sqrt{\frac{zx + 1}{2}}.
\]
\item (Added by the problem selecting committee) When does the equality hold?
\end{enumerate}","
oindent\textbf{Solution.}

\medskip

oindent a) We rewrite the inequality as
\[
\left(\sqrt{xy + 1} + \sqrt{yz + 1} + \sqrt{zx + 1}
ight)^2 \leq 2 \cdot (x + y + z)^2 \tag{1}
\]
and note that, from CBS,
\[
\text{LHS} \leq \left(\frac{xy + 1}{x} + \frac{yz + 1}{y} + \frac{zx + 1}{z}
ight)(x + y + z).
\]
But
\[
\frac{xy + 1}{x} + \frac{yz + 1}{y} + \frac{zx + 1}{z} = x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 2(x + y + z),
\]
which proves (1).

\medskip

oindent b) The equality occurs when we have equality in CBS, i.e.\ when
\[
\frac{xy + 1}{x^2} = \frac{yz + 1}{y^2} = \frac{zx + 1}{z^2} \left(= \frac{xy + yz + zx + 3}{x^2 + y^2 + z^2}
ight).
\]
Since we can also write
\[
\left(\sqrt{xy + 1} + \sqrt{yz + 1} + \sqrt{zx + 1}
ight)^2 \leq \left(\frac{xy + 1}{y} + \frac{yz + 1}{z} + \frac{zx + 1}{x}
ight)(y + z + x) = 2(x + y + z)^2,
\]
the equality implies also
\[
\frac{xy + 1}{y^2} = \frac{yz + 1}{z^2} = \frac{zx + 1}{x^2} \left(= \frac{xy + yz + zx + 3}{x^2 + y^2 + z^2}
ight).
\]
But then $x = y = z$, and since $x + y + z = \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}$, we conclude that $x = \dfrac{1}{x} = 1 = y = z$.

\medskip

oindent\textit{Alternative solution to b):} The equality condition
\[
\frac{xy + 1}{x^2} = \frac{yz + 1}{y^2} = \frac{zx + 1}{z^2}
\]
can be rewritten as
\[
\frac{y + \frac{1}{x}}{x} = \frac{z + \frac{1}{y}}{y} = \frac{x + \frac{1}{z}}{z} = \frac{x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}}{x + y + z} = 2
\]
and thus we obtain the system:
\[
\begin{cases}
y = 2x - \dfrac{1}{x}\\[6pt]
z = 2y - \dfrac{1}{y}\\[6pt]
x = 2z - \dfrac{1}{z}
\end{cases}.
\]
We show that $x = y = z$. Indeed, if for example $x > y$, then $2x - \dfrac{1}{x} > 2y - \dfrac{1}{y}$, that is,
$y > z$ and $z = 2y - \dfrac{1}{y} > 2z - \dfrac{1}{z} = x$, and we obtain the contradiction $x > y > z > x$.
Similarly, if $x < y$, we obtain $x < y < z < x$.

Hence, the numbers are equal, and as above we get $x = y = z = 1$.",359,1956,Algebra,5
118,shl_jbmo_2016_c1,shl_jbmo,2016,c,"Let $S_n$ be the sum of reciprocal values of non-zero digits of all positive integers up to
(and including) $n$. For instance, $S_{13} = \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} + \dfrac{1}{6} + \dfrac{1}{7} + \dfrac{1}{8} + \dfrac{1}{9} + \dfrac{1}{1} + \dfrac{1}{1} + \dfrac{1}{1} + \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{1} + \dfrac{1}{3}$.

Find the least positive integer $k$ making the number $k! \cdot S_{2016}$ an integer.","
oindent\textbf{Solution.}

We will first calculate $S_{999}$, then $S_{1999} - S_{999}$, and then $S_{2016} - S_{1999}$.

Writing the integers from 1 to 999 as 001 to 999, adding eventually also 000 (since 0 digits
actually do not matter), each digit appears exactly 100 times in each position (as unit, ten,
or hundred). Hence
\[
S_{999} = 300 \cdot \left(\frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{9}
ight)
\]
For the numbers in the interval $1000 \to 1999$, compared to $0 \to 999$, there are precisely
1000 more digits 1. We get
\[
S_{1999} - S_{999} = 1000 + S_{999} \implies S_{1999} = 1000 + 600 \cdot \left(\frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{9}
ight)
\]
Finally, in the interval $2000 \to 2016$, the digit 1 appears twice as unit and seven times as
a ten, the digit 2 twice as a unit and 17 times as a thousand, the digits 3, 4, 5, and 6 each
appear exactly twice as units, and the digits 7, 8, and 9 each appear exactly once as a unit.
Hence
\[
S_{2016} - S_{1999} = 9 \cdot 1 + 19 \cdot \frac{1}{2} + 2 \cdot \left(\frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6}
ight) + 1 \cdot \left(\frac{1}{7} + \frac{1}{8} + \frac{1}{9}
ight).
\]
In the end, we get
\[
S_{2016} = 1609 \cdot 1 + 619 \cdot \frac{1}{2} + 602 \cdot \left(\frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6}
ight) + 601 \cdot \left(\frac{1}{7} + \frac{1}{8} + \frac{1}{9}
ight)
\]
\[
= m + \frac{1}{2} + \frac{2}{3} + \frac{2}{4} + \frac{2}{5} + \frac{2}{6} + \frac{6}{7} + \frac{1}{8} + \frac{7}{9}
= n + \frac{p}{2^3 \cdot 3^2 \cdot 5 \cdot 7},
\]
where $m$, $n$, and $p$ are positive integers, $p$ coprime to $2^3 \cdot 3^2 \cdot 5 \cdot 7$. Then $k! \cdot S_{2016}$ is an
integer precisely when $k!$ is a multiple of $2^3 \cdot 3^2 \cdot 5 \cdot 7$. Since $7 \mid k!$, it follows that $k \geq 7$.
Also, $7! = 2^4 \cdot 3^2 \cdot 5 \cdot 7$, implying that the least $k$ satisfying $k! \cdot S_{2016} \in \mathbb{Z}$ is $k = 7$. $\square$",463,1940,Combinatorics,6
119,shl_jbmo_2016_c2,shl_jbmo,2016,c,"The natural numbers from 1 to 50 are written down on the blackboard. At least
how many of them should be deleted, in order that the sum of any two of the remaining
numbers is not a prime?","
oindent\textbf{Solution.} Notice that if the odd, respectively even, numbers are all deleted, then the
sum of any two remaining numbers is even and exceeds 2, so it is certainly not a prime.
We prove that 25 is the minimal number of deleted numbers. To this end, we group the
positive integers from 1 to 50 in 25 pairs, such that the sum of the numbers within each
pair is a prime:
\[
(1, 2),(3, 4),(5, 6),(7, 10),(8, 9),(11, 12),(13, 16),(14, 15),(17, 20),
\]
\[
(18, 19),(21, 22),(23, 24),(25, 28),(26, 27),(29, 30),(31, 36),(32, 35),
\]
\[
(33, 34),(37, 42),(38, 41),(39, 40),(43, 46),(44, 45),(47, 50),(48, 49).
\]
Since at least one number from each pair has to be deleted, the minimal number is 25.",187,705,Combinatorics,7
120,shl_jbmo_2016_c3,shl_jbmo,2016,c,"Consider any four pairwise distinct real numbers and write one of these numbers in
each cell of a $5 \times 5$ array so that each number occurs exactly once in every $2 \times 2$ subarray.
The sum over all entries of the array is called the total sum of that array. Determine the
maximum number of distinct total sums that may be obtained in this way.","
oindent\textbf{Solution.} We will prove that the maximum number of total sums is 60.

The proof is based on the following claim.

\medskip

oindent\textbf{Claim.} Either each row contains exactly two of the numbers, or each column contains
exactly two of the numbers.

\medskip

oindent\textit{Proof of the Claim.} Indeed, let $R$ be a row containing at least three of the numbers.
Then, in row $R$ we can find three of the numbers in consecutive positions, let $x, y, z$ be the
numbers in consecutive positions (where $\{x, y, s, z\} = \{a, b, c, d\}$). Due to our hypothesis
that in every $2 \times 2$ subarray each number is used exactly once, in the row above $R$ (if there
is such a row), precisely above the numbers $x, y, z$ will be the numbers $z, t, x$ in this order.
And above them will be the numbers $x, y, z$ in this order. The same happens in the rows
below $R$ (see at the following figure).

\[
\begin{pmatrix}
\bullet & x & y & z & \bullet \\
\bullet & z & t & x & \bullet \\
\bullet & x & y & z & \bullet \\
\bullet & z & t & x & \bullet \\
\bullet & x & y & z & \bullet
\end{pmatrix}
\]

Completing all the array, it easily follows that each column contains exactly two of the
numbers and our claim has been proven.

Rotating the matrix (if it is necessary), we may assume that each row contains exactly two
of the numbers. If we forget the first row and column from the array, we obtain a $4 \times 4$
array, that can be divided into four $2 \times 2$ subarrays, containing thus each number exactly
four times, with a total sum of $4(a + b + c + d)$. It suffices to find how many different ways
are there to put the numbers in the first row $R_1$ and the first column $C_1$.

Denoting by $a_1, b_1, c_1, d_1$ the number of appearances of $a, b, c$, and respectively $d$ in $R_1$
and $C_1$, the total sum of the numbers in the entire $5 \times 5$ array will be
\[
S = 4(a + b + c + d) + a_1 \cdot a + b_1 \cdot b + c_1 \cdot c + d_1 \cdot d.
\]
If the first, the third and the fifth row contain the numbers $x, y$, with $x$ denoting the
number at the entry $(1, 1)$, then the second and the fourth row will contain only the
numbers $z, t$, with $z$ denoting the number at the entry $(2, 1)$. Then $x_1 + y_1 = 7$ and $x_1 > 3$,
$y_1 > 2$, $z_1 + t_1 = 2$, and $z_1 > t_1$. Then $\{x_1, y_1\} = \{5, 2\}$ or $\{x_1, y_1\} = \{4, 3\}$, respectively
$\{z_1, t_1\} = \{2, 0\}$ or $\{z_1, t_1\} = \{1, 1\}$. Then $(a_1, b_1, c_1, d_1)$ is obtained by permuting one of
the following quadruples:
\[
(5, 2, 2, 0),(5, 2, 1, 1),(4, 3, 2, 0),(4, 3, 1, 1).
\]
There are a total of $\dfrac{4!}{2!} = 12$ permutations of $(5, 2, 2, 0)$, also 12 permutations of $(5, 2, 1, 1)$,
24 permutations of $(4, 3, 2, 0)$ and finally, there are 12 permutations of $(4, 3, 1, 1)$. Hence,
there are at most 60 different possible total sums.

We can obtain indeed each of these 60 combinations: take three rows $ababa$ alternating
with two rows $cdcdc$ to get $(5, 2, 2, 0)$; take three rows $ababa$ alternating with one row $cdcdc$
and a row $(dcdcd)$ to get $(5, 2, 1, 1)$; take three rows $ababc$ alternating with two rows $cdcda$
to get $(4, 3, 2, 0)$; take three rows $abcda$ alternating with two rows $cdabc$ to get $(4, 3, 1, 1)$.
By choosing for example $a = 10^3, b = 10^2, c = 10, d = 1$, we can make all these sums
different. Hence, 60 is indeed the maximum possible number of different sums.",351,3395,Combinatorics,8
121,shl_jbmo_2016_c4,shl_jbmo,2016,c,"A splitting of a planar polygon is a finite set of triangles whose interiors are pairwise
disjoint, and whose union is the polygon in question. Given an integer $n \geq 3$, determine
the largest integer $m$ such that no planar $n$-gon splits into less than $m$ triangles.","
oindent\textbf{Solution.} The required maximum is $\lceil n/3 
ceil$, the least integer greater than or equal to $n/3$.

To describe a planar $n$-gon splitting into this many triangles, write $n = 3m - r$, where $m$ is a
positive integer and $r = 0, 1, 2$, and consider $m$ coplanar equilateral triangles $A_{3i}A_{3i+1}A_{3i+2}$,
$i = 0, \ldots, m - 1$, where the $A_i$ are pairwise distinct, the $A$'s of rank congruent to $0$ or $2$
modulo $3$ are all collinear, $A_2, A_3, A_5, \ldots$, $A_{3m-3}, A_{3m-1}, A_0$, in order, and the line
$A_0A_2$ separates $A_1$ from the remaining $A$'s of rank congruent to 1 modulo 3. The polygon
$A_0A_1A_2A_3A_4A_5 \ldots A_{3m-3}A_{3m-2}A_{3m-1}$ settles the case $r = 0$; removal of $A_3$ from the list
settles the case $r = 1$; and removal of $A_3$ and $A_{3m-1}$ settles the case $r = 2$.

Next, we prove that no planar $n$-gon splits into less than $n/3$ triangles. Alternatively, but
equivalently, if a planar polygon splits into $t$ triangles, then its boundary has (combinatorial) length at most $3t$. Proceed by induction on $t$. The base case $t = 1$ is clear, so let
$t > 1$.

The vertices of the triangles in the splitting may subdivide the boundary of the polygon,
making it into a possibly combinatorially longer simple loop $\Omega$. Clearly, it is sufficient to
prove that the length of $\Omega$ does not exceed $3t$.

To this end, consider a triangle in the splitting whose boundary $\omega$ meets $\Omega$ along at least
one of its edges. Trace $\Omega$ counterclockwise and let $\alpha_1, \ldots$, $\alpha_k$, in order, be the connected
components of $\Omega - \omega$. Each $\alpha_i$ is a path along $\Omega$ with distinct end points, whose terminal
point is joined to the starting point of $\alpha_{i+1}$ by a (possibly constant) path $\beta_i$ along $\omega$. Trace
$\omega$ clockwise from the terminal point of $\alpha_i$ to its starting point to obtain a path $\alpha'_i$
of positive length, and notice that $\alpha_i + \alpha'_i$ is the boundary of a polygon split into $t_i < t$ triangles.
By the induction hypothesis, the length of $\alpha_i + \alpha'_i$ does not exceed $3t_i$, and since $\alpha'_i$ has
positive length, the length of $\alpha_i$ is at most $3t_i - 1$. Consequently, the length of $\Omega - \omega$ does
not exceed $\sum_{i=1}^k (3t_i - 1) = 3t - 3 - k$.

Finally, we prove that the total length of the $\beta_i$ does not exceed $k + 3$. Begin by noticing
that no $\beta_i$ has length greater than 4, at most one has length greater than 2, and at most
three have length 2. If some $\beta_i$ has length 4, then the remaining $k - 1$ are all of length at
most 1, so the total length of the $\beta$'s does not exceed $4 + (k -1) = k + 3$. Otherwise, either
some $\beta_i$ has length 3, in which case at most one other has length 2 and the remaining $k - 2$
all have length at most 1, or the $\beta_i$ all have length less than 3, in which case there are at
most three of length 2 and the remaining $k - 3$ all have length at most 1; in the former
case, the total length of the $\beta$'s does not exceed $3 + 2 + (k - 2) = k + 3$, and in the latter,
the total length of the $\beta$'s does not exceed $3 \cdot 2 + (k - 3) = k + 3$. The conclusion follows.",271,3237,Combinatorics,9
122,shl_jbmo_2016_g1,shl_jbmo,2016,g,"Let $ABC$ be an acute angled triangle, let $O$ be its circumcentre, and let $D, E, F$ be
points on the sides $BC, AC, AB$, respectively. The circle $(c_1)$ of radius $FA$, centred at $F$,
crosses the segment $(OA)$ at $A'$ and the circumcircle $(c)$ of the triangle $ABC$ again at $K$.
Similarly, the circle $(c_2)$ of radius $DB$, centred at $D$, crosses the segment $(OB)$ at $B'$ and
the circle $(c)$ again at $L$. Finally, the circle $(c_3)$ of radius $EC$, centred at $E$, crosses the
segment $(OC)$ at $C'$ and the circle $(c)$ again at $M$. Prove that the quadrilaterals $BKFA'$,
$CLDB'$ and $AMEC'$ are all cyclic, and their circumcircles share a common point.","
oindent\textbf{Solution.} We will prove that the quadrilateral $BKFA'$ is cyclic and its circumcircle
passes through the center $O$ of the circle $(c)$.

The triangle $AFK$ is isosceles, so $m(\widehat{KFB}) = 2m(\widehat{KAB}) = m(\widehat{KOB})$. It follows that the
quadrilateral $BKFO$ is cyclic. \hfill(1)

The triangles $OFK$ and $OFA$ are congruent (S.S.S.), hence $m(\widehat{OKF}) = m(\widehat{OAF})$. The
triangle $FAA'$ is isosceles, so $m(\widehat{FA'A}) = m(\widehat{OAF})$. Therefore $m(\widehat{FA'A}) = m(\widehat{OKF})$, so
the quadrilateral $OKFA'$ is cyclic. \hfill(2)

(1) and (2) prove the initial claim.

Along the same lines, we can prove that the points $C, D, L, B', O$ and $A, M, E, C', O$ are
concyclic, respectively, so their circumcircles also pass through $O$.",668,791,Geometry,10
123,shl_jbmo_2016_g2,shl_jbmo,2016,g,"Let $ABC$ be a triangle with $m(\widehat{BAC}) = 60^\circ$. Let $D$ and $E$ be the feet of the
perpendiculars from $A$ to the external angle bisectors of $\widehat{ABC}$ and $\widehat{ACB}$, respectively. Let
$O$ be the circumcenter of the triangle $ABC$. Prove that the circumcircles of the triangles
$\Delta ADE$ and $\Delta BOC$ are tangent to each other.","
oindent\textbf{Solution.} Let $X$ be the intersection of the lines $BD$ and $CE$.

We will prove that $X$ lies on the circumcircles of both triangles $\Delta ADE$ and $\Delta BOC$ and
then we will prove that the centers of these circles and the point $X$ are collinear, which is
enough for proving that the circles are tangent to each other.

In this proof we will use the notation $(MNP)$ to denote the circumcircle of the triangle
$\Delta MNP$.

Obviously, the quadrilateral $ADXE$ is cyclic, and the circle $(DAE)$ has $[AX]$ as diameter.
\hfill(1)

Let $I$ be the incenter of triangle $ABC$. So, the point $I$ lies on the segment $[AX]$ \hfill(2), and
the quadrilateral $XBIC$ is cyclic because $IC \perp XC$ and $IB \perp XB$. So, the circle $(BIC)$ has
$[IX]$ as diameter.

Finally, $m(\widehat{BIC}) = 90^\circ + \dfrac{1}{2}m(\widehat{BAC}) = 120^\circ$ and $m(\widehat{BOC}) = 2m(\widehat{BAC}) = 120^\circ$.

So, the quadrilateral $BOIC$ is cyclic and the circle $(BOC)$ has $[IX]$ as diameter. \hfill(3)

(1), (2), (3) imply the conclusion.",358,1052,Geometry,11
124,shl_jbmo_2016_g3,shl_jbmo,2016,g,"A trapezoid $ABCD$ ($AB \parallel CD$, $AB > CD$) is circumscribed. The incircle of triangle
$ABC$ touches the lines $AB$ and $AC$ at $M$ and $N$, respectively. Prove that the incenter of
the trapezoid lies on the line $MN$.","
oindent\textbf{Solution.} Let $I$ be the incenter of triangle $ABC$ and $R$ be the common point of the lines
$BI$ and $MN$. Since
\[
m(\widehat{ANM}) = 90^\circ - \frac{1}{2}m(\widehat{MAN}) \quad \text{and} \quad m(\widehat{BIC}) = 90^\circ + \frac{1}{2}m(\widehat{MAN})
\]
the quadrilateral $IRNC$ is cyclic.
It follows that $m(\widehat{BRC}) = 90^\circ$ and therefore
\[
m(\widehat{BCR}) = 90^\circ - m(\widehat{CBR}) = 90^\circ - \frac{1}{2}\left(180^\circ - m(\widehat{BCD})
ight) = \frac{1}{2}m(\widehat{BCD}).
\]
So, $CR$ is the angle bisector of $\widehat{DCB}$ and $R$ is the incenter of the trapezoid.",224,612,Geometry,12
125,shl_jbmo_2016_g4,shl_jbmo,2016,g,"Let $ABC$ be an acute angled triangle whose shortest side is $[BC]$. Consider a variable
point $P$ on the side $[BC]$, and let $D$ and $E$ be points on $(AB]$ and $(AC]$, respectively, such
that $BD = BP$ and $CP = CE$. Prove that, as $P$ traces $[BC]$, the circumcircle of the
triangle $ADE$ passes through a fixed point.","
oindent\textbf{Solution.} We claim that the fixed point is the center of the incircle of $ABC$.

Let $I$ be the center of the incircle of $ABC$. Since $BD = BP$ and $[BI$ is the bisector of
$\widehat{DBP}$, the line $BI$ is the perpendicular bisector of $[DP]$. This yields $DI = PI$. Analogously
we get $EI = PI$. So, the point $I$ is the circumcenter of the triangle $DEP$.
This means $m(\widehat{DIE}) = 2m(\widehat{DPE})$.

On the other hand
\begin{align*}
m(\widehat{DPE}) &= 180^\circ - m(\widehat{DPB}) - m(\widehat{EPC})\\
&= 180^\circ - \left(90^\circ - \frac{1}{2}m(\widehat{DBP})
ight) - \left(90^\circ - \frac{1}{2}m(\widehat{ECP})
ight)\\
&= 90^\circ - \frac{1}{2}m(\widehat{BAC}).
\end{align*}
So, $m(\widehat{DIE}) = 2m(\widehat{DPE}) = 180^\circ - m(\widehat{DAE})$, which means that the points $A$, $D$, $E$ and
$I$ are cocyclic.

\medskip

oindent\textbf{Remark.} The fact that the incentre $I$ of the triangle $ABC$ is the required fixed point
could be guessed by considering the two extremal positions of $P$. Thus, if $P = B$, then
$D = D_B = B$ as well, and $CE = CE_B = BC$, so $m(\angle AEB) = m(\angle C) + m(\angle EBC) =$
$m(\angle C) + \dfrac{180^\circ - m(\angle C)}{2} = 90^\circ + \dfrac{m(\angle C)}{2} = m(\angle AIB)$. Hence the points $A$, $E = E_B$, $I$, $D_B = B$
are cocyclic. Similarly, letting $P = C$, the points $A$, $D = D_C$, $I$, $E_C = C$ are cocyclic.
Consequently, the circles $AD_BE_B$ and $AD_CE_C$ meet again at $I$.",322,1468,Geometry,13
126,shl_jbmo_2016_g5,shl_jbmo,2016,g,"Let $ABC$ be an acute angled triangle with orthocenter $H$ and circumcenter $O$. Assume the circumcenter $X$ of $BHC$ lies on the circumcircle of $ABC$. Reflect $O$ across $X$ to
obtain $O'$, and let the lines $XH$ and $O'A$ meet at $K$. Let $L$, $M$ and $N$ be the midpoints
of $[XB]$, $[XC]$ and $[BC]$, respectively. Prove that the points $K$, $L$, $M$ and $N$ are cocyclic.","
oindent\textbf{Solution.} The circumcircles of $ABC$ and $BHC$ have the same radius. So, $XB =$
$XC = XH = XO = r$ (where $r$ is the radius of the circle $ABC$) and $O'$ lies on $C(X, r)$.
We conclude that $OX$ is the perpendicular bisector for $[BC]$. So, $BOX$ and $COX$ are
equilateral triangles.

It is known that $AH = 2ON = r$. So, $AHO'X$ is parallelogram, and $XK = KH = r/2$.
Finally, $XL = XK = XN = XM = r/2$. So, $K$, $L$, $M$ and $N$ lie on the circle $c(X, r/2)$.",377,478,Geometry,14
127,shl_jbmo_2016_g6,shl_jbmo,2016,g,"Given an acute triangle $ABC$, erect triangles $ABD$ and $ACE$ externally, so that
$m(\widehat{ADB}) = m(\widehat{AEC}) = 90^\circ$ and $\widehat{BAD} \equiv \widehat{CAE}$. Let $A_1 \in BC$, $B_1 \in AC$ and $C_1 \in AB$ be
the feet of the altitudes of the triangle $ABC$, and let $K$ and $L$ be the midpoints of $[BC_1]$
and $[CB_1]$, respectively. Prove that the circumcenters of the triangles $AKL$, $A_1B_1C_1$ and
$DEA_1$ are collinear.","
oindent\textbf{Solution.} Let $M$, $P$ and $Q$ be the midpoints of $[BC]$, $[CA]$ and $[AB]$, respectively.

The circumcircle of the triangle $A_1B_1C_1$ is the Euler's circle. So, the point $M$ lies on this
circle.

It is enough to prove now that $[A_1M]$ is a common chord of the three circles $(A_1B_1C_1)$,
$(AKL)$ and $(DEA_1)$.

The segments $[MK]$ and $[ML]$ are midlines for the triangles $BCC_1$ and $BCB_1$ respectively,
hence $MK \parallel CC_1 \perp AB$ and $ML \parallel BB_1 \perp AC$. So, the circle $(AKL)$ has diameter $[AM]$
and therefore passes through $M$.

Finally, we prove that the quadrilateral $DA_1ME$ is cyclic.
From the cyclic quadrilaterals $ADBA_1$ and $AECA_1$, $\widehat{AA_1D} \equiv \widehat{ABD}$ and $\widehat{AA_1E} \equiv \widehat{ACE} \equiv$
$\widehat{ABD}$, so $m(\widehat{DA_1E}) = 2m(\widehat{ABD}) = 180^\circ - 2m(\widehat{DAB})$.

We notice now that $DQ = AB/2 = MP$, $QM = AC/2 = PE$ and
\[
m(\widehat{DQM}) = m(\widehat{DQB}) + m(\widehat{BQM}) = 2m(\widehat{DAB}) + m(\widehat{BAC}),
\]
\[
m(\widehat{EPM}) = m(\widehat{EPC}) + m(\widehat{CPM}) = 2m(\widehat{EAC}) + m(\widehat{CAB}),
\]
so $\Delta MPE \equiv \Delta DQM$ (S.A.S.). This leads to $m(\widehat{DME}) = m(\widehat{DMQ}) + m(\widehat{QMP}) +$
$m(\widehat{PME}) = m(\widehat{DMQ}) + m(\widehat{BQM}) + m(\widehat{QDM}) = 180^\circ - m(\widehat{DQB}) = 180^\circ - 2m(\widehat{DAB})$.
Since $m(\widehat{DA_1E}) = m(\widehat{DME})$, the quadrilateral $DA_1ME$ is cyclic.",442,1479,Geometry,15
128,shl_jbmo_2016_g7,shl_jbmo,2016,g,"Let $[AB]$ be a chord of a circle $(c)$ centered at $O$, and let $K$ be a point on the segment
$(AB)$ such that $AK < BK$. Two circles through $K$, internally tangent to $(c)$ at $A$ and $B$,
respectively, meet again at $L$. Let $P$ be one of the points of intersection of the line $KL$
and the circle $(c)$, and let the lines $AB$ and $LO$ meet at $M$. Prove that the line $MP$ is
tangent to the circle $(c)$.","
oindent\textbf{Solution.} Let $(c_1)$ and $(c_2)$ be circles through $K$, internally tangent to $(c)$ at $A$ and $B$,
respectively, and meeting again at $L$, and let the common tangent to $(c_1)$ and $(c)$ meet the
common tangent to $(c_2)$ and $(c)$ at $Q$. Then the point $Q$ is the radical center of the circles
$(c_1)$, $(c_2)$ and $(c)$, and the line $KL$ passes through $Q$.

We have $m(\widehat{QLB}) = m(\widehat{QBK}) = m(\widehat{QBA}) = \frac{1}{2}m(\overset{\frown}{BA}) = m(\widehat{QOB})$. So, the quadrilateral
$OBQL$ is cyclic. We conclude that $m(\widehat{QLO}) = 90^\circ$ and the points $O, B, Q, A$ and $L$ are
cocyclic on a circle $(k)$.

In the sequel, we will denote $\mathcal{P}_\omega(X)$ the power of the point $X$ with respect of the circle $\omega$.

\medskip

oindent\textit{The first continuation.}

From $MO^2 - OP^2 = \mathcal{P}_c(M) = MA \cdot MB = \mathcal{P}_k(M) = ML \cdot MO = (MO - OL) \cdot MO =$
$MO^2 - OL \cdot MO$ follows that $OP^2 = OL \cdot OM$. Since $PL \perp OM$, this shows that the
triangle $MPO$ is right at point $P$. Thus, the line $MP$ is tangent to the circle $(c)$.

\medskip

oindent\textit{The second continuation.}

Let $R \in (c)$ be so that $BR \perp MO$. The triangle $LBR$ is isosceles with $LB = LR$, so
$\widehat{OLR} \equiv \widehat{OLB} \equiv \widehat{OQB} \equiv \widehat{OQA} \equiv \widehat{MLA}$. We conclude that the points $A$, $L$ and $R$ are
collinear.

Now $m(\widehat{AMR}) + m(\widehat{AOR}) = m(\widehat{AMR}) + 2m(\widehat{ABR}) = m(\widehat{AMR}) + m(\widehat{ABR}) + m(\widehat{MRB}) =$
$180^\circ$, since the triangle $MBR$ is isosceles. So, the quadrilateral $MAOR$ is cyclic.

This yields $LM \cdot LO = -\mathcal{P}_{(MAOR)}(L) = LA \cdot LR = -\mathcal{P}_c(L) = LP^2$, which as above, shows
that $OP \perp PM$.

\medskip

oindent\textit{The third continuation.}

$\widehat{KLA} \equiv \widehat{KAQ} \equiv \widehat{KLB}$ and $m(\widehat{MLK}) = 90^\circ$ show that $[LK$ and $[LM$ are the internal and
external bisectors of the angle $\widehat{ALB}$, so $(M, K)$ and $(A, B)$ are harmonic conjugates. So,
$LK$ is the polar line of $M$ in the circle $(c)$.",410,2148,Geometry,16
129,shl_jbmo_2016_n1,shl_jbmo,2016,n,Determine the largest positive integer $n$ that divides $p^6 - 1$ for all primes $p > 7$.,"
oindent\textbf{Solution.} Note that
\[
p^6 - 1 = (p - 1)(p + 1)(p^2 - p + 1)(p^2 + p + 1).
\]
For $p = 11$ we have
\[
p^6 - 1 = 1771560 = 2^3 \cdot 3^2 \cdot 5 \cdot 7 \cdot 19 \cdot 37.
\]
For $p = 13$ we have
\[
p^6 - 1 = 2^3 \cdot 3^2 \cdot 7 \cdot 61 \cdot 157.
\]
From the last two calculations we find evidence to try showing that $p^6 - 1$ is divisible by
$2^3 \cdot 3^2 \cdot 7 = 504$ and this would be the largest positive integer that divides $p^6 - 1$ for all
primes greater than 7.

By Fermat's theorem, $7 \mid p^6 - 1$.

Next, since $p$ is odd, $8 \mid p^2 - 1 = (p - 1)(p + 1)$, hence $8 \mid p^6 - 1$.

It remains to show that $9 \mid p^6 - 1$.

Any prime number $p$, $p > 3$ is $1$ or $-1$ modulo $3$.
In the first case both $p - 1$ and $p^2 + p + 1$ are divisible by 3, and in the second case, both
$p + 1$ and $p^2 - p + 1$ are divisible by 3.

Consequently, the required number is indeed 504.

\medskip

oindent\textit{Alternative solution}

Let $q$ be a (positive) prime factor of $n$. Then $q \leq 7$, as $q 
mid q^6 - 1$. Also, $q$ is not 5, as the
last digit of $13^6 - 1$ is 8.

Hence, the prime factors of $n$ are among 2, 3, and 7.

Next, from $11^6 - 1 = 2^3 \cdot 3^2 \cdot 5 \cdot 7 \cdot 19 \cdot 37$ it follows that the largest integer $n$ such that
$n \mid p^6 - 1$ for all primes $p$ greater than 7 is at most $2^3 \cdot 3^2 \cdot 7$, and it remains to prove that
504 divides $p^6 - 1$ for all primes greater than 7.",89,1451,Number Theory,17
130,shl_jbmo_2016_n2,shl_jbmo,2016,n,"Find the maximum number of natural numbers $x_1, x_2, \ldots, x_m$ satisfying the conditions:
\begin{enumerate}[label=\alph*)]
\item No $x_i - x_j$, $1 \leq i < j \leq m$ is divisible by 11; and
\item The sum $x_2 x_3 \ldots x_m + x_1 x_3 \ldots x_m + \cdots + x_1 x_2 \ldots x_{m-1}$ is divisible by 11.
\end{enumerate}","
oindent\textbf{Solution.} The required maximum is 10.

According to a), the numbers $x_i$, $1 \leq i \leq m$, are all different (mod 11) \hfill(1)

Hence, the number of natural numbers satisfying the conditions is at most 11.

If $x_j \equiv 0 \pmod{11}$ for some $j$, then
\[
x_2 x_3 \ldots x_m + x_1 x_3 \ldots x_m + \cdots + x_1 x_2 \ldots x_{m-1} \equiv x_1 \ldots x_{j-1} x_{j+1} \ldots x_m \pmod{11},
\]
which would lead to $x_i \equiv 0 \pmod{11}$ for some $i 
eq j$, contradicting (1).

We now prove that 10 is indeed the required maximum.

Consider $x_i = i$, for all $i \in \{1, 2, \ldots, 10\}$. The products $2 \cdot 3 \cdots 10$, $1 \cdot 3 \cdots 10$, $\ldots$,
$1 \cdot 2 \cdots 9$ are all different (mod 11), and so
\[
2 \cdot 3 \cdots 10 + 1 \cdot 3 \cdots 10 + \cdots + 1 \cdot 2 \cdots 9 \equiv 1 + 2 + \cdots + 10 \pmod{11},
\]
and condition b) is satisfied, since $1 + 2 + \cdots + 10 = 55 = 5 \cdot 11$.",320,926,Number Theory,18
131,shl_jbmo_2016_n3,shl_jbmo,2016,n,"Find all positive integers $n$ such that the number $A_n = \dfrac{2^{4n+2}+1}{65}$ is
\begin{enumerate}[label=\alph*)]
\item an integer;
\item a prime.
\end{enumerate}","
oindent\textbf{Solution.} a) Note that $65 = 5 \cdot 13$.

Obviously, $5 = 2^2 + 1$ is a divisor of $(2^2)^{2n+1} + 1 = 2^{4n+2} + 1$ for any positive integer $n$.

Since $2^{12} \equiv 1 \pmod{13}$, if $n \equiv r \pmod 3$, then $2^{4n+2} + 1 \equiv 2^{4r+2} + 1 \pmod{13}$. Now,
$2^{4 \cdot 0 + 2} + 1 = 5$, $2^{4 \cdot 1 + 2} + 1 = 65$, and $2^{4 \cdot 2 + 2} + 1 = 1025 = 13 \cdot 78 + 11$. Hence 13 is a divisor
of $2^{4n+2} + 1$ precisely when $n \equiv 1 \pmod 3$. Hence, $A_n$ is an integer iff $n \equiv 1 \pmod 3$.

b) Applying the identity $4x^4 + 1 = (2x^2 - 2x + 1)(2x^2 + 2x + 1)$, we have $2^{4n+2} + 1 =$
$(2^{2n+1} - 2^{n+1} + 1)(2^{2n+1} + 2^{n+1} + 1)$. For $n = 1$, $A_1 = 1$, which is not a prime. According
to a), if $n 
eq 1$, then $n \geq 4$. But then $2^{2n+1} + 2^{n+1} + 1 > 2^{2n+1} - 2^{n+1} + 1 > 65$, and $A_n$
has at least two factors. We conclude that $A_n$ can never be a prime.

\medskip

oindent\textit{Alternative Solution to b):} Knowing that $n = 3k + 1$ in order for $A_n$ to be an integer,
\[
2^{4n+2} + 1 = 2^{12k+6} + 1 = (2^{4k+2})^3 + 1 = (2^{4k+2} + 1)(2^{8k+4} - 2^{4k+2} + 1) \quad (*)
\]
As in the previous solution, if $k = 0$, then $A_1 = 1$, if $k = 1$, then $A_4 = 2^{12} - 2^6 + 1 = 4033 = 37 \cdot 109$,
and for $k \geq 2$ both factors in $(*)$ are larger than 65, so $A_{3k+1}$ is not a prime.",167,1350,Number Theory,19
132,shl_jbmo_2016_n4,shl_jbmo,2016,n,"Find all triples of integers $(a, b, c)$ such that the number
\[
N = \frac{(a - b)(b - c)(c - a)}{2} + 2
\]
is a power of 2016.","
oindent\textbf{Solution.} Let $z$ be a positive integer such that
\[
(a - b)(b - c)(c - a) + 4 = 2 \cdot 2016^z.
\]
We set $a - b = -x$, $b - c = -y$ and we rewrite the equation as
\[
xy(x + y) + 4 = 2 \cdot 2016^z.
\]
Note that the right hand side is divisible by 7, so we have that
\[
xy(x + y) + 4 \equiv 0 \pmod 7
\]
or
\[
3xy(x + y) \equiv 2 \pmod 7
\]
or
\[
(x + y)^3 - x^3 - y^3 \equiv 2 \pmod 7. \tag{4.1}
\]
Note that, by Fermat's Little Theorem, we have that for any integer $k$ the cubic residues
are $k^3 \equiv -1, 0, 1 \pmod 7$. It follows that in (4.1) some of $(x + y)^3$, $x^3$ and $y^3$
should be divisible by 7, but in this case, $xy(x + y)$ is divisible by 7 and this is a contradiction. So,
the only possibility is to have $z = 0$ and consequently, $xy(x + y) + 4 = 2$, or, equivalently,
$xy(x + y) = -2$. The only solution of the latter is $(x, y) = (-1, -1)$, so the required triples
are $(a, b, c) = (k + 2, k + 1, k)$, $k \in \mathbb{Z}$, and all their cyclic permutations.",127,999,Number Theory,20
133,shl_jbmo_2016_n5,shl_jbmo,2016,n,"Determine all four-digit numbers $\overline{abcd}$ such that
\[
(a + b)(a + c)(a + d)(b + c)(b + d)(c + d) = \overline{abcd}.
\]","
oindent\textbf{Solution.} Depending on the parity of $a, b, c, d$, at least two of the factors $(a + b)$, $(a + c)$, $(a + d)$, $(b + c)$, $(b + d)$, $(c + d)$ are even, so that $4 \mid \overline{abcd}$.

We claim that $3 \mid \overline{abcd}$.

Assume $a + b + c + d \equiv 2 \pmod 3$. Then $x + y \equiv 1 \pmod 3$, for all distinct $x, y \in \{a, b, c, d\}$.
But then the left hand side in the above equality is congruent to $1 \pmod 3$ and the right
hand side congruent to $2 \pmod 3$, contradiction.

Assume $a + b + c + d \equiv 1 \pmod 3$. Then $x + y \equiv 2 \pmod 3$, for all distinct $x, y \in \{a, b, c, d\}$,
and $x \equiv 1 \pmod 3$, for all $x, y \in \{a, b, c, d\}$. Hence, $a, b, c, d \in \{1, 4, 7\}$, and since $4 \mid \overline{abcd}$,
we have $c = d = 4$. Therefore, $8 \mid \overline{ab44}$, and since at least one more factor is even, it follows
that $16 \mid \overline{ab44}$. Then $b 
eq 4$, and the only possibilities are $b = 1$, implying $a = 4$, which is
impossible because $\overline{4144}$ is not divisible by $5 = 1 + 4$, or $b = 7$, implying $11 \mid \overline{a744}$, hence
$a = 7$, which is also impossible because $7744$ is not divisible by $14 = 7 + 7$.

We conclude that $3 \mid \overline{abcd}$, hence also $3 \mid a + b + c + d$. Then at least one factor $x + y$
of $(a + b)$, $(a + c)$, $(a + d)$, $(b + c)$, $(b + d)$, $(c + d)$ is a multiple of 3, implying that also
$3 \mid a + b + c + d - x - y$, so $9 \mid \overline{abcd}$. Then $9 \mid a + b + c + d$, and $a + b + c + d \in \{9, 18, 27, 36\}$.

Using the inequality $xy \geq x + y - 1$, valid for all $x, y \in \mathbb{N}^*$, if $a + b + c + d \in \{27, 36\}$, then
\[
\overline{abcd} = (a + b)(a + c)(a + d)(b + c)(b + d)(c + d) \geq 26^3 > 10^4,
\]
which is impossible.

Using the inequality $xy \geq 2(x + y) - 4$ for all $x, y \geq 2$, if $a + b + c + d = 18$ and all
two-digit sums are greater than 1, then $\overline{abcd} \geq 32^3 > 10^4$. Hence, if $a + b + c + d = 18$,
some two-digit sum must be 1, hence the complementary sum will be 17, and the digits
are $\{a, b, c, d\} = \{0, 1, 8, 9\}$. But then $\overline{abcd} = 1 \cdot 17 \cdot 8 \cdot 9^2 \cdot 10 > 10^4$.

We conclude that $a + b + c + d = 9$. Then among $a, b, c, d$ there are either three odd or
three even numbers, and $8 \mid \overline{abcd}$.

If three of the digits are odd, then $d$ is even and since $c$ is odd, divisibility by 8 implies
that $d \in \{2, 6\}$. If $d = 6$, then $a = b = c = 1$. But 1116 is not divisible by 7, so this is not
a solution. If $d = 2$, then $a, b, c$ are either $1, 1, 5$ or $1, 3, 3$ in some order. In the first case
$2 \cdot 6^2 \cdot 3^2 \cdot 7 = 4536 
eq \overline{abcd}$. The second case cannot hold because the resulting number is
not a multiple of 5.

Hence, there has to be one odd and three even digits. At least one of the two-digits sums
of even digits is a multiple of 4, and since there cannot be two zero digits, we have either
$x + y = 4$ and $z + t = 5$, or $x + y = 8$ and $z + t = 1$ for some ordering $x, y, z, t$ of $a, b, c, d$.

In the first case we have $d = 0$ and the digits are $0, 1, 4, 4$, or $0, 2, 3, 4$, or $0, 2, 2, 5$. None of
these is a solution because $1 \cdot 4^2 \cdot 5^2 \cdot 8 = 3200$, $2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 = 5040$ and $2^2 \cdot 5 \cdot 4 \cdot 7^2 = 3920$.

In the second case two of the digits are 0 and 1, and the other two have to be either 4 and
4, or 2 and 6. We already know that the first possibility fails. For the second, we get
\[
(0 + 1) \cdot (0 + 2) \cdot (0 + 6) \cdot (1 + 2) \cdot (1 + 6) \cdot (2 + 6) = 2016
\]
and $\overline{abcd} = 2016$ is the only solution.",128,3658,Number Theory,21
492,tst_jbmo_ro_2016_1_p1,tst_jbmo,2016,g,"Let $ABC$ be a non-equilateral triangle with $m(\angle A) = 60^\circ$. If the Euler line of triangle $ABC$ intersects the sides of angle $\angle BAC$ at points $D$ and $E$, show that triangle $ADE$ is equilateral.","\subsection*{Solution}

Let $H$ and $O$ be the orthocenter and circumcenter of triangle $ABC$, respectively, $R$ the circumradius, $D$ and $E$ the intersections of line $OH$ with $AB$ and $AC$ respectively, and $B'$, $C'$ the feet of the altitudes from $B$ and $C$.

Quadrilateral $BCC'B'$ is cyclic, so $\angle AB'C' \equiv \angle ABC$. It follows that $\triangle AB'C' \sim \triangle ABC$, with similarity ratio
\[
\frac{AC'}{AC} = \cos\!\left(\widehat{BAC}
ight) = \frac{1}{2}.
\]
Then the ratio of the circumradii of these triangles also equals the similarity ratio, that is
\[
\frac{AH}{2R} = \frac{1}{2}, \quad \text{so} \quad AH = R = AO. \tag{1}
\]
It is known that rays $(AH)$ and $(AO)$ are isogonal, i.e.\ $\angle BAO \equiv \angle CAH$. \hfill(2)

From (1) it follows that $\angle AOH \equiv \angle AHO$, hence $\angle AOD \equiv \angle AHE$. From (1) and (2) it follows that triangles $AOD$ and $AHE$ are congruent, so $AD = AE$ and the conclusion follows.

(The order of points on line $OH$ depends on which of the sides $[AB]$ and $[AC]$ is longer, but the statements hold in both cases.)",213,1103,Geometry,1
493,tst_jbmo_ro_2016_1_p2,tst_jbmo,2016,a,"Let $m, n$ be positive integers and let $x, y, z \in [0,1]$ be real numbers. Prove that
\[
0 \;\le\; x^{m+n} + y^{m+n} + z^{m+n} - x^m y^n - y^m z^n - z^m x^n \;\le\; 1
\]
and determine the cases of equality.","\subsection*{Solution}

Without loss of generality, we may assume $x$ is the largest among $x, y, z$, and then we can write
\[
x^{m+n}+y^{m+n}+z^{m+n}-x^m y^n - y^m z^n - z^m x^n
= (x^m - z^m)(x^n - y^n) + (y^m - z^m)(y^n - z^n) \ge 0.
\]
The minimum $0$ is thus achieved when $x = \max\{y,z\}$ and $y = z$, i.e.\ when $x = y = z$.

The maximum is achieved for $x = 1$, and has the value
\[
1 + y^{m+n} + z^{m+n} - y^n - y^m z^n - z^m
= 1 - y^n(1 - y^m) - z^m(1 - z^n) - y^m z^n \le 1,
\]
with equality only when one of $y, z$ is $0$ and the other is $0$ or $1$. Therefore there are $6$ cases of equality:
\[
(x,y,z) \in \{(1,0,0),\,(0,1,0),\,(0,0,1),\,(1,1,0),\,(1,0,1),\,(0,1,1)\}.
\]",208,686,Algebra,2
494,tst_jbmo_ro_2016_1_p3,tst_jbmo,2016,n,Let $M$ be the set of natural numbers $k$ for which there exists $n \in \mathbb{N}^*$ such that the remainder of the division of $3^n$ by $n$ is $k$. Prove that the set $M$ is infinite.,"\subsection*{Solution 1}

Let $j$ be a fixed positive integer, and let $p > 2$ be a prime such that $2^j p > 3^{2^j}$. We have $3^{2^j}(3^{2^j(p-1)} - 1) \equiv 0 \pmod{2^j p}$, since $2\varphi(2^j p) = 2^j(p-1)$. Then $3^{2^{jp}} \equiv 3^{2^j} \pmod{2^j p}$, and so for $n = 2^j p$ we have $r_n = 3^{2^j}$. Therefore $3^{2^j} \in M$ for all $j \in \mathbb{N}^*$, so $M$ is infinite.

\subsection*{Solution 2}

Choosing $n = 2 \cdot 3^j$, the remainder $r_n$ of the division of $3^n$ by $n$ satisfies: $r_n 
e 0$, $r_n \equiv 0 \pmod{3^j}$, so $r_n \ge 3^j$. (In fact $r_n = 3^j$.) Therefore $M$ contains arbitrarily large numbers, so $M$ is infinite.",185,652,Number Theory,3
495,tst_jbmo_ro_2016_1_p4,tst_jbmo,2016,g,"Let triangle $ABC$ be acute with $AB 
e AC$, and let $D, E, F$ be the points of tangency of the incircle $\omega$ of the triangle with sides $BC$, $CA$, and $AB$ respectively. The perpendicular to $BC$ at $C$ meets $EF$ at $M$, and the perpendicular to $BC$ at $B$ meets $EF$ at $N$. Line $DM$ meets $\omega$ again at $P$ and line $DN$ meets $\omega$ again at $Q$. Prove that $DP = DQ$.","\subsection*{Solution 1}

Let $\{T\} = EF \cap BC$. Applying Menelaus' theorem to triangle $ABC$ and transversal $E$-$F$-$T$:
\[
\frac{TB}{TC} \cdot \frac{EC}{EA} \cdot \frac{FA}{FB} = 1,
\quad \text{i.e.} \quad
\frac{TB}{TC} \cdot \frac{p-c}{p-a} \cdot \frac{p-a}{p-b} = 1,
\quad \text{so} \quad
\frac{TB}{TC} = \frac{p-b}{p-c}. \tag{1}
\]
It follows that triangles $TBN$ and $TCM$ are similar, so $\dfrac{TB}{TC} = \dfrac{BN}{CM}$. From (1):
\[
\frac{BN}{CM} = \frac{p-b}{p-c}, \quad \frac{BD}{CD} = \frac{p-b}{p-c},
\]
and $m(\angle DBN) = m(\angle DCM) = 90^\circ$, which shows that triangles $BDN$ and $CDM$ are similar, so angles $\angle BDN$ and $\angle CDM$ are congruent. We deduce that arcs $DQ$ and $DP$ are equal, so $DP = DQ$.

\subsection*{Solution 2}

Let $S$ be the intersection of the altitude from $A$ with line $EF$. Lines $BN$, $AS$, $CM$ are parallel, so triangles $BNF$ and $ASF$ are similar, and likewise triangles $ASE$ and $CME$. We obtain $\dfrac{BN}{AS} = \dfrac{BF}{FA}$ and $\dfrac{AS}{CM} = \dfrac{AE}{EC}$.

Multiplying these two relations:
\[
\frac{BN}{CM} = \frac{BF}{FA} \cdot \frac{AE}{EC} = \frac{BF}{EC} = \frac{BD}{DC}.
\]
(We used $AE = AF$, $BF = BD$, $CE = CD$.)

It follows that right triangles $BDN$ and $CDM$ are similar (SAS), which leads to the same conclusion as in Solution 1.",386,1324,Geometry,4
496,tst_jbmo_ro_2016_1_p5,tst_jbmo,2016,c,"The unit squares of an $n \times n$ board, $n \ge 2$, are colored either black or white such that every black square has at least $3$ white neighbors. (A neighbor of a unit square is a unit square that shares a side with it.) What is the maximum number of black unit squares on the board?","The answer is $\dfrac{n^2-1}{2}$ if $n$ is odd and $\dfrac{n^2-4}{2}$ if $n$ is even.

Observe that:
\begin{itemize}
  \item Corner unit squares, having fewer than $3$ neighbors, must be white.
  \item The other border squares have exactly $3$ neighbors, so no two adjacent border squares can both be black.
  \item In any $2 \times 2$ square we can have at most two black squares (otherwise a black square would already have two black neighbors, leaving at most two white ones).
\end{itemize}

\textbf{Case $n$ odd:} Color the board like a chessboard with white corners. This coloring satisfies the condition and contains $\dfrac{n^2-1}{2}$ black squares, so the maximum is at least $\dfrac{n^2-1}{2}$.

On the other hand, partition the board into: an $(n-1)\times(n-1)$ square in the top-left corner, a unit square in the bottom-right corner, an $(n-1)\times 1$ rectangle on the right side, and a $1\times(n-1)$ rectangle on the bottom. Tiling the $(n-1)\times(n-1)$ square with $2\times 2$ squares and each $(n-1)$-square strip with dominoes, by the observations above we can have at most
\[
\frac{(n-1)^2}{2} + \frac{n-1}{2} + \frac{n-1}{2} + 0 = \frac{n^2-1}{2}
\]
black squares. Hence the maximum for odd $n$ is $\dfrac{n^2-1}{2}$.

\textbf{Case $n$ even:} Color the board like a chessboard; this makes two corners black. Recolor those corners white. The resulting coloring satisfies the condition and has $\dfrac{n^2-4}{2}$ black squares, so the maximum is at least $\dfrac{n^2-4}{2}$.

Partition the board into: an $(n-2)\times(n-2)$ square in the center, $4$ unit squares at the corners, and $4$ rectangles of size $(n-2)\times 1$ or $1\times(n-2)$ along the sides. By the same argument, we can have at most
\[
\frac{(n-2)^2}{2} + 4 \cdot \frac{n-2}{2} + 4 \cdot 0 = \frac{n^2-4}{2}
\]
black squares. Hence the maximum for even $n$ is $\dfrac{n^2-4}{2}$.",288,1863,Combinatorics,5
497,tst_jbmo_ro_2016_2_p1,tst_jbmo,2016,g,"Triangle $ABC$, inscribed in a circle with center $O$ and radius $R$, has the property that the radius of the $A$-excircle equals $R$. If $D, E, F$ are the points of tangency of the $A$-excircle with lines $BC$, $CA$, and $AB$ respectively, prove that lines $OD$ and $EF$ are perpendicular.","\subsection*{Solution}

Let $T$ be the midpoint of arc $BC$ of the circumcircle that does not contain $A$. Let $I_a$ be the center of the $A$-excircle. Then $OT$ is the perpendicular bisector of $[BC]$, so $OT \perp BC$ and $I_a D \perp BC$. Also $OT = I_a D = R$, so $ODI_aT$ is a parallelogram. It follows that $OD \parallel TI_a$, or $OD \parallel AI_a$. But $AE = AF$ and $(AI_a)$ is the bisector of angle $\angle FAE$, so $AI_a \perp EF$. Therefore $OD \perp EF$.",290,468,Geometry,6
498,tst_jbmo_ro_2016_2_p2,tst_jbmo,2016,a,"Let $a, b, c > 0$ with $abc \ge 1$. Prove that
\[
\frac{1}{a^3 + 2b^3 + 6} + \frac{1}{b^3 + 2c^3 + 6} + \frac{1}{c^3 + 2a^3 + 6} \le \frac{1}{3}.
\]","\subsection*{Solution 1}

Using $a^3 + b^3 + 1 \ge 3ab$ and $b^3 + 1 + 1 \ge 3b$, we get $a^3 + 2b^3 + 6 \ge 3(ab + b + 1)$, so
\[
\frac{1}{a^3+2b^3+6} \le \frac{1}{3(ab+b+1)}.
\]
It suffices to prove
\[
\frac{1}{ab+b+1} + \frac{1}{bc+c+1} + \frac{1}{ca+a+1} \le 1.
\]
But
\[
\frac{1}{ab+b+1} + \frac{1}{bc+c+1} + \frac{1}{ca+a+1}
= \frac{1}{ab+b+1} + \frac{ab}{ab^2c+abc+ab} + \frac{b}{abc+ab+b}
\;\overset{abc\ge 1}{\le}\;
\frac{1}{ab+b+1} + \frac{ab}{b+1+ab} + \frac{b}{1+ab+b}
= \frac{1+ab+b}{ab+b+1} = 1.
\]
Equality holds if and only if $a = b = c = 1$.

\subsection*{Solution 2}

Subtract $\frac{1}{6}$ from each fraction on the left side. The inequality becomes successively:
\[
\sum_{\text{cyc}} \left(\frac{1}{a^3+2b^3+6} - \frac{1}{6}
ight) \le \frac{1}{3} - \frac{1}{2},
\]
\[
\sum_{\text{cyc}} \frac{-(a^3+2b^3)}{6(a^3+2b^3+6)} \le -\frac{1}{6},
\]
\[
\sum_{\text{cyc}} \frac{a^3+2b^3}{a^3+2b^3+6} \ge 1.
\]
We prove separately that
\[
\sum_{\text{cyc}} \frac{a^3}{a^3+2b^3+6} \ge \frac{1}{3}
\quad \text{and} \quad
\sum_{\text{cyc}} \frac{b^3}{a^3+2b^3+6} \ge \frac{1}{3},
\]
which combined imply the desired inequality.

We have
\[
\sum_{\text{cyc}} \frac{a^3}{a^3+2b^3+6}
= \sum_{\text{cyc}} \frac{a^4}{a^4+2ab^3+6a}
\overset{\text{CBS}}{\ge}
\frac{(a^2+b^2+c^2)^2}{a^4+b^4+c^4+2(ab^3+bc^3+ca^3)+6(a+b+c)}
\overset{?}{\ge} \frac{1}{3}.
\]
The last step $(?)$ is equivalent to
\[
3(a^4+b^4+c^4) + 6(a^2b^2+b^2c^2+c^2a^2) \ge a^4+b^4+c^4 + 2(ab^3+bc^3+ca^3) + 6(a+b+c),
\]
which follows from $2(a^4+b^4+c^4) \ge 2(ab^3+bc^3+ca^3)$ (by the rearrangement inequality) and
\[
6(a^2b^2+b^2c^2+c^2a^2) \ge 6abc(a+b+c) \ge 6(a+b+c).
\]
Similarly
\[
\sum_{\text{cyc}} \frac{b^3}{a^3+2b^3+6}
= \sum_{\text{cyc}} \frac{b^4}{ba^3+2b^4+6b}
\overset{\text{CBS}}{\ge}
\frac{(a^2+b^2+c^2)^2}{2(a^4+b^4+c^4)+(ba^3+cb^3+ac^3)+6(a+b+c)}
\ge \frac{1}{3},
\]
using $a^4+b^4+c^4 \ge ba^3+cb^3+ac^3$ (by rearrangement) and the same estimate as before.

\subsection*{Solution 3}

Using $a^3+2b^3 = a^3+b^3+b^3 \ge 3ab^2$ and the cyclic analogues, it suffices to prove
\[
\frac{1}{ab^2+2} + \frac{1}{bc^2+2} + \frac{1}{ca^2+2} \le 1.
\]
Subtracting $\frac{1}{2}$ from each fraction, this reduces to
\[
\frac{ab^2}{ab^2+2} + \frac{bc^2}{bc^2+2} + \frac{ca^2}{ca^2+2} \ge 1.
\]
Since $abc \ge 1$, it suffices to show
\[
\frac{ab^2}{ab^2+2abc} + \frac{bc^2}{bc^2+2abc} + \frac{ca^2}{ca^2+2abc} \ge 1,
\quad \text{i.e.} \quad
\frac{b}{b+2c} + \frac{c}{c+2a} + \frac{a}{a+2b} \ge 1.
\]
But
\[
\frac{b}{b+2c}+\frac{c}{c+2a}+\frac{a}{a+2b}
= \frac{b^2}{b^2+2bc}+\frac{c^2}{c^2+2ca}+\frac{a^2}{a^2+2ab}
\overset{\text{CBS}}{\ge}
\frac{(a+b+c)^2}{b^2+2bc+c^2+2ca+a^2+2ab} = 1.
\]
Equality holds when $a = b = c = 1$.",148,2697,Algebra,7
499,tst_jbmo_ro_2016_2_p3,tst_jbmo,2016,n,"Let $n$ be a natural number, $n \ge 3$, and $A = \{2^n - 1,\; 3^n - 1,\; \ldots,\; (n-1)^n - 1\}$. If none of the elements of $A$ is divisible by $n$, show that $n$ is squarefree. Is $n$ necessarily prime?","\subsection*{Solution}

Suppose no element of $A$ is divisible by $n$; we show $n$ is squarefree. Assume not; then $n = p \cdot a$ with $p$ prime and $p \mid a$, $a > 1$. We show $n \mid (a+1)^n - 1$.

Indeed $(a+1)^n - 1 = a \cdot \bigl[(a+1)^{n-1} + (a+1)^{n-2} + \cdots + 1\bigr]$. Since $a+1 \equiv 1 \pmod{p}$, we get $(a+1)^{n-1}+\cdots+1 \equiv n \equiv 0 \pmod{p}$. Hence $p^2 \mid (a+1)^n - 1$; similarly the full factor works out, giving a contradiction with the assumption.

The answer to the second question is \textbf{no}: $n$ need not be prime. For example, take $n = 3 \cdot 5 = 15$; one can verify that none of the elements of $A$ is divisible by $15$.

\begin{remark}
Every prime $n$ has the stated property. But $n = 15$ shows that non-primes can also have it. The problem shows that \emph{only} squarefree numbers can have the property. Finally, $n = 6$ shows that not all squarefree numbers have the property.
\end{remark}",205,942,Number Theory,8
500,tst_jbmo_ro_2016_2_p4,tst_jbmo,2016,c,"The unit squares of a $4 \times 4$ board are initially colored white. A move consists of choosing a $1 \times 3$ or $3 \times 1$ rectangle of unit squares and toggling the color of each of its three squares (white to black or black to white). Is it possible, after finitely many moves, to make the entire board black?","\subsection*{Solution}

Fill the unit squares with numbers $1$, $2$, $3$ as follows:
\[
\begin{array}{|c|c|c|c|}
\hline
1 & 2 & 3 & 1 \\
\hline
2 & 3 & 1 & 2 \\
\hline
3 & 1 & 2 & 3 \\
\hline
1 & 2 & 3 & 1 \\
\hline
\end{array}
\]
Every $1 \times 3$ or $3 \times 1$ rectangle contains exactly one square of each label. So each move affects exactly one square of each label. Initially we have $6$ white squares labeled $1$, $5$ labeled $2$, and $5$ labeled $3$, for a total of $11$ white squares labeled $1$ or $2$. A move does not change the parity of the total number of white squares labeled $1$ or $2$. Therefore it is impossible to reach a configuration with $0$ white squares labeled $1$ or $2$.

Hence the answer is \textbf{no}.",317,734,Combinatorics,9
501,tst_jbmo_ro_2016_3_p1,tst_jbmo,2016,n,"For $n \in \mathbb{N}$ consider the system
\[
(S_n):\quad \begin{cases} x^2 + ny^2 = z^2 \\ nx^2 + y^2 = t^2 \end{cases}, \quad x, y, z, t \in \mathbb{N}^*.
\]
Let $M_1 = \{n \in \mathbb{N} \mid \text{system } (S_n) \text{ has infinitely many solutions}\}$ and $M_2 = \{n \in \mathbb{N} \mid \text{system } (S_n) \text{ has no solutions}\}$. Prove that:
\begin{enumerate}[label=\alph*)]
  \item $7 \in M_1$,\quad $10 \in M_2$.
  \item The sets $M_1$ and $M_2$ are infinite.
\end{enumerate}","\subsection*{Solution}

\textbf{a)} Observe that $x=1,\, y=3,\, z=8,\, t=4$ is a solution of $(S_7)$, so $(k, 3k, 8k, 4k)$ for $k \in \mathbb{N}^*$ are all solutions, hence $7 \in M_1$.

If $(x,y,z,t)$ were a solution of $(S_{10})$, then $11(x^2+y^2) = z^2+t^2$. From $11 \mid z^2+t^2$ it follows that $11 \mid z$ and $11 \mid t$. Then $11 \mid x^2+y^2$, so $11 \mid x$ and $11 \mid y$. Thus there exist $x_1,y_1,z_1,t_1 \in \mathbb{N}^*$ with $x=11x_1$, $y=11y_1$, $z=11z_1$, $t=11t_1$. It follows that $11(x_1^2+y_1^2)=z_1^2+t_1^2$. Continuing this (infinite descent), $x$ must be divisible by every power of $11$, which contradicts $x 
e 0$. So $(S_{10})$ has no solutions, i.e.\ $10 \in M_2$.

\textbf{b)} Every natural number of the form $m^2 - 1$, $m \in \mathbb{N}^*$, belongs to $M_1$: choose $x=y$ and note that $(k,k,mk,mk)$ for $k \in \mathbb{N}^*$ are solutions of $(S_{m^2-1})$, so $m^2-1 \in M_1$ for all $m \in \mathbb{N}^*$.

The argument in part a) for $n=10$ works for any $n = p-1$ where $p$ is a prime of the form $4m+3$. All such numbers belong to $M_2$. Since there are infinitely many primes of the form $4m+3$, the set $M_2$ is infinite.",489,1161,Number Theory,10
502,tst_jbmo_ro_2016_3_p2,tst_jbmo,2016,a,Let $x$ and $y$ be nonzero real numbers such that $x^3 + y^3 + 3x^2y^2 = x^3y^3$. Determine all possible values of the expression $E = \dfrac{1}{x} + \dfrac{1}{y}$.,"\subsection*{Solution}

The given relation is rewritten successively:
\[
(x+y)^3 - 3xy(x+y) = x^3y^3 - 3x^2y^2,
\]
\[
(x+y)^3 - (xy)^3 = 3xy(x+y) - 3x^2y^2,
\]
\[
(x+y-xy)\bigl(x^2+2xy+y^2+x^2y+xy^2+x^2y^2\bigr) = 3xy(x+y-xy).
\]
So either $x+y = xy$, i.e.\ $E = 1$, or
\[
x^2+2xy+y^2+x^2y+xy^2+x^2y^2 = 3xy.
\]
The latter, multiplied by $2$, can be rewritten as
\[
x^2(y+1)^2 + y^2(x+1)^2 + (x-y)^2 = 0,
\]
which is possible only if $x=y=-1$ (since $x=y=0$ is excluded). In this case $E = -2$.

Therefore the set of possible values of $E$ is contained in $\{-2, 1\}$. Both values are attained: $E=1$ for $x=y=2$, and $E=-2$ for $x=y=-1$.",164,638,Algebra,11
503,tst_jbmo_ro_2016_3_p3,tst_jbmo,2016,g,"Let $ABCD$ be a cyclic quadrilateral in which the diagonals intersect at $X$ and are not perpendicular. Let $A', C'$ be the projections of $A$ and $C$ onto $BD$, and $B', D'$ the projections of $B$ and $D$ onto $AC$. Show that:
\begin{enumerate}[label=\alph*)]
  \item The perpendiculars from the midpoints of the sides to the opposite side are concurrent at a point $M$, called the \textit{Mathot point}.
  \item The points $A', B', C', D'$ are concyclic.
  \item If $O'$ is the circumcenter of triangle $A'B'C'$, then $O'$ is the midpoint of the segment determined by the orthocenters of triangles $XAB$ and $XCD$.
  \item $O'$ is the Mathot point of quadrilateral $ABCD$.
\end{enumerate}","\subsection*{Solution}

\textbf{a)} Let $O$ be the circumcenter of $ABCD$. The midpoints of the sides of $ABCD$ form a parallelogram, so the segments joining the midpoints of two opposite sides have the same midpoint $G$. The perpendiculars from $O$ to $AB$ and $CD$ pass through the midpoints of those sides; thus the perpendiculars from the midpoints (and from $O$) to two opposite sides form a parallelogram whose center is $G$. It follows that the perpendiculars from the midpoints of two opposite sides to those opposite sides meet at the reflection of $O$ over $G$. The other two perpendiculars also meet there.

\textbf{b)} We assume angle $AXB$ is acute; the other case is analogous. Since quadrilaterals $ABA'B'$, $CDC'D'$, and $ABCD$ are cyclic,
\[
\angle XD'C' \equiv \angle XDC \equiv \angle XAB \equiv \angle XA'B',
\]
so $A'B'C'D'$ is cyclic.

\textbf{c)} Let $H_1$ and $H_2$ be the orthocenters of triangles $XAB$ and $XCD$ respectively, and let $O''$ be the midpoint of $[H_1H_2]$. Since $O''$ lies on the midline of trapezoid $H_1B'H_2D'$, it lies on the perpendicular bisector of $[B'D']$. Similarly $O''$ lies on the midline of trapezoid $A'H_1C'H_2$, so it lies on the perpendicular bisector of $[A'C']$. Since $A'C'$ and $B'D'$ are not parallel, $O''$ is precisely the circumcenter of $A'B'C'D'$, i.e.\ $O'' = O'$.

\textbf{d)} We have $\angle A'B'X \equiv \angle ABX \equiv \angle DCX$, so $A'B' \parallel CD$. If $N$ is the midpoint of $[AB]$, then $NA' = NB'$ (medians in right triangles), so $N$ lies on the perpendicular bisector of $[A'B']$. Since $O'$ also lies on this bisector, $NO' \perp CD$. Analogously $O'$ lies on the perpendicular from the midpoint of $[CD]$ to $AB$, so $O'$ is the Mathot point of the quadrilateral.",690,1753,Geometry,12
504,tst_jbmo_ro_2016_3_p4,tst_jbmo,2016,c,"In the unit squares of an $n \times n$ board are written $n^2$ natural numbers with sum $S$. A move consists of choosing a $2 \times 2$ square and increasing exactly three of the four numbers in that square by one. We say that a natural number $n$ is \emph{good} if, for any $S$, there exists a sequence of moves that makes all numbers on the board equal.
\begin{enumerate}[label=\alph*)]
  \item Show that $n = 6$ is not good.
  \item Prove that $4$ and $1024$ are good.
\end{enumerate}","\subsection*{Solution}

\textbf{a)} Each move increases the sum of the $36$ numbers by $3$. Since the final sum must be a multiple of $36$, hence of $3$, if the initial sum $S$ is not divisible by $3$ we can never make all numbers equal. Therefore $6$ (and likewise any multiple of $3$) is not good.

\textbf{b)} We pick any one cell. It is possible to increase all other $15$ cells by one unit. For example, if the chosen cell is in the top-left $2\times 2$ square, we increase successively the cells labeled $B$, then $C$, $D$, $E$, and finally three of the four cells labeled $A$ (all except the chosen one):
\[
\begin{array}{|c|c|c|c|}
\hline
A & A & B & B \\
\hline
A & A & C & B \\
\hline
D & C & C & E \\
\hline
D & D & E & E \\
\hline
\end{array}
\]
Combining $5$ moves, we obtain a new \emph{compound move} that decreases the chosen cell by one. Applying this compound move repeatedly, always choosing the maximum value, all numbers will eventually become equal.

We prove by induction on $m \in \mathbb{N}$ that for any $n = 2^m$, $m \ge 1$, the $2^m \times 2^m$ board with one arbitrary cell removed can be tiled by \emph{trominoes} (L-shaped pieces formed by removing one unit square from a $2\times 2$ square).

\textit{Base case} $m=1$: obvious.

\textit{Inductive step}: Assume the statement holds for $n = 2^m$; prove it for $n = 2^{m+1}$. Divide the board into four $2^m \times 2^m$ quadrants. Assume (WLOG) the missing cell is in the top-left quadrant. Tile the rest of that quadrant by trominoes (inductive hypothesis). From each of the other three quadrants, remove the corner cell that lies at the center of the full $2^{m+1} \times 2^{m+1}$ board. By the inductive hypothesis, each of these three quadrants minus a corner cell can be tiled by trominoes. Finally, the three removed central unit squares can themselves be covered by a single tromino. The induction is complete.

For any cell, we compose the original moves into an \emph{atom-move} that increases by $1$ every cell on the board except the chosen one. Repeatedly applying this atom-move to a cell with the maximum value will eventually make all numbers equal.",487,2144,Combinatorics,13
505,tst_jbmo_ro_2016_4_p1,tst_jbmo,2016,g,"The altitudes $AA_1$, $BB_1$, $CC_1$ of acute triangle $ABC$ meet at $H$. Let $A_2$ be the reflection of $A$ over line $B_1C_1$, and let $O$ be the circumcenter of triangle $ABC$.
\begin{enumerate}[label=\alph*)]
  \item Prove that points $O, A_2, B_1, C$ are concyclic.
  \item Prove that points $O, H, A_1, A_2$ are concyclic.
\end{enumerate}","\subsection*{Solution 1}

\textbf{a)} Angles $\angle ABC$ and $\angle AB_1C_1$ are congruent, so their complements $\angle BAA_1$ and $\angle A_2AC$ are also congruent. It follows that rays $(AH)$ and $(AA_2)$ are isogonal, so $A_2 \in (AO)$. Since $AO = CO$, we have $\angle ACO \equiv \angle OAC \equiv \angle AA_2B_1$, so points $O, A_2, B_1, C$ are concyclic. (The arguments work both when $A_2 \in (AO)$ and when $O \in (AA_2)$.)

\textbf{b)} From the power of point $A$ with respect to the circles through $O, A_2, B_1, C$ and through $H, A_1, C, B_1$ respectively:
\[
AB_1 \cdot AC = AO \cdot AA_2 \quad \text{and} \quad AB_1 \cdot AC = AH \cdot AA_1.
\]
From $AO \cdot AA_2 = AH \cdot AA_1$, by the converse of the power of a point theorem, points $O, A_2, H, A_1$ are concyclic.

\begin{remark}
The result remains valid even when the triangle is not acute.
\end{remark}

\subsection*{Solution 2 (for part b)}

Let $O_1$ be the midpoint of $[AH]$ ($O_1$ is the circumcenter of triangle $AB_1C_1$), and let $\{A_3\} = (AA_2) \cap B_1C_1$. Since $[O_1A_3]$ is a midline in triangle $AHA_2$, we have $\angle HA_2A \equiv \angle O_1A_3A$. But the similarity of triangles $ABC$ and $AB_1C_1$ implies the equality of corresponding angles $\angle AA_1O$ and $\angle AA_3O_1$, from which $\angle AA_2H \equiv \angle AA_1O$ and the conclusion follows.",344,1350,Geometry,14
506,tst_jbmo_ro_2016_4_p2,tst_jbmo,2016,c,"Given three colors and an $m \times n$ rectangle divided into unit squares, we wish to color each segment that forms a side of a unit square with one of the three colors, such that every unit square has two sides of one color and two sides of another color. How many such colorings are there?","\subsection*{Solution}

Number the rows top to bottom and left to right. The left side of the unit square in the top-left corner can be colored in $3$ ways. There are $3$ ways to choose which other side of this square gets the same color. For the remaining two sides we have $2$ options (either the same color for both, chosen from the two remaining colors). In total, there are $18$ colorings for this first square.

We then color the squares along the first row, left to right. Each time, the left side is already colored, so there are $6$ variants for each such square. The same applies when we color the squares of the first column, top to bottom (starting from row 2): $6$ variants each.

For each square in rows $2, 3, \ldots, m$ and columns $2, 3, \ldots, n$, its left side and top side are already colored. If they have different colors, the remaining two sides must use exactly these two colors, giving $2$ arrangements. If the left and top sides have the same color, the remaining two sides must both have the same color (chosen in $2$ ways). So in either case there are $2$ variants.

In conclusion, the total number of colorings is
\[
18 \cdot 6^{m-1} \cdot 6^{n-1} \cdot 2^{(m-1)(n-1)} = 3^{m+n} \cdot 2^{mn}.
\]",292,1225,Combinatorics,15
507,tst_jbmo_ro_2016_4_p3,tst_jbmo,2016,a,"Let $a, b, c$ be real numbers with $a \ge b \ge 1 \ge c \ge 0$ and $a + b + c = 3$.
\begin{enumerate}[label=\alph*)]
  \item Show that $2 \le ab + bc + ca \le 3$.
  \item Prove the inequality
  \[
  \frac{24}{a^3+b^3+c^3} + \frac{25}{ab+bc+ca} \ge 14
  \]
  and determine the cases of equality.
\end{enumerate}","\subsection*{Solution}

\textbf{a)} Denote $ab+bc+ca = q$ and $abc = p$. By hypothesis $(1-a)(1-b)(1-c) \ge 0$, which gives $q - 2 \ge p$; but $p \ge 0$, so $q \ge 2$. Also $3(ab+bc+ca) \le (a+b+c)^2$, giving $q \le 3$.

\textbf{b)} We have
\[
a^3+b^3+c^3 = (a+b+c)^3 - 3(a+b+c)(ab+bc+ca)+3abc = 3(9-3q+p).
\]
From part a), $p \le q-2$, so $3(9-3q+p) \le 3(7-2q)$, hence
\[
\frac{24}{a^3+b^3+c^3} \ge \frac{8}{7-2q}.
\]
It suffices to show that
\[
\frac{8}{7-2q} + \frac{25}{q} \ge 14,
\]
which is equivalent to $7(2q-5)^2 \ge 0$, which is obviously true.

From the above, equality holds if and only if $p = q-2$ and $q = \tfrac{5}{2}$, i.e.\ $p = \tfrac{1}{2}$.

The condition $p = q-2$ means $(1-a)(1-b)(1-c) = 0$, so at least one of $a, b, c$ equals $1$.

Denoting the other two by $x$ and $y$, we get the system
\[
\begin{cases} x + y + xy = \tfrac{5}{2} \\ xy = \tfrac{1}{2}. \end{cases}
\]
Solving, we obtain
\[
a = 1 + \frac{1}{\sqrt{2}}, \quad b = 1, \quad c = 1 - \frac{1}{\sqrt{2}}.
\]",310,995,Algebra,16
508,tst_jbmo_ro_2016_4_p4,tst_jbmo,2016,g,"Let $ABCD$ be a cyclic quadrilateral, $E$ and $F$ the midpoints of diagonals $[AC]$ and $[BD]$ respectively, $\{G\} = AB \cap CD$, $\{H\} = AD \cap BC$. Show that:
\begin{enumerate}[label=\alph*)]
  \item The intersections of the bisectors of angles $\angle AHB$ and $\angle AGD$ with the sides of quadrilateral $ABCD$ are the vertices of a rhombus.
  \item The center of this rhombus lies on line $EF$.
\end{enumerate}","\subsection*{Solution}

We consider the case $C \in (GD)$ and $C \in (BH)$; the other cases are handled analogously.

\textbf{a)} Let $M$ and $N$ be the intersections of the bisector of $\angle G$ with sides $[BC]$ and $[AD]$ respectively, and let $I$ and $K$ be the intersections of the bisector of $\angle H$ with $[CD]$ and $[AB]$ respectively. Let $\{J\} = MN \cap IK$. We have
\[
m(\angle G) = 180^\circ - m(\angle GBC) - m(\angle GCB) = m(\angle B)+m(\angle C)-180^\circ,
\]
so $m(\angle CMJ) = 90^\circ + \tfrac{1}{2}(m(\angle B)-m(\angle C))$. Similarly $m(\angle CIJ) = 90^\circ + \tfrac{1}{2}(m(\angle D)-m(\angle C))$. From quadrilateral $CMJI$, using $m(\angle B)+m(\angle D)=180^\circ$, we get $m(\angle MJI) = 90^\circ$.

In triangles $GIK$ and $HMN$, the segments $[GJ]$ and $[HJ]$ are simultaneously bisectors and altitudes, hence also medians. Therefore the diagonals of quadrilateral $MKNI$ are perpendicular and bisect each other, making $MKNI$ a rhombus.

\textbf{b)} The sides of the rhombus are parallel to diagonals $AC$ and $BD$ respectively. Indeed, by the angle bisector theorem, $\dfrac{MB}{MC} = \dfrac{GB}{GC}$ and $\dfrac{BK}{KA} = \dfrac{HB}{HA}$. By the power of point $G$ with respect to the circle, $\dfrac{GB}{GC} = \dfrac{GD}{GA}$. To show $MK \parallel AC$, i.e.\ $\dfrac{BK}{KA} = \dfrac{BM}{MC}$, it suffices to prove $HA \cdot GD = HB \cdot GA$. This follows from
\[
[AHG] = \frac{HA \cdot GD \sin D}{2} = \frac{HB \cdot GA \sin B}{2} \quad \text{and} \quad \sin B = \sin D.
\]
The other sides follow analogously.

Let $\{L\} = MK \cap BF$, $\{P\} = AE \cap KN$, $\{O\} = DF \cap IN$, $\{Q\} = MI \cap CE$. The segments $AE$, $BF$, $CE$, $DF$ are medians in triangles $ABD$, $ABC$, $BCD$, $CDA$ respectively, so segments $LO$ and $PQ$ are the \emph{bimedians} of the rhombus (joining midpoints of pairs of opposite sides). Therefore $LO$ and $PQ$ are concurrent at the center of the rhombus.

Let $\{X\} = LO \cap EF$ and $\{Y\} = QP \cap EF$. We have
\[
\frac{XE}{XF} = \frac{LB}{LF} = \frac{BK}{KA} \quad \text{and} \quad \frac{YE}{YF} = \frac{QE}{QC} = \frac{BM}{MC}.
\]
Since $\dfrac{BK}{KA} = \dfrac{BM}{MC}$, we get $X = Y$, so lines $EF$, $LO$, and $PQ$ are concurrent at the center of the rhombus. \qed",419,2250,Geometry,17
243,jbmo_2017_p1,jbmo,2017,n,"Determine all the sets of six consecutive positive integers such that the product of some two of them, added to the product of some other two of them is equal to the product of the remaining two numbers.","\subsection*{Solution}

Exactly two of the six numbers are multiples of 3 and these two need to be multiplied together, otherwise two of the three terms of the equality are multiples of 3 but the third one is not. Let $n$ and $n + 3$ denote these multiples of 3. Two of the four remaining numbers give remainder 1 when divided by 3, while the other two give remainder 2, so the two other products are either $\equiv 1 \cdot 1 = 1 \pmod{3}$ and $\equiv 2 \cdot 2 \equiv 1 \pmod{3}$, or they are both $\equiv 1 \cdot 2 \equiv 2 \pmod{3}$. In conclusion, the term $n(n+3)$ needs to be on the right hand side of the equality.

Looking at parity, three of the numbers are odd, and three are even. One of $n$ and $n + 3$ is odd, the other even, so exactly two of the other numbers are odd. As $n(n + 3)$ is even, the two remaining odd numbers need to appear in different terms.

We distinguish the following cases:

\textbf{I.} The numbers are $n-2, n-1, n, n+1, n+2, n+3$.

The product of the two numbers on the RHS needs to be larger than $n(n+3)$. The only possibility is
\[
(n-2)(n-1) + n(n+3) = (n+1)(n+2)
\]
which leads to $n = 3$. Indeed, $1 \cdot 2 + 3 \cdot 6 = 4 \cdot 5$.

\textbf{II.} The numbers are $n-1, n, n+1, n+2, n+3, n+4$.

As $(n+4)(n-1) + n(n+3) = (n+1)(n+2)$ has no solutions, $n+4$ needs to be on the RHS, multiplied with a number having a different parity, so $n-1$ or $n+1$.

$(n+2)(n-1) + n(n+3) = (n+1)(n+4)$ leads to $n = 3$. Indeed, $2 \cdot 5 + 3 \cdot 6 = 4 \cdot 7$.

$(n+2)(n+1) + n(n+3) = (n-1)(n+4)$ has no solution.

\textbf{III.} The numbers are $n, n+1, n+2, n+3, n+4, n+5$.

We need to consider the following situations:

$(n+1)(n+2) + n(n+3) = (n+4)(n+5)$ which leads to $n = 6$; indeed $7 \cdot 8 + 6 \cdot 9 = 10 \cdot 11$;

$(n+2)(n+5) + n(n+3) = (n+1)(n+4)$ obviously without solutions, and

$(n+1)(n+4) + n(n+3) = (n+2)(n+5)$ which leads to $n = 2$ (not a multiple of 3).

In conclusion, the problem has three solutions:
\[
1 \cdot 2 + 3 \cdot 6 = 4 \cdot 5, \quad 2 \cdot 5 + 3 \cdot 6 = 4 \cdot 7, \quad \text{and} \quad 7 \cdot 8 + 6 \cdot 9 = 10 \cdot 11.
\]",203,2102,Number Theory,1
244,jbmo_2017_p2,jbmo,2017,a,"Let $x, y, z$ be positive integers such that $x 
eq y 
eq z 
eq x$. Prove that
\[
(x + y + z)(xy + yz + zx - 2) \geq 9xyz.
\]
When does the equality hold?","\subsection*{Solution}

Since $x, y, z$ are distinct positive integers, the required inequality is symmetric and WLOG we can suppose that $x \geq y + 1 \geq z + 2$. We consider 2 possible cases:

\textbf{Case 1.} $y \geq z + 2$. Since $x \geq y + 1 \geq z + 3$ it follows that
\[
(x - y)^2 \geq 1, \quad (y - z)^2 \geq 4, \quad (x - z)^2 \geq 9
\]
which are equivalent to
\[
x^2 + y^2 \geq 2xy + 1, \quad y^2 + z^2 \geq 2yz + 4, \quad x^2 + z^2 \geq 2xz + 9
\]
or otherwise
\[
zx^2 + zy^2 \geq 2xyz + z, \quad xy^2 + xz^2 \geq 2xyz + 4x, \quad yx^2 + yz^2 \geq 2xyz + 9y.
\]
Adding up the last three inequalities we have
\[
xy(x + y) + yz(y + z) + zx(z + x) \geq 6xyz + 4x + 9y + z
\]
which implies that $(x + y + z)(xy + yz + zx - 2) \geq 9xyz + 2x + 7y - z$.

Since $x \geq z + 3$ it follows that $2x + 7y - z \geq 0$ and our inequality follows.

\textbf{Case 2.} $y = z + 1$. Since $x \geq y + 1 = z + 2$ it follows that $x \geq z + 2$, and replacing $y = z + 1$ in the required inequality we have to prove
\[
(x + z + 1 + z)(x(z+1) + (z+1)z + zx - 2) \geq 9x(z+1)z
\]
which is equivalent to
\[
(x + 2z + 1)(z^2 + 2zx + z + x - 2) - 9x(z+1)z \geq 0.
\]
Doing easy algebraic manipulations, this is equivalent to prove
\[
(x - z - 2)(x - z + 1)(2z + 1) \geq 0
\]
which is satisfied since $x \geq z + 2$.

The equality is achieved only in Case 2 for $x = z + 2$, so we have equality when $(x, y, z) = (k+2, k+1, k)$ and all the permutations for any positive integer $k$.",154,1470,Algebra,2
245,jbmo_2017_p3,jbmo,2017,g,"Let $ABC$ be an acute triangle such that $AB 
eq AC$, with circumcircle $\Gamma$ and circumcenter $O$. Let $M$ be the midpoint of $BC$ and $D$ be a point on $\Gamma$ such that $AD \perp BC$. Let $T$ be a point such that $BDCT$ is a parallelogram and $Q$ a point on the same side of $BC$ as $A$ such that
\[
\angle BQM = \angle BCA \quad \text{and} \quad \angle CQM = \angle CBA.
\]
Let the line $AO$ intersect $\Gamma$ at $E$, ($E 
eq A$) and let the circumcircle of $\triangle ETQ$ intersect $\Gamma$ at point $X 
eq E$. Prove that the points $A$, $M$, and $X$ are collinear.","\subsection*{Solution}

Let $X'$ be symmetric point to $Q$ in line $BC$. Now since $\angle CBA = \angle CQM = \angle CX'M$, $\angle BCA = \angle BQM = \angle BX'M$, we have
\[
\angle BX'C = \angle BX'M + \angle CX'M = \angle CBA + \angle BCA = 180^\circ - \angle BAC
\]
we have that $X' \in \Gamma$. Now since $\angle AX'B = \angle ACB = \angle MX'B$ we have that $A, M, X'$ are collinear.

Note that since
\[
\angle DCB = \angle DAB = 90^\circ - \angle ABC = \angle OAC = \angle EAC
\]
we get that $DBCE$ is an isosceles trapezoid.

Since $BDCT$ is a parallelogram we have $MT = MD$, with $M, D, T$ being collinear, $BD = CT$, and since $BDEC$ is an isosceles trapezoid we have $BD = CE$ and $ME = MD$. Since
\[
\angle BTC = \angle BDC = \angle BED, \quad CE = BD = CT \quad \text{and} \quad ME = MT
\]
we have that $E$ and $T$ are symmetric with respect to the line $BC$. Now since $Q$ and $X'$ are symmetric with respect to the line $BC$ as well, this means that $QX'ET$ is an isosceles trapezoid which means that $Q, X', E, T$ are concyclic. Since $X' \in \Gamma$ this means that $X \equiv X'$ and therefore $A, M, X$ are collinear.

\subsection*{Alternative solution}

Denote by $H$ the orthocenter of $\triangle ABC$. We use the following well known properties:

\begin{enumerate}[label=(
oman*)]
    \item Point $D$ is the symmetric point of $H$ with respect to $BC$. Indeed, if $H_1$ is the symmetric point of $H$ with respect to $BC$ then $\angle BH_1C + \angle BAC = 180^\circ$ and therefore $H_1 \equiv D$.
    \item The symmetric point of $H$ with respect to $M$ is the point $E$. Indeed, if $H_2$ is the symmetric point of $H$ with respect to $M$ then $BH_2CH$ is parallelogram, $\angle BH_2C + \angle BAC = 180^\circ$ and since $EB \parallel CH$ we have $\angle EBA = 90^\circ$.
\end{enumerate}

Since $DETH$ is a parallelogram and $MH = MD$ we have that $DETH$ is a rectangle. Therefore $MT = ME$ and $TE \perp BC$ implying that $T$ and $E$ are symmetric with respect to $BC$. Denote by $Q'$ the symmetric point of $Q$ with respect to $BC$. Then $Q'ETQ$ is isosceles trapezoid, so $Q'$ is a point on the circumcircle of $\triangle ETQ$. Moreover $\angle BQ'C + \angle BAC = 180^\circ$ and we conclude that $Q' \in \Gamma$. Therefore $Q' \equiv X$.

It remains to observe that $\angle CXM = \angle CQM = \angle CBA$ and $\angle CXA = \angle CBA$ and we infer that $X$, $M$ and $A$ are collinear.",576,2409,Geometry,3
246,jbmo_2017_p4,jbmo,2017,c,"Consider a regular $2n$-gon $P$, $A_1A_2\ldots A_{2n}$ in the plane, where $n$ is a positive integer. We say that a point $S$ on one of the sides of $P$ \emph{can be seen} from a point $E$ that is external to $P$, if the line segment $SE$ contains no other points that lie on the sides of $P$ except $S$. We color the sides of $P$ in 3 different colors (ignore the vertices of $P$, we consider them colorless), such that every side is colored in exactly one color, and each color is used at least once. Moreover, from every point in the plane external to $P$, points of at most 2 different colors on $P$ can be seen. Find the number of distinct such colorings of $P$ (two colorings are considered distinct if at least one of the sides is colored differently).","\subsection*{Solution}

\textbf{Answer:} For $n = 2$, the answer is $36$; for $n = 3$, the answer is $30$ and for $n \geq 4$, the answer is $6n$.

\begin{lemma}
Given a regular $2n$-gon in the plane and a sequence of $n$ consecutive sides $s_1, s_2, \ldots, s_n$ there is an external point $Q$ in the plane, such that the color of each $s_i$ can be seen from $Q$, for $i = 1, 2, \ldots, n$.
\end{lemma}

\begin{proof}
It is obvious that for a semi-circle $S$, there is a point $R$ in the plane far enough on the bisector of its diameter such that almost the entire semi-circle can be seen from $R$.

Now, it is clear that looking at the circumscribed circle around the $2n$-gon, there is a semi-circle $S$ such that each $s_i$ either has both endpoints on it, or has an endpoint that's on the semi-circle, and is not on the semi-circle's end. So, take $Q$ to be a point in the plane from which almost all of $S$ can be seen, clearly, the color of each $s_i$ can be seen from $Q$.
\end{proof}

\begin{lemma}
Given a regular $2n$-gon in the plane, and a sequence of $n+1$ consecutive sides $s_1, s_2, \ldots, s_{n+1}$ there is no external point $Q$ in the plane, such that the color of each $s_i$ can be seen from $Q$, for $i = 1, 2, \ldots, n+1$.
\end{lemma}

\begin{proof}
Since $s_1$ and $s_{n+1}$ are parallel opposite sides of the $2n$-gon, they cannot be seen at the same time from an external point.
\end{proof}

For $n = 2$, we have a square, so all we have to do is make sure each color is used. Two sides will be of the same color, and we have to choose which are these 2 sides, and then assign colors according to this choice, so the answer is $\binom{4}{2} \cdot 3 \cdot 2 = 36$.

For $n = 3$, we have a hexagon. Denote the sides as $a_1, a_2, \ldots, a_6$, in that order. There must be 2 consecutive sides of different colors, say $a_1$ is red, $a_2$ is blue. We must have a green side, and only $a_4$ and $a_5$ can be green. We have 3 possibilities:
\begin{enumerate}
    \item $a_4$ is green, $a_5$ is not. So, $a_3$ must be blue and $a_5$ must be blue (by elimination) and $a_6$ must be blue, so we get a valid coloring.
    \item Both $a_4$ and $a_5$ are green, thus $a_6$ must be red and $a_5$ must be blue, and we get the coloring $rbbggr$.
    \item $a_5$ is green, $a_4$ is not. Then $a_6$ must be red. Subsequently, $a_4$ must be red (we assume it is not green). It remains that $a_3$ must be red, and the coloring is $rbrrgr$.
\end{enumerate}

Thus, we have 2 kinds of configurations:
\begin{enumerate}[label=
oman*)]
    \item 2 opposite sides have 2 opposite colors and all other sides are of the third color. This can happen in $3 \cdot (3 \cdot 2 \cdot 1) = 18$ ways (first choosing the pair of opposite sides, then assigning colors),
    \item 3 pairs of consecutive sides, each pair in one of the 3 colors. This can happen in $2 \cdot 6 = 12$ ways (2 partitionings into pairs of consecutive sides, for each partitioning, 6 ways to assign the colors).
\end{enumerate}

Thus, for $n = 3$, the answer is $18 + 12 = 30$.

Finally, let's address the case $n \geq 4$. The important thing now is that any 4 consecutive sides can be seen from an external point, by Lemma 1.

Denote the sides as $a_1, a_2, \ldots, a_{2n}$. Again, there must be 2 adjacent sides that are of different colors, say $a_1$ is blue and $a_2$ is red. We must have a green side, and by Lemma 1, that can only be $a_{n+1}$ or $a_{n+2}$. So, we have 2 cases:

\textbf{Case 1:} $a_{n+1}$ is green, so $a_n$ must be red (cannot be green due to Lemma 1 applied to $a_1, a_2, \ldots, a_n$, cannot be blue for the sake of $a_2, \ldots, a_{n+1}$). If $a_{n+2}$ is red, so are $a_{n+3}, \ldots, a_{2n}$, and we get a valid coloring: $a_1$ is blue, $a_{n+1}$ is green, and all the others are red.

If $a_{n+2}$ is green:
\begin{enumerate}[label=\alph*)]
    \item $a_{n+3}$ cannot be green, because of $a_2, a_1, a_{2n}, \ldots, a_{n+3}$.
    \item $a_{n+3}$ cannot be blue, because the 4 adjacent sides $a_n, \ldots, a_{n+3}$ can be seen (this is the case that makes the separate treatment of $n \geq 4$ necessary).
    \item $a_{n+3}$ cannot be red, because of $a_1, a_{2n}, \ldots, a_{n+2}$.
\end{enumerate}
So, in the case that $a_{n+2}$ is also green, we cannot get a valid coloring.

\textbf{Case 2:} $a_{n+2}$ is green is treated the same way as Case 1.

This means that the only valid configuration for $n \geq 4$ is having 2 opposite sides colored in 2 different colors, and all other sides colored in the third color. This can be done in $n \cdot 3 \cdot 2 = 6n$ ways.",759,4562,Combinatorics,4
96,shl_jbmo_2017_a1,shl_jbmo,2017,a,"Let $a, b, c$ be positive real numbers such that $a + b + c + ab + bc + ca + abc = 7$. Prove that
\[
\sqrt{a^2 + b^2 + 2} + \sqrt{b^2 + c^2 + 2} + \sqrt{c^2 + a^2 + 2} \geq 6.
\]","\begin{proof}
First we see that $x^2 + y^2 + 1 \geq xy + x + y$. Indeed, this is equivalent to
\[
(x - y)^2 + (x - 1)^2 + (y - 1)^2 \geq 0.
\]
Therefore
\begin{align*}
&\sqrt{a^2 + b^2 + 2} + \sqrt{b^2 + c^2 + 2} + \sqrt{c^2 + a^2 + 2} \\
&\geq \sqrt{ab + a + b + 1} + \sqrt{bc + b + c + 1} + \sqrt{ca + c + a + 1} \\
&= \sqrt{(a+1)(b+1)} + \sqrt{(b+1)(c+1)} + \sqrt{(c+1)(a+1)}.
\end{align*}
It follows from the AM-GM inequality that
\[
\sqrt{(a+1)(b+1)} + \sqrt{(b+1)(c+1)} + \sqrt{(c+1)(a+1)}
\geq 3\sqrt[3]{\sqrt{(a+1)(b+1)} \cdot \sqrt{(b+1)(c+1)} \cdot \sqrt{(c+1)(a+1)}}
= 3\sqrt[3]{(a+1)(b+1)(c+1)}.
\]
On the other hand, the given condition is equivalent to $(a+1)(b+1)(c+1) = 8$ and we get the desired inequality.

Equality is attained if and only if $a = b = c = 1$.
\end{proof}

\begin{remark}
The condition of positivity of $a, b, c$ is superfluous and the equality $\cdots = 7$ can be replaced by the inequality $\cdots \geq 7$. Indeed, the above proof and the triangle inequality imply that
\[
\sqrt{a^2+b^2+2}+\sqrt{b^2+c^2+2}+\sqrt{c^2+a^2+2}
\geq 3\sqrt[3]{(|a|+1)(|b|+1)(|c|+1)}
\geq 3\sqrt[3]{|a+1| \cdot |b+1| \cdot |c+1|} \geq 6.
\]
\end{remark}",178,1167,Algebra,1
97,shl_jbmo_2017_a2,shl_jbmo,2017,a,"Let $a$ and $b$ be positive real numbers such that $3a^2 + 2b^2 = 3a + 2b$. Find the minimum value of
\[
A = \sqrt{\frac{a}{b(3a+2)}} + \sqrt{\frac{b}{a(2b+3)}}.
\]","\begin{proof}[Solution]
By the Cauchy--Schwarz inequality we have that
\[
5(3a^2 + 2b^2) = 5(a^2 + a^2 + a^2 + b^2 + b^2) \geq (3a + 2b)^2
\]
(or use that the last inequality is equivalent to $(a-b)^2 \geq 0$).

So, with the help of the given condition we get that $3a + 2b \leq 5$. Now, by the AM-GM inequality we have that
\[
A \geq 2\sqrt[4]{\sqrt{\frac{a}{b(3a+2)}} \cdot \sqrt{\frac{b}{a(2b+3)}}} = \frac{2}{\sqrt[4]{(3a+2)(2b+3)}}.
\]
Finally, using again the AM-GM inequality, we get that
\[
(3a+2)(2b+3) \leq \left(\frac{3a+2b+5}{2}
ight)^2 \leq 25,
\]
so $A \geq 2/\sqrt{5}$ and the equality holds if and only if $a = b = 1$.
\end{proof}",164,646,Algebra,2
98,shl_jbmo_2017_a3,shl_jbmo,2017,a,"Let $a, b, c, d$ be real numbers such that $0 \leq a \leq b \leq c \leq d$. Prove the inequality
\[
ab^3 + bc^3 + cd^3 + da^3 \geq a^2b^2 + b^2c^2 + c^2d^2 + d^2a^2.
\]","\begin{proof}
The inequality is equivalent to
\[
(ab^3 + bc^3 + cd^3 + da^3)^2 \geq (a^2b^2 + b^2c^2 + c^2d^2 + d^2a^2)^2.
\]
By the Cauchy--Schwarz inequality,
\[
(ab^3 + bc^3 + cd^3 + da^3)(a^3b + b^3c + c^3d + d^3a) \geq (a^2b^2 + b^2c^2 + c^2d^2 + d^2a^2)^2.
\]
Hence it is sufficient to prove that
\[
(ab^3 + bc^3 + cd^3 + da^3)^2 \geq (ab^3 + bc^3 + cd^3 + da^3)(a^3b + b^3c + c^3d + d^3a),
\]
i.e.\ to prove $ab^3 + bc^3 + cd^3 + da^3 \geq a^3b + b^3c + c^3d + d^3a$.

This inequality can be written successively
\[
a(b^3 - d^3) + b(c^3 - a^3) + c(d^3 - b^3) + d(a^3 - c^3) \geq 0,
\]
or
\[
(a - c)(b^3 - d^3) - (b - d)(a^3 - c^3) \geq 0,
\]
which comes down to
\[
(a - c)(b - d)(b^2 + bd + d^2 - a^2 - ac - c^2) \geq 0.
\]
The last inequality is true because $a - c \leq 0$, $b - d \leq 0$, and $(b^2 - a^2) + (bd - ac) + (d^2 - c^2) \geq 0$ as a sum of three non-negative numbers.

The last inequality is satisfied with equality when $a = b$ and $c = d$. Combining this with the equality cases in the Cauchy--Schwarz inequality we obtain the equality cases for the initial inequality: $a = b = c = d$.
\end{proof}

\begin{remark}
Instead of using the Cauchy--Schwarz inequality, once the inequality $ab^3 + bc^3 + cd^3 + da^3 \geq a^3b + b^3c + c^3d + d^3a$ is established, we have
\begin{align*}
2(ab^3 + bc^3 + cd^3 + da^3)
&\geq (ab^3 + bc^3 + cd^3 + da^3) + (a^3b + b^3c + c^3d + d^3a) \\
&= (ab^3 + a^3b) + (bc^3 + b^3c) + (cd^3 + c^3d) + (da^3 + d^3a) \\
&\overset{\mathrm{AM\text{-}GM}}{\geq} 2a^2b^2 + 2b^2c^2 + 2c^2d^2 + 2d^2a^2,
\end{align*}
which gives the conclusion.
\end{remark}",168,1601,Algebra,3
99,shl_jbmo_2017_a4,shl_jbmo,2017,a,"Let $x, y, z$ be three distinct positive integers. Prove that
\[
(x + y + z)(xy + yz + zx - 2) \geq 9xyz.
\]
When does the equality hold?","\begin{proof}
Since $x, y, z$ are distinct positive integers, the required inequality is symmetric and WLOG we can suppose that $x \geq y + 1 \geq z + 2$. We consider 2 possible cases.

\medskip

oindent\textbf{Case 1.} $y \geq z + 2$. Since $x \geq y + 1 \geq z + 3$ it follows that
\[
(x-y)^2 \geq 1, \quad (y-z)^2 \geq 4, \quad (x-z)^2 \geq 9,
\]
which are equivalent to
\[
x^2 + y^2 \geq 2xy + 1, \quad y^2 + z^2 \geq 2yz + 4, \quad x^2 + z^2 \geq 2xz + 9,
\]
or otherwise
\[
zx^2 + zy^2 \geq 2xyz + z, \quad xy^2 + xz^2 \geq 2xyz + 4x, \quad yx^2 + yz^2 \geq 2xyz + 9y.
\]
Adding up the last three inequalities we have
\[
xy(x+y) + yz(y+z) + zx(z+x) \geq 6xyz + 4x + 9y + z,
\]
which implies that $(x+y+z)(xy+yz+zx-2) \geq 9xyz + 2x + 7y - z$.

Since $x \geq z + 3$ it follows that $2x + 7y - z \geq 0$ and our inequality follows.

\medskip

oindent\textbf{Case 2.} $y = z + 1$. Since $x \geq y + 1 = z + 2$, replacing $y = z + 1$ in the required inequality we have to prove
\[
(x + z + 1 + z)(x(z+1) + (z+1)z + zx - 2) \geq 9x(z+1)z,
\]
which is equivalent to
\[
(x + 2z + 1)(z^2 + 2zx + z + x - 2) - 9x(z+1)z \geq 0.
\]
After algebraic manipulation, this is equivalent to proving
\[
(x - z - 2)(x - z + 1)(2z + 1) \geq 0,
\]
which is satisfied since $x \geq z + 2$.

The equality is achieved only in Case 2 for $x = z + 2$, so we have equality when $(x, y, z) = (k+2, k+1, k)$ and all permutations, for any positive integer $k$.
\end{proof}",137,1447,Algebra,4
100,shl_jbmo_2017_c1,shl_jbmo,2017,c,"Consider a regular $(2n+1)$-gon $P$ in the plane, where $n$ is a positive integer. We say that a point $S$ on one of the sides of $P$ can be \emph{seen} from a point $E$ that is external to $P$, if the line segment $SE$ contains no other points that lie on the sides of $P$ except $S$. We want to color the sides of $P$ in 3 colors, such that every side is colored in exactly one color, and each color must be used at least once. Moreover, from every point in the plane external to $P$, at most 2 different colors on $P$ can be seen (ignore the vertices of $P$, we consider them colorless). Find the largest positive integer $n$ for which such a coloring is possible.","\begin{proof}[Solution]
\textbf{Answer: $n = 1$.}

$n = 1$ is clearly a solution: we can just color each side of the equilateral triangle in a different color, and the conditions are satisfied. We prove there is no larger $n$ that fulfills the requirements.

\begin{lemma*}[Lemma 1]
Given a regular $(2n+1)$-gon in the plane, and a sequence of $n+1$ consecutive sides $s_1, s_2, \ldots, s_{n+1}$, there is an external point $Q$ in the plane such that the color of each $s_i$ can be seen from $Q$, for $i = 1, 2, \ldots, n+1$.
\end{lemma*}

\begin{proof}
It is obvious that for a semi-circle $S$, there is a point $R$ in the plane far enough on the perpendicular bisector of the diameter of $S$ such that almost the entire semi-circle can be seen from $R$.

Now, it is clear that looking at the circumscribed circle around the $(2n+1)$-gon, there is a semi-circle $S$ such that each $s_i$ either has both endpoints on it, or has an endpoint that is on the semi-circle, and is not on the semi-circle's end. So, take $Q$ to be a point in the plane from which almost all of $S$ can be seen; clearly, the color of each $s_i$ can be seen from $Q$.
\end{proof}

Take $n \geq 2$, denote the sides $a_1, a_2, \ldots, a_{2n+1}$ in order, and suppose we have a coloring that satisfies the condition of the problem. Let us call the 3 colors red, green and blue. We must have 2 adjacent sides of different colors, say $a_1$ is red and $a_2$ is green. Then, by Lemma 1:
\begin{enumerate}[(i)]
  \item We cannot have a blue side among $a_1, a_2, \ldots, a_{n+1}$.
  \item We cannot have a blue side among $a_2, a_1, a_{2n+1}, \ldots, a_{n+3}$.
\end{enumerate}
We are required to have at least one blue side, and according to (i) and (ii), that can only be $a_{n+2}$, so $a_{n+2}$ is blue. Now, applying Lemma 1 on the sequence $a_2, a_3, \ldots, a_{n+2}$ we get that $a_2, a_3, \ldots, a_{n+1}$ are all green. Applying Lemma 1 on the sequence $a_1, a_{2n+1}, a_{2n}, \ldots, a_{n+2}$ we get that $a_{2n+1}, a_{2n}, \ldots, a_{n+3}$ are all red.

Therefore $a_{n+1}, a_{n+2}$ and $a_{n+3}$ are all of different colors, and for $n \geq 2$ they can all be seen from the same point according to Lemma 1, so we have a contradiction.
\end{proof}",667,2224,Combinatorics,5
101,shl_jbmo_2017_c2,shl_jbmo,2017,c,"Consider a regular $2n$-gon $P$ in the plane, where $n$ is a positive integer. We say that a point $S$ on one of the sides of $P$ can be \emph{seen} from a point $E$ that is external to $P$, if the line segment $SE$ contains no other points that lie on the sides of $P$ except $S$. We want to color the sides of $P$ in 3 colors, such that every side is colored in exactly one color, and each color must be used at least once. Moreover, from every point in the plane external to $P$, at most 2 different colors on $P$ can be seen (ignore the vertices of $P$, we consider them colorless). Find the number of distinct such colorings of $P$ (two colorings are considered distinct if at least one side is colored differently).","\begin{proof}[Solution]
\textbf{Answer:} For $n = 2$, the answer is $36$; for $n = 3$, the answer is $30$; and for $n \geq 4$, the answer is $6n$.

\begin{lemma*}[Lemma 1]
Given a regular $2n$-gon in the plane and a sequence of $n$ consecutive sides $s_1, s_2, \ldots, s_n$, there is an external point $Q$ in the plane such that the color of each $s_i$ can be seen from $Q$, for $i = 1, 2, \ldots, n$.
\end{lemma*}

\begin{proof}
It is obvious that for a semi-circle $S$, there is a point $R$ in the plane far enough on the bisector of its diameter such that almost the entire semi-circle can be seen from $R$. Now, it is clear that looking at the circumscribed circle around the $2n$-gon, there is a semi-circle $S$ such that each $s_i$ either has both endpoints on it, or has an endpoint on the semi-circle that is not at the semi-circle's end. So, take $Q$ to be a point in the plane from which almost all of $S$ can be seen; clearly, the color of each $s_i$ can be seen from $Q$.
\end{proof}

\begin{lemma*}[Lemma 2]
Given a regular $2n$-gon in the plane, and a sequence of $n + 1$ consecutive sides $s_1, s_2, \ldots, s_{n+1}$, there is no external point $Q$ in the plane such that the color of each $s_i$ can be seen from $Q$, for $i = 1, 2, \ldots, n+1$.
\end{lemma*}

\begin{proof}
Since $s_1$ and $s_{n+1}$ are parallel opposite sides of the $2n$-gon, they cannot be seen at the same time from an external point.
\end{proof}


oindent\textbf{Case $n = 2$ (square).} All we have to do is make sure each color is used. Two sides will be of the same color, and we have to choose which are these 2 sides, and then assign colors, so the answer is $\binom{4}{2} \cdot 3 \cdot 2 = 36$.

\medskip

oindent\textbf{Case $n = 3$ (hexagon).} Denote the sides as $a_1, a_2, \ldots, a_6$ in order. There must be 2 consecutive sides of different colors, say $a_1$ is red and $a_2$ is blue. We must have a green side, and only $a_4$ and $a_5$ can be green. We have 3 possibilities:
\begin{enumerate}
  \item $a_4$ is green, $a_5$ is not. So $a_3$ must be blue and $a_5$ must be blue (by elimination) and $a_6$ must be blue, giving coloring $\mathtt{rbbgbb}$.
  \item Both $a_4$ and $a_5$ are green, thus $a_6$ must be red and $a_3$ must be blue, giving coloring $\mathtt{rbbggr}$.
  \item $a_5$ is green, $a_4$ is not. Then $a_6$ must be red. Subsequently, $a_4$ and $a_3$ must be red, giving coloring $\mathtt{rbrrgr}$.
\end{enumerate}
Thus we have 2 kinds of configurations:
\begin{itemize}
  \item[(i)] 2 opposite sides have 2 opposite colors and all other sides are of the third color. This can happen in $3 \cdot (3 \cdot 2 \cdot 1) = 18$ ways.
  \item[(ii)] 3 pairs of consecutive sides, each pair in one of the 3 colors. This can happen in $2 \cdot 6 = 12$ ways.
\end{itemize}
Thus, for $n = 3$, the answer is $18 + 12 = 30$.

\medskip

oindent\textbf{Case $n \geq 4$.} The important thing now is that any 4 consecutive sides can be seen from an external point, by Lemma 1.

Denote the sides as $a_1, a_2, \ldots, a_{2n}$. Again, there must be 2 adjacent sides of different colors, say $a_1$ is blue and $a_2$ is red. We must have a green side, and by Lemma 1, that can only be $a_{n+1}$ or $a_{n+2}$. So we have 2 cases.

\medskip

oindent\textit{Case 1:} $a_{n+1}$ is green, so $a_n$ must be red. If $a_{n+2}$ is red, so are $a_{n+3}, \ldots, a_{2n}$, and we get a valid coloring: $a_1$ is blue, $a_{n+1}$ is green, and all the others are red.

If $a_{n+2}$ is green:
\begin{enumerate}[(a)]
  \item $a_{n+3}$ cannot be green, because of $a_2, a_1, a_{2n}, \ldots, a_{n+3}$.
  \item $a_{n+3}$ cannot be blue, because the 4 adjacent sides $a_n, \ldots, a_{n+3}$ can be seen (this is the case that makes the separate treatment of $n \geq 4$ necessary).
  \item $a_{n+3}$ cannot be red, because of $a_1, a_{2n}, \ldots, a_{n+2}$.
\end{enumerate}
So, if $a_{n+2}$ is also green, we cannot get a valid coloring.

\medskip

oindent\textit{Case 2:} $a_{n+2}$ is green, treated the same way as Case 1.

This means that the only valid configuration for $n \geq 4$ is having 2 opposite sides colored in 2 different colors, and all other sides colored in the third color. This can be done in $n \cdot 3 \cdot 2 = 6n$ ways.
\end{proof}",721,4224,Combinatorics,6
102,shl_jbmo_2017_c3,shl_jbmo,2017,c,"We have two piles with 2000 and 2017 coins respectively. Ann and Bob take alternate turns making the following moves: The player whose turn is to move picks a pile with at least two coins, removes from that pile $t$ coins for some $2 \leq t \leq 4$, and adds to the other pile 1 coin. The players can choose a different $t$ at each turn, and the player who cannot make a move loses. If Ann plays first, determine which player has a winning strategy.","\begin{proof}[Solution]
Denote the number of coins in the two piles by $X$ and $Y$. We say that the pair $(X, Y)$ is \emph{losing} if the player who begins the game loses and that the pair $(X, Y)$ is \emph{winning} otherwise. We shall prove that $(X, Y)$ is losing if $X - Y \equiv 0, 1, 7 \pmod{8}$, and winning if $X - Y \equiv 2, 3, 4, 5, 6 \pmod{8}$.

\begin{lemma*}[Lemma 1]
If we have a winning pair $(X, Y)$, then we can always play in such a way that the other player is then faced with a losing pair.
\end{lemma*}

\begin{proof}
Assume $X \geq Y$ and write $X = Y + 8k + \ell$ for some non-negative integer $k$ and some $\ell \in \{2, 3, 4, 5, 6\}$. If $\ell = 2, 3, 4$ then we remove two coins from the first pile and add one coin to the second pile. If $\ell = 5, 6$ then we remove four coins from the first pile and add one coin to the second pile. In each case we then obtain a losing pair.
\end{proof}

\begin{lemma*}[Lemma 2]
If we are faced with a losing distribution, then either we cannot play, or, however we play, the other player is faced with a winning distribution.
\end{lemma*}

\begin{proof}
Without loss of generality we may assume that we remove $k$ coins from the first pile. The following table shows the new difference for all possible values of $k$ and all possible differences $X - Y$. So however we move, the other player will be faced with a winning distribution.
\[
\begin{array}{c|ccc}
k \backslash X - Y & 0 & 1 & 7 \\ \hline
2 & 5 & 6 & 4 \\
3 & 4 & 5 & 3 \\
4 & 3 & 4 & 2
\end{array}
\]
\end{proof}

Since initially the coin difference is $1 \pmod{8}$, by Lemmas 1 and 2, Bob has a winning strategy: he can play so that he is always faced with a winning distribution while Ann is always faced with a losing distribution. So Bob cannot lose. On the other hand, the game finishes after at most 4017 moves, so Ann has to lose.
\end{proof}",449,1875,Combinatorics,7
103,shl_jbmo_2017_g1,shl_jbmo,2017,g,"Given a parallelogram $ABCD$. The line perpendicular to $AC$ passing through $C$ and the line perpendicular to $BD$ passing through $A$ intersect at point $P$. The circle centered at point $P$ and radius $PC$ intersects the line $BC$ at point $X$ ($X 
eq C$) and the line $DC$ at point $Y$ ($Y 
eq C$). Prove that the line $AX$ passes through the point $Y$.","\begin{proof}
Denote the feet of the perpendiculars from $P$ to the lines $BC$ and $DC$ by $M$ and $N$ respectively, and let $O = AC \cap BD$. Since the points $O$, $M$ and $N$ are midpoints of $CA$, $CX$ and $CY$ respectively, it suffices to prove that $M$, $N$ and $O$ are collinear. According to Menelaus's theorem for $\triangle BCD$ and points $M$, $N$ and $O$ we have to prove that
\[
\frac{BM}{MC} \cdot \frac{CN}{ND} \cdot \frac{DO}{OB} = 1.
\]
Since $DO = OB$ the above simplifies to $\dfrac{BM}{CM} = \dfrac{DN}{CN}$. It follows from $BM = BC + CM$ and $DN = DC - CN = AB - CN$ that the last equality is equivalent to:
\begin{equation}\label{eq:G1}
\frac{BC}{CM} + 2 = \frac{AB}{CN}.
\end{equation}

Denote by $S$ the foot of the perpendicular from $B$ to $AC$. Since $\angle BCS = \angle CPM = \varphi$ and $\angle BAC = \angle ACD = \angle CPN = \psi$, we conclude that $\triangle CBS \sim \triangle PCM$ and $\triangle ABS \sim \triangle PCN$. Therefore
\[
\frac{CM}{BS} = \frac{CP}{BC} \quad \text{and} \quad \frac{CN}{BS} = \frac{CP}{AB},
\]
and thus
\[
CM = \frac{CP \cdot BS}{BC} \quad \text{and} \quad CN = \frac{CP \cdot BS}{AB}.
\]
Now equality \eqref{eq:G1} becomes $AB^2 - BC^2 = 2CP \cdot BS$. It follows from
\[
AB^2 - BC^2 = AS^2 - CS^2 = (AS - CS)(AS + CS) = 2OS \cdot AC
\]
that
\[
DC^2 - BC^2 = 2CP \cdot BS \iff 2OS \cdot AC = 2CP \cdot BS \iff OS \cdot AC = CP \cdot BS.
\]
Since $\angle ACP = \angle BSO = 90^\circ$ and $\angle CAP = \angle SBO$, we conclude that $\triangle ACP \sim \triangle BSO$. This implies $OS \cdot AC = CP \cdot BS$, which completes the proof.
\end{proof}",357,1611,Geometry,8
104,shl_jbmo_2017_g2,shl_jbmo,2017,g,"Let $ABC$ be an acute triangle such that $AB$ is the shortest side of the triangle. Let $D$ be the midpoint of the side $AB$ and $P$ be an interior point of the triangle such that
\[
\angle CAP = \angle CBP = \angle ACB.
\]
Denote by $M$ and $N$ the feet of the perpendiculars from $P$ to $BC$ and $AC$, respectively. Let $p$ be the line through $M$ parallel to $AC$ and $q$ be the line through $N$ parallel to $BC$. If $p$ and $q$ intersect at $K$, prove that $D$ is the circumcenter of triangle $MNK$.","\begin{proof}
Let $\gamma = \angle ACB$, so $\angle CAP = \angle CBP = \angle ACB = \gamma$. Let $E = KN \cap AP$ and $F = KM \cap BP$. We show that $E$ and $F$ are midpoints of $AP$ and $BP$, respectively.

Indeed, consider triangle $AEN$. Since $KN \parallel BC$, we have $\angle ENA = \angle BCA = \gamma$. Moreover $\angle EAN = \gamma$, giving that triangle $AEN$ is isosceles, i.e.\ $AE = EN$. Next, consider triangle $ENP$. Since $\angle ENA = \gamma$ we find that
\[
\angle PNE = 90^\circ - \angle ENA = 90^\circ - \gamma.
\]
Now $\angle EPN = 90^\circ - \gamma$ implies that triangle $ENP$ is isosceles, i.e.\ $EN = EP$. Since $AE = EN = EP$, point $E$ is the midpoint of $AP$ and, analogously, $F$ is the midpoint of $BP$. Moreover, $D$ is also the midpoint of $AB$ and we conclude that $DFPE$ is a parallelogram.

It follows from $DE \parallel AP$ and $KE \parallel BC$ that $\angle DEK = \angle CBP = \gamma$ and analogously $\angle DFK = \gamma$.

We conclude that $\triangle EDN \cong \triangle FMD$ (since $ED = FP = FM$, $EN = EP = FD$ and $\angle DEN = \angle MFD = 180^\circ - \gamma$) and thus $ND = MD$. Therefore $D$ is a point on the perpendicular bisector of $MN$. Further,
\[
\angle FDE = \angle FPE = 360^\circ - \angle BPM - \angle MPN - \angle NPA
= 360^\circ - (90^\circ - \gamma) - (180^\circ - \gamma) - (90^\circ - \gamma) = 3\gamma.
\]
It follows that
\[
\angle MDN = \angle FDE - \angle FDM - \angle EDN = \angle FDE - (\angle END + \angle EDN) = 3\gamma - \gamma = 2\gamma.
\]
Finally, $KMCN$ is a parallelogram, i.e.\ $\angle MKN = \angle MCN = \gamma$. Therefore $D$ is a point on the perpendicular bisector of $MN$ and $\angle MDN = 2\angle MKN$, so $D$ is the circumcenter of $\triangle MNK$.
\end{proof}",503,1742,Geometry,9
105,shl_jbmo_2017_g3,shl_jbmo,2017,g,"Consider triangle $ABC$ such that $AB \leq AC$. Point $D$ on the arc $BC$ of the circumcircle of $ABC$ not containing point $A$, and point $E$ on side $BC$, are such that
\[
\angle BAD = \angle CAE < \tfrac{1}{2}\angle BAC.
\]
Let $S$ be the midpoint of segment $AD$. If $\angle ADE = \angle ABC - \angle ACB$, prove that
\[
\angle BSC = 2\angle BAC.
\]","\begin{proof}
Let the tangent to the circumcircle of $\triangle ABC$ at point $A$ intersect line $BC$ at $T$. Since $AB \leq AC$ we get that $B$ lies between $T$ and $C$. Since $\angle BAT = \angle ACB$ and $\angle ABT = 180^\circ - \angle ABC$, we get
\[
\angle ETA = \angle BTA = \angle ABC - \angle ACB = \angle ADE,
\]
which gives that $A, E, D, T$ are concyclic. Since
\[
\angle TDB + \angle BCA = \angle TDB + \angle BDA = \angle TDA = \angle AET = \angle ACB + \angle EAC,
\]
this means $\angle TDB = \angle EAC = \angle DAB$, which means that $TD$ is tangent to the circumcircle of $\triangle ABC$ at point $D$.

Using similar triangles $TAB$ and $TCA$ we get
\begin{equation}\label{eq:G3_1}
\frac{AB}{AC} = \frac{TA}{TC}.
\end{equation}

Using similar triangles $TBD$ and $TDC$ we get
\begin{equation}\label{eq:G3_2}
\frac{BD}{CD} = \frac{TD}{TC}.
\end{equation}

Using the fact that $TA = TD$ with \eqref{eq:G3_1} and \eqref{eq:G3_2} we get
\begin{equation}\label{eq:G3_3}
\frac{AB}{AC} = \frac{BD}{CD}.
\end{equation}

Now since $\angle DAB = \angle CAE$ and $\angle BDA = \angle ECA$, the triangles $DAB$ and $CAE$ are similar. Analogously, the triangles $CAD$ and $EAB$ are similar. These similarities give us
\[
\frac{DB}{CE} = \frac{AB}{AE} \quad \text{and} \quad \frac{CD}{EB} = \frac{CA}{EA},
\]
which, combined with \eqref{eq:G3_3}, give $BE = CE$, so $E$ is the midpoint of side $BC$.

Using the fact that triangles $DAB$ and $CAE$ are similar, and that $E$ is the midpoint of $BC$, we get:
\[
\frac{2DS}{CA} = \frac{DA}{CA} = \frac{DB}{CE} = \frac{DB}{\frac{CB}{2}} = \frac{2DB}{CB},
\]
implying that
\begin{equation}\label{eq:G3_4}
\frac{DS}{DB} = \frac{CA}{CB}.
\end{equation}

Since $\angle SDB = \angle ADB = \angle ACB$, we get from \eqref{eq:G3_4} that the triangles $SDB$ and $ACB$ are similar, giving us $\angle BSD = \angle BAC$. Analogously, $\triangle SDC \sim \triangle ABC$ gives $\angle CSD = \angle CAB$. Combining the last two equalities we get
\[
2\angle BAC = \angle BAC + \angle CAB = \angle CSD + \angle BSD = \angle CSB.
\]
This completes the proof.
\end{proof}

\begin{proof}[Alternative Solution]
\begin{lemma*}[Lemma 1]
A point $P$ is such that $\angle PXY = \angle PYZ$ and $\angle PZY = \angle PYX$. If $R$ is the midpoint of $XZ$, then $\angle XYP = \angle ZYR$.
\end{lemma*}

\begin{proof}
We consider the case when $P$ is inside triangle $XYZ$ (the other case is treated similarly). Let $Q$ be the isogonal conjugate of $P$ in $\triangle XYZ$ and let $YQ$ intersect $XZ$ at $S$.

Then $\angle QXZ = \angle QYX$ and $\angle QZX = \angle QYZ$, and therefore $\triangle SXY \sim \triangle SQX$ and $\triangle SZY \sim \triangle SQZ$. Thus $SX^2 = SQ \cdot SY = SZ^2$ and we conclude that $S \equiv R$. This completes the proof of the Lemma.
\end{proof}

For $\triangle DCA$ we have $\angle CDE = \angle ECA$ and $\angle EAC = \angle ECD$. By Lemma 1 for $\triangle DCA$ and point $E$, we have $\angle SCA = \angle DCE$. Therefore
\[
\angle DSC = \angle SAC + \angle SCA = \angle SAC + \angle DCE = \angle SAC + \angle BAD = \angle BAC.
\]
By analogy, Lemma 1 applied for $\triangle BDA$ and point $E$ gives $\angle BSD = \angle BAC$. Thus, $\angle BSC = 2\angle BAC$.
\end{proof}",353,3225,Geometry,10
106,shl_jbmo_2017_g4,shl_jbmo,2017,g,"Let $ABC$ be a scalene triangle with circumcircle $\Gamma$ and circumcenter $O$. Let $M$ be the midpoint of $BC$ and $D$ be a point on $\Gamma$ such that $AD \perp BC$. Let $T$ be a point such that $BDCT$ is a parallelogram and $Q$ a point on the same side of $BC$ as $A$ such that
\[
\angle BQM = \angle BCA \quad \text{and} \quad \angle CQM = \angle CBA.
\]
Let $AO$ intersect $\Gamma$ again at $E$ and let the circumcircle of $ETQ$ intersect $\Gamma$ at point $X 
eq E$. Prove that the points $A$, $M$, and $X$ are collinear.","\begin{proof}
Let $X'$ be the point symmetric to $Q$ in line $BC$. Now since $\angle CBA = \angle CQM = \angle CX'M$, $\angle BCA = \angle BQM = \angle BX'M$, we have
\[
\angle BX'C = \angle BX'M + \angle CX'M = \angle CBA + \angle BCA = 180^\circ - \angle BAC,
\]
so $X' \in \Gamma$. Now since $\angle AX'B = \angle ACB = \angle MX'B$, we have that $A, M, X'$ are collinear.

Note that since
\[
\angle DCB = \angle DAB = 90^\circ - \angle ABC = \angle OAC = \angle EAC,
\]
we get that $DBCE$ is an isosceles trapezoid.

Since $BDCT$ is a parallelogram we have $MT = MD$, with $M, D, T$ being collinear, $BD = CT$, and since $BDCE$ is an isosceles trapezoid we have $BD = CE$ and $ME = MD$. Since
\[
\angle BTC = \angle BDC = \angle BED, \quad CE = BD = CT, \quad \text{and} \quad ME = MT,
\]
we have that $E$ and $T$ are symmetric with respect to line $BC$. Now since $Q$ and $X'$ are symmetric with respect to $BC$ as well, this means that $QX'ET$ is an isosceles trapezoid, so $Q, X', E, T$ are concyclic. Since $X' \in \Gamma$, this means that $X \equiv X'$ and therefore $A, M, X$ are collinear.
\end{proof}

\begin{proof}[Alternative Solution]
Denote by $H$ the orthocenter of $\triangle ABC$. We use the following well-known properties:
\begin{enumerate}[(i)]
  \item Point $D$ is the symmetric point of $H$ with respect to $BC$. Indeed, if $H_1$ is the symmetric point of $H$ with respect to $BC$, then $\angle BH_1C + \angle BAC = 180^\circ$ and therefore $H_1 \equiv D$.
  \item The symmetric point of $H$ with respect to $M$ is the point $E$. Indeed, if $H_2$ is the symmetric point of $H$ with respect to $M$, then $BH_2CH$ is a parallelogram, $\angle BH_2C + \angle BAC = 180^\circ$, and since $EB \parallel CH$ we have $\angle EBA = 90^\circ$.
\end{enumerate}

Since $DETH$ is a parallelogram and $MH = MD$, we have that $DETH$ is a rectangle. Therefore $MT = ME$ and $TE \perp BC$, implying that $T$ and $E$ are symmetric with respect to $BC$. Denote by $Q'$ the symmetric point of $Q$ with respect to $BC$. Then $Q'ETQ$ is an isosceles trapezoid, so $Q'$ is a point on the circumcircle of $\triangle ETQ$. Moreover, $\angle BQ'C + \angle BAC = 180^\circ$ and we conclude that $Q' \in \Gamma$. Therefore $Q' \equiv X$.

It remains to observe that $\angle CXM = \angle CQM = \angle CBA$ and $\angle CXA = \angle CBA$, and we infer that $X$, $M$ and $A$ are collinear.
\end{proof}",528,2393,Geometry,11
107,shl_jbmo_2017_g5,shl_jbmo,2017,g,"A point $P$ lies in the interior of triangle $ABC$. The lines $AP$, $BP$, and $CP$ intersect $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$, respectively. Prove that if two of the quadrilaterals $ABDE$, $BCEF$, $CAFD$, $AEPF$, $BFPD$, and $CDPE$ are concyclic, then all six are concyclic.","\begin{proof}
We first prove the following lemma.

\begin{lemma*}[Lemma 1]
Let $ABCD$ be a convex quadrilateral and let $AB \cap CD = E$ and $BC \cap DA = F$. Then the circumcircles of triangles $ABF$, $CDF$, $BCE$ and $DAE$ all pass through a common point $P$. This point lies on line $EF$ if and only if $ABCD$ is concyclic.
\end{lemma*}

\begin{proof}
Let the circumcircles of $ABF$ and $BCF$ intersect at $P 
eq B$. We have
\begin{align*}
\angle FPC &= \angle FPB + \angle BPC = \angle BAD + \angle BEC = \angle EAD + \angle AED \\
&= 180^\circ - \angle ADE = 180^\circ - \angle FDC,
\end{align*}
which gives us that $F, P, C, D$ are concyclic. Similarly we have
\begin{align*}
\angle APE &= \angle APB + \angle BPE = \angle AFB + \angle BCD = \angle DFC + \angle FCD \\
&= 180^\circ - \angle FDC = 180^\circ - \angle ADE,
\end{align*}
which gives us that $E, P, A, D$ are concyclic. Since $\angle FPE = \angle FPB + \angle EPB = \angle BAD + \angle BCD$, we get that $\angle FPE = 180^\circ$ if and only if $\angle BAD + \angle BCD = 180^\circ$, which completes the lemma.
\end{proof}

We now divide the problem into cases.

\medskip

oindent\textbf{Case 1:} $AEPF$ and $BFEC$ are concyclic. Here we get
\[
180^\circ = \angle AEP + \angle AFP = 360^\circ - \angle CEB - \angle BFC = 360^\circ - 2\angle CEB,
\]
and here we get that $\angle CEB = \angle CFB = 90^\circ$. From here it follows that $P$ is the orthocenter of $\triangle ABC$ and that gives us $\angle ADB = \angle ADC = 90^\circ$. Now the quadrilaterals $CEPD$ and $BDPF$ are concyclic because
\[
\angle CEP = \angle CDP = \angle PDB = \angle PFB = 90^\circ.
\]
Quadrilaterals $ACDF$ and $ABDE$ are concyclic because
\[
\angle AEB = \angle ADB = \angle ADC = \angle AFC = 90^\circ.
\]

\medskip

oindent\textbf{Case 2:} $AEPF$ and $CEPD$ are concyclic. By Lemma 1 applied to quadrilateral $AEPF$, the circumcircles of $CEP$, $CAF$, $BPF$ and $BEA$ intersect at a point on $BC$. Since $D \in BC$ and $CEPD$ is concyclic, we get that $D$ is the desired point and it follows that $BDPF$, $BAED$, $CAFD$ are all concyclic. We can now finish as in Case 1 since $AEDB$ and $CEPD$ are concyclic.

\medskip

oindent\textbf{Case 3:} $AEPF$ and $AEDB$ are concyclic. We apply Lemma 1 as in Case 2 on quadrilateral $AEPF$. From the lemma we get that $BDPF$, $CEPD$ and $CAFD$ are concyclic, and we finish as in Case 1.

\medskip

oindent\textbf{Case 4:} $ACDF$ and $ABDE$ are concyclic. We apply Lemma 1 on quadrilateral $AEPF$ and get that the circumcircles of $ACF$, $ECP$, $PFB$ and $BAE$ intersect at one point. Since this point is $D$ (because $ACDF$ and $ABDE$ are concyclic), we get that $AEPF$, $CEPD$ and $BFPD$ are concyclic. We now finish as in Case 1.

These four cases prove the problem statement.
\end{proof}

\begin{remark}
A more natural approach is to solve each of the four cases by simple angle chasing.
\end{remark}",290,2892,Geometry,12
108,shl_jbmo_2017_n1,shl_jbmo,2017,n,"Determine all sets of six consecutive positive integers such that the product of two of them, added to the product of other two of them, is equal to the product of the remaining two numbers.","\begin{proof}[Solution]
Exactly two of the six numbers are multiples of 3, and these two need to be multiplied together; otherwise two of the three terms of the equality are multiples of 3 but the third one is not.

Let $n$ and $n+3$ denote these multiples of 3. Two of the four remaining numbers give remainder 1 when divided by 3, while the other two give remainder 2, so the two other products are either $\equiv 1 \cdot 1 = 1 \pmod{3}$ and $\equiv 2 \cdot 2 \equiv 1 \pmod{3}$, or they are both $\equiv 1 \cdot 2 \equiv 2 \pmod{3}$. In conclusion, the term $n(n+3)$ needs to be on the right-hand side of the equality.

Looking at parity, three of the numbers are odd and three are even. One of $n$ and $n+3$ is odd, the other even, so exactly two of the other numbers are odd. As $n(n+3)$ is even, the two remaining odd numbers need to appear in different terms.

We distinguish the following cases.

\medskip

oindent\textbf{Case I.} The numbers are $n-2, n-1, n, n+1, n+2, n+3$.

The product of the two numbers on the RHS needs to be larger than $n(n+3)$. The only possibility is $(n-2)(n-1) + n(n+3) = (n+1)(n+2)$, which leads to $n = 3$. Indeed,
\[
1 \cdot 2 + 3 \cdot 6 = 4 \cdot 5.
\]

\medskip

oindent\textbf{Case II.} The numbers are $n-1, n, n+1, n+2, n+3, n+4$.

As $(n+4)(n-1) + n(n+3) = (n+1)(n+2)$ has no solutions, $n+4$ needs to be on the RHS, multiplied with a number having a different parity, so $n-1$ or $n+1$.
\begin{itemize}
  \item $(n+2)(n-1) + n(n+3) = (n+1)(n+4)$ leads to $n = 3$. Indeed, $2 \cdot 5 + 3 \cdot 6 = 4 \cdot 7$.
  \item $(n+2)(n+1) + n(n+3) = (n-1)(n+4)$ has no solution.
\end{itemize}

\medskip

oindent\textbf{Case III.} The numbers are $n, n+1, n+2, n+3, n+4, n+5$.

We need to consider the following situations:
\begin{itemize}
  \item $(n+1)(n+2) + n(n+3) = (n+4)(n+5)$ which leads to $n = 6$; indeed $7 \cdot 8 + 6 \cdot 9 = 10 \cdot 11$.
  \item $(n+2)(n+5) + n(n+3) = (n+1)(n+4)$ obviously without solutions.
  \item $(n+1)(n+4) + n(n+3) = (n+2)(n+5)$ which leads to $n = 2$ (not a multiple of 3).
\end{itemize}

In conclusion, the problem has three solutions:
\[
1 \cdot 2 + 3 \cdot 6 = 4 \cdot 5, \qquad 2 \cdot 5 + 3 \cdot 6 = 4 \cdot 7, \qquad 7 \cdot 8 + 6 \cdot 9 = 10 \cdot 11.
\]
\end{proof}",190,2252,Number Theory,13
109,shl_jbmo_2017_n2,shl_jbmo,2017,n,Determine all positive integers $n$ such that $n^2 \mid (n-1)!$.,"\begin{proof}[First Solution]
This is true for all positive integers $n$ unless $n = 8$, $n = 9$, $n = p$, or $n = 2p$ for some prime $p$.

It is easy to check that $8^2 
mid 7!$ and $9^2 
mid 8!$ by determining the largest powers of 2 and 3 which divide the right-hand sides. It is also immediate that $p^2 
mid (p-1)!$ and $(2p)^2 
mid (2p-1)!$, as $(p-1)!$ is not divisible by $p$, while the largest power of $p$ dividing $(2p-1)!$ is 1.

The case $n = 1$ is also clearly true. So it remains to show that $n^2 \mid (n-1)!$ in all other cases. It is enough to show that in those cases, for every prime $p$ which divides $n$, the largest power of $p$ dividing $n^2$ is less than or equal to the largest power of $p$ dividing $(n-1)!$. So let us write $n = mp^r$ where $(m, p) = 1$. The largest power of $p$ dividing $(n-1)!$ is
\[
\left\lfloor \frac{n-1}{p} 
ight
floor + \left\lfloor \frac{n-1}{p^2} 
ight
floor + \cdots
> (mp^{r-1} - 1) + \cdots + (m - 1) = m\frac{p^r - 1}{p - 1} - r.
\]
So it is enough to prove that $m\dfrac{p^r - 1}{p - 1} > 3r$.

We distinguish between the cases $p = 2$, $p = 3$, and $p \geq 5$.

\medskip

oindent\textbf{Case 1: $p = 2$.}


oindent\textit{Case 1A: $r \geq 4$.} Then
\[
m\frac{p^r-1}{p-1} \geq 2^r - 1 = 8(1+1)^{r-3} - 1 \geq 8(1 + r - 3) - 1 = 3r + (5r - 17) > 3r.
\]
(Here we used Bernoulli's inequality.)


oindent\textit{Case 1B: $r \leq 3$.} Because $n 
eq 2, 4, 8$, then $n$ has another prime divisor and so $m \geq 3$. Then
\[
m\frac{p^r-1}{p-1} \geq 3(2^r - 1) \geq 3r,
\]
where the last inequality is easily verifiable for $r \leq 3$.

\medskip

oindent\textbf{Case 2: $p = 3$.}


oindent\textit{Case 2A: $r \geq 3$.} Then
\[
m\frac{p^r-1}{p-1} \geq \frac{3^r - 1}{2} = \frac{9(1+2)^{r-2} - 1}{2} \geq \frac{9(1+2(r-2)) - 1}{2} = 3r + (6r - 14) > 3r.
\]


oindent\textit{Case 2B: $r = 2$.} Because $n 
eq 9$, then $n$ has another prime divisor and so $m \geq 2$. Then
\[
m\frac{p^r-1}{p-1} \geq 8 > 6 = 3r.
\]


oindent\textit{Case 2C: $r = 1$.} Because $n 
eq 3, 6$, then $n$ has another divisor which is bigger than 2. So $m \geq 4$. Then
\[
m\frac{p^r-1}{p-1} \geq 4 > 3 = 3r.
\]

\medskip

oindent\textbf{Case 3: $p \geq 5$.}


oindent\textit{Case 3A: $r \geq 2$.} Then
\[
m\frac{p^r-1}{p-1} \geq \frac{5^r - 1}{4} = \frac{5(1+4)^{r-1} - 1}{4} \geq \frac{5(1+4(r-1)) - 1}{4} = 3r + 2(r-2) \geq 3r.
\]


oindent\textit{Case 3B: $r = 1$.} Because $n 
eq p, 2p$, then $n$ has another divisor which is bigger than 2. So $m \geq 3$. Then
\[
m\frac{p^r-1}{p-1} \geq 3 = 3r.
\]
\end{proof}

\begin{proof}[Second Solution]
Let $n 
eq 8, 9, p, 2p$, where $p$ is prime.

Let $n$ be odd and $p$ be the smallest prime divisor of $n$. If $n = p^2$, then $p \geq 5$, and $p < 2p < 3p < 4p$ all participate in $(n-1)!$, so $p^4 = n^2 \mid (n-1)!$. If $p < \frac{n}{p}$, then $p < 2p$ and $\frac{n}{p} < \frac{2n}{p}$ all are less than $n$ and therefore participate in $(n-1)!$. So $n^2 \mid 4n^2 = p \cdot 2p \cdot \frac{n}{p} \cdot \frac{2n}{p} \mid (n-1)!$.

Let $n$ be even and $n = 2^k m$, where $k$ is a positive integer and $m$ is odd. If $m = 1$, then $k \geq 4$ and $2 < 2^2 < 2^{k-2} < 2^{k-1}$ shows that $n^2 = 2^{2k} \mid (n-1)!$ for $k \geq 5$, and the case $k = 4$ is seen directly.

Let now $m > 1$. If $k \geq 2$, then the divisors $2 < m < 2^{k-1}m$ and $2^k$ of $n$ work. If $k = 1$, then $m$ is not prime, and let $p$ be the smallest prime divisor of $m$. Now $4, p < 2p$ and $\frac{m}{p} < \frac{2m}{p}$ work when $m 
eq p^2$, and $4, p < 2p < 3p < 4p$ work when $m = p^2$.
\end{proof}",64,3546,Number Theory,14
110,shl_jbmo_2017_n3,shl_jbmo,2017,n,"Find all pairs of positive integers $(x, y)$ such that $2^x + 3^y$ is a perfect square.","\begin{proof}[Solution]
In order for $2^x + 3^y$ to be a perfect square, a positive integer $t$ must exist such that $2^x + 3^y = t^2$.

\medskip

oindent\textbf{Case 1.} If $x$ is even, then there exists a positive integer $z$ such that $x = 2z$. Then
\[
(t - 2^z)(t + 2^z) = 3^y.
\]
Since $t + 2^z - (t - 2^z) = 2^{z+1}$, which implies $\gcd(t - 2^z, t + 2^z) \mid 2^{z+1}$, it follows that $\gcd(t - 2^z, t + 2^z) = 1$, hence $t - 2^z = 1$ and $t + 2^z = 3^y$, so we have $2^{z+1} + 1 = 3^y$.

For $z = 1$ we have $5 = 3^y$ which clearly has no solution. For $z \geq 2$ we have (modulo 4) that $y$ is even. Let $y = 2k$. Then $2^{z+1} = (3^k - 1)(3^k + 1)$, which is possible only when $3^k - 1 = 2$, i.e.\ $k = 1$, $y = 2$, which implies $t = 5$. So the pair $(4, 2)$ is a solution.

\medskip

oindent\textbf{Case 2.} If $y$ is even, then there exists a positive integer $w$ such that $y = 2w$, and
\[
(t - 3^w)(t + 3^w) = 2^x.
\]
Since $t + 3^w - (t - 3^w) = 2 \cdot 3^w$, we have $\gcd(t - 3^w, t + 3^w) = 2$. Hence $t - 3^w = 2$ and $t + 3^w = 2^{x-1}$. So we have
\[
2 \cdot 3^w + 2 = 2^{x-1} \implies 3^w + 1 = 2^{x-2}.
\]
Here we see modulo 3 that $x - 2$ is even. Let $x - 2 = 2m$; then $3^w = (2^m - 1)(2^m + 1)$, whence $m = 1$ since $\gcd(2^m - 1, 2^m + 1) = 1$. So we arrive again at the solution $(4, 2)$.

\medskip

oindent\textbf{Case 3.} Let $x$ and $y$ be odd. For $x \geq 3$ we have $2^x + 3^y \equiv 3 \pmod{4}$ while $t^2 \equiv 0, 1 \pmod{4}$, a contradiction. For $x = 1$ we have $2 + 3^y = t^2$. For $y \geq 2$ we have $2 + 3^y \equiv 2 \pmod{9}$ while $t^2 \equiv 0, 1, 4, 7 \pmod{9}$. For $y = 1$ we have $5 = 2 + 3 = t^2$, which clearly has no solution.

\medskip

oindent Hence the only solution is $(x, y) = (4, 2)$.
\end{proof}

\begin{note}
The proposer's solution used Zsygmondy's theorem in the final steps of Cases 1 and 2.
\end{note}",87,1870,Number Theory,15
111,shl_jbmo_2017_n4,shl_jbmo,2017,n,Solve in nonnegative integers the equation $5^t + 3 \cdot 4^y = z^2$.,"\begin{proof}[Solution]
If $x = 0$ we have
\[
z^2 - 2^{2y} = 5^t \iff (z + 2^y)(z - 2^y) = 5^t.
\]
Putting $z + 2^y = 5^a$ and $z - 2^y = 5^b$ with $a + b = t$ we get $5^a - 5^b = 2^{y+1}$. This gives us $b = 0$ and now we have $5^t - 1 = 2^{y+1}$. If $y \geq 2$, consideration modulo 8 gives $2 \mid t$. Putting $t = 2s$ we get $(5^s - 1)(5^s + 1) = 2^{y+1}$. This means $5^s - 1 = 2^c$ and $5^s + 1 = 2^d$ with $c + d = y + 1$. Subtracting we get $2 = 2^d - 2^c$, so $c = 1$, $d = 2$, but the equation $5^s - 1 = 2$ has no solutions over nonneg. integers. Therefore $y \geq 2$ gives no solutions. If $y = 0$ we get again $5^t - 1 = 2$ which has no solutions. If $y = 1$ we get $t = 1$ and $z = 3$, giving the solution $(t, x, y, z) = (1, 0, 1, 3)$.

Now if $x \geq 1$, then by modulo 3 we have $2 \mid t$. Putting $t = 2s$ we get
\[
3^x 4^y = z^2 - 5^{2s} \iff 3^x 4^y = (z + 5^s)(z - 5^s).
\]
Now we have $z + 5^s = 3^m 2^k$ and $z - 5^s = 3^n 2^l$, with $k + l = 2y$ and $m + n = x \geq 1$. Subtracting we get
\[
2 \cdot 5^s = 3^m 2^k - 3^n 2^l.
\]
Here we get that $\min\{m, n\} = 0$. We now have a couple of cases.

\medskip

oindent\textbf{Case 1: $k = l = 0$.} Now we have $n = 0$ and we get the equation $2 \cdot 5^s = 3^m - 1$. From modulo 4 we get that $m$ is odd. If $s \geq 1$ we get modulo 5 that $4 \mid m$, a contradiction. So $s = 0$ and we get $m = 1$. This gives us $t = 0$, $x = 1$, $y = 0$, $z = 2$.

\medskip

oindent\textbf{Case 2: $\min\{k,l\} = 1$.}


oindent\textit{Case 2a: $l > k = 1$.} We get $5^s = 3^m - 3^n 2^{l-1}$. Since $\min\{m,n\} = 0$, we get $n = 0$. Now the equation becomes $5^s = 3^m - 2^{l-1}$. Note that $l - 1 = 2y - 2$ is even. By modulo 3 we get that $s$ is odd and so $s \geq 1$. By modulo 5 we get $3^m \equiv 2^{2y-2} \equiv 1, -1 \pmod{5}$. Here we get that $m$ is even, so write $m = 2q$. Now we get $5^s = (3^q - 2^{y-1})(3^q + 2^{y-1})$.

Therefore $3^q - 2^{y-1} = 5^v$ and $3^q + 2^{y-1} = 5^u$ with $u + v = s$. Then $2^y = 5^u - 5^v$, whence $v = 0$ and we have $3^q - 2^{y-1} = 1$. Plugging in $y = 1, 2$ we get the solution $y = 2$, $q = 1$. This gives $m = 2$, $s = 1$, $n = 0$, $x = 2$, $t = 2$, and therefore $z = 13$. Thus we have the solution $(t, x, y, z) = (2, 2, 2, 13)$.

If $y \geq 3$ we get modulo 4 that $q$ is even; write $q = 2r$. Then $(3^r - 1)(3^r + 1) = 2^{y-1}$. Putting $3^r - 1 = 2^e$ and $3^r + 1 = 2^f$ with $e + f = y - 1$ and subtracting gives $2^{f-1} - 2^{e-1} = 1$, whence $e = 1$, $f = 2$. Therefore $r = 1$, $q = 2$, $y = 4$. Now since $2^4 = 5^u - 1$ does not have a solution, there are no more solutions in this case.


oindent\textit{Case 2b: $k > l = 1$.} We now get $5^s = 3^m 2^{k-1} - 3^n$. By modulo 4 (which we can use since $0 < k - 1 = 2y - 2$) we get $3^n \equiv -1 \pmod{4}$ and therefore $n$ is odd. Now since $\min\{m,n\} = 0$ we get $m = 0$, and $0 + n = m + n = x \geq 1$. The equation becomes $5^s = 2^{2y-2} - 3^x$. By modulo 3 we see that $s$ is even. Putting $s = 2g$ we obtain
\[
(2^{y-1} - 5^g)(5^g + 2^{y-1}) = 3^x.
\]
Putting $2^{y-1} - 5^g = 3^h$ and $2^{y-1} + 5^g = 3^i$ where $i + h = x$, and subtracting, we get $3^i - 3^h = 2^y$. This gives $h = 0$ and now we are solving the equation $3^x + 1 = 2^y$.

The solution $x = 0$, $y = 1$ gives $1 - 5^g = 1$ without solution. If $x \geq 1$, by modulo 3 we get $y$ is even. Putting $y = 2y_1$ we obtain $3^x = (2^{y_1} - 1)(2^{y_1} + 1)$. Putting $2^{y_1} - 1 = 3^{x_1}$ and $2^{y_1} + 1 = 3^{x_2}$ and subtracting we get $3^{x_2} - 3^{x_1} = 2$. This gives $x_1 = 0$, $x_2 = 1$. Then $y_1 = 1$, $x = 1$, $y = 2$ is the only solution to $3^x + 1 = 2^y$ with $x \geq 1$. Now from $2 - 5^g = 1$ we get $g = 0$, giving $t = 0$. This gives the solution $1 + 3 \cdot 16 = 49$ and $(t, x, y, z) = (0, 1, 2, 7)$.

\medskip
This completes all the cases, and thus the solutions are
\[
(t, x, y, z) = (1, 0, 1, 3),\ (0, 1, 0, 2),\ (2, 2, 2, 13),\ \text{and}\ (0, 1, 2, 7).
\]
\end{proof}

\begin{note}
The problem can be simplified by asking for solutions in positive integers (without significant loss in ideas).
\end{note}",69,4079,Number Theory,16
112,shl_jbmo_2017_n5,shl_jbmo,2017,n,"Find all positive integers $n$ such that there exists a prime number $p$ such that
\[
p^n - (p-1)^n
\]
is a power of 3.

(\emph{Note.} A power of 3 is a number of the form $3^a$ where $a$ is a positive integer.)","\begin{proof}[Solution]
Suppose that the positive integer $n$ is such that
\begin{equation}\label{eq:NT5}
p^n - (p-1)^n = 3^a \tag{1}
\end{equation}
for some prime $p$ and positive integer $a$.

If $p = 2$, then $2^n - 1 = 3^a$ by \eqref{eq:NT5}, whence $(-1)^n - 1 \equiv 0 \pmod{3}$, so $n$ must be even. Setting $n = 2s$ we obtain $(2^s - 1)(2^s + 1) = 3^a$. It follows that $2^s - 1$ and $2^s + 1$ are both powers of 3, but since they are both odd they are coprime, and we have $2^s - 1 = 1$, i.e.\ $s = 1$ and $n = 2$.

If $p = 3$, then \eqref{eq:NT5} gives $3 \mid 2^n$, which is impossible.

Let $p \geq 5$. Then it follows from \eqref{eq:NT5} that we cannot have $3 \mid p - 1$. This means that $p^n - (p-1)^n 
ot\equiv 0 \pmod{3}$ forces $n$ to be even; let $n = 2k$. Then
\[
p^{2k} - (p-1)^{2k} = 3^a \iff (p^k - (p-1)^k)(p^k + (p-1)^k) = 3^a.
\]
If $d = \gcd(p^k - (p-1)^k,\, p^k + (p-1)^k)$, then $d \mid 2p^k$. However, both numbers are powers of 3, so $d = 1$ and
\[
p^k - (p-1)^k = 1, \quad p^k + (p-1)^k = 3^a.
\]
If $k = 1$, then $n = 2$ and we can take $p = 5$. For $k \geq 2$ we have
\[
1 = p^k - (p-1)^k \geq p^2 - (p-1)^2 = 2p - 1 \geq 9,
\]
which is absurd (the inequality $p^k - (p-1)^k \geq p^2 - (p-1)^2$ is equivalent to $p^2(p^{k-2} - 1) \geq (p-1)^2((p-1)^{k-2} - 1)$, which is obviously true).

It follows that the only solution is $\boxed{n = 2}$.
\end{proof}",211,1389,Number Theory,17
472,tst_jbmo_ro_2017_1_p1,tst_jbmo,2017,g,"Let $P$ be a point in the interior of the acute-angled triangle $ABC$. Prove that if the reflections of $P$ with respect to the sides of the triangle lie on the circumcircle of the triangle, then $P$ is the orthocenter of $ABC$.","\textbf{Solution.}
Let $C_1$, $C_2$ and $C_3$ be the reflections of the circumcircle with respect to the sides of the triangle. Point $P$ lies on each of these circles. On the other hand, so does the orthocenter. But these three circles have only one common point, therefore $P$ is the orthocenter.

\textbf{Remark.} The fact that the reflections of $P$ with respect to two of the sides of the triangle lie on the circumcircle already makes $P$ the orthocenter.",228,461,Geometry,1
473,tst_jbmo_ro_2017_1_p2,tst_jbmo,2017,a,"\textbf{a)} Determine the set
\[
A = \{(a,b,c) \in \mathbb{R}^3 \mid a+b+c=3,\ (6a+b^2+c^2)(6b+c^2+a^2)(6c+a^2+b^2)
eq 0\}.
\]
\textbf{b)} Prove that for all $(a,b,c)\in A$ the following inequality holds:
\[
\frac{a}{6a+b^2+c^2} + \frac{b}{6b+c^2+a^2} + \frac{c}{6c+a^2+b^2} \leq \frac{3}{8}.
\]","\textbf{Solution.}

\textbf{a)} If $a+b+c=3$ and $6a+b^2+c^2=0$, then $0=b^2+c^2+6(3-b-c)=(b-3)^2+(c-3)^2$, hence $b=c=3$, $a=-3$. By proceeding similarly in the other cases, we conclude that the only triples not belonging to $A$ are $(3,3,-3)$, $(3,-3,3)$ and $(-3,3,3)$.

\textbf{b)} The inequality can be rewritten successively:
\[
\frac{6a}{6a+b^2+c^2} + \frac{6b}{6b+c^2+a^2} + \frac{6c}{6c+a^2+b^2} \leq \frac{9}{4},
\]
\[
\left(1 - \frac{b^2+c^2}{6a+b^2+c^2}
ight) + \left(1 - \frac{c^2+a^2}{6b+c^2+a^2}
ight) + \left(1 - \frac{a^2+b^2}{6c+a^2+b^2}
ight) \leq \frac{9}{4},
\]
\[
\frac{b^2+c^2}{6a+b^2+c^2} + \frac{c^2+a^2}{6b+c^2+a^2} + \frac{a^2+b^2}{6c+a^2+b^2} \geq \frac{3}{4},
\]
\[
\frac{b^2+c^2}{2a(a+b+c)+b^2+c^2} + \frac{c^2+a^2}{2b(a+b+c)+c^2+a^2} + \frac{a^2+b^2}{2c(a+b+c)+a^2+b^2} \geq \frac{3}{4},
\]
\[
\frac{2(b^2+c^2)}{(a+b)^2+(a+c)^2} + \frac{2(c^2+a^2)}{(b+c)^2+(b+a)^2} + \frac{2(a^2+b^2)}{(c+a)^2+(c+b)^2} \geq \frac{3}{2}.
\]
As $2(b^2+c^2) \geq (b+c)^2$ and its analogues hold, it is sufficient to prove that
\[
\frac{(b+c)^2}{(a+b)^2+(a+c)^2} + \frac{(c+a)^2}{(b+c)^2+(b+a)^2} + \frac{(a+b)^2}{(c+a)^2+(c+b)^2} \geq \frac{3}{2},
\]
which is Nesbitt's well-known inequality (which remains valid even if one of the variables is 0) written for the numbers $(a+b)^2$, $(b+c)^2$, $(c+a)^2$.

The equality case is when $a=b=c=1$.

\textbf{Remark.} Working on the denominators, the inequality
\[
\frac{2(b^2+c^2)}{(a+b)^2+(a+c)^2} + \frac{2(c^2+a^2)}{(b+c)^2+(b+a)^2} + \frac{2(a^2+b^2)}{(c+a)^2+(c+b)^2} \geq \frac{3}{2}
\]
also follows from Nesbitt's inequality for $a^2+b^2$, $b^2+c^2$, $c^2+a^2$.",295,1624,Algebra,2
474,tst_jbmo_ro_2017_1_p3,tst_jbmo,2017,n,Determine the integers $x$ and $y$ for which $\sqrt{4^x+5^y}$ is rational.,"\textbf{Solution.}
We treat four cases: I. $x,y\geq 0$, II. $x,y<0$, III. $x<0,\ y\geq 0$ and IV. $x\geq 0,\ y<0$.

\textbf{Case I.} $\sqrt{4^x+5^y}$ is rational if and only if $4^x+5^y$ is a perfect square, i.e.\ there exists $n\in\mathbb{N}$ such that $4^x+5^y=n^2$. Analyzing this equation modulo 3, we have $4^x\equiv 1\pmod{3}$, $5^y\equiv (-1)^y\pmod{3}$ and $n^2\equiv 0,1\pmod{3}$, hence $y$ needs to be odd. Let $z\in\mathbb{N}$ be such that $y=2z+1$. The previous equation comes to $4^x+5\cdot 25^z=n^2$. If $x\geq 2$, then $4^x\equiv 0\pmod{8}$, $5\cdot 25^z\equiv 5\pmod{8}$, while $n^2\equiv 0,1,4\pmod{8}$, which means the equation has no solutions in this case. We are left with $x=0$ and $x=1$. For $x=0$ we have $4^x+5^y\equiv 2\pmod{4}$, hence $4^x+5^y$ cannot be a perfect square. If $x=1$, then $5^y=(n-2)(n+2)$, hence there exist $a,b\in\mathbb{N}$ with $a+b=y$ such that $n-2=5^a$ and $n+2=5^b$. By subtraction, $5^b-5^a=4$. If $a,b\geq 1$ then $5\mid 5^b-5^a$, hence $5\mid 4$, contradiction. As $a<b$, it follows that $a=0$, then $b=1$, i.e.\ $y=1$. We obtain the solution $x=y=1$.

\textbf{Case II.} Let $u=-x$, $v=-y$, $u,v>0$. $\sqrt{4^x+5^y}$ is rational if and only if there exist $p,q\in\mathbb{N}^*$, coprime, such that $4^x+5^y=\frac{p^2}{q^2}$, i.e.\ $\frac{4^u+5^v}{4^u\cdot 5^v}=\frac{p^2}{q^2}$, which means $p^2\cdot 4^u\cdot 5^v=q^2(4^u+5^v)$. Numbers $4^u\cdot 5^v$ and $4^u+5^v$ are coprime, therefore every prime factor of $4^u+5^v$ is a prime factor of $p^2$, which means that it appears at an even exponent. It follows that $4^u+5^v$ is a perfect square, and, similarly, $4^u\cdot 5^v$ is a perfect square. From Case I it follows that $4^u+5^v$ is a perfect square if and only if $u=v=1$, but then $4^u\cdot 5^v=20$ is not a perfect square. We conclude that there are no solutions in this case.

\textbf{Case III.} Let $u=-x\in\mathbb{N}^*$. Then $\sqrt{4^x+5^y}=\frac{\sqrt{1+4^u\cdot 5^y}}{2^u}$ is rational if and only if $1+4^u\cdot 5^y$ is a perfect square, i.e.\ there exists $n\in\mathbb{N}$ such that $1+4^u\cdot 5^y=n^2$. Then $4^u\cdot 5^y=(n-1)(n+1)$. As $u>0$, $n$ is odd and $\gcd(n-1,n+1)=2$. We distinguish the following sub-cases:
\begin{itemize}
  \item[A.] $n-1=2\cdot 5^y$, $n+1=2^{2u-1}$;
  \item[B.] $n-1=2$, $n+1=2^{2u-1}\cdot 5^y$;
  \item[C.] $n-1=2^{2u-1}$, $n+1=2\cdot 5^y$;
  \item[D.] $n-1=2^{2u-1}\cdot 5^y$, $n+1=2$.
\end{itemize}

In sub-case A we obtain $2^{2u-2}-5^y=1$, i.e.\ $(2^{u-1}-1)(2^{u-1}+1)=5^y$. It follows that $2^{u-1}-1$ and $2^{u-1}+1$ should be powers of 5, but no two powers of 5 are at distance 2.

Sub-case B leads to $n=3$ and immediately to $4=2^{2u-1}\cdot 5^y$, with no solutions.

In sub-case C we get $5^y-2^{2u-2}=1$. Then $5^y\equiv(-1)^y\pmod{3}$ and $2^{2u-2}=4^{u-1}\equiv 1\pmod{3}$, therefore $y$ needs to be odd. It follows that $5^y\equiv 5\pmod{8}$, hence $2^{2u-2}\equiv 4\pmod{8}$, i.e.\ $u=2$. We obtain the solution $x=-2$, $y=1$.

Sub-case D leads to $n=1$ and then to $0=2^{2u-1}\cdot 5^y$, with no solutions.

\textbf{Case IV.} Let $v=-y\in\mathbb{N}^*$. Then $\sqrt{4^x+5^y}=\sqrt{\frac{1+4^x\cdot 5^v}{5^v}}$ is rational if and only if there exist $p,q\in\mathbb{N}^*$ coprime such that $\frac{1+4^x\cdot 5^v}{5^v}=\frac{p^2}{q^2}$, i.e.\ such that $p^2\cdot 5^v=q^2(1+4^x\cdot 5^v)$. Numbers $1+4^x\cdot 5^v$ and $5^v$ are coprime, hence, as in Case I, they need to be perfect squares. We have seen in the previous case that $1+4^x\cdot 5^v$ is a perfect square only when $x=2$ and $v=1$, but in this situation $5^v=5$ is not a perfect square. We conclude that in this case there are no solutions.

To summarize, the only solutions to the problem are $x=y=1$ and $x=-2$, $y=1$.",74,3698,Number Theory,3
475,tst_jbmo_ro_2017_1_p4,tst_jbmo,2017,c,"Two right isosceles triangles of legs equal to 1 are glued together to form either an isosceles triangle --- called \textit{t-shape} --- of leg $\sqrt{2}$, or a parallelogram --- called \textit{p-shape} --- of sides $1$ and $\sqrt{2}$. Find all integers $m$ and $n$, $m,n\geq 2$, such that a rectangle $m\times n$ can be tiled with t-shapes and p-shapes.","\textbf{Solution.}
We will prove that any rectangle with (at least) an even side length obeys the condition.

To this end, notice that 4 t-shapes can be glued to produce a $2\times 2$ square, which is sufficient for tiling a rectangle with both sides even. If $m+n$ is odd, a $2\times 3$ rectangle can be obtained by a suitable arrangement. Starting from this rectangle and using several $2\times 2$ squares one can tile any given $m\times n$ rectangle with precisely one even length side.

Alternatively, we can tile any rectangle $2\times n$, with $n\geq 2$ tiles as follows: place three t-shapes on the left, three on the right, and fill the middle with p-shapes.

Obviously a $1\times n$ rectangle cannot be tiled for any $n$. Any way one would position the tiles, the one that covers one of the vertices of the rectangle renders impossible the covering of the closest vertex.

It is left to prove that any odd-sided rectangle cannot be tessellated. For this, color the unit squares of an $m\times n$ rectangle with $m,n$ odd in a chessboard pattern, with the corners being black. On one hand, we have more black squares than white squares; on the other hand each tile covers a surface area that is half black, half white, so the total surface area covered by the tiles is half black, half white. In conclusion, the surface of a rectangle with both dimensions odd cannot be tiled.

In conclusion, the tiling is possible if and only if $m,n\geq 2$ and at least one dimension is even.",354,1486,Combinatorics,4
476,tst_jbmo_ro_2017_2_p1,tst_jbmo,2017,n,Determine the integers $x$ such that $2^x+x^2+25$ is the cube of a prime number.,"\textbf{Solution.}
Let $y$ be a prime such that $y^3=2^x+x^2+25$; clearly, $y$ is odd. Moreover, $x$ cannot be a negative integer nor can it belong to the set $\{0,1,2,3\}$. Hence $x\geq 4$ is even.

\textbf{1.} If $x=6k$, with $k\in\mathbb{N}^*$, we get $64^k+36k^2+25=y^3$.

For $k=1$ we get $x=6$ and $y=5$.

For $k\geq 2$ we have no solutions because of the following inequalities:
\[
\left(\frac{4^k}{3}
ight)^3 < y^3 = 2^{6k}+(6k)^2+25 < \left(\frac{4^k+1}{3}
ight)^3,
\]
the second inequality being equivalent to $3k^2+2<4^{k-1}(4^k+1)$, which is true (induction).

\textbf{2.} If $x=6k+2$ or $x=6k+4$ then $2^x\equiv 1\pmod{3}$, $x^2\equiv 1\pmod{3}$, hence $2^x+x^2+25\equiv 0\pmod{3}$, i.e.\ $y^3\equiv 0\pmod{3}$, which means $y\equiv 0\pmod{3}$ and, as $y$ is prime, we get $y=3$, which does not fulfill the condition.

In conclusion, the only solution is $x=6$.",80,874,Number Theory,5
477,tst_jbmo_ro_2017_2_p2,tst_jbmo,2017,c,"Determine the smallest positive integer $n$ such that, for any coloring of the elements of the set $\{2,3,\ldots,n\}$ with two colors, the equation $x+y=z$ has a monochrome solution with $x
eq y$.

(We say that the equation $x+y=z$ has a monochrome solution if there exist $a,b,c$ distinct, having the same color, such that $a+b=c$.)","\textbf{Solution.}
We prove that $n=13$.

For $n=12$ there exists a coloring of the numbers from $\{2,3,\ldots,12\}$ with two colors such that the equation $x+y=z$ has no monochrome solution: we color the numbers from $A=\{2,3,4,11,12\}$ with one color, and the elements of $B=\{5,6,7,8,9,10\}$ with the second color.

For $n<12$ we color with the first color the numbers from $\{2,3,\ldots,n\}\cap A$ and with the second one the elements of $\{2,3,\ldots,n\}\cap B$. Again, the equation $x+y=z$ has no monochrome solution. It follows that $n\geq 13$.

We prove that, for any coloring with two colors of the elements of the set $\{2,3,\ldots,13\}$, the equation $x+y=z$ has monochrome solutions. Assume the contrary: there exists a coloring such that there is no monochrome solution to the equation.

\textbf{Case 1:} If $2,3,4$ have color 1, then $5=2+3$, $6=2+4$, $7=3+4$ need to have color 2, hence $11=5+6$, $12=5+7$, $13=6+7$ need to be of color 1. But then $2+11=13$, and the equation has a monochrome solution (of color 1).

\textbf{Case 2:} If $2$ and $3$ are of color 1 while $4$ has color 2, then $5=2+3$ has color 2, and $9=4+5$ has color 1. But $3+6=9$, and numbers $3$ and $9$ have color 1, hence $6$ needs to be of color 2. Similarly, as $2+7=9$ and $2,9$ have color 1, it follows that $7$ has color 2. If $11$ has color 1, then $2+9=11$ leads to a monochrome solution. If $11$ has color 2, then $5+6=11$ is a monochrome solution.

\textbf{Case 3:} If $2$ and $4$ have color 1 while $3$ has color 2, then $6$ has color 2, $9$ has color 1. As $4$ and $9$ have color 1, we need $5$ to have color 2. Similarly, $2$ and $9$ are of color 1, therefore $7$ has color 2, $8=3+5$ has color 1. But $12=4+8=5+7$, hence we have a monochrome solution (of color 1 or color 2, depending on the color of 12).

\textbf{Case 4:} $2$ has color 1, $3$ and $4$ have color 2. Then $7$ has color 1. But $2$ and $7$ having color 1 means that $5$ needs to be of color 2. This leads to $9=2+7=4+5$ and, again, regardless of the color of $9$, the equation has a monochrome solution.",333,2069,Combinatorics,6
478,tst_jbmo_ro_2017_2_p3,tst_jbmo,2017,g,"The incircle of triangle $ABC$ touches the sides $BC$, $CA$, and $AB$ at $D$, $E$, and $F$ respectively. On the line segments $EF$, $FD$, and $DE$, consider the points $M$, $N$, and $P$ respectively such that the sums $BM+MC$, $CN+NA$, and $AP+PB$ are minimum.
\begin{itemize}
  \item[a)] Prove that the lines $AM$, $BN$, and $CP$ are concurrent.
  \item[b)] Prove that $DM$, $EN$ and $FP$ are the altitudes of triangle $DEF$.
\end{itemize}","\textbf{Solution.}

\textbf{a)} Let $S$ be the reflection of $C$ across the line $EF$. As $AE=AF$, it follows that $\angle AFE=\angle AEF$, hence $\angle BFE=\angle FEC=\angle FES$, therefore $FB$ and $ES$ are parallel. $FBES$ is a trapezoid (or a parallelogram); let $M$ be the intersection point of its diagonals. According to the billiards problem, $M$ is the point of the line $EF$ for which the sum $BM+MC$ is minimum.

Triangles $BFM$ and $SEM$ are similar, therefore $\dfrac{FM}{EM}=\dfrac{FB}{ES}=\dfrac{FB}{EC}=\dfrac{s-b}{s-c}$. It follows that
\[
\frac{\sin(\angle FAM)}{\sin(\angle EAM)} = \frac{[FAM]}{[EAM]} = \frac{FM}{EM} = \frac{s-b}{s-c}.
\]
Multiplying this with the two analogous relations we get
\[
\frac{\sin(\angle FAM)}{\sin(\angle EAM)}\cdot\frac{\sin(\angle ECP)}{\sin(\angle DCP)}\cdot\frac{\sin(\angle DBN)}{\sin(\angle NBF)} = 1.
\]
By the converse of the trigonometric form of Ceva's theorem, it follows that the lines $AM$, $BN$, and $CP$ are concurrent.

\textbf{b)} We have $\dfrac{FM}{EM}=\dfrac{p-b}{p-c}=\dfrac{BD}{CD}=\dfrac{BF}{CE}$, hence, from the ``Gliding Bisector Theorem'', it follows that $DM$ is parallel to the bisector of angle $\angle BAC$. The bisector is perpendicular to $EF$, therefore $DM$ is also perpendicular to $EF$.

\textbf{Remark.} One can avoid the trigonometric form of Ceva's theorem by considering $A'$ the intersection point of $AM$ with $BC$, computing $\dfrac{BA'}{CA'}=\dfrac{c(s-b)}{b(s-c)}$, then applying the classical form of Ceva's theorem.

\textbf{Remark.} Another approach: one could start by proving first point b) (possibly by using the Billiard's Problem and then the similarity of triangles $BFM$ and $SEM$), then point a) follows from:

\textbf{Theorem (Cevian Nest).} Let $ABC$ be a triangle, $A_1\in BC$, $B_1\in CA$, $C_1\in AB$, $A_2\in B_1C_1$, $B_2\in C_1A_1$ and $C_2\in A_1B_1$ such that the lines $AA_1$, $BB_1$ and $CC_1$ are concurrent and $A_1A_2$, $B_1B_2$ and $C_1C_2$ are also concurrent. Then the lines $AA_2$, $BB_2$ and $CC_2$ are also concurrent.

Returning to our problem, lines $AD$, $BE$ and $CF$ meet in Gergonne's Point of triangle $ABC$, while $DM$, $EN$ and $FP$ meet in the orthocenter of triangle $DEF$. From the Cevian Nest Theorem it follows that $AD$, $BE$, $CF$ are concurrent.

\textbf{Further remarks} (Mircea Fianu): Triangles $ABC$ and $MNP$ are homothetic, the lines $AM$, $BN$ and $CN$ being concurrent in the center of homothety, $X$. Point $X$ lies on the Euler line of triangle $DEF$.

\textit{Proof:} The quadrilateral $DEMN$ is cyclic, hence $\angle FMN=\angle FDE=\dfrac{\angle B+\angle C}{2}=\angle AFE$, and $MN\parallel AB$. Similarly, $NP\parallel BC$ and $PM\parallel CA$, which proves the homothety of triangles $ABC$ and $MNP$.

The homothety transforming triangle $ABC$ into triangle $MNP$ takes the incenter of $ABC$ to the incenter of $MNP$. But $MNP$ is the orthic triangle of $DEF$, hence the angle bisectors of $MNP$ are the altitudes of $DEF$. It follows that the incenter of $MNP$ is the orthocenter of $DEF$, while the incenter of $ABC$ is the circumcenter of $DEF$. As $X$ lies on the line determined by the two incenters which are the orthocenter and the circumcenter of $DEF$, we conclude that $X$ is on the Euler line of $DEF$.",440,3273,Geometry,7
479,tst_jbmo_ro_2017_2_p4,tst_jbmo,2017,a,"Let $a,b,c,d$ be non-negative real numbers satisfying $a+b+c+d=3$. Prove that
\[
\frac{a}{1+2b^3} + \frac{b}{1+2c^3} + \frac{c}{1+2d^3} + \frac{d}{1+2a^3} \geq \frac{a^2+b^2+c^2+d^2}{3}.
\]
When does the equality hold?","\textbf{Solution 1.} We notice the equality case $a=3$, $b=c=d=0$ and we try to mix the variables as follows:
\[
f(a,b,c,d) \geq f(a+b+c+d,0,0,0) = a+b+c+d - \frac{(a+b+c+d)^2}{3} = 0,
\]
where
\[
f(a,b,c,d) = \frac{a}{1+2b^3}+\frac{b}{1+2c^3}+\frac{c}{1+2d^3}+\frac{d}{1+2a^3} - \frac{a^2+b^2+c^2+d^2}{3}.
\]
Then
\begin{align*}
&f(a,b,c,d) - a - b - c - d + \frac{(a+b+c+d)^2}{3} \\
&= \frac{a}{1+2b^3}+\frac{b}{1+2c^3}+\frac{c}{1+2d^3}+\frac{d}{1+2a^3} - a - b - c - d + \frac{2(ab+bc+cd+da+ac+bd)}{3} \\
&\geq \frac{a}{1+2b^3}+\frac{b}{1+2c^3}+\frac{c}{1+2d^3}+\frac{d}{1+2a^3} - a - b - c - d + \frac{2(ab+bc+cd+da)}{3} \\
&= a\!\left(\frac{1}{1+2b^3}-1+\frac{2b}{3}
ight) + b\!\left(\frac{1}{1+2c^3}-1+\frac{2c}{3}
ight) + c\!\left(\frac{1}{1+2d^3}-1+\frac{2d}{3}
ight) + d\!\left(\frac{1}{1+2a^3}-1+\frac{2a}{3}
ight)\\
&= \frac{2ab(2b^3-3b^2+1)}{3(1+2b^3)} + \frac{2bc(2c^3-3c^2+1)}{3(1+2c^3)} + \frac{2cd(2d^3-3d^2+1)}{3(1+2d^3)} + \frac{2ad(2a^3-3a^2+1)}{3(1+2a^3)} \\
&= \frac{2ab(2b+1)(b-1)^2}{3(1+2b^3)} + \frac{2bc(2c+1)(c-1)^2}{3(1+2c^3)} + \frac{2cd(2d+1)(d-1)^2}{3(1+2d^3)} + \frac{2ad(2a+1)(a-1)^2}{3(1+2a^3)} \geq 0.
\end{align*}

We have equality when $ac+bd=0$ and also $a=3$, $b=c=d=0$ or $a=2$, $b=1$, $c=d=0$ as well as for their cyclic permutations.

\textbf{Solution 2.} (Ionu\c{t} Nicolae) We use the inequality $\dfrac{1}{1+2a^3}\geq 1-\dfrac{2a}{3}$ which, after computations, comes to $a(a-1)^2(2a+1)\geq 0$. The last inequality is obviously true and its equality cases are $a=0$ and $a=1$. We have
\[
\frac{d}{1+2a^3} \geq d - \frac{2ad}{3},\quad\text{with equality if }d=0\text{ or }a\in\{0,1\}.
\]
Then $\sum_{\text{cycl}}\dfrac{a}{1+2b^3} \geq \sum_{\text{cycl}}\!\left(a-\dfrac{2ab}{3}
ight) = 3 - \dfrac{2}{3}\sum_{\text{cycl}}ab$. It remains to be proven that $a^2+b^2+c^2+d^2+2(ab+bc+cd+da)\leq 9$, which follows from $a^2+b^2+c^2+d^2+2(ab+bc+cd+da+ac+bd)=9$.

We have equality if $ac+bd=0$ and each non-zero variable is followed in cyclic order by one that is either 0 or 1. From $ac+bd=0$ we see that two variables need to be 0. If another one is 0 then the last one is 3 and we have indeed equality; if the other two are positive, the second one needs to be 1, hence the first one must be 2.

\textbf{Solution 3.} (given in the contest by Andrei Pantea) From the AM-GM inequality we have $1+2x^3=1+x^3+x^3\geq 3x^2$, $\forall x>0$.

We treat the following cases: 1.\ $a,b,c,d>0$; 2.\ $a,b,c>0$, $d=0$; 3.\ $a,b>0$, $c=d=0$; 4.\ $a,c>0$, $b=d=0$; 5.\ $a=b=c=0$, $d=3$. All the other cases follow by cyclic permutations.

\textbf{Case 1.} The inequality is equivalent to
\[
\sum_{\text{cycl}}\left(a - \frac{a}{1+2b^3}
ight) \leq \frac{1}{3}\left[(a+b+c+d)^2 - \sum_{\text{cycl}}a^2
ight].
\]
But $\sum_{\text{cycl}}\dfrac{2ab^3}{1+2b^3}\leq\sum_{\text{cycl}}\dfrac{2ab^3}{3b^2}=\dfrac{2}{3}\sum_{\text{cycl}}ab < \dfrac{2}{3}\sum_{\text{sym}}ab=\dfrac{1}{3}\left[(a+b+c+d)^2-\sum_{\text{cycl}}a^2
ight]$, so in this case the inequality is satisfied without equality.

\textbf{Case 2.} We have $a+b+c=3$ and we need to prove $\dfrac{a}{1+2b^3}+\dfrac{b}{1+2c^3}+c\geq\dfrac{a^2+b^2+c^2}{3}$. But
\[
\frac{2}{3}(ab+bc+ca) = 3 - \frac{a^2+b^2+c^2}{3} \geq 3 - \frac{a}{1+2b^3} - \frac{b}{1+2c^3} - c = \frac{2ab^3}{1+2b^3}+\frac{2bc^3}{1+2c^3}.
\]
The inequality follows from $\dfrac{2ab^3}{1+2b^3}+\dfrac{2bc^3}{1+2c^3}\leq\dfrac{2ab^3}{3b^2}+\dfrac{2bc^3}{3c^2}=\dfrac{2}{3}(ab+bc)<\dfrac{2}{3}(ab+bc+ca)$. Again, there is no equality in this case.

\textbf{Case 3.} We have $a+b=3$ and we need to prove that $\dfrac{a}{1+2b^3}+b\geq\dfrac{a^2+b^2}{3}$, i.e.\ $3-\dfrac{a^2+b^2}{3}\geq 3-\dfrac{a}{1+2b^3}-b$, or $\dfrac{2}{3}ab\geq\dfrac{2ab^3}{1+2b^3}$. We have equality if $b=1$, which leads to $a=2$.

\textbf{Case 4.} We have $a+c=3$ and we need to prove that $a+c\geq\dfrac{a^2+c^2}{3}$, i.e.\ $a^2+c^2\leq 9=a^2+2ac+c^2$, which is true (no equality in this case).

\textbf{Case 5.} The inequality is fulfilled with equality.

In conclusion, the inequality is true in each of the cases and we have equality if $(a,b,c,d)\in\{(3,0,0,0),(2,1,0,0)\}$ and their cyclic permutations.",218,4123,Algebra,8
480,tst_jbmo_ro_2017_3_p1,tst_jbmo,2017,n,"Let $n$ and $k$ be two positive integers such that $1\leq n\leq k$. Prove that, if $d^k+k$ is a prime number for each positive divisor $d$ of $n$, then $n+k$ is a prime number.","\textbf{Solution.}
For $d=1$ it follows that $1+k$ is a prime.

For $d=n$ it follows that $n^k+k$ is a prime. As $k+1$ does not divide $n$ (being larger than $n$), from Fermat's Theorem we get $n^k\equiv 1\pmod{k+1}$, hence $n^k+k\equiv 0\pmod{k+1}$. It follows that $n^k+k=k+1$, hence $n=1$. We have seen at the beginning that $1+k$ is a prime.",176,345,Number Theory,9
481,tst_jbmo_ro_2017_3_p2,tst_jbmo,2017,a,"Given $x_1, x_2, \ldots, x_n$ real numbers, prove that there exists a real number $y$ such that
\[
\{y-x_1\} + \{y-x_2\} + \cdots + \{y-x_n\} \leq \frac{n-1}{2}.
\]","\textbf{Solution.}
As $\{a\}+\{-a\}\leq 1$, $\forall a\in\mathbb{R}$ (with equality if $a$ is not an integer), we have
\[
\sum_{1\leq i
eq j\leq n}\{x_i-x_j\} \leq \frac{n(n-1)}{2},
\]
which means that at least one of the sums $S_i=\displaystyle\sum_{j=1}^n\{x_i-x_j\}$ is $\leq\dfrac{n-1}{2}$ and we can choose $y=x_i$.",164,320,Algebra,10
482,tst_jbmo_ro_2017_3_p3,tst_jbmo,2017,g,"Let $I$ be the incenter of the scalene triangle $ABC$, with $AB<AC$, and let $I'$ be the reflection of point $I$ in the line $BC$. The angle bisector $AI$ meets the side $BC$ at $D$ and the circumcircle of $ABC$ at $E$. The line $EI'$ meets the circumcircle at $F$. Prove that $\dfrac{AI}{IE}=\dfrac{ID}{DE}$ and $IA=IF$.","\textbf{Solution 1.}
We have $\dfrac{AI}{ID}=\dfrac{AB}{BD}=\dfrac{EC}{DE}=\dfrac{BE}{DE}=\dfrac{IE}{DE}$, which proves the first relation.

Let $\{G\}=EF\cap BC$. We know that $GD$ is the bisector of the angle $\angle EGI$, hence $\dfrac{GI}{GE}=\dfrac{ID}{DE}=\dfrac{AI}{IE}$. \hfill(1)

Triangles $FBE$ and $BGE$ are similar (AA), therefore $EB^2=EF\cdot EG=EI^2$. This means that triangles $FIE$ and $IGE$ are also similar (SAS), which leads to $\dfrac{FI}{IE}=\dfrac{GI}{GE}$. \hfill(2)

From (1) and (2) we obtain $FI=AI$, and the conclusion.

\textbf{Solution 2.} (for the second task, given in the contest by Paul B\'ecsi)

Let $O$ be the circumcenter of triangle $ABC$. We show that triangles $AIO$ and $II'E$ are similar, which proves that the points $I$, $O$, $F$, $E$ are co-cyclic.

$OE$ and $II'$ are both perpendicular to $BC$, hence they are parallel. It follows that $\angle I'IE\equiv\angle OEI\equiv\angle OAI$.

We prove that $\dfrac{AI}{II'}=\dfrac{AO}{IE}$. From the angle bisector theorem we obtain
\[
\frac{AI}{ID}=\frac{AB}{BD}=\frac{AC}{CD}=\frac{AB+AC}{BC}=\frac{b+c}{a},
\]
hence $\dfrac{AI}{AD}=\dfrac{b+c}{a+b+c}$. As $AD=\ell_a=\dfrac{2bc\cos\frac{A}{2}}{b+c}$, we obtain $AI=\dfrac{2bc\cos\frac{A}{2}}{a+b+c}$.

But $IE=EB$ and $\dfrac{EB}{BC}=\dfrac{\sin\frac{A}{2}}{\sin A}=\dfrac{1}{2\cos\frac{A}{2}}$. Moreover, $EI=EB=\dfrac{a}{2\cos\frac{A}{2}}$. It follows that
\[
AI\cdot IE = \frac{abc}{a+b+c}.
\]
On the other hand, $AO\cdot II'=R\cdot 2r=\dfrac{2RS}{p}=\dfrac{abc}{a+b+c}$, hence $AO\cdot II'=AI\cdot IE$, which, together with $\angle I'IE\equiv\angle OAI$, proves the similarity of the triangles $AIO$ and $II'E$.

We deduce that $\angle EIO\equiv\angle II'F\equiv\angle OEF\equiv\angle OFE$, hence $IOEF$ is cyclic (the order of the points and the arguments depend slightly on the configuration).

It follows that $\angle IOF\equiv\angle IEI'\equiv\angle AOI$. We also have $\angle IAO\equiv\angle IEO\equiv\angle IFO$, which means that triangles $IAO$ and $IFO$ are congruent (SAA), therefore $IF=IA$.",321,2047,Geometry,11
483,tst_jbmo_ro_2017_3_p4,tst_jbmo,2017,c,"The sides of an equilateral triangle are divided into $n$ equal parts by $n-1$ points on each side. Through these points one draws parallel lines to the sides of the triangle. Thus, the initial triangle is divided into $n^2$ equal equilateral triangles. In every vertex of such a triangle there is a beetle. The beetles start crawling simultaneously, with equal speed, along the sides of the small triangles. When they reach a vertex, the beetles change the direction of their movement by $60^\circ$ or by $120^\circ$.

\begin{itemize}
  \item[a)] Prove that, if $n\geq 7$, the beetles can move indefinitely on the sides of the small triangles without two beetles ever meeting in a vertex of a small triangle.
  \item[b)] Determine all the values of $n\geq 1$ for which the beetles can move along the sides of the small triangles without meeting in their vertices.
\end{itemize}","\textbf{Solution.}
It is easy to see (by induction or otherwise) that, for $n\geq 3$ odd, the set of the vertices of the small triangle can be partitioned into groups of 3 and 4 vertices that form either an equilateral triangle or a rhombus formed by gluing together two such triangles. On each of these polygonal lines, the beetles can move in a circuit, changing at each step their direction.

Alternatively, one can provide a direct example in this case: we call a \emph{strip} an isosceles trapezoid whose legs are part of the original triangle's sides and whose bases are horizontal and we label the strips from top to bottom, strip number $k$ consisting of $2k-1$ small equilateral triangles. Then we can choose the triangles and the rhombi as follows: on the strip $k$ with $k$ odd we place a triangle, then $k-1$ rhombi.

In general, if for a given $n$ the beetles can move without meeting, then they can move indefinitely for $n+2$ as well. For $n=1$ and $n=8$ one can give examples.

The inductive step for the case when $n$ is even: if the triangle whose sides have been divided into $n\geq 2$ equal parts can be travelled by the beetles without intersecting, then so can the triangle whose sides have been divided into $n+2$ equal parts: it is enough to split the triangle with side length $n+2$ into an equilateral triangle of side length $n$ plus two strips.

Why can the beetles not move indefinitely without meeting in the case $n=6$? We color red all the vertices situated on odd position in rows with an odd number, the numbering starting from top to bottom. We color the remaining points blue. A beetle cannot get in one step from a red vertex to another red one, nor can it achieve this in two steps because it needs to change direction after the first step. After two steps, those 10 beetles initially positioned in a red vertex need to be in blue vertices, but so do the 10 beetles that were in red vertices after the first step. But there are only 18 vertices to accommodate these 20 beetles, so some beetles need to meet after two steps.

If for $n=2$ the beetles would be able to move without meeting, from the inductive step it would follow that the same would be true for $n=4$ and then also for $n=6$, which is not happening.

In conclusion, the beetles can move indefinitely without ever meeting if and only if $n\in\mathbb{N}\setminus\{2,4,6\}$.",878,2375,Combinatorics,12
484,tst_jbmo_ro_2017_4_p1,tst_jbmo,2017,c,"Alina and Bogdan play a game on a $2\times n$ rectangular grid ($n\geq 2$) whose sides of length 2 are glued together to form a cylinder. Alternating moves, each player cuts out a unit square of the grid. A player loses if his/her move causes the grid to lose circular connection (two unit squares that only touch at a corner are considered to be disconnected). Suppose Alina makes the first move. Which player has a winning strategy?","\textbf{Solution.}
\textbf{Answer:} Alina if $n$ is odd, Bogdan if $n$ is even.

If $n=2j+1$ is odd, Alina's strategy is the following: she cuts out a unit square and labels the columns from $-j$ to $j$, the column from which the first unit square has been removed receiving the label 0. Starting at this point of the game, whenever Bogdan removes a square from column number $k\in\{-j,\ldots,-2,-1,1,2,\ldots,j\}$, Alina removes the unit square positioned in column number $-k$ and on the same row as the unit square removed by Bogdan in his last move. (If Bogdan removes the square remaining in column 0 he loses instantly.) If on Bogdan's move the cylinder didn't lose its circular connection, it will not lose it after Alina's move either. Hence, Alina will never destroy the cylinder and, as the game is bound to finish sooner or later, Bogdan is the one who will lose the game. Alina wins.

If $n=2j$ is even, Bogdan wins by adopting the following strategy: he labels the columns from $-j+1$ to $j$, column 0 being the one from which Alina has removed a square in her initial move. Bogdan removes a square from column $j$ (the one lying opposite to column 0). From now on, if Alina removes a square from column number $k\in\{-j+1,\ldots,-2,-1,1,2,\ldots,j-1\}$, Bogdan removes the square situated in the same row, but in the opposite column, namely $-k$. (If Alina removes the remaining square from column 0 or from column $j$, she loses instantly.) If Alina's move didn't dismantle the surface of the cylinder, Bogdan's move won't do it either. Eventually, Alina will dismantle the cylindric surface and Bogdan will win.",434,1627,Combinatorics,13
485,tst_jbmo_ro_2017_4_p2,tst_jbmo,2017,n,"Let $n$ be a positive integer. For each of the numbers $1,2,\ldots,n$ we compute the difference between the number of its odd positive divisors and its even positive divisors. Prove that the sum of these differences is at least 0 and at most $n$.","\textbf{Solution.}
We count how many times a number $d$ contributes to the sum of the differences. It appears $\left\lfloor\dfrac{n}{d}
ight
floor$ times; if $d$ is odd, then this term is to be added, while if $d$ is even this term is to be subtracted. Thus, the desired sum is $\displaystyle\sum_{k=1}^n (-1)^{k+1}\left\lfloor\frac{n}{k}
ight
floor$.

Adding, if necessary, the term $\left\lfloor\dfrac{n}{n+1}
ight
floor=0$ at the end, we can group the terms of the sum into pairs having the sum $\left\lfloor\dfrac{n}{2k-1}
ight
floor-\left\lfloor\dfrac{n}{2k}
ight
floor\geq 0$, hence the sum is non-negative. Similarly, separating the first term (which is equal to $n$), adding, if necessary, a 0 term to the end, we can again group the terms into pairs having the sum $-\left\lfloor\dfrac{n}{2k}
ight
floor+\left\lfloor\dfrac{n}{2k+1}
ight
floor\leq 0$, hence the sum is at most $n$.",246,889,Number Theory,14
486,tst_jbmo_ro_2017_4_p3,tst_jbmo,2017,a,"Prove that if $a,b,c,d\in[1,2]$, then
\[
\frac{a+b}{b+c} + \frac{c+d}{d+a} \leq 4\cdot\frac{a+c}{b+d}.
\]
When does the equality hold?","\textbf{Solution 1.}
Swapping, if necessary, the roles of $a$ and $c$ and those of $b$ and $d$, we may assume that $a\leq c$. In this case, the function $f(x)=\dfrac{a+x}{c+x}$ is increasing on $[1,2]$, while $g(x)=\dfrac{c+x}{a+x}$ is decreasing on $[1,2]$, hence
\[
\frac{a+b}{b+c} + \frac{c+d}{d+a} \leq \frac{a+2}{c+2}+\frac{c+1}{a+1}.
\]
On the other hand, $4\cdot\dfrac{a+c}{b+d}\geq a+c$, therefore it is sufficient to prove that
\[
\frac{a+2}{c+2}+\frac{c+1}{a+1} \leq a+c,
\]
which reduces to $a+c+4\leq a^2c+ac^2+a^2+3ac$. From $(a-1)(c-1)\geq 0$ and $a^2c+ac^2+a^2+2ac\geq 5$ the conclusion follows, with equality if $a=c=1$ and $b=d=2$.

\textbf{Solution 2.}
We show that $\dfrac{a+b}{b+c}\leq 2\cdot\dfrac{a+c}{b+2}\leq 2\cdot\dfrac{a+c}{b+d}$ and $\dfrac{c+d}{d+a}\leq 2\cdot\dfrac{a+c}{b+d}$, inequalities which give the conclusion. The second inequality can be obtained from the first one by swapping $a$ with $c$ and $b$ with $d$, therefore it is sufficient to prove the first one. This inequality reduces to $b^2+ab+2b+2a\leq 2c^2+2ac+2bc+2ab$. But $2a\leq 2ac$, $2b\leq 2bc$ and $b^2\leq b+2\leq ab+2c^2$, the inequality $b^2\leq b+2$ being equivalent to $(b+1)(b-2)\leq 0$. We obtain the equality case $a=1$, $b=2$, $c=1$ and $d=2$.

\textbf{Solution 3.}
As $0<b+d\leq 4$, it is sufficient to prove that the left hand side is at most $a+c$. Without loss of generality, we may assume that $a\leq c$. In this case,
\[
\frac{a+b}{b+c}+\frac{c+d}{d+a} \leq 1+1+\frac{c-a}{d+a} \leq 2+c-a \leq c+a,
\]
with equality if $a=1$, $b=2$, $c=1$ and $d=2$.",134,1564,Algebra,15
487,tst_jbmo_ro_2017_4_p4,tst_jbmo,2017,g,"Let $ABC$ be a right triangle, with the right angle at $A$. The altitude from $A$ meets $BC$ at $H$ and $M$ is the midpoint of the hypotenuse $[BC]$. On the legs, in the exterior of the triangle, equilateral triangles $BAP$ and $ACQ$ are constructed. If $N$ is the intersection point of the lines $AM$ and $PQ$, prove that the angles $\angle NHP$ and $\angle AHQ$ are equal.","\textbf{Solution 1.} 

If $AB=AC$, the statement is obvious. In the following, we assume $AB<AC$, the other case being similar.

\begin{itemize}
  \item Triangles $PAM$ and $PBM$ are congruent (SSS), hence $\angle PMA\equiv\angle PMB$. Similarly, $\angle QMA\equiv\angle QMC$, and this leads rapidly to $\angle PMQ=90^\circ$.
  \item As $\tan B=\dfrac{AH}{BH}=\dfrac{AC}{AB}=\dfrac{AQ}{BP}$ and $\angle HAQ=\angle HBP=60^\circ+\angle B$, triangles $HAQ$ and $HBP$ are similar (SAS). It follows that $\angle QHA\equiv\angle PHB$, hence $\angle QHP=\angle QHA+\angle AHP=\angle PHB+\angle AHP=\angle AHB=90^\circ$.
  \item In conclusion, the points $P,Q,M$ and $H$ are co-cyclic, hence $\angle HPQ\equiv\angle CMQ\equiv\angle AMQ$.
  \item Let $\{D\}=HQ\cap MP$. The angle $\angle MNQ$ is exterior to the triangle $PMN$, therefore $\angle MNQ=\angle MPN+\angle NMP=\angle MHQ+\angle BMP=\angle MDQ$. \hfill(1)
  \item It follows that the quadrilateral $NDMQ$ is cyclic, hence $\angle DNQ=90^\circ$. This means that $NDHP$ is also cyclic, therefore $\angle PNH\equiv\angle PDH$. \hfill(2)
  \item From (1) and (2) it follows that $\angle PNH\equiv\angle PDH\equiv\angle MDQ\equiv\angle MNQ$, i.e.\ $ND$ is the bisector of angle $\angle HNM$. For the triangle $HMN$ ray $NP$ is the external bisector, while $MP$ is an internal bisector, which means that $P$ is the excenter opposite to the vertex $M$. Then $HP$ is the internal bisector of angle $\angle NHB$.
  \item Finally, $\angle NHP\equiv\angle PHB\equiv\angle AHQ$.
\end{itemize}

\textbf{Solution 2.} 
Let $\{S\}=AH\cap PQ$. An easy computation shows that $\angle SAQ=\angle NAP=120^\circ-\angle B$, which shows that the rays $AN$ and $AS$ are isogonal in the angle $\angle PAQ$. From Steiner's Theorem it follows that
\[
\frac{PN}{NQ}\cdot\frac{PS}{SQ} = \left(\frac{PA}{AQ}
ight)^2.
\]
The conclusion means the rays $HN$ and $HS$ are isogonal in the angle $\angle PHQ$, which, according to Steiner's Theorem, is equivalent to
\[
\frac{PN}{NQ}\cdot\frac{PS}{SQ} = \left(\frac{PH}{HQ}
ight)^2.
\]
In conclusion, we need to prove that $\dfrac{PA}{AQ}=\dfrac{PH}{HQ}$.

But $\triangle BHA\sim\triangle BAC$ leads to $\dfrac{BH}{HA}=\dfrac{BA}{CA}=\dfrac{BP}{CQ}$. As $\angle PBH=60^\circ+\angle B=\angle QAH$, it follows that $\triangle PBH\sim\triangle QAH$, i.e.\ $\dfrac{PH}{QH}=\dfrac{PB}{AQ}=\dfrac{PA}{AQ}$ and the conclusion.",374,2384,Geometry,16
488,tst_jbmo_ro_2017_5_p1,tst_jbmo,2017,a,"If $a,b,c\in[-1,1]$ satisfy $a+b+c+abc=0$, prove that
\[
a^2+b^2+c^2 \geq 3(a+b+c).
\]
When does the equality hold?","\textbf{Solution 1.}
If $a+b+c\leq 0$, the inequality is obviously satisfied, with equality occurring if and only if $a=b=c=0$. If $a+b+c>0$, then $abc<0$. It is not possible for all the variables to be negative (their sum would be negative), therefore one of them is negative and the other two are positive. We may assume that $a<0<b,c$.

Put $x=-a$. Then $-x+b+c=xbc>0$ and
\[
(a^2+b^2+c^2)^2 = (x^2+b^2+c^2)^2 \geq (xb+xc+bc)^2 \geq 3(xb\cdot xc+xc\cdot bc+bc\cdot xb)
\]
\[
= 3xbc(x+b+c) = 3(-x+b+c)(x+b+c) \geq 9(-x+b+c)^2.
\]
As $-x+b+c>0$, the last inequality comes to $x+b+c\geq 3(-x+b+c)$, i.e.\ to $x\geq\dfrac{b+c}{2}$. But $x=\dfrac{b+c}{1+bc}\geq\dfrac{b+c}{2}$ because $bc\leq 1$. In conclusion, we obtain $a^2+b^2+c^2\geq 3(-x+b+c)=3(a+b+c)$, with equality (in the case $a<0<b,c$) if and only if $x=b=c$ and $bc=1$, i.e.\ if $a=-1$, $b=c=1$. To conclude, we have equality if $(a,b,c)\in\{(0,0,0),(-1,1,1),(1,-1,1),(1,1,-1)\}$.

\textbf{Solution 2.}
If $a+b+c\leq 0$, the inequality is obviously satisfied, with equality occurring if and only if $a=b=c=0$. If $a+b+c>0$, then $abc<0$.

From the AM-GM inequality, $a^2+b^2+c^2\geq 3\sqrt[3]{(abc)^2}\geq 3|abc|=-3abc=3(a+b+c)$, with equality if $a^2=b^2=c^2$, $abc\leq 0$ and $|abc|=1$, which leads to $(a,b,c)\in\{(0,0,0),(-1,1,1),(1,-1,1),(1,1,-1)\}$.

The inequality $3\sqrt[3]{(abc)^2}\geq 3|abc|$ follows from $|abc|\leq 1$.",115,1392,Algebra,17
489,tst_jbmo_ro_2017_5_p2,tst_jbmo,2017,g,"Let $A$ be a point outside the circle $\mathscr{C}$. The tangents from $A$ touch the circle at $B$ and $C$. Let $P$ be an arbitrary point on $AC$ produced, $Q$ the projection of $C$ onto $PB$ and $E$ the second intersection point of the circumcircle of $ABP$ with the circle $\mathscr{C}$. Prove that $\angle PEQ=2\angle APB$.","\textbf{Solution.}
Let $\{D\}=BP\cap\mathscr{C}$ and $\{M\}=DE\cap AC$.

Then $\angle BDE=\angle BCE=180^\circ-\angle ACB-\angle PCE=180^\circ-\angle ABC-\angle CBE=\angle APE$, hence $\triangle MPD\sim\triangle MEP$ (AA), which leads to $\angle MPD\equiv\angle MEP$ and $MP^2=MD\cdot ME$. On the other hand, from the power of the point $M$ with respect to the circle $\mathscr{C}$ it follows that $MC^2=MD\cdot ME$, i.e.\ $MP^2=MC^2$, which means that $M$ is the midpoint of $[PC]$. It follows that $\angle MEP\equiv\angle MPD\equiv\angle MQP$, i.e.\ the quadrilateral $MQEP$ is cyclic. We obtain that $\angle MEQ\equiv\angle MPQ\equiv\angle MEP$, and the conclusion.",326,668,Geometry,18
490,tst_jbmo_ro_2017_5_p3,tst_jbmo,2017,n,"Let $n\geq 2$ be a positive integer. Prove that the following assertions are equivalent:
\begin{itemize}
  \item[a)] for all integer $x$ coprime with $n$ the congruence $x^6\equiv 1\pmod{n}$ holds;
  \item[b)] $n$ divides 504.
\end{itemize}","\textbf{Solution.}
We have $504=2^3\cdot 3^2\cdot 7$.

\textbf{a) $\Rightarrow$ b).} First we prove that $\gcd(n,5)=1$. If $\gcd(n,5)
eq 1$ then $n=5k$.

The sequence $(5m+2)_{m\geq 1}$ is an arithmetic sequence with ratio 5. Its first term being coprime with the ratio, from Dirichlet's Theorem it follows that the sequence contains infinitely many primes. Then, there exists a prime $p$ such that $p=5m+2$ and $p>n$. As $\gcd(p,n)=1$ it follows that $p^6\equiv 1\pmod{n}$, i.e.\ $n\mid(5m+2)^6-1$. As $5\mid n$ and $(5m+2)^6=M_5+2^6$, we get that 5 divides 63, absurd!

As $\gcd(n,5)=1$, according to the hypothesis $5^6\equiv 1\pmod{n}$, i.e.\ $n\mid(5^6-1)$.

But $5^6-1=(5^3-1)(5^3+1)=(5-1)(5^2+5+1)(5+1)(5^2-5+1)=4\cdot 31\cdot 6\cdot 21=2^3\cdot 3^2\cdot 7\cdot 31$, hence $n\mid 2^3\cdot 3^2\cdot 7\cdot 31$. \hfill(1)

From (1) one can notice that $\gcd(n,11)=1$, hence $11^6\equiv 1\pmod{n}$, i.e.\ $n\mid(11^6-1)$.

But $11^6-1=(11^3-1)(11^3+1)=(11-1)(11^2+11+1)(11+1)(11^2-11+1)=10\cdot 133\cdot 12\cdot 111$, therefore
\[
n\mid 2^3\cdot 3^2\cdot 5\cdot 7\cdot 19\cdot 37. \hfill(2)
\]
From (1) and (2) it follows that $n\mid 2^3\cdot 3^2\cdot 7$, i.e.\ $n\mid 504$.

\textbf{b) $\Rightarrow$ a).} Let $x\in\mathbb{Z}$. Then $x^6-1=(x^2-1)(x^4+x^2+1)$. We have:

\begin{enumerate}
  \item if $\gcd(x,2)=1$ then $x-1$ and $x+1$ are consecutive even numbers, hence $2^3\mid x^6-1$;
  \item if $\gcd(x,7)=1$, from Fermat's Theorem we obtain $7\mid x^6-1$;
  \item if $\gcd(x,3)=1$ then $x=3k\pm 1$. Obviously $3\mid x-1$ or $3\mid x+1$, hence $3\mid x^2-1$. We also have $x^4+x^2+1=(3k\pm 1)^4+(3k\pm 1)^2+1\equiv (M_3+1)+(M_3+1)+1=M_3$, i.e.\ $3\mid x^4+x^2+1$. It follows that for $\gcd(x,3)=1$ we have $3^2\mid x^6-1$.
\end{enumerate}

If $n\in\mathbb{N}$, $n\geq 2$ and $n\mid 504$, then $n=2^a\cdot 3^b\cdot 7^c$ with $a\in\{0,1,2,3\}$, $b\in\{0,1,2\}$, $c\in\{0,1\}$ and $a,b,c$ not all of them 0. From the three statements above it follows that for all integer $x$ coprime with $n$ we have $n\mid x^6-1$, i.e.\ $x^6\equiv 1\pmod{n}$.

\textbf{Remark:} Case 3 above can be proven more directly as follows: if $\gcd(x,3)=1$ then $\gcd(x,9)=1$ and, from Euler's Theorem, we obtain $9\mid x^{\varphi(9)}-1$, i.e.\ $3^2\mid x^6-1$.",240,2242,Number Theory,19
491,tst_jbmo_ro_2017_5_p4,tst_jbmo,2017,c,"Consider an $m\times n$ board where $m,n\geq 3$ are positive integers, divided into unit squares. Initially all the squares are white. What is the minimum number of squares that need to be painted red such that each $3\times 3$ square contains at least two red squares?","\textbf{Solution.}
We label the rows from 1 to $m$ and the columns from 1 to $n$.

If $m$ and $n$ are not congruent to 2 modulo 3, i.e.\ $m=3a+r_1$ and $n=3b+r_2$ with $r_1,r_2\in\{0,1\}$, we can tile the rectangle formed by the first $3a$ rows and $3b$ columns with $a\cdot b$ disjoint $3\times 3$ squares. Each of these squares needs to contain at least 2 red squares, therefore one needs to paint red at least $2ab=2\left\lfloor\dfrac{m}{3}
ight
floor\cdot\left\lfloor\dfrac{n}{3}
ight
floor$ squares. On the other hand, this number is sufficient: paint red those unit squares whose row number is a multiple of 3 (i.e.\ $3,6,\ldots,3a$) and whose column number belongs to the set $\{2,3,5,6,\ldots,3b-1,3b\}$. It is easy to check that each $3\times 3$ square contains exactly two red squares.

The same answer (with the exact same arguments) remains valid if $m\equiv 2\pmod{3}$ and $n\equiv 0,1\pmod{3}$ (the answer remains the same in the case $n\equiv 2\pmod{3}$ and $m\equiv 0,1\pmod{3}$).

Let us now tackle the case $m=3a+2$, $n=3b+2$. We prove that in this case the answer is $2ab+\min\{a,b\}$.

We begin with the case when $m=n=3a+2$. In this case the answer is $2a^2+a$. On one hand, the above coloring uses exactly $2a^2+a$ red squares, thus this number is sufficient. We prove by induction after $a$ that this number of red squares, $2a^2+a$, is also necessary.

Let us call a ``zone'' a $5\times 5$ square from which we have removed a $2\times 2$ square situated in one of its corners. We start by noticing that in any zone we need to have at least three red squares. Indeed, examining the two $3\times 3$ squares that contain the two remaining opposite corners of the $5\times 5$ square, they contain each at least two red squares and they only share one unit square (the one in the center), hence they contain together at least 3 red squares. With this remark, we move on to the promised induction:

For $a=1$: we can paint red the squares situated in column number 3, on the rows 1, 3 and 4. Any $3\times 3$ square contains exactly two of these red squares.

Assuming the assertion true for a $(3a-1)\times(3a-1)$ board, let us prove it for the $(3a+2)\times(3a+2)$ board. From the inductive hypothesis, in the $(3a-1)\times(3a-1)$ square situated in the top-left corner we must have at least $2(a-1)^2+a-1$ red squares. We cover the remaining part of the $(3a+2)\times(3a+2)$ square with $3\times 3$ squares, plus one zone placed in the bottom-right corner. In this remaining part we have at least $2(a-1)+2(a-1)+3$ red squares, hence we have in total at least $2(a-1)^2+a-1+4(a-1)+3=2a^2+a$ red squares.

For $m<n$ (the case when $m>n$ is similar), it is sufficient to notice that the rectangle $(3a+2)\times(3b+2)$ can be obtained from the $(3a+2)\times(3a+2)$ square by gluing a $(3a+2)\times(3b-3a)$ rectangle next to it. In the square there are at least $2a^2+a$ red squares, while in the rectangle one can fit $a\cdot(b-a)$ disjoint $3\times 3$ squares, therefore it contains at least $2a(b-a)$ red squares. In total, the rectangle has at least $2a^2+a+2a(b-a)=2ab+a$ red squares. We have already seen that this number can actually be achieved, therefore the statement is proven.",269,3204,Combinatorics,20
247,jbmo_2018_p1,jbmo,2018,n,"Find all the pairs $(m, n)$ of integers which satisfy the equation
\[
m^5 - n^5 = 16mn.
\]","\textbf{Solution.} If one of $m$, $n$ is 0, the other has to be 0 too, and $(m, n) = (0, 0)$ is one solution. If $mn 
eq 0$, let $d = \gcd(m, n)$ and we write $m = da$, $n = db$, $a, b \in \mathbb{Z}$ with $(a, b) = 1$. Then, the given equation is transformed into
\begin{equation}
    d^3 a^5 - d^3 b^5 = 16ab \tag{1}
\end{equation}

So, by the above equation, we conclude that $a \mid d^3 b^5$ and thus $a \mid d^3$. Similarly $b \mid d^3$. Since $(a, b) = 1$, we get that $ab \mid d^3$, so we can write $d^3 = abr$ with $r \in \mathbb{Z}$. Then, equation (1) becomes
\[
abra^5 - abrb^5 = 16ab \Rightarrow
\]
\[
r(a^5 - b^5) = 16.
\]

Therefore, the difference $a^5 - b^5$ must divide 16. Therefore, the difference $a^5 - b^5$ must divide 16. This means that
\[
a^5 - b^5 = \pm 1, \pm 2, \pm 4, \pm 8, \pm 16.
\]

The smaller values of $|a^5 - b^5|$ are 1 or 2. Indeed, if $|a^5 - b^5| = 1$ then $a = \pm 1$ and $b = 0$ or $a = 0$ and $b = \pm 1$, a contradiction. If $|a^5 - b^5| = 2$, then $a = 1$ and $b = -1$ or $a = -1$ and $b = 1$. Then $r = -8$, and $d^3 = -8$ or $d = -2$. Therefore, $(m, n) = (-2, 2)$. If $|a^5 - b^5| > 2$ then, without loss of generality, let $a > b$ and $a \geq 2$. Putting $a = x + 1$ with $x \geq 1$, we have

\begin{align*}
|a^5 - b^5| &= |(x+1)^5 - b^5| \\
&\geq |(x+1)^5 - x^5| \\
&= |5x^4 + 10x^3 + 10x^2 + 5x + 1| \geq 31
\end{align*}

which is impossible. Thus, the only solutions are $(m, n) = (0, 0)$ or $(-2, 2)$.",90,1455,Number Theory,1
248,jbmo_2018_p2,jbmo,2018,c,"Let $n$ three-digit numbers satisfy the following properties:
\begin{enumerate}
    \item No number contains the digit 0.
    \item The sum of the digits of each number is 9.
    \item The units digits of any two numbers are different.
    \item The tens digits of any two numbers are different.
    \item The hundreds digits of any two numbers are different.
\end{enumerate}
Find the largest possible value of $n$.","\textbf{Solution.} Let $S$ denote the set of three-digit numbers that have digit sum equal to 9 and no digit equal to 0. We will first find the cardinality of $S$. We start from the number 111 and each element of $S$ can be obtained from 111 by a string of 6 A's (which means that we add 1 to the current digit) and 2 G's (which means go to the next digit). Then for example 324 can be obtained from 111 by the string AAGAGAAA. There are in total
\[
\frac{8!}{6! \cdot 2!} = 28
\]
such words, so $S$ contains 28 numbers. Now, from the conditions (3),(4),(5), if $abc$ is in $T$ then each of the other numbers of the form $**c$ cannot be in $T$, neither $*b*$ can be, nor $a**$. Since there are $a + b - 2$ numbers of the first category, $a + c - 2$ from the second and $b + c - 2$ from the third one. In these three categories there are
\[
(a + b - 2) + (b + c - 2) + (c + a - 2) = 2(a + b + c) - 6 = 2 \cdot 9 - 6 = 12
\]
distinct numbers that cannot be in $T$ if $abc$ is in $T$. So, if $T$ has $n$ numbers, then $12n$ are the forbidden ones that are in $S$, but each number from $S$ can be a forbidden number no more than three times, once for each of its digits, so
\[
n + \frac{12n}{3} \leq 28 \iff n \leq \frac{28}{5},
\]
and since $n$ is an integer, we get $n \leq 5$. A possible example for $n = 5$ is
\[
T = \{144, 252, 315, 423, 531\}.
\]

\textbf{Comment by PSC.} It is classical to compute the cardinality of $S$ and this can be done in many ways. In general, the number of solutions of the equation
\[
x_1 + x_2 + \cdots + x_k = n
\]
in positive integers, where the order of $x_i$ matters, is well known that equals to $\binom{n-1}{k-1}$. In our case, we want to count the number of positive solutions to $a + b + c = 9$. By the above, this equals to $\binom{9-1}{3-1} = 28$. Using the general result above, we can also find that there are $a + b - 2$ numbers of the form $**c$.",415,1891,Combinatorics,2
249,jbmo_2018_p3,jbmo,2018,a,"Let $k > 1$ be a positive integer and $n > 2018$ be an odd positive integer. The nonzero rational numbers $x_1, x_2, \ldots, x_n$ are not all equal and satisfy
\[
x_1 + \frac{k}{x_2} = x_2 + \frac{k}{x_3} = x_3 + \frac{k}{x_4} = \cdots = x_{n-1} + \frac{k}{x_n} = x_n + \frac{k}{x_1}\cdot
\]
Find:
\begin{itemize}
    \item[a)] the product $x_1 x_2 \cdots x_n$ as a function of $k$ and $n$
    \item[b)] the least value of $k$, such that there exist $n, x_1, x_2, \ldots, x_n$ satisfying the given conditions.
\end{itemize}","\textbf{a)} If $x_i = x_{i+1}$ for some $i$ (assuming $x_{n+1} = x_1$), then by the given identity all $x_i$ will be equal, a contradiction. Thus $x_1 
eq x_2$ and
\[
x_1 - x_2 = k\,\frac{x_2 - x_3}{x_2 x_3}.
\]
Analogously
\[
x_1 - x_2 = k\,\frac{x_2 - x_3}{x_2 x_3} = k^2\,\frac{x_3 - x_4}{(x_2 x_3)(x_3 x_4)} = \cdots = k^n\,\frac{x_1 - x_2}{(x_2 x_3)(x_3 x_4)\cdots(x_1 x_2)}\cdot
\]
Since $x_1 
eq x_2$ we get
\[
x_1 x_2 \cdots x_n = \pm\sqrt{k^n} = \pm k^{\frac{n-1}{2}}\sqrt{k}.
\]
If one among these two values, positive or negative, is obtained, then the other one will be also obtained by changing the sign of all $x_i$ since $n$ is odd.

\textbf{b)} From the above result, as $n$ is odd, we conclude that $k$ is a perfect square, so $k \geq 4$. For $k = 4$ let $n = 2019$ and $x_{3j} = 4$, $x_{3j-1} = 1$, $x_{3j-2} = -2$ for $j = 1, 2, \ldots, 673$. So the required least value is $k = 4$.

\textbf{Comment by PSC.} There are many ways to construct the example when $k = 4$ and $n = 2019$. Since $3 \mid 2019$, the idea is to find three numbers $x_1, x_2, x_3$ satisfying the given equations, not all equal, and repeat them as values for the rest of the $x_i$'s. So, we want to find $x_1, x_2, x_3$ such that
\[
x_1 + \frac{4}{x_2} = x_2 + \frac{4}{x_3} = x_3 + \frac{4}{x_1}\cdot
\]
As above, $x_1 x_2 x_3 = \pm 8$. Suppose without loss of generality that $x_1 x_2 x_3 = -8$. Then, solving the above system we see that if $x_1 
eq 2$, then
\[
x_2 = -\frac{4}{x_1 - 2} \quad \text{and} \quad x_3 = 2 - \frac{4}{x_1},
\]
leading to infinitely many solutions. The example in the official solution is obtained by choosing $x_1 = -2$.",523,1642,Algebra,3
250,jbmo_2018_p4,jbmo,2018,g,"Let $ABC$ be an acute triangle, $A'$, $B'$ and $C'$ be the reflections of the vertices $A$, $B$ and $C$ with respect to $BC$, $CA$, and $AB$, respectively, and let the circumcircles of triangles $ABB'$ and $ACC'$ meet again at $A_1$. Points $B_1$ and $C_1$ are defined similarly. Prove that the lines $AA_1$, $BB_1$ and $CC_1$ have a common point.","\textbf{Solution.} Let $O_1$, $O_2$ and $O$ be the circumcenters of triangles $ABB'$, $ACC'$ and $ABC$ respectively. As $AB$ is the perpendicular bisector of the line segment $CC'$, $O_2$ is the intersection of the perpendicular bisector of $AC$ with $AB$. Similarly, $O_1$ is the intersection of the perpendicular bisector of $AB$ with $AC$. It follows that $O$ is the orthocenter of triangle $AO_1O_2$. This means that $AO$ is perpendicular to $O_1O_2$. On the other hand, the segment $AA_1$ is the common chord of the two circles, thus it is perpendicular to $O_1O_2$. As a result, $AA_1$ passes through $O$. Similarly, $BB_1$ and $CC_1$ pass through $O$, so the three lines are concurrent at $O$.

\bigskip

\textbf{Comment by PSC.} We present here a different approach.

We first prove that $A_1$, $B$ and $C'$ are collinear. Indeed, since $\angle BAB' = \angle CAC' = 2\angle BAC$, then from the circles $(ABB')$, $(ACC')$ we get
\[
\angle AA_1B = \frac{\angle BA_1B'}{2} = \frac{180^\circ - \angle BAB'}{2} = 90^\circ - \angle BAC = \angle AA_1C'.
\]
It follows that
\begin{equation}
\angle A_1AC = \angle A_1C'C = \angle BC'C = 90^\circ - \angle ABC \tag{1}
\end{equation}
On the other hand, if $O$ is the circumcenter of $ABC$, then
\begin{equation}
\angle OAC = 90^\circ - \angle ABC. \tag{2}
\end{equation}
From (1) and (2) we conclude that $A_1$, $A$ and $O$ are collinear. Similarly, $BB_1$ and $CC_1$ pass through $O$, so the three lines are concurrent in $O$.",347,1474,Geometry,4
323,shl_jbmo_2018_a1,shl_jbmo,2018,a,"Let $x,y,z$ be positive real numbers . Prove:
$\frac{x}{\sqrt{\sqrt[4]{y}+\sqrt[4]{z}}}+\frac{y}{\sqrt{\sqrt[4]{z}+\sqrt[4]{x}}}+\frac{z}{\sqrt{\sqrt[4]{x}+\sqrt[4]{y}}}\geq \frac{\sqrt[4]{(\sqrt{x}+\sqrt{y}+\sqrt{z})^7}}{\sqrt{2\sqrt{27}}}$","Let $x=u^4,y=v^4,z=w^4$, then by Titu's Lemma,\[\sum_{cyc}\frac{u^4}{\sqrt{v+w}}\geq \frac{(u^2+v^2+w^2)^2}{\sum_{cyc}\sqrt{v+w}}.\]Now it is suffices to prove that\[\frac{(u^2+v^2+w^2)^2}{\sum_{cyc}\sqrt{v+w}}\geq \frac{\sqrt[4]{(u^2+v^2+w^2)^7}}{\sqrt{2\sqrt{27}}}.\]Taking both sides to the power of $4$, we have to prove\[108(u^2+v^2+w^2)\geq \left(\sum_{cyc} \sqrt{v+w}
ight)^4\]which by Cauchy we have\[\left(\sum_{cyc}u+v
ight)\left(\sum_{cyc}1
ight)\geq \left(\sum_{cyc}\sqrt{u+v}
ight)^2\]so we are left to prove\[108(u^2+v^2+w^2)\geq 36(u+v+w)^2\]which is just\[u^2+v^2+w^2\geq uv+vw+wu\]by AM-GM. Equality holds at $x=y=z=9$.",241,636,Algebra,1
324,shl_jbmo_2018_a2,shl_jbmo,2018,a,"Find the maximum positive integer $k$ such that for any positive integers $m,n$ such that $m^3+n^3>(m+n)^2$, we have

$$m^3+n^3\geq (m+n)^2+k$$","Let $x=u^4,y=v^4,z=w^4$, then by Titu's Lemma,\[\sum_{cyc}\frac{u^4}{\sqrt{v+w}}\geq \frac{(u^2+v^2+w^2)^2}{\sum_{cyc}\sqrt{v+w}}.\]Now it is suffices to prove that\[\frac{(u^2+v^2+w^2)^2}{\sum_{cyc}\sqrt{v+w}}\geq \frac{\sqrt[4]{(u^2+v^2+w^2)^7}}{\sqrt{2\sqrt{27}}}.\]Taking both sides to the power of $4$, we have to prove\[108(u^2+v^2+w^2)\geq \left(\sum_{cyc} \sqrt{v+w}
ight)^4\]which by Cauchy we have\[\left(\sum_{cyc}u+v
ight)\left(\sum_{cyc}1
ight)\geq \left(\sum_{cyc}\sqrt{u+v}
ight)^2\]so we are left to prove\[108(u^2+v^2+w^2)\geq 36(u+v+w)^2\]which is just\[u^2+v^2+w^2\geq uv+vw+wu\]by AM-GM. Equality holds at $x=y=z=9$.",143,636,Algebra,2
325,shl_jbmo_2018_a3,shl_jbmo,2018,a,"Let $a,b,c$ be positive real numbers . Prove that$$ \frac{1}{ab(b+1)(c+1)}+\frac{1}{bc(c+1)(a+1)}+\frac{1}{ca(a+1)(b+1)}\geq\frac{3}{(1+abc)^2}.$$","Let $abc=k^3$ and set $a=k\dfrac{x}{y}$, $b=k\dfrac{y}{z}$, $c=k\dfrac{z}{x}$, where $k,x,y,z>0$. Then, the inequality can be rewritten as
$$\frac{z^2}{(ky+z)(kz+x)}+\frac{x^2}{(kz+x)(kx+y)}+\frac{y^2}{(kx+y)(ky+z)}\geq\frac{3k^2}{(1+k^3)^2}.$$Using the Cauchy-Schwarz inequality we have that
$$\sum_{cyclic}\frac{z^2}{(ky+z)(kz+x)}\geq
\frac{(x+y+z)^2}{(ky+z)(kz+x)+(kz+x)(kx+y)+(kx+y)(ky+z)},$$therefore it suffices to prove that
$$\frac{(x+y+z)^2}{(ky+z)(kz+x)+(kz+x)(kx+y)+(kx+y)(ky+z)}\geq \frac{3k^2}{(1+k^3)^2}$$or
$$\left((1+k^3)^2-3k^3
ight)(x^2+y^2+z^2)\geq \left(3k^2(k^2+k+1)-2(1+k^3)^2
ight)(xy+yz+zx).$$Since $x^2+y^2+z^2\geq xy+yz+zx$ and $(1+k^3)^2-3k^3>0$, it is enough to prove that
$$(1+k^3)^2-3k^3\geq 3k^2(k^2+k+1)-2(1+k^3)^2,$$or
$$(k-1)^2(k^2+1)(k+1)^2\geq 0,$$which is true.",146,798,Algebra,3
326,shl_jbmo_2018_a4,shl_jbmo,2018,a,"Let $k > 1, n > 2018$ be positive integers, and let $n$ be odd. The nonzero rational numbers $x_1,x_2,\ldots,x_n$ are not all equal and satisfy$$x_1+\frac{k}{x_2}=x_2+\frac{k}{x_3}=x_3+\frac{k}{x_4}=\ldots=x_{n-1}+\frac{k}{x_n}=x_n+\frac{k}{x_1}$$Find:
a) the product $x_1 x_2 \ldots x_n$ as a function of $k$ and $n$
b) the least value of $k$, such that there exist $n,x_1,x_2,\ldots,x_n$ satisfying the given conditions.","a)\[ x_1 - x_2 = k \left( \frac{1}{x_3} - \frac{1}{x_2} 
ight) \Rightarrow \frac{x_1 - x_2}{x_2 - x_3} = \frac{k}{x_2 x_3} \]Multiply cyclically, we get $x_1 x_2 \dots x_n = k^{\frac{n}{2}}$.
b) $k > 1$. Since $n$ is odd, $k$ must be a square. Thus, $k \ge 4$. This works. Take $n = 2019$, $(2,-1,4, 2, -1, 4, \dots)$.",422,318,Algebra,4
327,shl_jbmo_2018_a5,shl_jbmo,2018,a,"Let a$,b,c,d$ and $x,y,z,t$ be real numbers such that
$0\le a,b,c,d \le 1$ , $x,y,z,t \ge 1$ and $a+b+c+d +x+y+z+t=8$.
Prove that $a^2+b^2+c^2+d^2+x^2+y^2+z^2+t^2\le 28$","Let
$$L(a,b,c,d,x,y,z,t) = a^2+b^2+c^2+d^2+x^2+y^2+z^2+t^2 $$First of all, we see that:
$$L(a,b,c,d,x,y,z,t) \leq L(0,0,0,0,x+a,y+b,z+c,t+d)$$Therefore, it is enough to check the case $a=b=c=d=0$. In this case, we have $x,y,z,t \geq 1$ with $x+y+z+t=8$ and we need to maximise:
$$L(x,y,z,t) = x^2+y^2+z^2+t^2$$WLOG, let's assume $x \geq y \geq z \geq t$. We have:
$$L(x-t+1,y,z,1)-L(x,y,z,t) = 2(x-1)(t-1) \geq 0 $$Therefore, we conclude that:
$$L(x,y,z,t) \leq L(x+t-1,y,z,1) \leq L(5,1,1,1) = 28$$Therfore:
$$\boxed{L_{max} = 28}$$achieved at $a=b=c=d=0$, $y=z=t=1$, $x=5$.",169,575,Algebra,5
328,shl_jbmo_2018_a6,shl_jbmo,2018,a,"For $a,b,c$ positive real numbers such that $ab+bc+ca=3$, prove:
$ \frac{a}{\sqrt{a^3+5}}+\frac{b}{\sqrt{b^3+5}}+\frac{c}{\sqrt{c^3+5}} \leq \frac{\sqrt{6}}{2}$","Observe that $2(a^3+5)=(a^3+a^3+1)+9\geq 3a^2+9=3(a+b)(a+c)$.
Hence, we have$$\sum_{cyc} \frac{a}{\sqrt{a^3+5}}\leq \sqrt{\frac{2}{3}}\sum_{cyc} \frac{a}{\sqrt{(a+b)(a+c)}}$$
$$=\sqrt{\frac{2}{3}\cdot \frac{\displaystyle{\left(\sum_{cyc} \sqrt{a^2(b+c)}
ight)^2}}{(a+b)(b+c)(c+a)}}$$
(By CS.)$$\leq \sqrt{\frac{2}{3}\cdot \frac{(\sum a)(\sum (ab+ac))}{(a+b)(b+c)(c+a))}}\leq \frac{\sqrt{6}}{2}$$",160,395,Algebra,6
329,shl_jbmo_2018_a7,shl_jbmo,2018,a,"Let $A$ be a set of positive integers satisfying the following :

$a.)$ If $n \in A$ , then $n \le 2018$.

$b.)$ If $S \subset A$ such that $|S|=3$, then there exists $m,n \in S$ such that $|n-m| \ge \sqrt{n}+\sqrt{m}$

What is the maximum cardinality of $A$ ?","Condition (b) rewrites to $|\sqrt{m}-\sqrt{n}| \ge 1$. Now, partition the set $T = \{ 1,2, \dots, 2018\}$ according to the floor of the square of its elements, i.e. $T_1 = \{ 1,2,3 \}, T_2 = \{ 4,5,6,7,8 \}, \dots, T_{44} = \{ 1936, 1937, \dots, 2018\}$. If there are three elements $a_1,a_2,a_3$ of $A$ who are in the same partition, then the maximal value of $|\sqrt{a_i}-\sqrt{a_j}|$ would be less than 1, violating (b). Therefore, there are at most two elements of $A$ in each partition. So, $|A| \le 2 \times 44 = 88$.

For the construction, pick $A = \{ t^2, t^2+t | 1\le t \le 44, t \in \mathbb{Z} \}$. Take arbitrary $a_1<a_2<a_3\in A$. To show it works:
If all of them are in different partition, then $|\sqrt{a_3}-\sqrt{a_1}| > 1$.
If two of them are in one partition, bash cases and the observations $|\sqrt{(t+1)^2}-\sqrt{t^2}| \ge 1$ and $|\sqrt{(t+1)^2+t+1}-\sqrt{t^2+t}| \ge 1$ will do the work.",260,910,Algebra,7
330,shl_jbmo_2018_c1,shl_jbmo,2018,c,"A set $S$ is called neighbouring if it has the following two properties:
a) $S$ has exactly four elements
b) for every element $x$ of $S$, at least one of the numbers $x - 1$ or $x+1$ belongs to $S$.
Find the number of all neighbouring subsets of the set $\{1,2,... ,n\}$.","Let be $S=\{a,b,c,d\}$, where $a,b,c,d\in\{1,2,\dots,n\}, a<b<c<d$.
$b-a\ge1; c-b\ge 1; d-c\ge1\Longrightarrow d-a\ge3$.
$a\in S$ and $a-1
otin S\Longrightarrow a+1\in S\Longrightarrow b=a+1$.
$d\in S$ and $d+1
otin S\Longrightarrow d-1\in S\Longrightarrow c=d-1$.
Hence: $S=\{a,a+1,d-1,d\}; d-a\ge 3$.
The number of the sets $S$ is equal to the number of pairs $(a,d)$, where $a,d\in\{1,2,\dots,n\}, d-a\ge3$.

Obviously, $n\ge 4$.
For $a=1$ results $d\in\{4,5,\dots,n\}$, hence $n-3$ possible values of $d$.
For $a=2$ results $d\in\{5,6,\dots,n\}$, hence $n-4$ possible values of $d$.
......
For $a=n-3$ results $d=n$, hence $1$ possible value of $d$.
Results the number of possible sets $S$:
$N_S=(n-3)+(n-4)+\dots+1=\dfrac{(n-3)(n-2)}{2}$.",272,743,Combinatorics,8
331,shl_jbmo_2018_c2,shl_jbmo,2018,c,"Find max number $n$ of numbers of three digits such that :
1. Each has digit sum $9$
2. No one contains digit $0$
3. Each $2$ have different unit digits
4. Each $2$ have different decimal digits
5. Each $2$ have different hundreds digits","We use first and second conditions, we get digits are $1,2,3,4,5,6,7$.So, maximum number of this form numbers are less than 7.
$$711$$$$621,612$$$$531,513,522$$$$441,414,432,423$$$$351,315,342,324,333$$$$261,216,252,225,243,234$$$$171,117,162,126,153,135,144$$Sum of digits are 9.

Case 1:$ n=7 $.
So,we must choose exactly one number from each group.This means we must use 711. But if we use 711, then we don't use 621 and 612. Because, we have third and fourth conditions.

Case 2: $ n=6 $.

Subcase 1: We use 711. We don't use 621,612,531,513,441,414,351,315,261,216,171,117.So, From third group we must choose 522, then we mustn't use 432 or 423.This means $n \le 4$.

Subcase 2: We don't use 711. This means we use 621 or 612. If we choose 621, then we must choose 513. And we must choose 432.This means in fifth we don't choose 351,315,342,324,333.But this is not true.
Case 612 is similar with case 621.

Case 3: $ n=5 $.
In this case we have construction.
513,432,351,225,144.",237,984,Combinatorics,9
332,shl_jbmo_2018_c3,shl_jbmo,2018,c,The cells of a $8 \times 8$ table are initially white. Alice and Bob play a game. First Alice paints $n$ of the fields in red. Then Bob chooses $4$ rows and $4$ columns from the table and paints all fields in them in black. Alice wins if there is at least one red field left. Find the least value of $n$ such that Alice can win the game no matter how Bob plays.,"1. We claim that the least value of $n$ should be greater than $12$.
1.1 Consider $n=12$.
Suppose $a_i$ is the number of red fields in row $i$, $1 \leq i \leq 8$. So $\sum_{i=1}^{8} a_i=12$.
Consider the sum of the largest $4$ elements of $\{a_i\}$, let it be $m$.
(1) We claim $m \geq 8$. If not, suppose the largest $4$ rows be set $A$ and the other $4$ rows be set $B$.
The sum of red fields number in $A$, that is $m$, is less than $8$; so there is at least one row in $A$ including at most only $1$ red field, or else $m \geq 8$ (Pigeonhole principle).
However, the sum of red fields number in $B$, that is $12-m$, is greater than 4; so there is at least one row in $B$ including at least $2$ red fields (Pigeonhole principle again).
So there would be at least one row in $A$, the red field number of which is less than one row in $B$. This leads to contradiction since $A$ includes the largest $4$ red field number rows!
(2) Select those $4$ rows to cover, and thus there would be less than $12-8=4$ red fields left. Obviously we could use no more than $4$ columns to cover the left red fields.
So $n=12$ could always be covered by $4$ rows and $4$ columns.
1.2 If $n<12$, then randomly paint some additional fields red to $n=12$ and due to 1.1, it could be covered by $4$ rows and $4$ columns.

2. A construction of $n=13$.
Consider the matrix below, where $1$ is the red fields and $0$ is white fields. There are $5$ rows and $5$ columns that include $2$ red fields. It can not be covered by $4$ rows and $4$ columns.
$$\begin{matrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 1\\
0 & 0 & 0 & 1 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 1 & 1 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 1 & 1 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 1
\end{matrix} $$So the least value of $n$ is $\boxed{13}$.",361,1854,Combinatorics,10
333,shl_jbmo_2018_g1,shl_jbmo,2018,g,"Let $H$ be the orthocentre of an acute triangle $ABC$ with $BC > AC$, inscribed in a circle $\Gamma$. The circle with centre $C$ and radius $CB$ intersects $\Gamma$ at the point $D$, which is on the arc $AB$ not containing $C$. The circle with centre $C$ and radius $CA$ intersects the segment $CD$ at the point $K$. The line parallel to $BD$ through $K$, intersects $AB$ at point $L$. If $M$ is the midpoint of $AB$ and $N$ is the foot of the perpendicular from $H$ to $CL$, prove that the line $MN$ bisects the segment $CH$.","et $X,Y,$ and $Z$ be the foot of perpendiculars from $C,A,$ and $B$ to $AB, BC,$ and $AC$ respectively. Also, let $P$ and $E$ be the midpoint of $CH$ and $BC$ respectively, which results in $P$ and $E$ are the circumcenters of $(CHN)$ and $(CXB)$ respectively, along with $PE||BZ$ and $EM||AC$

Claim 1: $ALKC$ is cyclic
since $LK||BD$, we have that $\angle CAB = \angle CDB = \angle LKD = 180 -\angle LKC$, as desired. Furthermore, since $AC=CK$, then $\angle ALC = \angle CKA = \angle CAK = \angle CLK$, making $\angle CLA = \frac{\angle KLA}{2}$

Claim 2: $PXME$ is cyclic
From $\angle PEM = 180 - \angle MEB - \angle PEC = 180 - \angle ACB -\angle ZBC = 90$, we have that $\angle PXM = \angle PEM = 90$, therefore $PXME$ is cyclic

For finishing up, just notice that $\angle XPN = 2\angle HCN = 2(90 - \angle HLC) = 180 - \angle ALK = \angle KLM = \angle ABD$, and $\angle XPM =\angle XEM = \angle XEB - \angle MEB = 2\angle XCB - \angle ACB = 2(90 -\angle XBC) - (180 - \angle CAB - \angle CBA) = 180 -2(\angle CBD - \angle ABD) - (180 - \angle CDB - (\angle (CBD - \angle (ABD)) = 180 -2\angle CBD +2\angle ABD - 180 + \angle CBD + \angle CBD -\angle ABD = \angle ABD$, which reveals that $\angle XPM = \angle XPN$, so $M,N,P$ are colinear, thus done",526,1256,Geometry,11
334,shl_jbmo_2018_g2,shl_jbmo,2018,g,"Let $ABC$ be a right angled triangle with $\angle A = 90^o$ and $AD$ its altitude. We draw parallel lines from $D$ to the vertical sides of the triangle and we call $E, Z$ their points of intersection with $AB$ and $AC$ respectively. The parallel line from $C$ to $EZ$ intersects the line $AB$ at the point $N$. Let $A' $ be the symmetric of $A$ with respect to the line $EZ$ and $I, K$ the projections of $A'$ onto $AB$ and $AC$ respectively. If $T$ is the point of intersection of the lines $IK$ and $DE$, prove that $\angle NA'T = \angle  ADT$.","First, let us denote $\angle B=\alpha, AD\cap EZ={P}, IK\cap AA'={R}$ and $AA'\cap DE={Q}$. Since $DE||AC$, we have $\angle DEA=90^o$. Similarly, $\angle DZA=90^o$.

Now, since $\angle ADB=90^o$,
we have $\angle DAB = 180^o-\angle ADB  -\angle ABD=180^o-90^o-\alpha=90^o-\alpha$.
From here, $\angle ADE=180^o-\angle DEA-\angle DAE=180^o-90^o-(90^o - \alpha) = \alpha$.
Because $AZDE$ is a rectangle, $\angle ADE=\angle AZE \equiv \angle AZR$.

$ZE$ and $CN$ are parallel, so $\angle AZE=\angle ACN=\alpha$.
From here we derive
$\angle CNA=90^o-\angle ACN=90^o-\alpha$
and
$\angle INC=180^o-\angle CNA=90^o+\alpha$

$A'$ is symmetric of $A$ with respect to $ZE$, so we must have $AA'\perp ZE$. Thus,
$\angle ZAR=90^o-\angle AZR=90^o-\alpha$.
Now,
$\angle AA'K=90^o-\angle KAA' \equiv 90^o-\angle ZAR=\alpha$
Because $AKA'I$ is a rectangle, we get$\angle AA'K=\angle AIK=\alpha$
So,
$\angle ETI=90^o-\angle EIT \equiv 90^o-\angle AIK=90^o-\alpha$
and
$\angle ITQ=180^o-\angle ETI=90^o+\alpha$

Lemma 1: $CN, AA'$ and $DE$ are concurrent
Proof:
1) Let us suppose a line $p$ through $Q$, such that $p\perp AA'$. Let $C'$ be the intersection of $p$ with $AC$. Since $\angle CAA'=90^o-\alpha$, and therefore $\angle C'AA'=90^o-\alpha$,
$\angle QC'A=\alpha$
Because of this, and because $\angle ADQ \equiv \angle ADE=\alpha$, $\angle QC'A=\angle QDA, C'AQD$ is cyclic, a.e. the excircle of $\triangle AQD$ contains $C'$.

2) We have $\angle A'AI \equiv \angle QAE=\alpha$. Since $\angle DEA \equiv \angle QEA=90^o$, we obtain
$\angle DQA=\angle QAE+\angle QEA=90^o+\alpha$
By definition, $\angle ACB \equiv \angle DCA=90^o-\alpha$.
Now we see that $\angle DCA+\angle DQA=90^o-\alpha+90^o+\alpha=180^o$, so the excircle of $\triangle AQD$ must contain $C$.

Combining 1) and 2), we see that the excircle of $\triangle AQD$ intersects the line $AC$ at three points: $A, C$ and $C'$. It is impossible for all three to be distinct points, since then the excircle of $\triangle AQD$ will have three common points with $AC$, which is impossible. It follows that two of $A, C$ and $C'$ must be the same point. $A$ and $C'$ cannot be equivalent because, in that case $QC$' couldn't be perpendicular to $AA'$. So $C\equiv C'$.

Now, $CQ$ is perpendicular to $AA'$, which means that $CQ\parallel ZE$, and that $CQ$ contains $N$. Thus, the proof of the lemma is complete.

Lemma 2: $A'QNI$ is cyclic
Proof:
We have $\angle A'QN=90^o$ and $\angle A'IN=90^o$.
Case 1: $N$ is between $A$ and $I$
-- $\angle A'IN+\angle A'QN=180^o$, so the lemma is true.
Case 2: $N$ is between $B$ and $I$
-- $\angle A'QN=\angle A'IN=90^o$, so the lemma is true.

So, $\angle CNI \equiv \angle QNI=\angle DTI \equiv \angle QTI$, or $\angle QTI=\angle QNI$. From here we obtain that $QTNI$ is cyclic. Because $QNIA'$ is also cyclic, $Q, T, N, I$ and $A'$ all lie on the same circle.

*Note: When $N$ is between $I$ and $B$, the proof is done by using the fact that $\angle QNI \equiv\angle CNA=90^o -\alpha$

Case 1: $N$ is between $A$ and $I$
-- $\angle AIK \equiv \angle NIT=\alpha$, and by peripheral angles, we obtain $\angle NA'T=\angle NIT=\alpha$, a.e. $\angle NA'T=\angle ADT=\alpha$ . Thus, the proof is complete.

Case 2: $N$ is between $B$ and $I$
-- $\angle NA'T=180^o-\angle TIN=\alpha$, so $\angle NA'T=\alpha=\angle ADT$. Thus, the proof is complete.

*Note: When $N$ and $I$ are equivalent, the proof follows directly from the fact that $QNIA'$ is cyclic and $\angle CQD=\alpha$ .",547,3457,Geometry,12
335,shl_jbmo_2018_g3,shl_jbmo,2018,g,"Let $\triangle ABC$ and $A'$,$B'$,$C'$ the symmetrics of vertex over opposite sides.The intersection of the circumcircles of $\triangle ABB'$ and $\triangle ACC'$ is $A_1$.$B_1$ and $C_1$ are defined similarly.Prove that lines $AA_1$,$BB_1$ and $CC_1$ are concurent.","$\angle AA_1C=\angle AC'C=\angle ACC'=90-\alpha$.
Similarly, $\angle AA_1B=90-\alpha$
Let $A_1C \cap (ABB'A_1)=X$, then $\angle AA_1X=90-\alpha=\angle AA_1B$. Therefore, $AB=AB'=AX$, i.e. $B'\equiv X$, i.e. $A_1-C-B'$ are collinear.
Now, $\angle CAH=90-\gamma=\angle B'BC=\angle BB'C=\angle BB'A_1=\angle BAA_1$ and since the circumcenter $O$ and the orthocenter $H$ are isogonal conjugates, it follows that $O$ lies on $AA_1$. Similarly, $O$ lies on $BB_1$ and $CC_1$.",266,469,Geometry,13
336,shl_jbmo_2018_g4,shl_jbmo,2018,g,"Let $ABC$ be a triangle with side-lengths $a, b, c$, inscribed in a circle with radius $R$ and let $I$ be ir's incenter. Let $P_1, P_2$ and $P_3$ be the areas of the triangles $ABI, BCI$ and $CAI$, respectively. Prove that$$\frac{R^4}{P_1^2}+\frac{R^4}{P_2^2}+\frac{R^4}{P_3^2}\ge 16$$","It's
$$\sum_{cyc}\frac{1}{a^2}\geq\frac{(a+b+c)\prod\limits_{cyc}(a+b-c)^3}{a^4b^4c^4}$$and since by Schur
$$abc\geq\prod_{cyc}(a+b-c),$$it's enough to prove that
$$\sum_{cyc}a^2b^2\geq\sum_{cyc}(2a^2b^2-a^4)$$or
$$\sum_{cyc}(a^2-b^2)^2\geq0.$$",285,244,Geometry,14
337,shl_jbmo_2018_g5,shl_jbmo,2018,g,"Given a rectangle $ABCD$ such that $AB = b > 2a = BC$, let $E$ be the midpoint of $AD$. On a line parallel to $AB$ through point $E$, a point $G$ is chosen such that the area of $GCE$ is
$$(GCE)= \frac12 \left(\frac{a^3}{b}+ab
ight)$$Point $H$ is the foot of the perpendicular from $E$ to $GD$ and a point $I$ is taken on the diagonal $AC$ such that the triangles $ACE$ and $AEI$ are similar. The lines $BH$ and $IE$ intersect at $K$ and the lines $CA$ and $EH$ intersect at $J$. Prove that $KJ \perp AB$.","Because $(GCE)=\frac{1}{2}\cdot a\cdot \frac{a^2+b^2}{b},$ we get $EG=\frac{a^2+b^2}{b}=\frac{EC^2}{EF},$ and therefore $\angle ECG=90^\circ$ and $EHCG$ is cyclic with $EG$ being a diameter of the corresponding circle. But $DE\perp EG,$ and thus $DEA$ is tangent to this circle, and by construction of the location of $I,$ we have that $I$ is also on this circle (because $AE^2=AI\cdot AC$). So is point $B,$ by symmetry (wrt to $C$ and the diameter $EG$); now, $\angle IKH=(\overarc {BI}-\overarc {EH})/2=(\overarc {HC}-\overarc {IE})/2=\angle HJI,$ and therefore $KJIH$ is cyclic. Finally, $\angle KJH=\angle KIH=\angle EGH=\angle HED=\angle JEA,$ and therefore $KJ\parallel DA$ and $KJ\perp AB.$",505,698,Geometry,15
338,shl_jbmo_2018_g6,shl_jbmo,2018,g,"Let $XY$ be a chord of a circle $\Omega$, with center $O$, which is not a diameter. Let $P, Q$ be two distinct points inside the segment $XY$, where $Q$ lies between $P$ and $X$. Let $\ell$ the perpendicular line drawn from $P$ to the diameter which passes through $Q$. Let $M$ be the intersection point of $\ell$ and $\Omega$, which is closer to $P$. Prove that$$ MP \cdot XY \ge 2 \cdot QX \cdot PY$$","Let $\ell’$ be the line through $Q$ parallel to $\ell$. Let $U, V = \ell’ \cap \Omega$ and $R = \ell’ \cap PM$.

Claim : $XY \cdot QU \geq 2 \cdot QX \cdot QY$

Proof : Note that $UV \leq XY$ as $XY$ is farther from the centre than $UV$. Also, $Q$ is midpoint of $UV$. So, $XY \cdot QU \geq 2 \cdot QU^2 = 2 \cdot QX \cdot QY$. $\square$

Since $QR \geq QU$, we have $\frac{XY}{2 \cdot QX} \geq \frac{QY}{QU} \geq \frac{QY}{QR} = \frac{PY}{PM}$ which gives the desired result.",402,476,Geometry,16
339,shl_jbmo_2018_n1,shl_jbmo,2018,n,Find all integers $m$ and $n$ such that the fifth power of $m$ minus the fifth power of $n$ is equal to $16mn$.,"$m^5-n^5=16mn$
Case 1: $m,n>0$
Let $(m,n)=d$, so that $m=dx, n=dy$. note that this implies $d^3(x^5-y^5)=16xy$. Then $d^3y^5$ is divisible by $x$ which is impossible unless $x=1$. Similarly $y=1$. However putting in the values we see that there's no solution in this case.
Case 2: $m>0, n<0$
Let $n=-a, a >0$. Notice that $m^5+a^5=-16am$. Since the $LHS > 0$, there is no solution in this case.
Case 3 : $m<0, n>0$
Let $m=-a, a>0$. Notice that $a^5+n^5=16an$. Let $(a,n)=d, a=dx, n=dy$. Notice that $d^3(x^5+y^5)=16xy$. This implies $d^3x^5$ is divisible by $y$, but because $(d^3x^5,y)=1, y=1$. Similarly $x=1$. Put in the values to get $d^3 \cdot 2 = 16 \implies d=2$. So, $m=-2, n=2$
Case 4: $m<0, n<0$
Notice that in this case the signs of the LHS and RHS don't match; no solutions here.
Case 5: WLOG $m=0$
Then we get $n=0$ as well. So $(m,n)=(0,0)$ is a solution as well

Combining these cases we get $(m,n)=(-2,2),(0,0)$",111,927,Number Theory,17
340,shl_jbmo_2018_n2,shl_jbmo,2018,n,"Find all ordered pairs of positive integers $(m,n)$ such that :
$125*2^n-3^m=271$","The only solution in natural numbers of the Diophantine equation

$(1) \;\; 125 \cdot 2^n - 3^m = 271$

is $(m,n) = (6,3)$.

Proof: Equation (1) implies $5 | 3^m + 1$, yielding $4 | m+2$. Hence $m \geq 2$, which according to equation (1) means $9 | 2^n + 1$, which give us $6 | n - 3$. Thus there exist four non-negative integers $q_1,q_2,r_1,r_2$ s.t.

$(2) \;\; m = 12q_1 + 4r_1 + 2, \: r_1 \leq 2,$

and

$(3) \;\; n = 12q_2 + 6r_2 + 3, \: r_2 \leq 1$.

Combining (1)-(3) we obtain

$(4) \;\; (-1)^{r_2+1} - 3^{4r_1+2} \equiv 11 \pmod{13}$.

Checking the six combination of $r_1$ and $r_2$, we find that there are two combinations which satisfies congruence (4), namely $(r_1,r_2)=(1,0)$ and $(r_1,r_2) = (2,1)$. Thus we have the following two cases:

Case 1: $(r_1,r_2) = (1,0)$. Inserting this in (2)-(3), the results are

$(5) \;\; (m,n) = (12q_1+6,12q_2+3)$.

Therefore by equation (1)

$(6) \;\; x^3 - y^3 = 271$,

where

$(7) \;\; (x,y) = (5 \cdot 2^{4q_2+1}, 3^{4q_1+2})$.

Equation (6) is equivalent to $(x - y)(x^2 + xy + y^2) = 271$, yielding $(x-y,x^2+xy+y^2) = (1,271)$ (since 271 is a prime), which give us $(x,y) = (10,9)$. Consequently $q_1=q_2=0$ by (7), which according to (5) give us $(m,n) = (6,3)$.

Case 2: $(r_1,r_2) = (2,1)$. Inserting this in (2)-(3), we obtain

$(8) \;\; (m,n) = (12q_1+10,12q_2+9)$.

Using the fact that $2^6 \equiv 3^6 \equiv 1 \pmod{7}$ combined with (1) and (8) we obtain

$125 \cdot 2^{9} - 3^{10} \equiv 271 \pmod{7}$,

i.e. $7 | 4680$. This contradiction implies there is no solution of equation (1).

Conclusion: The only solution of equation (1) in natural number is $(m,n) = (6,3)$. q.e.d.",81,1644,Number Theory,18
341,shl_jbmo_2018_n3,shl_jbmo,2018,n,"Find all four-digit positive integers $\overline{abcd}=10^3a+10^2b+10c+d$ $(a
e0)$ such that:
$\overline{abcd} =a^{a+b+c+d}-a^{-a+b-c+d}+a$","The only solution of the Diophantine equation

$(1) \;\; \overline{abcd} = a^{a + b + c + d} - a^{-a + b - c + d} + a$,

(where $\overline{abcd}$ is a four digit number) is $(a,b,c,d) = (2,0,1,8)$.

Proof: Clearly the LHS of equation (1) is greater or equal to 1000, implying $a 
eq 1$. Hence $a \geq 2$, which means the exponent $-a + b - c + d$ in equation (1) is non-negative, i.e.

$(2) \;\; b + d \geq a + c$

Moreover by equation (1)

$1000a \leq \overline{abcd} = a^{a+b+c+d} - a^{-a+b-c+d} + a < a^{a+b+c+d} + a$,

yielding

$(3) \;\; a^{a+b+c+d-1} > 999$.

Futhermore

$1000(a + 1) > \overline{abcd} = a^{a+b+c+d} - a^{-a+b-c+d} + a > a^{a+b+c+d} - a^{b+c+d} + a$

i.e.

$(4) \;\; (a^a - 1) \cdot a^{b+c+d} \leq 999(a+1)$.

By inequality (2) we obtain $a + b + c + d \geq 2(a + c) \geq 2a$, which combined with inequality (4) give us

$(5) \;\; a^{2a} \leq 999(a+1)$.

Hence $a \leq 3$ by inequality (5).

Assume $a=3$. Then according to inequalities (3) and (4) we obtain $3^{b+c+d+2} > 999 > 3^6$ and ${\textstyle 3^{b+c+d} \leq \frac{3996}{26}} < 3^5$, yielding $b+c+d \geq 5$ and $b+c+d \leq 4$ respectively. This contradiction implies $a=2$. Applying the inequalities (3) and (4) we obtain $2^{b+c+d+1} > 999 > 2^9$ and $2^{b+c+d} \leq 999 < 2^{10}$, yielding $b+c+d \geq 9$ and $b+c+d \leq 9$ respectively. In other words, $b+c+d=9$. This means

$-a+b-c+d = -(2 - c) + (9 - c) = 7 - 2c$,

which inserted in equation (1) result in

$\overline{2bcd} = 2000 + \overline{bcd} = 2^{2+9} - 2^{7-2c} + 2$,

i.e.

$(6) \;\; \overline{bcd} = 50 - 2^{7-2c}$.

Thus we have $\overline{bcd} < 100$ by equation (6), which give us $b=0$. Consequently by equation (6)

$(7) \;\; \overline{cd} = 50 - 2^{7-2c}$

with

$(8) \;\; c+d=9$.

According to equation (7) the digit $d$ is even. Thus $c$ is odd by equation (8).

We also have $7 - 2c \geq 0$ by equation (7), i.e. $c \leq 3$. Therefore according to equation (7)

$2^{7-2c} = 50 - \overline{cd} \geq 50 - 39 = 11 > 2^3$,

yielding $7-2c>3$. Hence $c<2$, which means $c=1$ (since $c$ is odd) and $d=9-c=9-1=8$ by equation (8).

Conclusion: The only solution of equation (1) is $(a,b,c,d) = (2,0,1,8)$. q.e.d.",139,2162,Number Theory,19
342,shl_jbmo_2018_n4,shl_jbmo,2018,n,Prove that there exist infinitely many positive integers $n$ such that $\frac{4^n+2^n+1}{n^2+n+1}$ is a positive integer.,"The form of $n$ which works is highly motivated by looking at small examples, and trying constructive ideas.

We claim that all integers of the form $n=2^{2^r}$ satisfy the given condition. We first make the observation that for all positive integers $m$,
\[m^3 - 1 = (m-1)(m^2+m+1)\]Thus, the given fraction can be rewritten as,
\[\frac{2^{3n}-1}{2^{n-1}}\cdot \frac{n-1}{n^3-1}\]Thus we wish to show that for all positive integers $r$,
\[\frac{2^{3\cdot 2^{2^r}}-1}{2^{2^{2^r}}-1}\cdot \frac{2^{2^r}-1}{2^{3\cdot 2^r}-1}\]is an integer. Now, we consider prime divisors $p$ of the denominator. Note that since the denominator is clearly odd, $p
e 2$. We have two cases.

Case 1 : For a prime $p$ divisor of the denominator, $p$ divides exactly one of $2^{2^{2^r}}-1$ and $2^{3\cdot 2^r}-1$. We sort out the case where $p\mid 2^{2^{2^r}}-1$. The argument can be copied to the other case. Note that, since $2^{2^r} \mid 3 \cdot 2^{2^r}$ $p\mid 2^{3\cdot 2^{2^r}}-1$ as well. Further by the Lifting the Exponent Lemma we have,
\[
u_p(2^{3\cdot 2^{2^r}}-1)=
u_p(2^{2^{2^r}}-1)+
u_p(3) \ge 
u_p(2^{2^{2^r}}-1)\]
Case 2 : For a prime $p$ divisor of the denominator, $p$ divides both $2^{2^{2^r}}-1$ and $2^{3\cdot 2^r}-1$. Then,
\[\text{ord}_p(2) \mid 2^{2^r} \text{ and } \text{ord}_p(2) \mid 3\cdot 2^r\]from which it is not hard to see that,
\[\text{ord}_p(2) \mid 2^r\]But then, $p\mid 2^{2^r}-1$ as well. We now repeat the argument using the Lifting the Exponent Lemma. Note,
\[
u_p(2^{3\cdot 2^{2^r}}-1)=
u_p(2^{2^r}-1)+
u_p(3\cdot 2^{2^{2^r-r}}) =
u_p(2^{2^r}-1)+
u_p(3) + 
u_p(2^{2^{2^r-r}}) \]\[
u_p(2^{2^{2^r}}-1)=
u_p(2^{2^r}-1)+
u_p(2^{{2^{2^r}-r}})\]and
\[
u_p(2^{3\cdot 2^r}-1) = 
u_p(2^{2^r}-1)+
u_p(3)\]From which we have that
\[
u_p\left((2^{3\cdot 2^{2^r}}-1)(2^{2^r}-1)
ight) \ge 
u_p\left((2^{2^{2^r}}-1)(2^{3\cdot 2^r}-1)
ight)\]which finishes the proof.",121,1870,Number Theory,20
448,tst_jbmo_ro_2018_1_p1,tst_jbmo,2018,n,Prove that the equation $x^2 + y^2 + z^2 = x + y + z + 1$ has no rational solutions.,"\begin{solution}
The equation can be written equivalently $(2x-1)^2 + (2y-1)^2 + (2z-1)^2 = 7$. If $(x, y, z)$ would be a solution of this equation with rational components, denoting $2x - 1 = \dfrac{a_1}{b_1}$, $2y - 1 = \dfrac{a_2}{b_2}$, $2z - 1 = \dfrac{a_3}{b_3}$, one would have integers $a_1, b_1, a_2, b_2, a_3, b_3$ satisfying the equality
\[
  (a_1 b_2 b_3)^2 + (b_1 a_2 b_3)^2 + (b_1 b_2 a_3)^2 = 7(b_1 b_2 b_3)^2.
\]
This would lead to the existence of four integers $a, b, c, d$ such that $a^2 + b^2 + c^2 = 7d^2$. If the greatest common divisor of $a, b, c, d$ is $k$, dividing by $k^2$ one would obtain a solution $(a, b, c, d) \in \mathbb{Z}^4$ of the equation $a^2 + b^2 + c^2 = 7d^2$ with $\gcd(a, b, c, d) = 1$. Then $a, b, c, d$ cannot all be even. But a perfect square is congruent to either $0$, $1$, or $4$ modulo $8$, so the left-hand side can only be $1, 2, 3, 5$ or $6$ modulo $8$, while the right-hand side, $7d^2$, is $0$, $4$ or $7$ modulo $8$. In conclusion, the equality cannot hold.
\end{solution}",84,1029,Number Theory,1
449,tst_jbmo_ro_2018_1_p2,tst_jbmo,2018,a,"Let $a, b, c$ be positive real numbers such that $a^2 + b^2 + c^2 = 3$. Prove that
\[
  \frac{1}{a} + \frac{3}{b} + \frac{5}{c} \geq 4a^2 + 3b^2 + 2c^2.
\]
When does the equality hold?","\begin{solution}[Solution 1]
The inequality can be written
\[
  \frac{1}{a} + \frac{3}{b} + \frac{5}{c} + b^2 + 2c^2 \geq 4(a^2 + b^2 + c^2)
\]
or
\[
  \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{2}{b} + b^2 + \frac{4}{c} + 2c^2 \geq 12.
\]
This follows from the following inequalities by using the AM-GM inequality:
\[
  \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq \frac{3}{\sqrt[3]{abc}} \geq 3
\]
because
\[
  3 = a^2 + b^2 + c^2 \geq 3\sqrt[3]{a^2 b^2 c^2} \implies abc \leq 1.
\]
On the other hand,
\[
  \frac{2}{b} + b^2 = \frac{1}{b} + \frac{1}{b} + b^2 \geq 3\sqrt[3]{\frac{1}{b} \cdot \frac{1}{b} \cdot b^2} = 3,
\]
\[
  \frac{4}{c} + 2c^2 = \frac{2}{c} + \frac{2}{c} + 2c^2 \geq 3\sqrt[3]{\frac{2}{c} \cdot \frac{2}{c} \cdot 2c^2} = 6.
\]
The equality holds if $a = b = c = 1$.

\textit{Remark.} One can apply directly the AM-GM inequality for the following 12 numbers:
\[
  \frac{1}{a},\; \frac{1}{b},\; \frac{1}{b},\; \frac{1}{b},\; \frac{1}{c},\; \frac{1}{c},\; \frac{1}{c},\; \frac{1}{c},\; \frac{1}{c},\; b^2,\; c^2,\; c^2.
\]
\end{solution}

\begin{solution}[Solution 2]
As $x^3 - 3x + 2 \geq 0$, $\forall\, x \geq 0$ (equivalent to $(x-1)^2(x+2) \geq 0$; alternatively, the previous inequality follows from AM-GM: $x^3 + 1 + 1 \geq 3\sqrt[3]{x^3 \cdot 1 \cdot 1} = 3x$; the equality holds when $x = 1$), we deduce that $\dfrac{1}{x} \geq \dfrac{3}{2} - \dfrac{x^2}{2}$.

Writing this for $a$, $b$, $c$ and multiplying these inequalities by $1$, $3$, and $5$, respectively, we obtain by summing
\[
  \frac{1}{a} + \frac{3}{b} + \frac{5}{c} \geq \frac{27}{2} - \frac{a^2 + 3b^2 + 5c^2}{2}.
\]
Using $27 = 9(a^2 + b^2 + c^2)$ one obtains the conclusion.

The equality holds if and only if $a = b = c = 1$.
\end{solution}",184,1739,Algebra,2
450,tst_jbmo_ro_2018_1_p3,tst_jbmo,2018,g,Let $ABC$ be a triangle with $AB > AC$. Point $P \in (AB)$ is such that $\angle ACP = \angle ABC$. Let $D$ be the reflection of $P$ into the line $AC$ and let $E$ be the point in which the circumcircle of $BCD$ meets again the line $AC$. Prove that $AE = AC$.,"\begin{solution}[Solution 1]
Let $Q$ be the point in which the circumcircle of $BCD$ meets again the line $AB$. Then $\angle QEA = \angle QBC = \angle ECP$, hence $EQ \parallel PC$. Moreover, $\angle ECP = \angle ECD$ implies $QD$ is parallel to $EC$, hence $EQ = CD = CP$. It follows that $EQCP$ is a parallelogram, which leads to the conclusion.
\end{solution}

\begin{solution}[Solution 2 (given in the contest)]
If $\{O\} = BD \cap AC$, it is easy to prove that $BA$ and $BO$ are isogonal in the angle $\angle EBC$, therefore, in order to show that $BA$ is the median, one has to prove that $BO$ is the symmedian. This follows readily by computation, using Steiner's Theorem.

Indeed, $\angle EBA = \angle BAC - \angle BEA = 180^\circ - \angle ABC - \angle ACB - \angle BDC = 180^\circ - \angle ECD - \angle ECB - \angle BDC = \angle CBD$.

Triangles $OEB$ and $ODC$ are similar, hence $\dfrac{EO}{OD} = \dfrac{EB}{CD}$. \hfill (1)

Triangles $OCB$ and $ODE$ are also similar, therefore $\dfrac{OC}{OD} = \dfrac{BC}{ED}$. \hfill (2)

We have $\angle ADC = \angle APC = \angle PBC + \angle PCB = \angle ACB = \angle EDB$, which shows that the triangles $ACD$ and $EBD$ are similar. Hence $ED = \dfrac{EB \cdot AD}{AC}$ and, as $AD = \dfrac{CD \cdot AC}{BC}$ (from $\triangle ACB \sim \triangle APC \sim \triangle ADC$), it follows that $ED = \dfrac{EB \cdot CD}{BC}$.

From (1) and (2) we obtain $\dfrac{EO}{OC} = \dfrac{EB \cdot ED}{CD \cdot BC} = \left(\dfrac{EB}{BC}
ight)^2$, which shows that $BO$ is indeed the symmedian.
\end{solution}

\begin{solution}[Solution 3 (given in the contest by Andrei M\u{a}rginean)]
Let $\alpha = \angle EBA$, $x = \angle ABC$. We have that $\angle DCE = \angle ACP = x$ and $\angle EBC = \alpha + x$. As $B, C, D, E$ are concyclic, it follows that $\angle CDE = 180^\circ - \alpha - x$, $\angle DEC = 180^\circ - \angle EDC - \angle DCE = \alpha = \angle EBA$. But $AE$ is the perpendicular bisector of $[PD]$, hence $\angle PEA = \angle DEC = \angle EBA$, which shows that the triangles $APE$ and $AEB$ are similar. It follows that $AE^2 = AB \cdot AP$. But triangles $ACP$ and $ABC$ are also similar, hence $AC^2 = AP \cdot AB = AE^2$, and the conclusion follows immediately.
\end{solution}",259,2232,Geometry,3
451,tst_jbmo_ro_2018_1_p4,tst_jbmo,2018,c,What is the maximum number of rooks one can place on a chessboard such that any rook attacks exactly two other rooks? (We say that two rooks attack each other if they are on the same line or on the same column and between them there are no other rooks.),"\begin{solution}[Solution 1]
We say that there can be two types of rooks on the chessboard: a rook of type $T_1$ is a rook that is attacked from two perpendicular lines; a rook is of type $T_2$ if it is attacked by two rooks situated on the same line, but in opposite directions.

Suppose one can place $m$ rooks of type $T_1$ and $n$ of type $T_2$, with $m + n = x$. Each rook of type $T_1$ determines two lines from which it is attacked (out of the 16 of the chessboard: 8 horizontal and 8 vertical ones), while exactly one other rook of type $T_1$ is situated on each of these two lines. Notice that a rook of type $T_2$ cannot be attacked from any other line than the one it is already attacked from, hence, for each of these $m$ rooks, there are $m$ lines on which there can be no other rooks. In total, we could have at most $\dfrac{2n}{2} + m = n + m$ lines. Hence $16 \geq m + n = x$. To prove that 16 is indeed the desired maximum, it is sufficient to exhibit an example of 16 rooks that attack exactly two other ones.
\end{solution}

\begin{solution}[Solution 2 (given in the contest by Andrei M\u{a}rginean)]
We prove by induction after $n \geq 2$ that on an $n \times n$ board one can place at most $2n$ rooks with the above restrictions.

For $n = 2$ the statement is obvious.

For the inductive step, assume the statement above to hold for an arbitrary $n \geq 2$ and let us prove it for $n + 1$.

Assume that on an $(n+1) \times (n+1)$ one could place at least $2n + 3$ rooks such that each of them attacks exactly two other ones. From the Pigeonhole Principle it follows that there exists at least one horizontal line containing at least 3 rooks and, also, a vertical line with 3 or more rooks. On a horizontal line with 3 or more rooks there exists at least one rook that is attacked by two rooks situated on the same horizontal line with it. Therefore this rook must stand alone on the vertical line it occupies. Similarly, there must be a horizontal line containing exactly one rook. Eliminating these two lines, we obtain an $n \times n$ board with $2n + 1$ rooks that still satisfies the condition that each rook attacks exactly two other rooks. This contradicts our inductive hypothesis for $n$.

Again, an example with 16 rooks finishes the proof.

\textit{Remark.} By induction after $n + m$ one can prove that the maximum number of rooks one can place on an $m \times n$ board such that each rook attacks exactly two other rooks is $m + n$.
\end{solution}

\begin{solution}[Solution 3 (given in the contest by Iustinian Constantinescu)]
Let us consider a configuration of rooks attacking each other. A rook can attack on 4 rays, 2 by 2 opposite. On exactly 2 of these rays one must have rooks; on each of the other two rays the rook attacks one of the 32 segments that constitute the margins of the chessboard. Each of these 32 segments is attacked by a different rook, and each rook attacks two such segments, therefore one can have at most 16 rooks on the board.

An example with 16 rooks on the board finishes the proof. (Iustinian placed the rooks on the two diagonals.)
\end{solution}",253,3114,Combinatorics,4
452,tst_jbmo_ro_2018_2_p1,tst_jbmo,2018,a,"Determine the positive integers $n \geq 3$ such that, for every integer $m \geq 0$, there exist integers $a_1, a_2, \ldots, a_n$ such that $a_1 + a_2 + \cdots + a_n = 0$ and $a_1 a_2 + a_2 a_3 + \cdots + a_{n-1} a_n + a_n a_1 = -m$.","\begin{solution}
Any $n \geq 5$ has the desired property: one can choose $a_1 = 1 - m$, $a_2 = a_3 = \cdots = a_{n-3} = 0$, $a_{n-2} = -1$, $a_{n-1} = m$, $a_n = 0$.

Numbers $n = 3$ and $n = 4$ do not have the property.

For $n = 3$,
\[
  -2m = 2a_1 a_2 + 2a_2 a_3 + 2a_3 a_1 = (a_1 + a_2 + a_3)^2 - a_1^2 - a_2^2 - a_3^2
\]
comes to $a_1^2 + a_2^2 + a_3^2 = 2m$, which does not hold for $m = 14$. Indeed, 28 cannot be written as a sum of three perfect squares.

For $n = 4$,
\[
  -m = a_1 a_2 + a_2 a_3 + a_3 a_4 + a_4 a_1 = (a_1 + a_3)(a_2 + a_4) = -(a_1 + a_3)^2
\]
can hold only if $m$ is a perfect square, i.e.\ it does not hold for all $m \geq 0$.
\end{solution}",232,669,Algebra,5
453,tst_jbmo_ro_2018_2_p2,tst_jbmo,2018,a,"Let $x, y, z$ be positive real numbers satisfying $2x^2 + 3y^2 + 6z^2 + 12(x + y + z) = 108$. Find the maximum value of $x^3 y^2 z$.","\begin{solution}
Guessing that the maximum is obtained when $x = 3$, $y = 2$, $z = 1$, we multiply the given equality by 6, and apply AM-GM:
\begin{align*}
  6 \cdot 108 &= 4x^2 + 4x^2 + 4x^2 + 9y^2 + 9y^2 + 36z^2\\
  &\quad + 12x + 12x + 12x + 12x + 12x + 12x + 18y + 18y + 18y + 18y + 36z + 36z\\
  &\geq 18\sqrt[18]{2^8 \cdot 3^4 \cdot 2^4 \cdot x^{12} \cdot y^8 \cdot z^4},
\end{align*}
from which it follows immediately that $108$ is the desired maximum and this value is obtained when all the numbers are equal, i.e.\ when $x = 3$, $y = 2$, $z = 1$.
\end{solution}",132,570,Algebra,6
454,tst_jbmo_ro_2018_2_p3,tst_jbmo,2018,g,Triangle $ABC$ has the property that there exists a unique point $X$ on the line segment $BC$ such that $AX^2 = BX \cdot CX$. Prove that $AB + AC = BC\sqrt{2}$.,"\begin{solution}
Let $T$ be the reflection of $A$ with respect to $X$. By the converse of the Power of a Point Theorem, it follows that the quadrilateral $ABTC$ is cyclic. If the line parallel to $BC$ through $T$ intersects the circumcircle of $ABC$ again at $U$, and lines $AU$ and $BC$ meet at $Y$, then $Y \in [BC]$ and $BY \cdot CY = AY \cdot UY = AY^2$, thus, in order for the point $X$ to be unique, it is necessary that $T = U$, i.e.\ $T$ is the midpoint of the arc $BC$, in other words $X$ is the foot of the angle bisector from $A$.

From here one can finish the proof in several ways. One can use the formula for the length of the angle bisector in a triangle, or Stewart's Theorem.

If $AX = \ell$, $BX = x$, $CX = y$, then $b^2 x + c^2 y = \ell^2 a + axy$. It is known that $x = \dfrac{ac}{b+c}$, $y = \dfrac{ab}{b+c}$. It follows that
\[
  \ell^2 = xy \iff b^2 x + c^2 y = 2axy \iff (b+c)^2 = 2a^2 \iff b + c = a\sqrt{2}.
\]

\medskip

oindent\textit{Comment:} Writing Stewart's Theorem and substituting $AX^2 = BX \cdot CX$ and $CX = a - BX$, leads to the following equation in the unknown $x = BX$:
\[
  f(x) := 2ax^2 - (c^2 + 2a^2 - b^2)x + c^2 a = 0.
\]
As $f(0) \cdot f(a) > 0$, this equation has a unique solution in $[0, a]$ if and only if $\Delta = 0$ (it is easy to check that this solution is indeed in $[0, a]$ when $\Delta = 0$). But
\[
  \Delta = (a\sqrt{2} + b + c)(a\sqrt{2} + b - c)(a\sqrt{2} - b + c)(a\sqrt{2} - b - c).
\]
The only factor that can be $0$ is $a\sqrt{2} - b - c$.
\end{solution}",160,1524,Geometry,7
455,tst_jbmo_ro_2018_2_p4,tst_jbmo,2018,c,"For $n \geq 2$, consider $n$ boxes aligned from left to right. In each box, one puts a ball that can be red, blue or white such that the following condition is fulfilled: each box is neighboring at least one box containing a ball of the same color. We denote by $I_n$ the number of such configurations.

\begin{enumerate}[label=\alph*)]
  \item Determine $I_{11}$. Justify your answer.
  \item Find, with proof, the general formula for $I_n$.
\end{enumerate}","\begin{solution}
Let $a_n$ be the number of configurations that have a red ball in the first box. Clearly, $I_n = 3a_n$. Obviously $a_2 = 1$, $a_3 = 1$. For $n \geq 4$, if the first box contains a red ball, then so does the second one. Now we distinguish two types of configurations: those containing a red ball in the third box, and those with a blue or a white ball in the third box. In order to count the configurations of the first type, simply forget about the first box. We can easily see that there are $a_{n-1}$ configurations of this type. For the second type of configurations, we omit the first two boxes and obtain $2a_{n-2}$ configurations. Thus, $a_n = a_{n-1} + 2a_{n-2}$.

By induction, it is easy to prove that $a_n = \dfrac{2^{n-1} + (-1)^n}{3}$, hence
\[
  I_n = 2^{n-1} + (-1)^n.
\]
For $n = 11$ we get $I_{11} = 1023$.
\end{solution}",458,854,Combinatorics,8
456,tst_jbmo_ro_2018_3_p1,tst_jbmo,2018,a,"Determine all triples of real numbers $(a, b, c)$ that satisfy simultaneously the equations:
\[
  \begin{cases}
    a(b^2 + c) = c(c + ab),\\
    b(c^2 + a) = a(a + bc),\\
    c(a^2 + b) = b(b + ca).
  \end{cases}
\]","\begin{solution}
Let $(a, b, c)$ be a solution of the system. If one of the numbers $a, b, c$ is $0$, e.g.\ if $c = 0$, then $c(a^2 + b) = a(b + ca)$ leads to $a = 0$, and similarly one gets $b = 0$. Thus $abc = 0$ leads to $a = b = c = 0$, which is indeed a solution. We now look for solutions with $abc 
eq 0$.

We rewrite the equations:
\[
  ab(b - c) = c(c - a), \quad bc(c - a) = a(a - b), \quad ca(a - b) = b(b - c).
\]
It follows that $a^2 b^2 c^2 (a-b)(b-c)(c-a) = abc(a-b)(b-c)(c-a)$.

\textbf{Case 1:} Among the numbers $a, b, c$ there exist (at least) two equal ones; say $a = b$. We have
\[
  a^2 c + bc = b^2 + abc \iff bc = b^2 \iff c = b,
\]
therefore in this case we obtain $a = b = c$. Conversely, any such triple is a solution to the system.

\textbf{Case 2:} If $a 
eq b$, $b 
eq c$, $c 
eq a$, then $abc = 1$. We obtain
\[
  ab(b - c) = c(c - a) \iff (b - c) = c^2(c - a)
\]
and two more equations, similar to this one. It follows that
\[
  a^3 + b^3 + c^3 = ac^2 + ba^2 + cb^2.
\]
If $a, b, c > 0$, from AM-GM it follows that $a^3 + a^3 + b^3 \geq 3ba^2$ (with equality if and only if $a = b$) which, added with two similar inequalities, leads to $a^3 + b^3 + c^3 \geq ac^2 + ba^2 + cb^2$. We thus have equality in the previous inequality, hence $a = b = c$.

If one of the variables is positive and the other two are negative, say $a > 0$, $b, c < 0$, then
\[
  b(c^2 + a) = a(a + bc) = a^2 + 1 > 0 \implies a < 0,
\]
contradiction.

In conclusion, the only solutions are $(a, b, c) = (x, x, x)$, $\forall\, x \in \mathbb{R}$.
\end{solution}",216,1563,Algebra,9
457,tst_jbmo_ro_2018_3_p2,tst_jbmo,2018,g,"In an acute triangle $ABC$ with $AB < BC$ let $BB'$ be an altitude, and let $O$ be the circumcenter. A line through $B'$ parallel to $CO$ meets $BO$ at $X$. Prove that $X$ and the midpoints of $AB$ and $AC$ are collinear.","\begin{solution}[Solution 1]
Let $M$ be the midpoint of the side $AB$. Then $\angle OBC = \angle OCB = 90^\circ - \angle A$ and $MX$ parallel to $BC$ comes to $\angle MXB = 90^\circ - \angle A$. But $\angle B'XB = \angle XOC = 2\angle OBC = 180^\circ - 2\angle A$. In the triangle $ABB'$ we have $MA = MB = MB'$ and $\angle BMB' = \angle MAB' + \angle MB'A = 2\angle A$. It follows that the quadrilateral $MBXB'$ is cyclic, hence $\angle MXB = \angle MB'B = \angle ABB' = \angle OBC$, which leads to the conclusion.
\end{solution}

\begin{solution}[Solution 2]
Let $K$ be the projection of $B$ onto the parallel through $A$ to the line $BC$. The quadrilateral $AKBB'$ is cyclic, hence $\angle AB'K = \angle ABK = 90^\circ - \angle B = \angle OCA = \angle CB'X$. It follows that points $K$, $B'$ and $X$ are collinear. Finally, the triangle $BXK$ is isosceles, with $BX = KX$, therefore $X$ lies on the perpendicular bisector of the line segment $BK$, i.e.\ $X$ is on the midline of triangle $ABC$ that is parallel to $BC$. We have $\angle XBK = 90^\circ - \angle OBC = \angle A = \angle XKB$ (from the cyclic quadrilateral $AKBB'$). The conclusion follows readily.
\end{solution}

\begin{solution}[Solution 3 (Alexandru Mihalcu)]
Let $S$ be the midpoint of the line segment $AC$ and $T$ be the intersection point of lines $BX$ and $AC$. As $OS$ is parallel to $BB'$, we have $\dfrac{B'S}{ST} = \dfrac{OB}{OT} = \dfrac{OC}{OT} = \dfrac{B'X}{XT}$. From the converse of the Angle Bisector Theorem it follows that $XS$ is the angle bisector of $\angle B'XT$. Then $\angle SXT = \dfrac{\angle B'XT}{2} = \dfrac{\angle XOC}{2} = \angle OBC$, and the conclusion follows.
\end{solution}",221,1678,Geometry,10
458,tst_jbmo_ro_2018_3_p3,tst_jbmo,2018,a,"Let
\[
  A = \left\{ a = q + \frac{1}{q} \;\middle|\; q \in \mathbb{Q}^*,\, q > 0 
ight\},
\]
\[
  A + A = \{a + b \mid a, b \in A\}, \quad A \cdot A = \{a \cdot b \mid a, b \in A\}.
\]
Prove that:
\begin{enumerate}[label=
oman*)]
  \item $A + A 
eq A \cdot A$;
  \item $(A + A) \cap \mathbb{N} = (A \cdot A) \cap \mathbb{N}$.
\end{enumerate}","\begin{solution}
\textbf{i)} Let $a = 1 + \frac{1}{1} = 2 \in A$, $b = 2 + \frac{1}{2} \in A$. Then
\[
  a + b = 2 + 2 + \frac{1}{2} = \frac{9}{2} \in A + A.
\]
We show that $\frac{9}{2} 
otin A \cdot A$. Assume that
\[
  \frac{9}{2} = \left(\frac{x}{y} + \frac{y}{x}
ight)\left(\frac{z}{t} + \frac{t}{z}
ight) = \frac{(x^2 + y^2)(z^2 + t^2)}{xyzt},
\]
with $(x, y) = (z, t) = 1$, $x, y \in \mathbb{N}^*$. It follows that $3 \mid x^2 + y^2$ or $3 \mid z^2 + t^2$. This means that $3 \mid x$ and $3 \mid y$, or $3 \mid z$ and $3 \mid t$. But then either $(x, y) 
eq 1$ or $(z, t) 
eq 1$, contradiction.

\textbf{ii)} First, we prove that $A \cdot A \subset A + A$. Indeed,
\[
  \left(q_1 + \frac{1}{q_1}
ight)\left(q_2 + \frac{1}{q_2}
ight) = \left(q_1 q_2 + \frac{1}{q_1 q_2}
ight) + \left(\frac{q_1}{q_2} + \frac{q_2}{q_1}
ight) \in A + A,
\]
hence $(A \cdot A) \cap \mathbb{N} \subset (A + A) \cap \mathbb{N}$.

Let us prove that $(A + A) \cap \mathbb{N} \subset (A \cdot A) \cap \mathbb{N}$.

If
\[
  n = \left(\frac{x}{y} + \frac{y}{x}
ight) + \left(\frac{z}{t} + \frac{t}{z}
ight) = \frac{(x^2 + y^2)zt + (z^2 + t^2)xy}{xyzt} \in \mathbb{N},
\]
with $x, y, z, t \in \mathbb{N}$, $(x, y) = (z, t) = 1$, then $(x^2 + y^2, xy) = 1$ and $(z^2 + t^2, zt) = 1$, hence $xy \mid zt$ and $zt \mid xy$. It follows that $xy = zt$.

Take $q_1 = \frac{z}{y}$, $q_2 = \frac{x}{z}$ and we have:
\[
  \left(q_1 + \frac{1}{q_1}
ight)\left(q_2 + \frac{1}{q_2}
ight) = \left(\frac{z}{y} + \frac{y}{z}
ight)\left(\frac{x}{z} + \frac{z}{x}
ight) = \frac{x}{y} + \frac{y}{x} + \frac{z^2}{xy} + \frac{xy}{z^2}
\]
\[
  = \frac{x}{y} + \frac{y}{x} + \frac{z^2}{zt} + \frac{zt}{z^2} = \left(\frac{x}{y} + \frac{y}{x}
ight) + \left(\frac{z}{t} + \frac{t}{z}
ight) = n \in A + A.
\]
\end{solution}",342,1774,Algebra,11
459,tst_jbmo_ro_2018_3_p4,tst_jbmo,2018,c,"Consider a $2018 \times 2018$ board. An ``LC-tile'' is a tile consisting of 9 unit squares, having the shape as in the figure below.

What is the maximum number of ``LC-tiles'' that can be placed on the board without superposing them? (Each of the 9 unit squares of the tile must cover one of the unit squares of the board; a tile may be rotated, turned upside down, etc.)","\begin{solution}
We label the rows and columns of the board from 1 to 2018. We color black the unit squares that have both coordinates multiples of 3.

There are $672^2$ black squares. Notice that each ``LC-tile'' covers exactly one black square, irrespective of its orientation. Therefore, one can place at most $672^2$ tiles.

On the other hand, the example below shows that it is indeed possible to place $672^2$ tiles on the board. Thus the required maximum is $\boxed{672^2}$.
\end{solution}",372,496,Combinatorics,12
460,tst_jbmo_ro_2018_4_p1,tst_jbmo,2018,n,Determine the prime numbers $p$ for which the number $a = 7^p - p - 16$ is a perfect square.,"\begin{solution}
$p = 2$ does not fulfill the requirement, but $p = 3$ does: $a = 7^3 - 3 - 16 = 324 = 18^2$. We show that there are no other solutions. Let $p \geq 5$ be a prime number. If $p \equiv 1 \pmod{4}$, then $a \equiv 2 \pmod{4}$, which shows that $a$ is not a perfect square. If $p > 3$ is a prime of the form $4k + 3$, then, from Fermat's Little Theorem it follows that $7^p \equiv 7 \pmod{p}$, hence $a \equiv -9 \pmod{p}$, i.e.\ $p \mid a + 9$. If $a$ is a perfect square, then $p$ divides $a + 9 = b^2 + 3^2$ would lead to $p$ divides $b$ and $p$ divides $3$, which is not possible. In conclusion, the only solution to the problem is $p = 3$.
\end{solution}",92,672,Number Theory,13
461,tst_jbmo_ro_2018_4_p2,tst_jbmo,2018,g,"Let $ABC$ be an acute triangle, with $AB 
eq AC$. Let $D$ be the midpoint of the line segment $BC$, and let $E$ and $F$ be the projections of $D$ onto the sides $AB$ and $AC$, respectively. If $M$ is the midpoint of the line segment $EF$, and $O$ is the circumcenter of triangle $ABC$, prove that the lines $DM$ and $AO$ are parallel.","\begin{solution}
Let $S$ be the intersection point of the lines $AO$ and $BC$. Let $T$ be the point in which the line $AD$ meets again the circumcircle of triangle $ABC$. The quadrilaterals $ABTC$ and $AEDF$ are cyclic, hence $\angle TBD = \angle TAC = \angle DEF$ and $\angle TCD = \angle TAB = \angle DFE$. It follows that triangles $DEF$ and $TBC$ are similar. Then we have
\[
  \frac{TB}{DE} = \frac{BC}{EF} = \frac{BC/2}{EF/2} = \frac{BD}{EM}.
\]
This shows that the triangles $TBD$ and $DEM$ are also similar, hence $\angle EDM = \angle BTD = \angle ACB$, which leads to
\[
  \angle BDM = \angle BDE + \angle EDM = 90^\circ - \angle ABC + \angle ACB.
\]
From $\angle OAC = 90^\circ - \angle ABC$, it follows that
\[
  \angle ASB = \angle SAC + \angle ACB = 90^\circ - \angle ABC + \angle ACB = \angle MDB,
\]
which means that $AS$ is parallel to $MD$.
\end{solution}",334,872,Geometry,14
462,tst_jbmo_ro_2018_4_p3,tst_jbmo,2018,g,"Let $ABCD$ be a cyclic quadrilateral. The line parallel to $BD$ passing through $A$ meets the line parallel to $AC$ passing through $B$ at $E$. The circumcircle of triangle $ABE$ meets the lines $EC$ and $ED$, again, at $F$ and $G$, respectively. Prove that the lines $AB$, $CD$ and $FG$ are either parallel or concurrent.","\begin{solution}
Angles $\angle ACB$ and $\angle ADB$ are equal, therefore points $C$ and $D$ are either both in the interior of the circumcircle of triangle $ABE$, or both on this circle, or both outside this circle. Thus we distinguish three cases:
\begin{enumerate}
  \item $F \in (EC)$ and $G \in (ED)$,
  \item $F = C$, $G = D$ (in this case the statement of the problem is obvious),
  \item $C \in (EF)$, $D \in (EG)$.
\end{enumerate}
We only treat the first case, the last one being similar. We have
\[
  \angle GDC = \angle GDB + \angle BDC = \angle GEA + \angle BAC = \frac{\overset{\frown}{AG}}{2} + \angle ABE = \frac{\overset{\frown}{AG} + \overset{\frown}{AE}}{2} = \angle GFE,
\]
which shows that the quadrilateral $FGDC$ is cyclic. We notice that line $AB$ is the radical axis of the circumcircles of triangle $ABE$ and quadrilateral $ABCD$, line $FG$ is the radical axis of the circumcircles of triangle $ABE$ and quadrilateral $FGDC$, and line $CD$ is the radical axis of the circumcircles of the quadrilaterals $ABCD$ and $FGDC$. It is well known that the radical axes of three circles are either parallel (if the three centers are collinear) or concurrent (in the radical center of the three circles).
\end{solution}",322,1235,Geometry,15
463,tst_jbmo_ro_2018_4_p4,tst_jbmo,2018,c,"Consider $n$ weights, $n \geq 2$, of masses $m_1, m_2, \ldots, m_n$, where $m_k$ are positive integers such that $1 \leq m_k \leq k$ for all $k \in \{1, 2, \ldots, n\}$. Prove that we can place the weights on the two pans of a balance such that the pans stay in equilibrium if and only if the number $m_1 + m_2 + \cdots + m_n$ is even.","\begin{solution}[Solution 1]
Obviously, if the two pans can stay in equilibrium, the sum of the masses must be even (namely twice the sum of the masses of the weights placed on one of the pans of the balance).

Conversely, we prove by ``strong'' induction after $n$ that, if the sum of the masses is even, then one can place the weights conveniently. Assume the statement to be true for any $k < n$ weights whose total mass is even. We distinguish two cases:
\begin{itemize}
  \item If $m_{n-1} = m_n$, then $m_1 + m_2 + \cdots + m_{n-2}$ is even, hence, according to the inductive hypothesis for $k = n-2$, we can place these weights on the two pans of the balance such that the balance stays in equilibrium. Now we simply add the two weights of mass $m_n$, one in each pan, and the balance remains in equilibrium.
  \item If $m_n 
eq m_{n-1}$, instead of the two weights we produce a new one of mass $m = |m_n - m_{n-1}|$. Then $1 \leq m \leq n-1$ and the sum of the masses of these $n-1$ weights,
  \[
    m_1 + m_2 + \cdots + m_{n-2} + m = m_1 + m_2 + \cdots + m_n - 2 \cdot \min\{m_{n-1}, m_n\},
  \]
  is even. From the inductive hypothesis for $k = n-1$ it follows that we can place these $n-1$ weights on the pans of the balance such that the balance stays in equilibrium. Now, instead of the weight of mass $|m_n - m_{n-1}|$ we put the weight $\max\{m_{n-1}, m_n\}$, and on the other pan we put the weight of mass $\min\{m_{n-1}, m_n\}$; the balance stays in equilibrium.
\end{itemize}
The statement is thus proven.
\end{solution}

\begin{solution}[Solution 2]
Obviously, if the two pans can stay in equilibrium, the sum of the masses must be even (namely twice the sum of the masses of the weights placed in one of the pans of the balance).

Conversely, assume that the total mass is even. We place the weight of mass $m_n$ on one of the pans. The modulus of the difference of the total weights of the two pans is $m_n \leq n$. We place successively the weights of masses $m_{n-1}, m_{n-2}, \ldots, m_1$ on the pan whose total mass is smaller. (In case the two pans are in equilibrium, we place the weight on any of the two pans.) After placing the weight of mass $m_k$, the difference between the total masses on the two pans is at most $k$. Proceeding this way, we ensure that in the end, after placing the weight of mass $m_1 = 1$, the modulus of the difference between the two pans is at most $1$. But the total mass being even, it follows that the difference is $0$, which is what we wanted.
\end{solution}",335,2521,Combinatorics,16
464,tst_jbmo_ro_2018_5_p1,tst_jbmo,2018,n,"Prove that a positive integer $A$ is a perfect square if and only if, for all positive integers $n$, at least one of the numbers
\[
  (A+1)^2 - A,\; (A+2)^2 - A,\; (A+3)^2 - A,\; \ldots,\; (A+n)^2 - A
\]
is a multiple of $n$.","\begin{solution}
If $A$ is a perfect square, i.e.\ there exists $B \in \mathbb{N}$ such that $A = B^2$, then
\[
  (A+k)^2 - A = (B^2 + k)^2 - B^2 = (B^2 + B + k)(B^2 - B + k)
\]
for all $k = 1, \ldots, n$, and exactly one of the (consecutive) numbers $B^2 + B + 1$, $B^2 + B + 2$, \ldots, $B^2 + B + n$ is a multiple of $n$.

Conversely, if $A$ is not a perfect square, then it has a prime factor that occurs in the prime factorization of $A$ at an odd exponent. Let $p$ be such a prime and $j \in \mathbb{N}$ such that $p^{2j-1} \mid A$, but $p^{2j} 
mid A$. We choose $n = p^{2j} \in \mathbb{N}$ and show that none of the numbers $(A+1)^2 - A$, $(A+2)^2 - A$, $(A+3)^2 - A$, \ldots, $(A+n)^2 - A$ is a multiple of $n$. Indeed, if $n \mid (A+m)^2 - A$, for some $m \in \{1, 2, \ldots, n\}$, i.e.\ $p^{2j} \mid A^2 + 2Am + m^2 - A$, from $p^{2j-1} \mid A$ it follows that $p^{2j-1} \mid m^2$, hence $p^j \mid m$. But then $p^{2j} \mid (A+m)^2$ and $p^{2j} \mid (A+m)^2 - A$, hence $p^{2j} \mid A$, which is a contradiction.
\end{solution}",225,1038,Number Theory,17
465,tst_jbmo_ro_2018_5_p2,tst_jbmo,2018,a,"If $a$, $b$, $c$ are positive real numbers, prove that
\[
  \frac{a}{\sqrt{(a+2b)^3}} + \frac{b}{\sqrt{(b+2c)^3}} + \frac{c}{\sqrt{(c+2a)^3}} \geq \frac{1}{\sqrt{a+b+c}}.
\]","\begin{solution}
From H\""{o}lder's inequality we have:
\[
  \left(\sum_{\mathrm{cyc}} \frac{a}{\sqrt{(a+2b)^3}}
ight) \cdot \left(\sum_{\mathrm{cyc}} a\sqrt{a+2b}
ight) \cdot \left(\sum_{\mathrm{cyc}} a\sqrt{a+2b}
ight) \cdot \left(\sum_{\mathrm{cyc}} a\sqrt{a+2b}
ight) \geq (a+b+c)^4,
\]
and from the Cauchy--Bunyakovsky--Schwarz inequality it follows that
\[
  \left(\sum_{\mathrm{cyc}} a\sqrt{a+2b}
ight)^2 = \left(\sum_{\mathrm{cyc}} \sqrt{a} \cdot \sqrt{a(a+2b)}
ight)^2 \leq (a+b+c) \cdot \sum_{\mathrm{cyc}} (a^2 + 2ab) = (a+b+c)^3,
\]
and we are done.

Equality holds if and only if $a = b = c$.

\medskip
Instead of H\""{o}lder's inequality one could have used the Cauchy--Bunyakovsky--Schwarz inequality twice:
\[
  \left(\sum_{\mathrm{cyc}} \frac{a}{\sqrt{(a+2b)^3}}
ight) \cdot \left(\sum_{\mathrm{cyc}} a\sqrt{a+2b}
ight) \cdot \left(\sum_{\mathrm{cyc}} a\sqrt{a+2b}
ight)^2
  \overset{\mathrm{CBS}}{\geq} \left(\sum_{\mathrm{cyc}} \frac{a}{a+2b}
ight)^2 \cdot \left(\sum_{\mathrm{cyc}} a\sqrt{a+2b}
ight)^2
  \overset{\mathrm{CBS}}{\geq} \left(\sum_{\mathrm{cyc}} a
ight)^4. \tag{1}
\]
On the other hand,
\[
  \left(\sum_{\mathrm{cyc}} a\sqrt{a+2b}
ight)^2 = \left(\sum_{\mathrm{cyc}} \sqrt{a} \cdot \sqrt{a(a+2b)}
ight)^2 \overset{\mathrm{CBS}}{\leq} (a+b+c)(a(a+2b)+b(b+2c)+c(c+2a)) = (a+b+c)^3,
\]
hence
\[
  \left(\sum_{\mathrm{cyc}} a\sqrt{a+2b}
ight)^3 \leq \sqrt{(a+b+c)^9}. \tag{2}
\]
From (1) and (2) the conclusion follows immediately. Equality holds when $a = b = c$.

\medskip

oindent\textit{Remark:} This inequality is a direct consequence of Jensen's inequality for the convex function $f(x) = \dfrac{1}{\sqrt{x^3}}$, the numbers $x_1 = a+2b$, $x_2 = b+2c$, $x_3 = c+2a$ and the weights $a_1 = a$, $a_2 = b$, $a_3 = c$.
\end{solution}",173,1763,Algebra,18
466,tst_jbmo_ro_2018_5_p3,tst_jbmo,2018,c,"Alina and Bogdan play the following game. They have a heap and 330 stones in it. They take turns. In one turn it is allowed to take from the heap exactly $1$, exactly $n$ or exactly $m$ stones. The player who takes the last stone wins. Before the beginning Alina says the number $n$, ($1 < n < 10$). After that Bogdan says the number $m$, ($m 
eq n$, $1 < m < 10$). Alina goes first. Which of the two players has a winning strategy? What if initially there are 2018 stones in the heap?","\begin{solution}
\textbf{For the heap initially containing 330 stones:}

Bogdan has a winning strategy. One such strategy is the following:
\begin{enumerate}
  \item If Alina chooses a number $n$ which is not a multiple of $3$, then Bogdan chooses $m = 2$ (or any other number that is not a multiple of $3$, in case Alina has chosen $n = 2$).
  \item If Alina chooses $3$ or $9$, Bogdan chooses $5$.
  \item If Alina chooses $6$, Bogdan chooses $4$.
\end{enumerate}

\textbf{Case 1.} The number of stones Alina leaves in the heap is not a multiple of $3$; Bogdan will take $1$ or $2$ stones, leaving a heap with a number of stones multiple of $3$. Thus, at the end, Bogdan is the one that leaves $0$ stones in the heap.

\textbf{Case 2.} If Alina chooses an odd number, in particular $3$ or $9$, Bogdan can choose $m$ odd and whatever Alina moves, he can always take one stone from the heap. After Alina's moves, the heap will always contain an odd number of stones, so she cannot win.

\textbf{Case 3.} If Alina chooses $6$, Bogdan chooses $4$ and he will move as follows: if Alina takes $1$ or $6$ stones, Bogdan takes $4$, and if Alina takes $4$, Bogdan takes $1$. Thus, after Bogdan's moves the number of stones in the heap will always be a multiple of $5$, while after Alina's moves it will not be a multiple of $5$. Bogdan wins again.

\medskip
\textbf{For the heap initially containing 2018 stones:}

Alina has a winning strategy. One such strategy is the following: she chooses $n = 2$.

\begin{itemize}
  \item If Bogdan chooses $m \in \{4, 5, 7, 8\}$, Alina moves such that the number of stones she leaves in the heap is always a multiple of $3$ (initially she takes $2$ stones).
  \item If Bogdan chooses $m = 3$, Alina moves such that the number of stones she leaves in the heap is always a multiple of $4$ (initially she takes $2$ stones). (If Bogdan takes $1$, $2$ or $3$ stones, Alina takes $3$, $2$, and $1$ stone(s), respectively, leaving a number of stones that is a multiple of $4$.)
  \item If Bogdan chooses $m = 6$, Alina moves such that the number of stones she leaves in the heap gives one of the remainders $0$ or $3$ when divided by $7$ (initially she takes $2$ stones). Later, if Bogdan finds $7k$ stones in the heap and takes $1$, $2$ or $6$ stones, then Alina takes $6$, $2$, or $1$ stones, respectively, while if Bogdan finds $7k+3$ stones in the heap and takes $1$, $2$ or $6$ stones, Alina takes $2$, $1$, or $1$ stones, respectively.
  \item If Bogdan chooses $m = 9$, Alina moves such that the number of stones she leaves in the heap gives one of the remainders $0$, $3$ or $6$ when divided by $10$ (initially she takes $2$ stones). Later, if Bogdan finds $10k$ stones in the heap and takes $1$, $2$ or $9$ stones, then Alina takes $9$, $2$, or $1$ stones, respectively, leaving $10(k-1)$ or $10(k-1)+6$ stones. If Bogdan finds $10k+3$ stones in the heap and takes $1$, $2$ or $9$ stones, Alina takes $2$, $1$, or $1$ stones, respectively, leaving $10k$ or $10(k-1)+3$ stones (the last situation is possible only if $k 
eq 0$). If Bogdan finds $10k+6$ stones in the heap and takes $1$, $2$ or $9$ stones, Alina takes $2$, $1$, or $1$ stones, respectively, leaving $10k+3$ or $10(k-1)+6$ stones (the last situation is possible only if $k 
eq 0$).
\end{itemize}

Thus, irrespective of Bogdan's choice of $m$, Alina wins by choosing $n = 2$ and taking initially $2$ stones from the heap.
\end{solution}",485,3433,Combinatorics,19
467,tst_jbmo_ro_2018_5_p4,tst_jbmo,2018,g,"Let $ABC$ be a triangle, and let $E$ and $F$ be two arbitrary points on the sides $AB$ and $AC$, respectively. The circumcircle of triangle $AEF$ meets the circumcircle of triangle $ABC$ again at point $M$. Let $D$ be the reflection of point $M$ across the line $EF$ and let $O$ be the circumcenter of triangle $ABC$. Prove that $D$ is on $BC$ if and only if $O$ belongs to the circumcircle of triangle $AEF$.","\begin{solution}[Solution 1]
Assume $M$ lies on the small arc $AB$.

Then $\angle AEF = \angle AMF < \angle AMC = \angle ABC$, therefore the line $EF$ does meet the line $BC$ at a point $G$ such that $B$ is between $C$ and $G$. It is known that the circumcircles of triangles $ABC$, $AEF$, $EBG$ and $FCG$ have a common point, the Miquel point of the complete quadrilateral $BCFEAG$. The circumcircles of triangles $ABC$ and $AEF$ meet (again) at $M$, hence the quadrilaterals $MGBE$ and $MFCG$ are cyclic. It follows that:
\[
  D \in BC \iff \angle MGE = \angle CGE \iff \angle AEM = 2\angle ABM = \angle AOM \iff O \text{ belongs to the circumcircle of triangle } AEF.
\]

The fact that the quadrilateral $MGBE$ is cyclic can be proven easily:
\[
  \angle EGB = \angle EBC - \angle BEG = \angle ABC - \angle AEF = \angle AMC - \angle AMF = \angle FMC,
\]
hence $MFCG$ is cyclic.

Then, $\angle BMC = \angle BAC = \angle EAF = \angle EMF$, which leads to $\angle EGB = \angle BME$ and $\angle BME = \angle BGE$, which shows that the quadrilateral $GBEM$ is indeed cyclic.
\end{solution}

\begin{solution}[Solution 2 (given in the contest by Ioana Popescu)]
\textbf{($\Leftarrow$)} If $O$ lies on the circumcircle of $AEF$, then $\angle AEM = \angle AOM = 2 \cdot \angle ABM$, hence $\angle MEB = 180^\circ - 2 \cdot \angle ABM$, which shows that the triangle $MEB$ is isosceles. Similarly, triangle $MFC$ is also isosceles. Moreover, the two triangles are similar.

Notice that we have a spiral similarity centered at $M$. Consider the point $T$ such that triangles $MTD$ and $MEB$ are similar. Then $T$ belongs to the line $EF$ (because $EF$ is the perpendicular bisector of the line segment $DM$), so it follows that $D \in BC$. (For a spiral similarity, if one of the points glides on a line, like in this case $T$ glides on $EF$, then its image glides on the line ``similar'' to $EF$. In our case, $E \mapsto B$, $F \mapsto C$, $T \mapsto D$ and $T \in EF$, therefore $D \in BC$.)

\textbf{($\Rightarrow$)} Assume that $D$ is on $BC$. Triangles $MEB$ and $MFC$ remain similar (AA) (but we no longer know that they are isosceles) and, again, a spiral similarity centered at $M$ takes $E \mapsto B$, $F \mapsto C$. There exists a point $T \in EF$ that is taken by the similarity into $D$. Triangles $MTD$ and $MEB$ are similar, but $EF$ is the perpendicular bisector of the line segment $MD$, hence triangle $MTD$ is isosceles. It follows that triangle $MEB$ is also isosceles, hence $\angle EMB = \angle EBM$, which leads to $\angle AEM = 2 \cdot \angle ABM$ and $\angle AEM = \angle AOM$. This shows that $O$ is on the circumcircle of triangle $AEF$.

\medskip

oindent\textit{Remark:} The result remains valid in the case the circumcircles of triangles $AEF$ and $ABC$ are tangent, in which case we consider $M = A$. Indeed, the homothety centered at $A$ transforming the circumcircle of triangle $AEF$ into the circumcircle of triangle $ABC$ takes the line segment $EF$ into a parallel line segment, $BC$. Then $D \in BC$ if and only if $EF$ is a midline, which is equivalent to $AEOF$ being cyclic.
\end{solution}",409,3121,Geometry,20
468,tst_jbmo_ro_2018_6_p1,tst_jbmo,2018,n,"Let $p > 5$ be a prime number and $S = \{p - n^2 \mid n \in \mathbb{N},\, n^2 < p\}$. Prove that $S$ contains two elements $a$ and $b$ such that $1 < a < b$ and $a$ divides $b$.","\begin{solution}
We show that the smallest element of $S$ that is greater than $1$ divides a larger element of $S$. If $p$ is of the form $m^2 + 1$, with $m \in \mathbb{N}$, we show that $p - (m-1)^2 = 2m$ divides $p - 0^2 = m^2$. Indeed, as $m$ is even, it follows that $2m \mid m^2$.

If $p$ cannot be written as $m^2 + 1$, $m \in \mathbb{N}$, $p$ cannot be written as $m^2 + 2m$ either (this is a composite number because $m > 1$), hence $m^2 + 1 < p < m^2 + 2m$ for some $m \in \mathbb{N}$ ($m \geq 2$). We show that $p - m^2$, which is an element of $S$ that is larger than $1$, divides another element of $S$. The condition that $p - m^2$ divides one of the numbers $p - n^2$ with $n \in \{0, 1, 2, \ldots, m-1\}$ is equivalent to $p - m^2$ dividing one of the numbers $m^2$, $m^2 - 1^2$, $m^2 - 2^2$, \ldots, $m^2 - (m-1)^2$. Being smaller than $2m$, $p - m^2$ divides one of the following $2m - 1$ consecutive numbers: $1, 2, \ldots, m-1, m, m+1, \ldots, 2m-1$, hence it divides one of the differences $m^2 - 0^2$, $m^2 - 1^2$, \ldots, $m^2 - (m-1)^2$. Moreover, it does not divide $m^2$, because it would divide $p$.
\end{solution}",177,1140,Number Theory,21
469,tst_jbmo_ro_2018_6_p2,tst_jbmo,2018,a,"Let $k > 2$ be a real number.

\begin{enumerate}[label=\alph*)]
  \item Prove that for all positive real numbers $x$, $y$ and $z$ the following inequality holds:
  \[
    \sqrt{x+y} + \sqrt{y+z} + \sqrt{z+x} > 2\sqrt{\frac{(x+y)(y+z)(z+x)}{xy+yz+zx}}.
  \]
  \item Prove that there exist positive real numbers $x$, $y$ and $z$ such that
  \[
    \sqrt{x+y} + \sqrt{y+z} + \sqrt{z+x} < k\sqrt{\frac{(x+y)(y+z)(z+x)}{xy+yz+zx}}.
  \]
\end{enumerate}","\begin{solution}
\textbf{a)} We have
\[
  \sqrt{x+y} + \sqrt{y+z} + \sqrt{z+x} > 2\sqrt{\frac{(x+y)(y+z)(z+x)}{xy+yz+zx}}
\]
\[
  \iff x + y + z + \sqrt{x^2+xy+yz+zx} + \sqrt{y^2+xy+yz+zx} + \sqrt{z^2+xy+yz+zx}
  > 2 \cdot \frac{(x+y)(y+z)(z+x)}{xy+yz+zx}.
\]
But $\sqrt{x^2 + xy + yz + zx} > x$, $\sqrt{y^2 + xy + yz + zx} > y$, and $\sqrt{z^2 + xy + yz + zx} > z$, hence
\[
  x + y + z + \sqrt{x^2+xy+yz+zx} + \sqrt{y^2+xy+yz+zx} + \sqrt{z^2+xy+yz+zx} > 2(x+y+z).
\]
It is sufficient to show that
\[
  x + y + z \geq \frac{(x+y)(y+z)(z+x)}{xy+yz+zx},
\]
i.e.\
\[
  (x+y+z)(xy+yz+zx) \geq (x+y)(y+z)(z+x).
\]
After some computations, the previous inequality comes to $xyz \geq 0$, which is obviously true.

\textbf{b)} Fix $z = 1$. We look for $x, y > 0$, with $y = x$, such that
\[
  \sqrt{x+y} + \sqrt{y+z} + \sqrt{z+x} < k\sqrt{\frac{(x+y)(y+z)(z+x)}{xy+yz+zx}},
\]
i.e.\ $2\sqrt{x} + 1 + \sqrt{2x} < k\sqrt{\dfrac{2(x+1)^2}{x+2}}$, or $\sqrt{\dfrac{x+2}{2(x+1)}}\left(2 + \sqrt{\dfrac{2x}{x+1}}
ight) < k$.

As $\sqrt{\dfrac{x+2}{2(x+1)}} < 1$, it is sufficient to find $x$ such that $2 + \sqrt{\dfrac{2x}{x+1}} < k$, i.e.\ $\sqrt{\dfrac{2x}{x+1}} < k - 2$, or, equivalently, $\dfrac{2x}{x+1} < (k-2)^2$. Putting $t = (k-2)^2 > 0$, the previous condition is fulfilled by any $x > 0$ if $t \geq 2$, while in the case that $t < 2$, it reduces to $x < \dfrac{t}{2-t}$. Since $\dfrac{t}{2-t} > 0$, there exist positive numbers $x, y, z$ that fulfill the conditions of the statement.
\end{solution}",447,1498,Algebra,22
470,tst_jbmo_ro_2018_6_p3,tst_jbmo,2018,g,"Let $ABC$ be an acute triangle with $AB < AC$, let $G$ be its centroid and $D$ the foot of the altitude from $A$. The line $DG$ meets the small arc $BC$ of the circumcircle of triangle $ABC$ at point $E$. Prove that the line $AB$ is tangent to the circumcircle of triangle $BDE$.","\begin{solution}
Let $F$ be the point in which the parallel through $A$ to $BC$ intersects again the circumcircle of triangle $ABC$. We prove that the points $D$, $G$ and $F$ are collinear. $AFCB$ is a cyclic trapezoid, hence a cyclic one. If $T$ is the orthogonal projection of point $F$ onto $BC$, then $AFTD$ is a rectangle. It is easy to prove that triangles $ABD$ and $FCT$ are equal; it follows that $BD = CT$, i.e.\ $M$ is the midpoint of the line segment $DT$. It follows that
\[
  \frac{DM}{FA} = \frac{DM}{DT} = \frac{1}{2} = \frac{GM}{GA}
\]
and, since $\angle GMD = \angle GAF$, triangles $GMD$ and $GAF$ are similar. We deduce that $\angle DGM = \angle FGA$, i.e.\ points $D$, $G$, $F$ are collinear.

Then $\angle BED = \angle BEF = \angle BCF = \angle ABC = \angle ABD$, which shows that the line $AB$ is tangent to the circumcircle of triangle $BDE$.
\end{solution}",279,881,Geometry,23
471,tst_jbmo_ro_2018_6_p4,tst_jbmo,2018,c,"In $n$ transparent boxes there are red balls and blue balls. One needs to choose 50 boxes such that, together, they contain at least half of the red balls and at least half of the blue balls. Is such a choice possible irrespective of the number of balls and of the way they are distributed in the boxes, if:

\begin{enumerate}[label=\alph*)]
  \item $n = 100$;
  \item $n = 99$?
\end{enumerate}","\begin{solution}
\textbf{a)} The answer is in the negative. If we have 100 boxes, and 25 of the boxes only contain one red ball each, while the other 75 boxes contain just one blue ball each, then one needs to choose at least 13 of the boxes containing red balls and at least 38 of the boxes containing blue balls, which means one would need to choose at least 51 boxes. Thus, such a choice is not always possible.

\textbf{b)} We prove by induction that from $2n+1$ boxes one can always choose $n+1$ that contain at least half of the balls of both colors.

\textit{Base case} $n = 1$: if choosing two boxes would not be possible, then the remaining box must contain the majority of the red balls, or the majority of the blue balls. But there is only one such choice for each color, and there are 3 ways of choosing the boxes, so at least one of the choices must work.

\textit{Inductive step.} We say two boxes are comparable if one of them has at least as many red balls as the other one and at least as many blue balls as the other one. If two comparable boxes exist, we eliminate them and apply the inductive hypothesis to the remaining $2n - 1$ boxes. Then, to the $n$ boxes thus chosen, we add that box of the two comparable ones that contains more red and more blue balls. If no two boxes are comparable, we denote by $a_i$ and $b_i$ the number of red balls and blue balls in box no.\ $i$, respectively. Changing the order of the boxes, one may assume that $a_1 > a_2 > \cdots > a_{99}$ and $b_1 < b_2 < \cdots < b_{99}$. In this case, simply choose the boxes whose ranks are odd (boxes $1, 3, 5, \ldots, 99$).

\medskip

oindent\textit{An easier solution to point b) (given in the contest by Dinu Iosifescu).}

Denoting by $r_i$ the number of red balls in box $i$, we may assume, without loss of generality, that $r_1 \geq r_2 \geq \cdots \geq r_{99}$. Then we can choose box no.\ 1, then, from box no.\ 2 and box no.\ 3 we choose the one containing more blue balls, from box no.\ 4 and box no.\ 5 we choose the one containing more blue balls, and so on from box no.\ 98 and box no.\ 99 we choose the one containing more blue balls. It is clear that this choice satisfies the conditions in the statement.
\end{solution}",394,2227,Combinatorics,24
251,jbmo_2019_p1,jbmo,2019,n,"Find all prime numbers $p$ for which there exist positive integers $x$, $y$ and $z$ such that the number
\[
x^p + y^p + z^p - x - y - z
\]
is a product of exactly three distinct prime numbers.","\textbf{Solution.} Let $A = x^p + y^p + z^p - x - y - z$. For $p = 2$, we take $x = y = 4$ and $z = 3$. Then $A = 30 = 2 \cdot 3 \cdot 5$. For $p = 3$ we can take $x = 3$ and $y = 2$ and $z = 1$. Then again $A = 30 = 2 \cdot 3 \cdot 5$. For $p = 5$ we can take $x = 2$ and $y = 1$ and $z = 1$. Again $A = 30 = 2 \cdot 3 \cdot 5$.

Assume now that $p \geqslant 7$. Working modulo $2$ and modulo $3$ we see that $A$ is divisible by both $2$ and $3$. Moreover, by Fermat's Little Theorem, we have
\[
x^p + y^p + z^p - x - y - z \equiv x + y + z - x - y - z = 0 \pmod{p}.
\]
Therefore, by the given condition, we have to solve the equation
\[
x^p + y^p + z^p - x - y - z = 6p.
\]
If one of the numbers $x$, $y$ and $z$ is bigger than or equal to $2$, let's say $x \geqslant 2$, then
\[
6p \geqslant x^p - x = x(x^{p-1} - 1) \geqslant 2(2^{p-1} - 1) = 2^p - 2.
\]
It is easy to check by induction that $2^n - 2 > 6n$ for all natural numbers $n \geqslant 6$. This contradiction shows that there are no more values of $p$ which satisfy the required property.

\begin{remark}
There are a couple of other ways to prove that $2^p - 2 > 6p$ for $p \geqslant 7$. For example, we can use the Binomial Theorem as follows:
\[
2^p - 2 \geqslant 1 + p + \frac{p(p-1)}{2} + \frac{p(p-1)(p-2)}{6} - 2 \geqslant 1 + p + 3p + 5p - 2 > 6p.
\]
We can also use Bernoulli's Inequality as follows:
\[
2^p - 2 = 8(1+1)^{p-3} - 2 \geqslant 8(1 + (p-3)) - 2 = 8p - 18 > 6p
\]
The last inequality is true for $p \geqslant 11$. For $p = 7$ we can see directly that $2^p - 2 > 6p$.

One can also use calculus to show that $f(x) = 2^x - 6x$ is increasing for $x \geqslant 5$.
\end{remark}",192,1655,Number Theory,1
252,jbmo_2019_p2,jbmo,2019,a,"Let $a$, $b$ be two distinct real numbers and let $c$ be a positive real number such that
\[
a^4 - 2019a = b^4 - 2019b = c.
\]
Prove that $-\sqrt{c} < ab < 0$.","\textbf{Solution.} Firstly, we see that
\[
2019(a - b) = a^4 - b^4 = (a-b)(a+b)(a^2+b^2).
\]
Since $a 
eq b$, we get $(a+b)(a^2+b^2) = 2019$, so $a + b 
eq 0$. Thus
\begin{align*}
2c &= a^4 - 2019a + b^4 - 2019b \\
   &= a^4 + b^4 - 2019(a+b) \\
   &= a^4 + b^4 - (a+b)^2(a^2+b^2) \\
   &= -2ab(a^2 + ab + b^2).
\end{align*}
Hence $ab(a^2 + ab + b^2) = -c < 0$. Note that
\[
a^2 + ab + b^2 = \frac{1}{2}\!\left(a^2 + b^2 + (a+b)^2
ight) > 0,
\]
thus $ab < 0$. Finally, $a^2 + ab + b^2 = (a+b)^2 - ab > -ab$ (the equality does not occur since $a + b 
eq 0$). So
\[
-c = ab(a^2 + ab + b^2) < -(ab)^2 \implies (ab)^2 < c \implies -\sqrt{c} < ab < \sqrt{c}.
\]
Therefore, we have $-\sqrt{c} < ab < 0$.

\begin{remark}
We can get $c = -ab(a^2 + ab + b^2)$ in several other ways. For example using that,
\[
(a-b)c = a(b^4 - 2019b) - b(a^4 - 2019a) = ab(b^3 - a^3) = ab(b-a)(a^2+ab+b^2).
\]
We can also divide $f(x) = x^4 - 2019x - c$ by $(x-a)(x-b)$ and look at the constant term of the remainder.
\end{remark}

\textbf{Alternative Solution.} By Descartes' Rule of Signs, the polynomial $p(x) = x^4 - 2019x - c$ has exactly one positive root and exactly one negative root. So $a$, $b$ must be its two real roots. Since one of them is positive and the other is negative, then $ab < 0$. Let $r \pm is$ be the two non-real roots of $p(x)$.

By Vieta, we have
\begin{align}
ab(r^2 + s^2) &= -c, \label{eq1} \\
a + b + 2r &= 0, \label{eq2} \\
ab + 2ar + 2br + r^2 + s^2 &= 0. \label{eq3}
\end{align}
Using \eqref{eq2} and \eqref{eq3}, we have
\begin{equation}
r^2 + s^2 = -2r(a+b) - ab = (a+b)^2 - ab \geqslant -ab. \label{eq4}
\end{equation}
If in the last inequality we actually have an equality, then $a + b = 0$. Then \eqref{eq2} gives $r = 0$ and \eqref{eq3} gives $s^2 = -ab$. Thus the roots of $p(x)$ are $a,\, -a,\, ia,\, -ia$. This would give that $p(x) = x^4 + a^4$, a contradiction.

So the inequality in \eqref{eq4} is strict and now from \eqref{eq1} we get
\[
c = -(r^2 + s^2)ab > (ab)^2,
\]
which gives that $ab > -\sqrt{c}$.

\begin{remark}
One can get that $x^4 - 2019x - c$ has only two real roots by showing (e.g.\ by calculus) that it is initially decreasing and then increasing.

Also, instead of Vieta one can also proceed by factorising the polynomial as:
\[
x^4 - 2019x - c = \left(x^2 - (a+b)x + ab
ight)\!\left(x^2 + (a+b)x - \frac{c}{ab}
ight).
\]
Since the second quadratic has no real roots, its discriminant is negative which gives that $c > (ab)^2$.
\end{remark}",159,2481,Algebra,2
253,jbmo_2019_p3,jbmo,2019,g,"Triangle $ABC$ is such that $AB < AC$. The perpendicular bisector of side $BC$ intersects lines $AB$ and $AC$ at points $P$ and $Q$, respectively. Let $H$ be the orthocentre of triangle $ABC$, and let $M$ and $N$ be the midpoints of segments $BC$ and $PQ$, respectively. Prove that lines $HM$ and $AN$ meet on the circumcircle of $ABC$.","\textbf{Solution.} We have
\[
\angle APQ = \angle BPM = 90^\circ - \angle MBP = 90^\circ - \angle CBA = \angle HCB,
\]
and
\[
\angle AQP = \angle MQC = 90^\circ - \angle QCM = 90^\circ - \angle ACB = \angle CBH.
\]
From these two equalities, we see that the triangles $APQ$ and $HCB$ are similar. Moreover, since $M$ and $N$ are the midpoints of the segments $BC$ and $PQ$ respectively, then the triangles $AQN$ and $HBM$ are also similar. Therefore, we have $\angle ANQ = \angle HMB$.

Let $L$ be the intersection of $AN$ and $HM$. We have
\[
\angle MLN = 180^\circ - \angle LNM - \angle NML = 180^\circ - \angle LMB - \angle NML = 180^\circ - \angle NMB = 90^\circ.
\]
Now let $D$ be the point on the circumcircle of $ABC$ diametrically opposite to $A$. It is known that $D$ is also the reflection of point $H$ over the point $M$. Therefore, we have that $D$ belongs on $MH$ and that $\angle DLA = \angle MLA = \angle MLN = 90^\circ$. But, as $DA$ is the diameter of the circumcircle of $ABC$, the condition that $\angle DLA = 90^\circ$ is enough to conclude that $L$ belongs on the circumcircle of $ABC$.

\begin{remark}
There is a spiral similarity mapping $AQP$ to $HBC$. Since the similarity maps $AN$ to $HM$, it also maps $AH$ to $NM$, and since these two lines are parallel, the centre of the similarity is $L = AN \cap HM$. Since the similarity maps $BC$ to $QP$, its centre belongs on the circumcircle of $BCX$, where $X = BQ \cap PC$. But $X$ is the reflection of $A$ on $QM$ and so it must belong on the circumcircle of $ABC$. Hence so must $L$.
\end{remark}

\begin{remark}
Students have also submitted correct proofs using radical axes, harmonic quadruples, coordinate geometry and complex numbers.
\end{remark}",336,1726,Geometry,3
254,jbmo_2019_p4,jbmo,2019,c,"A $5 \times 100$ table is divided into $500$ unit square cells, where $n$ of them are coloured black and the rest are coloured white. Two unit square cells are called \emph{adjacent} if they share a common side. Each of the unit square cells has at most two adjacent black unit square cells. Find the largest possible value of $n$.","\textbf{Solution.} If we colour all the cells along all edges of the board together with the entire middle row except the second and the last-but-one cell, the condition is satisfied and there are $302$ black cells. The figure below exhibits this colouring for the $5 \times 8$ case.

We can cover the table by one fragment like the first one on the figure below, $24$ fragments like the middle one, and one fragment like the third one.

\[
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
d & e & d & e & j & k & j & k \\
\hline
c & d & e & f & g & j & k & m \\
\hline
c & f & g & f & g & m & & \\
\hline
c & a & b & f & g & h & i & m \\
\hline
a & b & a & b & h & i & h & i \\
\hline
\end{array}
\]

In each fragment, among the cells with the same letter, there are at most two coloured black, so the total number of coloured cells is at most $(5 + 24 \cdot 6 + 1) \cdot 2 + 2 = 302$.

\textbf{Alternative Solution.} Consider the cells adjacent to all cells of the second and fourth row. Counting multiplicity, each cell in the first and fifth row is counted once, each cell in the third row twice, while each cell in the second and fourth row is also counted twice apart from their first and last cells which are counted only once.

So there are $204$ cells counted once and $296$ cells counted twice. Those cells contain, counting multiplicity, at most $400$ black cells. Suppose $a$ of the cells have multiplicity one and $b$ of them have multiplicity $2$. Then $a + 2b \leqslant 400$ and $a \leqslant 204$. Thus
\[
2a + 2b \leqslant 400 + a \leqslant 604,
\]
and so $a + b \leqslant 302$ as required.

\begin{remark}
The alternative solution shows that if we have equality, then all cells in the perimeter of the table except perhaps the two cells of the third row must be coloured black. No other cell in the second or fourth row can be coloured black as this will give a cell in the first or fifth row with at least three neighbouring black cells. For similar reasons we cannot colour black the second and last-but-one cell of the third row. So we must colour black all other cells of the third row and therefore the colouring is unique.
\end{remark}

\textbf{Alternative Solution.} Suppose we have $a$ black corner cells, $b$ black side cells, and $c$ black interior cells. Let $N$ be the number of pairs of cells $(c_1, c_2)$ such that $c_1$ is black and $c_2$ is a neighbour of $c_1$. Then $N = 2a + 3b + 4c$ and $N \leqslant 1000$. Since also $a \leqslant 4$ and $b \leqslant 202$ we get
\[
1000 \geqslant 4(a + b + c) - 2a - b \geqslant 4(a + b + c) - 210,
\]
giving $a + b + c \leqslant 302.5$.",331,2599,Combinatorics,4
70,shl_jbmo_2019_a1,shl_jbmo,2019,a,Real numbers $a$ and $b$ satisfy $a^3 + b^3 - 6ab = -11$. Prove that $-\dfrac{7}{3} < a + b < -2$.,"\textbf{Solution.} Using the identity
\[
x^3 + y^3 + z^3 - 3xyz = \frac{1}{2}(x+y+z)\left((x-y)^2 + (y-z)^2 + (z-x)^2
ight),
\]
we get
\[
-3 = a^3 + b^3 + 2^3 - 6ab = \frac{1}{2}(a+b+2)\left((a-b)^2 + (a-2)^2 + (b-2)^2
ight).
\]
Since $S = (a-b)^2 + (a-2)^2 + (b-2)^2$ must be positive, we conclude that $a+b+2 < 0$, i.e.\ that $a+b < -2$. Now $S$ can be bounded by
\[
S \geqslant (a-2)^2 + (b-2)^2 = a^2 + b^2 - 4(a+b) + 8 \geqslant \frac{(a+b)^2}{2} - 4(a+b) + 8 > 18.
\]
Here, we have used the fact that $a+b < -2$, which we have proved earlier. Since $a+b+2$ is negative, it immediately implies that $a+b+2 < \dfrac{-2 \cdot 3}{18} = -\dfrac{1}{3}$, i.e.\ $a+b < -\dfrac{7}{3}$ which we wanted.

\medskip

\textbf{Alternative Solution by PSC.} Writing $s = a+b$ and $p = ab$ we have
\[
a^3 + b^3 - 6ab = (a+b)(a^2 - ab + b^2) - 6ab = s(s^2 - 3p) - 6p = s^3 - 3ps - 6p.
\]
This gives $3p(s+2) = s^3 + 11$. Thus $s 
eq -2$ and using the fact that $s^2 \geqslant 4p$ we get
\[
p = \frac{s^3+11}{3(s+2)} \leqslant \frac{s^2}{4}. \tag{1}
\]
If $s > -2$, then (1) gives $s^3 - 6s^2 + 44 \leqslant 0$. This is impossible as
\[
s^3 - 6s^2 + 44 = (s+2)(s-4)^2 + 8 > 0.
\]
So $s < -2$. Then from (1) we get $s^3 - 6s^2 + 44 \geqslant 0$. If $s < -\dfrac{7}{3}$ this is again impossible as $s^3 - 6s^2 = s^2(s-6) < -\dfrac{49}{9} \cdot \dfrac{25}{3} < -44$. (Since $49 \cdot 25 = 1225 > 1188 = 44 \cdot 27$.) So $-\dfrac{7}{3} < s < -2$ as required.",98,1442,Algebra,1
71,shl_jbmo_2019_a2,shl_jbmo,2019,a,"Let $a, b, c$ be positive real numbers such that $abc = \dfrac{2}{3}$. Prove that
\[
\frac{ab}{a+b} + \frac{bc}{b+c} + \frac{ca}{c+a} \leqslant \frac{a+b+c}{a^3+b^3+c^3}.
\]","\textbf{Solution.} The given inequality is equivalent to
\[
(a^3+b^3+c^3)\left(\frac{ab}{a+b} + \frac{bc}{b+c} + \frac{ca}{c+a}
ight) \leqslant a+b+c. \tag{1}
\]
By the AM-GM Inequality it follows that
\[
a^3 + b^3 = \frac{a^3+a^3+b^3}{3} + \frac{b^3+b^3+a^3}{3} \geqslant a^2b + b^2a = ab(a+b).
\]
Similarly we have
\[
b^3 + c^3 \geqslant bc(b+c) \quad\text{and}\quad c^3 + a^3 \geqslant ca(c+a).
\]
Summing the three inequalities we get
\[
2(a^3+b^3+c^3) \geqslant ab(a+b) + bc(b+c) + ca(c+a). \tag{2}
\]
From the Cauchy-Schwarz Inequality we have
\[
\bigl(ab(a+b) + bc(b+c) + ca(c+a)\bigr)\left(\frac{ab}{a+b} + \frac{bc}{b+c} + \frac{ca}{c+a}
ight) \geqslant (ab+bc+ca)^2. \tag{3}
\]
We also have
\[
(ab+bc+ca)^2 \geqslant 3(ab \cdot bc + bc \cdot ca + ca \cdot ab) = 3abc(a+b+c) = 2(a+b+c). \tag{4}
\]
Combining together (2), (3) and (4) we obtain (1) which is the required inequality.

\medskip

\textbf{Alternative Solution by PSC.} By the Power Mean Inequality we have
\[
\frac{a^3+b^3+c^3}{3} \geqslant \left(\frac{a+b+c}{3}
ight)^3.
\]
So it is enough to prove that
\[
(a+b+c)^2\left(\frac{ab}{a+b} + \frac{bc}{b+c} + \frac{ca}{c+a}
ight) \geqslant 9,
\]
or equivalently, that
\[
(a+b+c)^2\left(\frac{1}{ac+bc} + \frac{1}{ba+ca} + \frac{1}{cb+ab}
ight) \geqslant \frac{27}{2}. \tag{5}
\]
Since $(a+b+c)^2 \geqslant 3(ab+bc+ca) = \dfrac{3}{2}\bigl((ac+bc)+(ba+ca)+(cb+ac)\bigr)$, then (5) follows by the Cauchy-Schwarz Inequality.

\medskip

\textbf{Alternative Solution by PSC.} We have
\[
(a^3+b^3+c^3)\frac{ab}{a+b} = ab(a^2-ab+b^2) + \frac{abc^3}{a+b} \geqslant a^2b^2 + \frac{2c^2}{3(a+b)}.
\]
So the required inequality follows from
\[
(a^2b^2+b^2c^2+c^2a^2) + \frac{2}{3}\left(\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b}
ight) \geqslant a+b+c. \tag{6}
\]
By applying the AM-GM Inequality three times we get
\[
a^2b^2 + b^2c^2 + c^2a^2 \geqslant abc(a+b+c) = \frac{2}{3}(a+b+c). \tag{7}
\]
By the Cauchy-Schwarz Inequality we also have
\[
\bigl((b+c)+(c+a)+(a+b)\bigr)\left(\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b}
ight) \geqslant (a+b+c)^2,
\]
which gives
\[
\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} \geqslant \frac{a+b+c}{2}. \tag{8}
\]
Combining (7) and (8) we get (6) as required.",173,2229,Algebra,2
72,shl_jbmo_2019_a3,shl_jbmo,2019,a,"Let $A$ and $B$ be two non-empty subsets of $X = \{1, 2, \ldots, 11\}$ with $A \cup B = X$. Let $P_A$ be the product of all elements of $A$ and let $P_B$ be the product of all elements of $B$. Find the minimum and maximum possible value of $P_A + P_B$ and find all possible equality cases.","\textbf{Solution.} For the maximum, we use the fact that $(P_A - 1)(P_B - 1) \geqslant 0$, to get that $P_A + P_B \leqslant P_A P_B + 1 = 11! + 1$. Equality holds if and only if $A = \{1\}$ or $B = \{1\}$.

For the minimum observe, first that $P_A \cdot P_B = 11! = c$. Without loss of generality let $P_A \leqslant P_B$. In this case $P_A \leqslant \sqrt{c}$. We write $P_A + P_B = P_A + \dfrac{c}{P_A}$ and consider the function $f(x) = x + \dfrac{c}{x}$ for $x \leqslant \sqrt{c}$. Since
\[
f(x) - f(y) = x - y + \frac{c(y-x)}{yx} = \frac{(x-y)(xy - c)}{xy},
\]
then $f$ is decreasing for $x \in (0, c]$.

Since $x$ is an integer and cannot be equal with $\sqrt{c}$, the minimum is attained at the closest integer to $\sqrt{c}$. We have $\lfloor\sqrt{11!}
floor = \lfloor\sqrt{2^8 \cdot 3^4 \cdot 5^2 \cdot 7 \cdot 11}
floor = \lfloor 720\sqrt{77}
floor = 6317$ and the closest integer which can be a product of elements of $X$ is $6300 = 2 \cdot 5 \cdot 7 \cdot 9 \cdot 10$. Therefore the minimum is $f(6300) = 6300 + 6336 = 12636$ and it is achieved for example for $A = \{2, 5, 7, 9, 10\}$, $B = \{1, 3, 4, 6, 8, 11\}$.

Suppose now that there are different sets $A$ and $B$ such that $P_A + P_B = 402$. Then the pairs of numbers $(6300, 6336)$ and $(P_A, P_B)$ have the same sum and the same product, thus the equality case is unique for the numbers 6300 and 6336. It remains to find all possible subsets $A$ with product $6300 = 2^2 \cdot 3^2 \cdot 5^2 \cdot 7$. It is immediate that $5, 7, 10 \in A$ and from here it is easy to see that all possibilities are $A = \{2, 5, 7, 9, 10\}$, $\{1, 2, 5, 7, 9, 10\}$, $\{3, 5, 6, 7, 10\}$ and $\{1, 3, 5, 6, 7, 10\}$.

\medskip

\textbf{Alternative Solution by PSC.} We have $P_A + P_B \geqslant 2\sqrt{P_A P_B} = 2\sqrt{11!} = 1440\sqrt{77}$. Since $P_A + P_B$ is an integer, we have $P_A + P_B \geqslant \lceil 1440\sqrt{77}
ceil = 12636$. One can then follow the approach of the first solution to find all equality cases.

\medskip

\textbf{Remark by PSC.} We can increase the difficulty of the alternative solution by taking $X = \{1, 2, \ldots, 9\}$. Following the first solution we have $\lfloor\sqrt{9!}
floor = \lfloor 72\sqrt{70}
floor = 602$ and the closest integer which can be a product of elements of $X$ is $2 \cdot 4 \cdot 8 \cdot 9 = 576$. The minimum is $f(576) = 576 + 630 = 1206$ achieved by $A = \{1, 2, 4, 8, 9\}$ and $B = \{3, 5, 6, 7\}$. For equality, the set with product 630 must contain 5 and 7, either 2 and 9 or 3 and 6, and finally it is allowed to either contain 1 or not.

Our alternative solution would give $P_A + P_B \geqslant \lceil 144\sqrt{70}
ceil = 1205$. One would then need to find a way to show that $P_A + P_B 
eq 1205$. To do this we can assume without loss of generality that $5 \in A$. Then the last digit of $P_A$ is either 5 or 0. In the first case the last digit of $P_B$ would be 0 and so $P_B$ would also be a multiple of 5 which is impossible. The second case is analogous.

The computation of the expressions here might be a bit simpler. For example $9! = 362880$ so one expects $\sqrt{9!}$ to be slightly larger than 600.",289,3126,Algebra,3
73,shl_jbmo_2019_a4,shl_jbmo,2019,a,"Let $a, b$ be two distinct real numbers and let $c$ be a positive real number such that
\[
a^4 - 2019a = b^4 - 2019b = c.
\]
Prove that $-\sqrt{c} < ab < 0$.","\textbf{Solution.} Firstly, we see that
\[
2019(a-b) = a^4 - b^4 = (a-b)(a+b)(a^2+b^2).
\]
Since $a 
eq b$, we get $(a+b)(a^2+b^2) = 2019$, so $a+b 
eq 0$. Thus
\begin{align*}
2c &= a^4 - 2019a + b^4 - 2019b \\
&= a^4 + b^4 - 2019(a+b) \\
&= a^4 + b^4 - (a+b)^2(a^2+b^2) \\
&= -2ab(a^2 + ab + b^2).
\end{align*}
Hence $ab(a^2 + ab + b^2) = -c < 0$. Note that
\[
a^2 + ab + b^2 = \frac{1}{2}\left(a^2 + b^2 + (a+b)^2
ight) \geqslant 0,
\]
thus $ab < 0$. Finally, $a^2 + ab + b^2 = (a+b)^2 - ab > -ab$ (the equality does not occur since $a+b 
eq 0$). So
\[
-c = ab(a^2+ab+b^2) < -(ab)^2 \implies (ab)^2 < c \implies -\sqrt{c} < ab < \sqrt{c}.
\]
Therefore, we have $-\sqrt{c} < ab < 0$.

\medskip

\textbf{Alternative Solution by PSC.} By Descartes' Rule of Signs, the polynomial $p(x) = x^4 - 2019x - c$ has exactly one positive root and exactly one negative root. So $a, b$ must be its two real roots. Since one of them is positive and the other is negative, then $ab < 0$. Let $r \pm is$ be the two non-real roots of $p(x)$.

By Vieta, we have
\[
ab(r^2 + s^2) = -c, \tag{1}
\]
\[
a + b + 2r = 0, \tag{2}
\]
\[
ab + 2ar + 2br + r^2 + s^2 = 0. \tag{3}
\]
Using (2) and (3), we have
\[
r^2 + s^2 = -2r(a+b) - ab = (a+b)^2 - ab \geqslant -ab. \tag{4}
\]
If in the last inequality we actually have an equality, then $a+b = 0$. Then (2) gives $r = 0$ and (3) gives $s^2 = -ab$. Thus the roots of $p(x)$ are $a, -a, ia, -ia$. This would give that $p(x) = x^4 + a^4$, a contradiction.

So the inequality in (4) is strict and now from (1) we get
\[
c = -(r^2+s^2)ab > (ab)^2,
\]
which gives that $ab > -\sqrt{c}$.",157,1606,Algebra,4
74,shl_jbmo_2019_a5,shl_jbmo,2019,a,"Let $a, b, c, d$ be positive real numbers such that $abcd = 1$. Prove the inequality
\[
\frac{1}{a^3+b+c+d} + \frac{1}{a+b^3+c+d} + \frac{1}{a+b+c^3+d} + \frac{1}{a+b+c+d^3} \leqslant \frac{a+b+c+d}{4}.
\]","\textbf{Solution.} From the Cauchy-Schwarz Inequality, we obtain
\[
(a+b+c+d)^2 \leqslant (a^3+b+c+d)\left(\frac{1}{a}+b+c+d
ight).
\]
Using this, together with the other three analogous inequalities, we get
\[
\frac{1}{a^3+b+c+d} + \frac{1}{a+b^3+c+d} + \frac{1}{a+b+c^3+d} + \frac{1}{a+b+c+d^3}
\leqslant \frac{3(a+b+c+d) + \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}
ight)}{(a+b+c+d)^2}.
\]
So it suffices to prove that
\[
(a+b+c+d)^3 \geqslant 12(a+b+c+d) + 4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}
ight),
\]
or equivalently, that
\[
(a^3+b^3+c^3+d^3) + 3\sum a^2b + 6(abc+abd+acd+bcd) \geqslant 12(a+b+c+d) + 4(abc+abd+acd+bcd).
\]
(Here, the sum is over all possible $x^2y$ with $x, y \in \{a,b,c,d\}$ and $x 
eq y$.) From the AM-GM Inequality we have
\[
a^3+a^2b+a^2b+a^2c+a^2c+a^2d+a^2d+b^2a+c^2a+d^2a+bcd+bcd \geqslant 12\sqrt[12]{a^{18}b^6c^6d^6} = 12a.
\]
Similarly, we get three more inequalities. Adding them together gives the inequality we wanted. Equality holds if and only if $a = b = c = d = 1$.

\medskip

\textbf{Remark by PSC.} Alternatively, we can finish off the proof by using the following two inequalities: Firstly, we have $a+b+c+d \geqslant 4\sqrt[4]{abcd} = 4$ by the AM-GM Inequality, giving
\[
\frac{3}{4}(a+b+c+d)^3 \geqslant 12(a+b+c+d).
\]
Secondly, by McLaurin's Inequality, we have
\[
\left(\frac{a+b+c+d}{4}
ight)^3 \geqslant \frac{bcd+acd+abd+abc}{4},
\]
giving
\[
\frac{1}{4}(a+b+c+d)^3 \geqslant 4(bcd+acd+abd+abc).
\]
Adding those inequalities we get the required result.",205,1536,Algebra,5
75,shl_jbmo_2019_a6,shl_jbmo,2019,a,"Let $a, b, c$ be positive real numbers. Prove the inequality
\[
(a^2+ac+c^2)\left(\frac{1}{a+b+c}+\frac{1}{a+c}
ight) + b^2\left(\frac{1}{b+c}+\frac{1}{a+b}
ight) > a+b+c.
\]","\textbf{Solution.} By the Cauchy-Schwarz Inequality, we have
\[
\frac{1}{a+b+c} + \frac{1}{a+c} \geqslant \frac{4}{2a+b+2c},
\]
and
\[
\frac{1}{b+c} + \frac{1}{a+b} \geqslant \frac{4}{a+2b+c}.
\]
Since
\[
a^2+ac+c^2 = \frac{3}{4}(a+c)^2 + \frac{1}{4}(a-c)^2 \geqslant \frac{3}{4}(a+c)^2,
\]
then, writing $L$ for the Left Hand Side of the required inequality, we get
\[
L \geqslant \frac{3(a+c)^2}{2a+b+2c} + \frac{4b^2}{a+2b+c}.
\]
Using again the Cauchy-Schwarz Inequality, we have:
\[
L \geqslant \frac{(\sqrt{3}(a+c)+2b)^2}{3a+3b+3c} > \frac{(\sqrt{3}(a+c)+\sqrt{3}b)^2}{3a+3b+3c} = a+b+c.
\]

\medskip

\textbf{Alternative Question by Proposers.} Let $a, b, c$ be positive real numbers. Prove the inequality
\[
\frac{a^2}{a+c} + \frac{b^2}{b+c} > \frac{ab-c^2}{a+b+c} + \frac{ab}{a+b}.
\]
Note that both this inequality and the original one are equivalent to
\[
\left(c+\frac{a^2}{a+c}
ight) + \left(a - \frac{ab-c^2}{a+b+c}
ight) + \frac{b^2}{b+c} + \left(b - \frac{ab}{a+b}
ight) > a+b+c.
\]

\medskip

\textbf{Alternative Solution by PSC.} The required inequality is equivalent to
\[
\left(\frac{b^2}{a+b}-(b-a)
ight) + \frac{b^2}{b+c} + \left(\frac{a^2+ac+c^2}{a+c}-a
ight) + \left(\frac{a^2+ac+c^2}{a+b+c}-(a+c)
ight) > 0,
\]
or equivalently, to
\[
\frac{a^2}{a+b} + \frac{b^2}{b+c} + \frac{c^2}{c+a} > \frac{ab+bc+ca}{a+b+c}.
\]
However, by the Cauchy-Schwarz Inequality we have
\[
\frac{a^2}{a+b} + \frac{b^2}{b+c} + \frac{c^2}{c+a} \geqslant \frac{(a+b+c)^2}{2(a+b+c)} \geqslant \frac{3(ab+bc+ca)}{2(a+b+c)} > \frac{ab+bc+ca}{a+b+c}.
\]",174,1549,Algebra,6
76,shl_jbmo_2019_a7,shl_jbmo,2019,a,"Show that for any positive real numbers $a, b, c$ such that $a + b + c = ab + bc + ca$, the following inequality holds
\[
3 + \sqrt[3]{\frac{a^3+1}{2}} + \sqrt[3]{\frac{b^3+1}{2}} + \sqrt[3]{\frac{c^3+1}{2}} \leqslant 2(a+b+c).
\]","\textbf{Solution.} Using the condition we have
\[
a^2 - a + 1 = a^2 - a + 1 + ab + bc + ca - a - b - c = (c+a-1)(a+b-1).
\]
Hence we have
\[
\sqrt[3]{\frac{a^3+1}{2}} = \sqrt[3]{\frac{(a+1)(a^2-a+1)}{2}} = \sqrt[3]{\left(\frac{a+1}{2}
ight)(c+a-1)(a+b-1)}.
\]
Using the last equality together with the AM-GM Inequality, we have
\begin{align*}
\sum_{\text{cyc}} \sqrt[3]{\frac{a^3+1}{2}} &= \sum_{\text{cyc}} \sqrt[3]{\left(\frac{a+1}{2}
ight)(c+a-1)(a+b-1)} \\
&\leqslant \sum_{\text{cyc}} \frac{\frac{a+1}{2} + c + a - 1 + a + b - 1}{3} \\
&= \sum_{\text{cyc}} \frac{5a+2b+2c-3}{6} \\
&= \frac{3(a+b+c-1)}{2}.
\end{align*}
Hence it is enough to prove that
\[
3 + \frac{3(a+b+c-1)}{2} \leqslant 2(a+b+c),
\]
or equivalently, that $a+b+c \geqslant 3$. From a well-known inequality and the condition, we have
\[
(a+b+c)^2 \geqslant 3(ab+bc+ca) = 3(a+b+c),
\]
thus $a+b+c \geqslant 3$ as desired.

\medskip

\textbf{Alternative Proof by PSC.} Since $f(x) = \sqrt[3]{x}$ is concave for $x \geqslant 0$, by Jensen's Inequality we have
\[
\sqrt[3]{\frac{a^3+1}{2}} + \sqrt[3]{\frac{b^3+1}{2}} + \sqrt[3]{\frac{c^3+1}{2}} \leqslant 3\sqrt[3]{\frac{a^3+b^3+c^3+3}{6}}.
\]
So it is enough to prove that
\[
\sqrt[3]{\frac{a^3+b^3+c^3+3}{6}} \leqslant \frac{2(a+b+c)-3}{3}. \tag{1}
\]
We now write $s = a+b+c = ab+bc+ca$ and $p = abc$. We have
\[
a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ca) = s^2 - 2s,
\]
and
\[
r = a^2b + ab^2 + b^2c + bc^2 + c^2a + ca^2 = (ab+bc+ca)(a+b+c) - 3abc = s^2 - 3p.
\]
Thus,
\[
a^3+b^3+c^3 = (a+b+c)^3 - 3r - 6abc = s^3 - 3s^2 + 3p.
\]
So to prove (1), it is enough to show that
\[
\frac{s^3 - 3s^2 + 3p + 3}{6} \leqslant \frac{(2s-3)^3}{27}.
\]
Expanding, this is equivalent to
\[
7s^3 - 45s^2 + 108s - 27p - 81 \geqslant 0.
\]
By the AM-GM Inequality we have $s^3 \geqslant 27p$. So it is enough to prove that $p(s) \geqslant 0$, where
\[
p(s) = 6s^3 - 45s^2 + 108s - 81 = 3(s-3)^2(2s-3).
\]
It is easy to show that $s \geqslant 3$ (e.g.\ as in the first solution) so $p(s) \geqslant 0$ as required.",230,2014,Algebra,7
77,shl_jbmo_2019_c1,shl_jbmo,2019,c,"Let $S$ be a set of 100 positive integers having the following property:
\begin{quote}
``Among every four numbers of $S$, there is a number which divides each of the other three or there is a number which is equal to the sum of the other three.''
\end{quote}
Prove that the set $S$ contains a number which divides each of the other 99 numbers of $S$.","\textbf{Solution.} Let $a < b$ be the two smallest numbers of $S$ and let $d$ be the largest number of $S$. Consider any two other numbers $x < y$ of $S$. For the quadruples $(a,b,x,d)$ and $(a,b,y,d)$ we cannot get both of $d = a+b+x$ and $d = a+b+y$, since $a+b+x < a+b+y$. From here, we get $a \mid b$ and $a \mid d$.

Consider any number $s$ of $S$ different from $a, b, d$. From the condition of the problem, we get $d = a+b+s$ or $a$ divides $b$, $s$ and $d$. But since we already know that $a$ divides $b$ and $d$ anyway, we also get that $a \mid s$, as in the first case we have $s = d - a - b$. This means that $a$ divides all other numbers of $S$.

\medskip

\textbf{Alternative Solution by PSC.} Order the elements of $S$ as $x_1 < x_2 < \cdots < x_{100}$. For $2 \leqslant k \leqslant 97$, looking at the quadruples $(x_1, x_k, x_{k+1}, x_{k+2})$ and $(x_1, x_k, x_{k+1}, x_{k+3})$, we get that $x_1 \mid x_k$ as alternatively, we would have $x_{k+2} = x_1 + x_k + x_{k+1} = x_{k+3}$, a contradiction.

For $5 \leqslant k \leqslant 100$, looking at the quadruples $(x_1, x_{k-2}, x_{k-1}, x_k)$ and $(x_1, x_{k-3}, x_{k-1}, x_k)$ we get that $x_1 \mid x_k$ as alternatively, we would have $x_k = x_1 + x_{k-2} + x_{k-1} = x_1 + x_{k-3} + x_{k-1}$, a contradiction.

So $x_1$ divides all other elements of $S$.

\medskip

\textbf{Alternative Solution by PSC.} The condition that one element is the sum of the other three cannot be satisfied by all quadruples. So we have four elements such that one divides the other three. Suppose inductively that we have a subset $S'$ of $S$ with $|S'| = k \geqslant 4$ such that there is $x \in S'$ with $x \mid y$ for every $y \in S'$. Pick $s \in S \setminus S'$ and $y, z \in S'$ different from $x$. Considering $(s, x, y, z)$ either $s \mid x$, or $x \mid s$ or one of the four is a sum of the other three. In the last case we have $s = \pm x \pm y \pm z$ and so $x \mid s$. In any case either $x$ or $s$ divides all elements of $S' \cup \{s\}$.

\medskip

\textbf{Remark by PSC.} The last solution shows that the condition that the elements of $S$ are positive can be ignored.",350,2129,Combinatorics,8
78,shl_jbmo_2019_c2,shl_jbmo,2019,c,"In a certain city there are $n$ straight streets, such that every two streets intersect, and no three streets pass through the same intersection. The City Council wants to organize the city by designating the main and the side street on every intersection. Prove that this can be done in such way that if one goes along one of the streets, from its beginning to its end, the intersections where this street is the main street, and the ones where it is not, will appear in alternating order.","\textbf{Solution.} Pick any street $s$ and organize the intersections along $s$ such that the intersections of the two types alternate, as in the statement of the problem.

On every other street $s_1$, exactly one intersection has been organized, namely the one where $s_1$ intersects $s$. Call this intersection $I_1$. We want to organize the intersections along $s_1$ such that they alternate between the two types. Note that, as $I_1$ is already organized, we have exactly one way to organize the remaining intersections along $s_1$.

For every street $s_1 
eq s$, we can apply the procedure described above. Now, we only need to show that every intersection not on $s$ is well-organized. More precisely, this means that for every two streets $s_1, s_2 
eq s$ intersecting at $s_1 \cap s_2 = A$, $s_1$ is the main street on $A$ if and only if $s_2$ is the side street on $A$.

Consider also the intersections $I_1 = s_1 \cap s$ and $I_2 = s_2 \cap s$. Now, we will define the ``role'' of the street $t$ at the intersection $X$ as ``main'' if this street $t$ is the main street on $X$, and ``side'' otherwise. We will prove that the roles of $s_1$ and $s_2$ at $A$ are different.

Consider the path $A \to I_1 \to I_2 \to A$. Let the number of intersections between $A$ and $I_1$ be $u_1$, the number of these between $A$ and $I_2$ be $u_2$, and the number of these between $I_1$ and $I_2$ be $v$. Now, if we go from $A$ to $I_1$, we will change our role $u_1 + 1$ times, as we will encounter $u_1 + 1$ new intersections. Then, we will change our street from $s_1$ to $s$, changing our role once more. Then, on the segment $I_1 \to I_2$, we have $v+1$ new role changes, and after that one more when we change our street from $s_1$ to $s_2$. The journey from $I_2$ to $A$ will induce $u_2 + 1$ new role changes, so in total we have changed our role
\[
u_1 + 1 + 1 + v + 1 + 1 + u_2 + 1 = u_1 + v + u_2 + 5.
\]
As we try to show that roles of $s_1$ and $s_2$ differ, we need to show that the number of role changes is odd, i.e.\ that $u_1 + v + u_2 + 5$ is odd.

Obviously, this claim is equivalent to $2 \mid u_1 + v + u_2$. But $u_1$, $v$ and $u_2$ count the number of intersections of the triangle $AI_1I_2$ with streets other than $s, s_1, s_2$. Since every street other than $s, s_1, s_2$ intersects the sides of $AI_1I_2$ in exactly two points, the total number of intersections is even. As a consequence, $2 \mid u_1 + v + u_2$ as required.",490,2447,Combinatorics,9
79,shl_jbmo_2019_c3,shl_jbmo,2019,c,In a $5 \times 100$ table we have coloured black $n$ of its cells. Each of the 500 cells has at most two adjacent (by side) cells coloured black. Find the largest possible value of $n$.,"\textbf{Solution.} If we colour all the cells along all edges of the board together with the entire middle row except the second and the last-but-one cell, the condition is satisfied and there are 302 black cells.

We can cover the table by one fragment like the first one, 24 fragments like the middle one, and one fragment like the third one. In each fragment, among the cells with the same letter, there are at most two coloured black, so the total number of coloured cells is at most $(5 + 24 \cdot 6 + 1) \cdot 2 + 2 = 302$.

\medskip

\textbf{Alternative Solution by PSC.} Consider the cells adjacent to all cells of the second and fourth row. Counting multiplicity, each cell in the first and fifth row is counted once, each cell in the third row twice, while each cell in the second and fourth row is also counted twice apart from their first and last cells which are counted only once.

So there are 204 cells counted once and 296 cells counted twice. Those cells contain, counting multiplicity, at most 400 black cells. Suppose $a$ of the cells have multiplicity one and $b$ of them have multiplicity 2. Then $a + 2b \leqslant 400$ and $a \leqslant 204$. Thus
\[
2a + 2b \leqslant 400 + a \leqslant 604,
\]
and so $a + b \leqslant 302$ as required.

\medskip

\textbf{Remark by PSC.} The alternative solution shows that if we have equality, then all cells in the perimeter of the table except perhaps the two cells of the third row must be coloured black. No other cell in the second or fourth row can be coloured black as this will give a cell in the first or fifth row with at least three neighbouring black cells. For similar reasons we cannot colour black the second and last-but-one cell of the third row. So we must colour black all other cells of the third row and therefore the colouring is unique.",185,1816,Combinatorics,10
80,shl_jbmo_2019_c4,shl_jbmo,2019,c,"We have a group of $n$ kids. For each pair of kids, at least one has sent a message to the other one. For each kid $A$, among the kids to whom $A$ has sent a message, exactly 25\% have sent a message to $A$. How many possible two-digit values of $n$ are there?","\textbf{Solution.} If the number of pairs of kids with two-way communication is $k$, then by the given condition the total number of messages is $4k + 4k = 8k$. Thus the number of pairs of kids is $\dfrac{n(n-1)}{2} = 7k$. This is possible only if $n \equiv 0, 1 \pmod{7}$.

\begin{itemize}
    \item In order to obtain $n = 7m+1$, arrange the kids in a circle and let each kid send a message to the first $4m$ kids to its right and hence receive a message from the first $4m$ kids to its left. Thus there are exactly $m$ kids to which it has both sent and received messages.
    \item In order to obtain $n = 7m$, let kid $X$ send no messages (and receive from every other kid). Arrange the remaining $7m-1$ kids in a circle and let each kid on the circle send a message to the first $4m-1$ kids to its right and hence receive a message from the first $4m-1$ kids to its left. Thus there are exactly $m$ kids to which it has both sent and received messages.
\end{itemize}

There are 26 two-digit numbers with remainder 0 or 1 modulo 7. (All numbers of the form $7m$ and $7m+1$ with $2 \leqslant m \leqslant 14$.)

\medskip

\textbf{Alternative Solution by PSC.} Suppose kid $x_i$ sent $4d_i$ messages. (Guaranteed by the conditions to be a multiple of 4.) Then it received $d_i$ messages from the kids that it has sent a message to, and another $n - 1 - 4d_i$ messages from the rest of the kids. So it received a total of $n - 1 - 3d_i$ messages. Since the total number of messages sent is equal to the total number of messages received, we must have:
\[
d_1 + \cdots + d_n = (n - 1 - 3d_1) + \cdots + (n - 1 - 3d_n).
\]
This gives $7(d_1 + \cdots + d_n) = n(n-1)$ from which we get $n \equiv 0, 1 \pmod{7}$ as in the first solution.

We also present an alternative inductive construction (which turns out to be different from the construction in the first solution).

For the case $n \equiv 0 \pmod{7}$, we start with a construction for $7k$ kids, say $x_1, \ldots, x_{7k}$, and another construction with 7 kids, say $y_1, \ldots, y_7$. We merge them by demanding that in addition, each kid $x_i$ sends and receives gifts according to the following table:

\begin{center}
\begin{tabular}{|c|c|c|}
\hline
$i \bmod 7$ & Sends & Receives \\
\hline
0 & $y_1, y_2, y_3, y_4$ & $y_4, y_5, y_6, y_7$ \\
1 & $y_2, y_3, y_4, y_5$ & $y_5, y_6, y_7, y_1$ \\
2 & $y_3, y_4, y_5, y_6$ & $y_6, y_7, y_1, y_2$ \\
3 & $y_4, y_5, y_6, y_7$ & $y_7, y_1, y_2, y_3$ \\
4 & $y_5, y_6, y_7, y_1$ & $y_1, y_2, y_3, y_4$ \\
5 & $y_6, y_7, y_1, y_2$ & $y_2, y_3, y_4, y_5$ \\
6 & $y_7, y_1, y_2, y_3$ & $y_3, y_4, y_5, y_6$ \\
\hline
\end{tabular}
\end{center}

So each kid $x_i$ sends an additional four messages and receives a message from only one of those four additional kids. Also, each kid $y_j$ sends an additional $4k$ messages and receives from exactly $k$ of those additional kids. So this is a valid construction for $7(k+1)$ kids.

For the case $n \equiv 1 \pmod{7}$, we start with a construction for $7k+1$ kids, say $x_1, \ldots, x_{7k+1}$, and we take another 7 kids, say $y_1, \ldots, y_7$ for which we do not yet mention how they exchange gifts. The kids $x_1, \ldots, x_{7k+1}$ exchange gifts with the kids $y_1, \ldots, y_7$ according to the previous table. As before, each kid $x_i$ satisfies the conditions. We now put $y_1, \ldots, y_7$ on a circle and demand that each of $y_1, \ldots, y_3$ sends gifts to the next four kids on the circle and each of $y_4, \ldots, y_7$ sends gifts to the next three kids on the circle. It is easy to check that the condition is satisfied by each $y_i$ as well.",260,3595,Combinatorics,11
81,shl_jbmo_2019_c5,shl_jbmo,2019,c,"An economist and a statistician play a game on a calculator which does only one operation. The calculator displays only positive integers and it is used in the following way: Denote by $n$ an integer that is shown on the calculator. A person types an integer, $m$, chosen from the set $\{1, 2, \ldots, 99\}$ of the first 99 positive integers, and if $m\%$ of the number $n$ is again a positive integer, then the calculator displays $m\%$ of $n$. Otherwise, the calculator shows an error message and this operation is not allowed. The game consists of doing alternatively these operations and the player that cannot do the operation loses. How many numbers from $\{1, 2, \ldots, 2019\}$ guarantee the winning strategy for the statistician, who plays second?","\textbf{Solution.} First of all, the game finishes because the number on the calculator always decreases. By picking $m\%$ of a positive integer $n$, players get the number
\[
\frac{m \cdot n}{100} = \frac{m \cdot n}{2^2 5^2}.
\]
We see that at least one of the powers of 2 and 5 that divide $n$ decreases after one move, as $m$ is not allowed to be 100, or a multiple of it. These prime divisors of $n$ are the only ones that can decrease, so we conclude that all the other prime factors of $n$ are not important for this game. Therefore, it is enough to consider numbers of the form $n = 2^k 5^\ell$ where $k, \ell \in \mathbb{N}_0$, and to draw conclusions from these numbers.

We will describe all possible changes of $k$ and $\ell$ in one move. Since $5^3 > 100$, then $\ell$ cannot increase, so all possible changes are from $\ell$ to $\ell + b$, where $b \in \{0, -1, -2\}$. For $k$, we note that $2^6 = 64$ is the biggest power of 2 less than 100, so $k$ can be changed to $k + a$, where $a \in \{-2, -1, 0, 1, 2, 3, 4\}$. But the changes of $k$ and $\ell$ are not independent. For example, if $\ell$ stays the same, then $m$ has to be divisible by 25, giving only two possibilities for a change $(k, \ell) \to (k-2, \ell)$, when $m = 25$ or $m = 75$, or $(k, \ell) \to (k-1, \ell)$, when $m = 50$. Similarly, if $\ell$ decreases by 1, then $m$ is divisible exactly by 5 and then the different changes are given by $(k, \ell) \to (k+a, \ell-1)$, where $a \in \{-2,-1,0,1,2\}$, depending on the power of 2 that divides $m$ and it can be from $2^0$ to $2^4$. If $\ell$ decreases by 2, then $m$ is not divisible by 5, so it is enough to consider when $m$ is a power of two, giving changes $(k, \ell) \to (k+a, \ell-2)$, where $a \in \{-2,-1,0,1,2,3,4\}$.

We have translated the starting game into another game with changing (the starting pair of non-negative integers) $(k, \ell)$ by moves described above and the player who cannot make the move loses, i.e.\ the player who manages to play the move $(k, \ell) \to (0, 0)$ wins. We claim that the second player wins if and only if $3 \mid k$ and $3 \mid \ell$.

We notice that all moves have their inverse modulo 3, namely after the move $(k, \ell) \to (k+a, \ell+b)$, the other player plays $(k+a, \ell+b) \to (k+a+c, \ell+b+d)$, where
\[
(c, d) \in \{(0,-1),(0,-2),(-1,0),(-1,-1),(-1,-2),(-2,0),(-2,-1),(-2,-2)\}
\]
is chosen such that $3 \mid a+c$ and $3 \mid b+d$. Such $(c,d)$ can be chosen as all possible residues different from $(0,0)$ modulo 3 are contained in the set above and there is no move that keeps $k$ and $\ell$ the same modulo 3. If the starting numbers $(k, \ell)$ are divisible by 3, then after the move of the first player at least one of $k$ and $\ell$ will not be divisible by 3, and then the second player will play the move so that $k$ and $\ell$ become divisible by 3 again. In this way, the first player can never finish the game, so the second player wins. In all other cases, the first player will make such a move to make $k$ and $\ell$ divisible by 3 and then he becomes the second player in the game, and by previous reasoning, wins.

The remaining part of the problem is to compute the number of positive integers $n \leqslant 2019$ which are winning for the second player. Those are the $n$ which are divisible by exactly $2^{3k} 5^{3\ell}$, $k, \ell \in \mathbb{N}_0$. Here, exact divisibility by $2^{3k} 5^{3\ell}$ in this context means that $2^{3k} \| n$ and $5^{3\ell} \| n$, even for $\ell = 0$, or $k = 0$. For example, if we say that $n$ is exactly divisible by 8, it means that $8 \mid n$, $16 
mid n$ and $5 
mid n$. We start by noting that for each ten consecutive numbers, exactly four of them are coprime to 10. Then we find the desired amount by dividing 2019 by numbers $2^{3k} 5^{3\ell}$ which are less than 2019, and then computing the number of numbers no bigger than $\left\lfloor\dfrac{2019}{2^{3k}5^{3\ell}}
ight
floor$ which are coprime to 10.

First, there are $4 \cdot 201 + 4 = 808$ numbers (out of positive integers $n \leqslant 2019$) coprime to 10. Then, there are $\left\lfloor\dfrac{2019}{8}
ight
floor = 252$ numbers divisible by 8, and $25 \cdot 4 + 1 = 101$ among them are exactly divisible by 8. There are $\left\lfloor\dfrac{2019}{64}
ight
floor = 31$ numbers divisible by 64, giving $3 \cdot 4 + 1 = 13$ divisible exactly by 64. And there are two numbers, $512$ and $3 \cdot 512$, which are divisible by exactly 512. Similarly, there are $\left\lfloor\dfrac{2019}{125}
ight
floor = 16$ numbers divisible by 125, implying that $4 + 2 = 6$ of them are exactly divisible by 125. Finally, there is only one number divisible by exactly 1000, and this is 1000 itself. All other numbers that are divisible by exactly $2^{3k} 5^{3\ell}$ are greater than 2019. So, we obtain that $808 + 101 + 13 + 2 + 6 + 1 = 931$ numbers not bigger than 2019 are winning for the statistician.

\medskip

\textbf{Alternative Solution by PSC.} Let us call a positive integer $n$ \emph{losing} if $n = 2^r 5^s k$ where $r \equiv s \equiv 0 \pmod{3}$ and $(k, 10) = 1$. We call all other positive integers \emph{winning}.

\medskip


oindent\textbf{Lemma 1.} \textit{If $n$ is losing, then $\dfrac{mn}{100}$ is winning for all $m \in \{1, 2, \ldots, 99\}$ such that $100 \mid mn$.}

\medskip


oindent\textit{Proof of Lemma 1.} Let $m = 2^t 5^u k'$. For $\dfrac{mn}{100}$ to be losing, we would need $t \equiv u \equiv 2 \pmod{3}$. But then $m \geqslant 100$, a contradiction. $\square$

\medskip


oindent\textbf{Lemma 2.} \textit{If $n$ is winning, then there is an $m \in \{1, 2, \ldots, 99\}$ such that $100 \mid mn$ and $\dfrac{mn}{100}$ is losing.}

\medskip


oindent\textit{Proof of Lemma 2.} Let $n = 2^r 5^s k$ where $(k, 10) = 1$. Pick $t, u \in \{0, 1, 2\}$ such that $t \equiv (2-r) \pmod{3}$ and $u \equiv (2-s) \pmod{3}$ and let $m = 2^t 5^u$. Then $100 \mid mn$ and $\dfrac{mn}{100}$ is winning. Furthermore $m < 100$ as otherwise $m = 100$, $t = u = 2$ giving $r \equiv s \equiv 0 \pmod{3}$ contradicting the fact that $n$ was winning. $\square$

\medskip

Combining Lemmas 1 and 2 we obtain that the second player wins if and only if the game starts from a losing number.",756,6187,Combinatorics,12
82,shl_jbmo_2019_g1,shl_jbmo,2019,g,Let $ABC$ be a right-angled triangle with $\hat{A} = 90^\circ$ and $\hat{B} = 30^\circ$. The perpendicular at the midpoint $M$ of $BC$ meets the bisector $BK$ of the angle $\hat{B}$ at the point $E$. The perpendicular bisector of $EK$ meets $AB$ at $D$. Prove that $KD$ is perpendicular to $DE$.,"\textbf{Solution.} Let $I$ be the incenter of $ABC$ and let $Z$ be the foot of the perpendicular from $K$ on $EC$. Since $KB$ is the bisector of $\hat{B}$, then $\angle EBC = 15^\circ$ and since $EM$ is the perpendicular bisector of $BC$, then $\angle ECB = \angle EBC = 15^\circ$. Therefore $\angle KEC = 30^\circ$. Moreover, $\angle ECK = 60^\circ - 15^\circ = 45^\circ$. This means that $KZC$ is isosceles and thus $Z$ is on the perpendicular bisector of $KC$.

Since $\angle KIC$ is the external angle of triangle $IBC$, and $I$ is the incenter of triangle $ABC$, then $\angle KIC = 15^\circ + 30^\circ = 45^\circ$. Thus, $\angle KIC = \dfrac{\angle KZC}{2}$. Since also $Z$ is on the perpendicular bisector of $KC$, then $Z$ is the circumcenter of $IKC$. This means that $ZK = ZI = ZC$. Since also $\angle EKZ = 60^\circ$, then the triangle $ZKI$ is equilateral. Moreover, since $\angle KEZ = 30^\circ$, we have that $ZK = \dfrac{EK}{2}$, so $ZK = IK = IE$.

Therefore $DI$ is perpendicular to $EK$ and this means that $DIKA$ is cyclic. So $\angle KDI = \angle IAK = 45^\circ$ and $\angle IKD = \angle IAD = 45^\circ$. Thus $ID = IK = IE$ and so $KD$ is perpendicular to $DE$ as required.

\medskip

\textbf{Alternative Question by Proposers.} We can instead ask to prove that $ED = 2AD$. (After proving $KD \perp DE$ we have that the triangle $EDK$ is right angled and isosceles, therefore $ED = DK = 2AD$.) This alternative is probably more difficult because the perpendicular relation is hidden.

\medskip

\textbf{Alternative Solution by PSC.} Let $P$ be the point of intersection of $EM$ with $AC$. The triangles $ABC$ and $MPC$ are equal since they have equal angles and $MC = \dfrac{BC}{2} = AC$. They also share the angle $\hat{C}$, so they must have identical incenter.

Let $I$ be the midpoint of $EK$. We have $\angle PEI = \angle BEM = 75^\circ = \angle EKP$. So the triangle $PEK$ is isosceles and therefore $PI$ is a bisector of $\angle CPM$. So the incenter of $MPC$ belongs on $PI$. Since it shares the same incentre with $ABC$, then $I$ is the common incenter. We can now finish the proof as in the first solution.

\medskip

\textbf{Alternative Solution by PSC.} Let $P$ be the point of intersection of $EM$ with $AC$ and let $I$ be the midpoint of $EK$. Then the triangle $PBC$ is equilateral. We also have $\angle PEI = \angle BEM = 75^\circ$ and $\angle PKE = 75^\circ$, so $PEK$ is isosceles. We also have $PI \perp EK$ and $DI \perp EK$, so the points $P, D, I$ are collinear.

Furthermore, $\angle PBI = \angle BPI = 45^\circ$, and therefore $BI = PI$.

We have $\angle DPA = \angle EBM = 15^\circ$ and also $BM = \dfrac{AB}{2} = AC = PA$. So the right-angled triangles $PDA$ and $BEM$ are equal. Thus $PD = BE$.

So
\[
EI = BI - BE = PI - PD = DI.
\]
Therefore $\angle DEI = \angle IDE = 45^\circ$. Since $DE = DK$, we also have $\angle DEI = \angle DKI = \angle KDI = 45^\circ$. So finally, $\angle EDK = 90^\circ$.

\medskip

\textbf{Coordinate Geometry Solution by PSC.} We may assume that $A = (0,0)$, $B = (0, \sqrt{3})$ and $C = (1, 0)$. Since $m_{BC} = -\sqrt{3}$, then $m_{EM} = \dfrac{\sqrt{3}}{3}$. Since also $M = \left(\dfrac{1}{2}, \dfrac{\sqrt{3}}{2}
ight)$, then the equation of $EM$ is $y = \dfrac{\sqrt{3}}{3}x + \dfrac{\sqrt{3}}{3}$. The slope of $BK$ is
\[
m_{BK} = \tan(105^\circ) = \frac{\tan(60^\circ)+\tan(45^\circ)}{1-\tan(60^\circ)\tan(45^\circ)} = -(2+\sqrt{3}).
\]
So the equation of $BK$ is $y = -(2+\sqrt{3})x + \sqrt{3}$ which gives $K = (2\sqrt{3}-3, 0)$ and $E = (2-\sqrt{3}, \sqrt{3}-1)$. Letting $I$ be the midpoint of $EK$ we get $I = \left(\dfrac{\sqrt{3}-1}{2}, \dfrac{\sqrt{3}-1}{2}
ight)$. Thus $I$ is equidistant from the sides $AB$, $AC$, so $AI$ is the bisector of $\hat{A}$, and thus $I$ is the incenter of triangle $ABC$. We can now finish the proof as in the first solution.

\medskip

\textbf{Metric Solution by PSC.} We can assume that $AC = 1$. Then $AB = \sqrt{3}$ and $BC = 2$. So $BM = MC = 1$. From triangle $BEM$ we get $BE = EC = \sec(15^\circ)$ and $EM = \tan(15^\circ)$. From triangle $BAK$ we get $BK = \sqrt{3}\sec(15^\circ)$. So $EK = BK - BE = (\sqrt{3}-1)\sec(15^\circ)$. Thus, if $N$ is the midpoint of $EK$, then $EN = NK = \dfrac{\sqrt{3}-1}{2}\sec(15^\circ)$ and $BN = BE + EN = \dfrac{\sqrt{3}+1}{2}\sec(15^\circ)$. From triangle $BDN$ we get $DN = BN\tan(15^\circ) = \dfrac{\sqrt{3}+1}{2}\tan(15^\circ)\sec(15^\circ)$. It is easy to check that $\tan(15^\circ) = 2-\sqrt{3}$. Thus $DN = \dfrac{\sqrt{3}-1}{2}\sec(15^\circ) = EN$. So $DN = EN = EK$ and therefore $\angle EDN = \angle KDN = 45^\circ$ and $\angle KDE = 90^\circ$ as required.",295,4635,Geometry,13
83,shl_jbmo_2019_g2,shl_jbmo,2019,g,"Let $ABC$ be a triangle and let $\omega$ be its circumcircle. Let $\ell_B$ and $\ell_C$ be two parallel lines passing through $B$ and $C$ respectively. The lines $\ell_B$ and $\ell_C$ intersect with $\omega$ for the second time at the points $D$ and $E$ respectively, with $D$ belonging on the arc $AB$, and $E$ on the arc $AC$. Suppose that $DA$ intersects $\ell_C$ at $F$, and $EA$ intersects $\ell_B$ at $G$. If $O$, $O_1$ and $O_2$ are the circumcenters of the triangles $ABC$, $ADG$ and $AEF$ respectively, and $P$ is the center of the circumcircle of the triangle $OO_1O_2$, prove that $OP$ is parallel to $\ell_B$ and $\ell_C$.","\textbf{Solution.} We write $\omega_1$, $\omega_2$ and $\omega'$ for the circumcircles of $AGD$, $AEF$ and $OO_1O_2$ respectively. Since $O_1$ and $O_2$ are the centers of $\omega_1$ and $\omega_2$, and because $DG$ and $EF$ are parallel, we get that
\[
\angle GAO_1 = 90^\circ - \frac{\angle GO_1A}{2} = 90^\circ - \angle GDA = 90^\circ - \angle EFA = 90^\circ - \frac{\angle EO_2A}{2} = \angle EAO_2.
\]
So, because $G$, $A$ and $E$ are collinear, we come to the conclusion that $O_1$, $A$ and $O_2$ are also collinear.

Let $\angle DFE = \varphi$. Then, as a central angle $\angle AO_2E = 2\varphi$. Because $AE$ is a common chord of both $\omega$ and $\omega_2$, the line $OO_2$ that passes through their centers bisects $\angle AO_2E$, thus $\angle AO_2O = \varphi$. By the collinearity of $O_1$, $A$, $O_2$, we get that $\angle O_1O_2O = \angle AO_2O = \varphi$. As a central angle in $\omega'$, we have $\angle O_1PO = 2\varphi$, so $\angle POO_1 = 90^\circ - \varphi$. Let $Q$ be the point of intersection of $DF$ and $OP$. Because $AD$ is a common chord of $\omega$ and $\omega_1$, we have that $OO_1$ is perpendicular to $DA$ and so $\angle DQP = 90^\circ - \angle POO_1 = \varphi$. Thus, $OP$ is parallel to $\ell_C$ and so to $\ell_B$ as well.

\medskip

\textbf{Alternative Solution by PSC.} Let us write $\alpha, \beta, \gamma$ for the angles of $ABC$. Since $ADBC$ is cyclic, we have $\angle GDA = 180^\circ - \angle BDA = \gamma$. Similarly, we have
\[
\angle GAD = 180^\circ - \angle DAE = \angle EBD = \angle BEC = \angle BAC = \alpha,
\]
where we have also used the fact that $\ell_B$ and $\ell_C$ are parallel.

Thus, the triangles $ABC$ and $AGD$ are similar. Analogously, $AEF$ is also similar to them.

Since $AD$ is a common chord of $\omega$ and $\omega_1$ then $AD$ is perpendicular to $OO_1$. Thus,
\[
\angle OO_1A = \frac{1}{2}\angle DO_1A = \angle DGA = \beta.
\]
Similarly, we have $\angle OO_2A = \gamma$. Since $O_1$, $A$, $O_2$ are collinear (as in the first solution) we get that $OO_1O_2$ is also similar to $ABC$. Their circumcentres are $P$ and $O$ respectively, thus $\angle POO_1 = \angle OAB = 90^\circ - \gamma$.

Since $OO_1$ is perpendicular to $AD$, letting $X$ be the point of intersection of $OO_1$ with $GD$, we get that $\angle DXO_1 = 90^\circ - \gamma$. Thus $OP$ is parallel to $\ell_B$ and therefore to $\ell_C$ as well.",634,2372,Geometry,14
84,shl_jbmo_2019_g3,shl_jbmo,2019,g,"Let $ABC$ be a triangle with incenter $I$. The points $D$ and $E$ lie on the segments $CA$ and $BC$ respectively, such that $CD = CE$. Let $F$ be a point on the segment $CD$. Prove that the quadrilateral $ABEF$ is circumscribable if and only if the quadrilateral $DIEF$ is cyclic.","\textbf{Solution.} Since $CD = CE$ it means that $E$ is the reflection of $D$ on the bisector of $\angle ACB$, i.e.\ the line $CI$. Let $G$ be the reflection of $F$ on $CI$. Then $G$ lies on the segment $CE$, the segment $EG$ is the reflection of the segment $DF$ on the line $CI$. Also, the quadrilateral $DEGF$ is cyclic since $\angle DFE = \angle EGD$.

Suppose that the quadrilateral $ABEF$ is circumscribable. Since $\angle FAI = \angle BAI$ and $\angle EBI = \angle ABI$, then $I$ is the centre of its inscribed circle. Then $\angle DFI = \angle EFI$ and since segment $EG$ is the reflection of segment $DF$ on the line $CI$, we have $\angle EFI = \angle DGI$. So $\angle DFI = \angle DGI$ which means that quadrilateral $DIGF$ is cyclic. Since the quadrilateral $DEGF$ is also cyclic, we have that the quadrilateral $DIEF$ is cyclic.

Suppose that the quadrilateral $DIEF$ is cyclic. Since quadrilateral $DEGF$ is also cyclic, we have that the pentagon $DIEGF$ is cyclic. So $\angle IEB = 180^\circ - \angle IEG = \angle IDG$ and since segment $EG$ is the reflection of segment $DF$ on the line $CI$, we have $\angle IDG = \angle IEF$. Hence $\angle IEB = \angle IEF$, which means that $EI$ is the angle bisector of $\angle BEF$. Since $\angle IFA = \angle IFD = \angle IGD$ and since the segment $EG$ is the reflection of segment $DF$ on the line $CI$, we have $\angle IGD = \angle IFE$, hence $\angle IFA = \angle IFE$, which means that $FI$ is the angle bisector of $\angle EFA$. We also know that $AI$ and $BI$ are the angle bisectors of $\angle FAB$ and $\angle ABE$. So all angle bisectors of the quadrilateral $ABEF$ intersect at $I$, which means that it is circumscribable.

\medskip

\textbf{Comment by PSC.} There is no need for introducing the point $G$. One can show that triangles $CID$ and $CIE$ are equal and also that the triangles $CDM$ and $CEM$ are equal, where $M$ is the midpoint of $DE$. From these, one can deduce that $\angle CDI = \angle CEI$ and $\angle IDE = \angle IED$ and proceed with similar reasoning as in the solution.",280,2059,Geometry,15
85,shl_jbmo_2019_g4,shl_jbmo,2019,g,"Let $ABC$ be a triangle such that $AB 
eq AC$, and let the perpendicular bisector of the side $BC$ intersect lines $AB$ and $AC$ at points $P$ and $Q$, respectively. If $H$ is the orthocenter of the triangle $ABC$, and $M$ and $N$ are the midpoints of the segments $BC$ and $PQ$ respectively, prove that $HM$ and $AN$ meet on the circumcircle of $ABC$.","\textbf{Solution.} We have
\[
\angle APQ = \angle BPM = 90^\circ - \angle MBP = 90^\circ - \angle CBA = \angle HCB,
\]
and
\[
\angle AQP = \angle MQC = 90^\circ - \angle QCM = 90^\circ - \angle ACB = \angle CBH.
\]
From these two equalities, we see that the triangles $APQ$ and $HCB$ are similar. Moreover, since $M$ and $N$ are the midpoints of the segments $BC$ and $PQ$ respectively, then the triangles $AQN$ and $HBM$ are also similar. Therefore, we have $\angle ANQ = \angle HMB$.

Let $L$ be the intersection of $AN$ and $HM$. We have
\[
\angle MLN = 180^\circ - \angle LNM - \angle NML = 180^\circ - \angle LMB - \angle NML = 180^\circ - \angle NMB = 90^\circ.
\]
Now let $D$ be the point on the circumcircle of $ABC$ diametrically opposite to $A$. It is known that $D$ is also the reflection of point $H$ over the point $M$. Therefore, we have that $D$ belongs on $MH$ and that $\angle DLA = \angle MLA = \angle MLN = 90^\circ$. But, as $DA$ is the diameter of the circumcircle of $ABC$, the condition that $\angle DLA = 90^\circ$ is enough to conclude that $L$ belongs on the circumcircle of $ABC$.

\medskip

\textbf{Remark by PSC.} There is a spiral similarity mapping $AQP$ to $HBC$. Since the similarity maps $AN$ to $HM$, it also maps $AH$ to $NM$, and since these two lines are parallel, the centre of the similarity is $L = AN \cap HM$. Since the similarity maps $BC$ to $QP$, its centre belongs on the circumcircle of $BCX$, where $X = BQ \cap PC$. But $X$ is the reflection of $A$ on $QM$ and so it must belong on the circumcircle of $ABC$. Hence so must $L$.",352,1577,Geometry,16
86,shl_jbmo_2019_g5,shl_jbmo,2019,g,"Let $P$ be a point in the interior of a triangle $ABC$. The lines $AP$, $BP$ and $CP$ intersect again the circumcircles of the triangles $PBC$, $PCA$, and $PAB$ at $D$, $E$ and $F$ respectively. Prove that $P$ is the orthocenter of the triangle $DEF$ if and only if $P$ is the incenter of the triangle $ABC$.","\textbf{Solution.} If $P$ is the incenter of $ABC$, then $\angle BPD = \angle ABP + \angle BAP = \dfrac{\hat{A}+\hat{B}}{2}$, and $\angle BDP = \angle BCP = \dfrac{\hat{C}}{2}$. From triangle $BDP$, it follows that $\angle PBD = 90^\circ$, i.e.\ that $EB$ is one of the altitudes of the triangle $DEF$. Similarly, $AD$ and $CF$ are altitudes, which means that $P$ is the orthocenter of $DEF$.

Notice that $AP$ separates $B$ from $C$, $B$ from $E$ and $C$ from $F$. Therefore $AP$ separates $E$ from $F$, which means that $P$ belongs to the interior of $\angle EDF$. It follows that $P \in \mathrm{Int}(\triangle DEF)$.

If $P$ is the orthocenter of $DEF$, then clearly $DEF$ must be acute. Let $A' \in EF$, $B' \in DF$ and $C' \in DE$ be the feet of the altitudes. Then the quadrilaterals $B'PA'F$, $C'PB'D$, and $A'PC'E$ are cyclic, which means that
\[
\angle B'FA' = 180^\circ - \angle B'PA' = 180^\circ - \angle BPA = \angle BFA.
\]
Similarly, one obtains that $\angle C'DB' = \angle CDB$, and $\angle A'EC' = \angle AEC$.

\begin{itemize}
    \item If $B \in \mathrm{Ext}(\triangle FPD)$, then $A \in \mathrm{Int}(\triangle EPF)$, $C \in \mathrm{Ext}(\triangle DPE)$, and thus $B \in \mathrm{Int}(\triangle FPD)$, contradiction.
    \item If $B \in \mathrm{Int}(\triangle FPD)$, then $A \in \mathrm{Ext}(\triangle EPF)$, $C \in \mathrm{Int}(\triangle DPE)$, and thus $B \in \mathrm{Ext}(\triangle FPD)$, contradiction.
\end{itemize}

This leaves us with $B \in FD$. Then we must have $A \in EF$, $C \in DE$, which means that $A = A'$, $B = B'$, $C = C'$. Thus $ABC$ is the orthic triangle of triangle $DEF$ and it is well known that the orthocenter of an acute triangle $DEF$ is the incenter of its orthic triangle.",308,1720,Geometry,17
87,shl_jbmo_2019_g6,shl_jbmo,2019,g,"Let $ABC$ be a non-isosceles triangle with incenter $I$. Let $D$ be a point on the segment $BC$ such that the circumcircle of $BID$ intersects the segment $AB$ at $E 
eq B$, and the circumcircle of $CID$ intersects the segment $AC$ at $F 
eq C$. The circumcircle of $DEF$ intersects $AB$ and $AC$ at the second points $M$ and $N$ respectively. Let $P$ be the point of intersection of $IB$ and $DE$, and let $Q$ be the point of intersection of $IC$ and $DF$. Prove that the three lines $EN$, $FM$ and $PQ$ are parallel.","\textbf{Solution.} Since $BDIE$ is cyclic, and $BI$ is the bisector of $\angle DBE$, then $ID = IE$. Similarly, $ID = IF$, so $I$ is the circumcenter of the triangle $DEF$. We also have
\[
\angle IEA = \angle IDB = \angle IFC,
\]
which implies that $AEIF$ is cyclic. We can assume that $A, E, M$ and $A, N, F$ are collinear in that order. Then $\angle IEM = \angle IFN$. Since also $IM = IE = IN = IF$, the two isosceles triangles $IEM$ and $INF$ are congruent, thus $EM = FN$ and therefore $EN$ is parallel to $FM$. From that, we can also see that the two triangles $IEA$ and $INA$ are congruent, which implies that $AI$ is the perpendicular bisector of $EN$ and $MF$.

Note that $\angle IDP = \angle IDE = \angle IBE = \angle IBD$, so the triangles $IPD$ and $IDB$ are similar, which implies that $\dfrac{ID}{IB} = \dfrac{IP}{ID}$ and $IP \cdot IB = ID^2$. Similarly, we have $IQ \cdot IC = ID^2$, thus $IP \cdot IB = IQ \cdot IC$. This implies that $BPQC$ is cyclic, which leads to
\[
\angle IPQ = \angle ICB = \frac{\hat{C}}{2}.
\]
But $\angle AIB = 90^\circ + \dfrac{\hat{C}}{2}$, so $AI$ is perpendicular to $PQ$. Hence, $PQ$ is parallel to $EN$ and $FM$.",518,1161,Geometry,18
88,shl_jbmo_2019_g7,shl_jbmo,2019,g,"Let $ABC$ be a right-angled triangle with $\hat{A} = 90^\circ$. Let $K$ be the midpoint of $BC$, and let $AKLM$ be a parallelogram with centre $C$. Let $T$ be the intersection of the line $AC$ and the perpendicular bisector of $BM$. Let $\omega_1$ be the circle with centre $C$ and radius $CA$ and let $\omega_2$ be the circle with centre $T$ and radius $TB$. Prove that one of the points of intersection of $\omega_1$ and $\omega_2$ is on the line $LM$.","\textbf{Solution.} Let $M'$ be the symmetric point of $M$ with respect to $T$. Observe that $T$ is equidistant from $B$ and $M$, therefore $M$ belongs on $\omega_2$ and $M'M$ is a diameter of $\omega_2$. It suffices to prove that $M'A$ is perpendicular to $LM$, or equivalently, to $AK$. To see this, let $S$ be the point of intersection of $M'A$ with $LM$. We will then have $\angle M'SM = 90^\circ$ which shows that $S$ belongs on $\omega_2$ as $M'M$ is a diameter of $\omega_2$. We also have that $S$ belongs on $\omega_1$ as $AL$ is a diameter of $\omega_1$.

Since $T$ and $C$ are the midpoints of $M'M$ and $KM$ respectively, then $TC$ is parallel to $M'K$ and so $M'K$ is perpendicular to $AB$. Since $KA = KB$, then $KM'$ is the perpendicular bisector of $AB$. But then the triangles $KBM'$ and $KAM'$ are equal, showing that $\angle M'AK = \angle M'BK = \angle M'BM = 90^\circ$ as required.

\medskip

\textbf{Alternative Solution by Proposers.} Since $CA = CL$, then $L$ belongs on $\omega_1$. Let $S$ be the other point of intersection of $\omega_1$ with the line $LM$. We need to show that $S$ belongs on $\omega_2$. Since $TB = TM$ ($T$ is on the perpendicular bisector of $BM$) it is enough to show that $TS = TM$.

Let $N$, $T'$ be points on the lines $AL$ and $LM$ respectively, such that $MN \perp LM$ and $TT' \perp LM$. It is enough to prove that $T'$ is the midpoint of $SM$. Since $AL$ is diameter of $\omega_1$ we have that $AS \perp LS$. Thus, it is enough to show that $T$ is the midpoint of $AN$.

We have
\[
AT = \frac{AN}{2} \iff AC - CT = \frac{AL - LN}{2} \iff 2AC - 2CT = AL - LN \iff LN = 2CT
\]
as $AL = 2AC$. So it suffices to prove that $LN = 2CT$.

Let $D$ be the midpoint of $BM$. Since $BK = KC = CM$, then $D$ is also the midpoint of $KC$. The triangles $LMN$ and $CTD$ are similar since they are right-angled with $\angle TCD = \angle CAK = \angle MLN$. ($AK = KC$ and $AK$ is parallel to $LM$.) So we have
\[
\frac{LN}{CT} = \frac{LM}{CD} = \frac{AK}{CD} = \frac{CK}{CD} = 2,
\]
as required.",454,2031,Geometry,19
89,shl_jbmo_2019_n1,shl_jbmo,2019,n,"Find all prime numbers $p$ for which there are non-negative integers $x$, $y$ and $z$ such that the number
\[
A = x^p + y^p + z^p - x - y - z
\]
is a product of exactly three distinct prime numbers.","\textbf{Solution.} For $p = 2$, we take $x = y = 4$ and $z = 3$. Then $A = 30 = 2 \cdot 3 \cdot 5$. For $p = 3$ we can take $x = 3$ and $y = 2$ and $z = 1$. Then again $A = 30 = 2 \cdot 3 \cdot 5$. For $p = 5$ we can take $x = 2$ and $y = 1$ and $z = 1$. Again $A = 30 = 2 \cdot 3 \cdot 5$.

Assume now that $p \geqslant 7$. Working modulo 2 and modulo 3 we see that $A$ is divisible by both 2 and 3. Moreover, by Fermat's Little Theorem, we have
\[
x^p + y^p + z^p - x - y - z \equiv x + y + z - x - y - z = 0 \pmod{p}.
\]
Therefore, by the given condition, we have to solve the equation
\[
x^p + y^p + z^p - x - y - z = 6p.
\]
If one of the numbers $x, y$ and $z$ is bigger than or equal to 2, let's say $x \geqslant 2$, then
\[
6p \geqslant x^p - x = x(x^{p-1}-1) \geqslant 2(2^{p-1}-1) = 2^p - 2.
\]
It is easy to check by induction that $2^p - 2 > 6p$ for all primes $p \geqslant 7$. This contradiction shows that there are no more values of $p$ which satisfy the required property.

\medskip

\textbf{Remark by PSC.} There are a couple of other ways to prove that $2^p - 2 > 6p$ for $p \geqslant 7$. For example, we can use the Binomial Theorem as follows:
\[
2^p - 2 \geqslant 1 + p + \frac{p(p-1)}{2} + \frac{p(p-1)(p-2)}{6} - 2 \geqslant 1 + p + 3p + 5p - 2 > 6p.
\]
We can also use Bernoulli's Inequality as follows:
\[
2^p - 2 = 8(1+1)^{p-3} - 2 \geqslant 8(1+(p-3)) - 2 = 8p - 18 > 6p.
\]
The last inequality is true for $p \geqslant 11$. For $p = 7$ we can see directly that $2^p - 2 > 6p$.",198,1503,Number Theory,20
90,shl_jbmo_2019_n2,shl_jbmo,2019,n,"Find all triples $(p, q, r)$ of prime numbers such that all of the following numbers are integers
\[
\frac{p^2+2q}{q+r},\quad \frac{q^2+9r}{r+p},\quad \frac{r^2+3p}{p+q}.
\]","\textbf{Solution.} We consider the following cases:

\textbf{1st Case:} If $r = 2$, then $\dfrac{r^2+3p}{p+q} = \dfrac{4+3p}{p+q}$. If $p$ is odd, then $4+3p$ is odd and therefore $p+q$ must be odd. From here, $q = 2$ and $\dfrac{r^2+3p}{p+q} = \dfrac{4+3p}{p+2} = 3 - \dfrac{2}{p+2}$ which is not an integer. Thus $p = 2$ and $\dfrac{r^2+3p}{p+q} = \dfrac{10}{q+2}$ which gives $q = 3$. But then $\dfrac{q^2+9r}{r+p} = \dfrac{27}{4}$ which is not an integer. Therefore $r$ is an odd prime.

\textbf{2nd Case:} If $q = 2$, then $\dfrac{q^2+9r}{r+p} = \dfrac{4+9r}{r+p}$. Since $r$ is odd, then $4+9r$ is odd and therefore $r+p$ must be odd. From here $p = 2$, but then $\dfrac{r^2+3p}{p+q} = \dfrac{r^2+6}{4}$ which is not an integer. Therefore $q$ is an odd prime.

Since $q$ and $r$ are odd primes, then $q+r$ is even. From the number $\dfrac{p^2+2q}{q+r}$ we get that $p = 2$. Since $\dfrac{p^2+2q}{q+r} = \dfrac{4+2q}{q+r} < 2$, then $4+2q = q+r$ or $r = q+4$. Since
\[
\frac{r^2+3p}{p+q} = \frac{(q+4)^2+6}{2+q} = q + 6 + \frac{10}{2+q},
\]
is an integer, then $q = 3$ and $r = 7$. It is easy to check that this triple works. So the only answer is $(p, q, r) = (2, 3, 7)$.",173,1177,Number Theory,21
91,shl_jbmo_2019_n3,shl_jbmo,2019,n,"Find all prime numbers $p$ and nonnegative integers $x 
eq y$ such that $x^4 - y^4 = p(x^3 - y^3)$.","\textbf{Solution.} If $x = 0$ then $y = p$ and if $y = 0$ then $x = p$. We will show that there are no other solutions.

Suppose $x, y > 0$. Since $x 
eq y$, we have
\[
p(x^2+xy+y^2) = (x+y)(x^2+y^2). \tag{$*$}
\]
If $p$ divides $x+y$, then $x^2+y^2$ must divide $x^2+xy+y^2$ and so it must also divide $xy$. This is a contradiction as $x^2+y^2 \geqslant 2xy > xy$.

Thus $p$ divides $x^2+y^2$, so $x+y$ divides $x^2+xy+y^2$. As $x+y$ divides $x^2+xy$ and $y^2+xy$, it also divides $x^2$, $xy$ and $y^2$. Suppose $x^2 = a(x+y)$, $y^2 = b(x+y)$ and $xy = c(x+y)$. Then $x^2+xy+y^2 = (a+b+c)(x+y)$, $x^2+y^2 = (a+b)(x+y)$, while $(x+y)^2 = x^2+y^2+2xy = (a+b+2c)(x+y)$ yields $x+y = a+b+2c$.

Substituting into $(*)$ gives
\[
p(a+b+c) = (a+b+2c)(a+b).
\]
Now let $a+b = dm$ and $c = dc_1$, where $\gcd(m, c_1) = 1$. Then
\[
p(m+c_1) = (m+2c_1)dm.
\]
If $m+c_1$ and $m$ had a common divisor, it would divide $c_1$, a contradiction. So $\gcd(m, m+c_1) = 1$ and similarly, $\gcd(m+c_1, m+2c_1) = 1$. Thus $m+2c_1$ and $m$ divide $p$, so $m+2c_1 = p$ and $m = 1$. Then $m+c_1 = d$ so $c \leqslant d = a+b$. Now
\[
xy = c(x+y) \leqslant (a+b)(x+y) = x^2+y^2,
\]
again a contradiction.

\medskip

\textbf{Alternative Solution by PSC.} Let $d = \gcd(x, y)$. Then $x = da$ and $y = db$ for some $a, b$ such that $\gcd(a,b) = 1$. Then
\[
d^4(a^4 - b^4) = pd^3(a^3 - b^3),
\]
which gives
\[
d(a+b)(a^2+b^2) = p(a^2+ab+b^2). \tag{$*$}
\]
If a prime $q$ divides both $a+b$ and $a^2+ab+b^2$, then it also divides $(a+b)^2 - (a^2+ab+b^2) = ab$. So $q$ divides $a$ or $q$ divides $b$. Since $q$ also divides $a+b$, it must divide both $a$ and $b$. This is impossible as $\gcd(a,b) = 1$. So $\gcd(a+b, a^2+ab+b^2) = 1$ and similarly $\gcd(a^2+b^2, a^2+ab+b^2) = 1$. Then $(a+b)(a^2+b^2)$ divides $p$ and since $a+b \leqslant a^2+b^2$, then $a+b = 1$.

If $a = 0$, $b = 1$ then $(*)$ gives $d = p$ and so $x = 0$, $y = p$ which is obviously a solution. If $a = 1$, $b = 0$ we similarly get the solution $x = p$, $y = 0$. These are the only solutions.",99,2031,Number Theory,22
92,shl_jbmo_2019_n4,shl_jbmo,2019,n,"Find all integers $x$, $y$ such that
\[
x^3(y+1) + y^3(x+1) = 19.
\]","\textbf{Solution.} Substituting $s = x+y$ and $p = xy$ we get
\[
2p^2 - (s^2 - 3s)p + 19 - s^3 = 0. \tag{1}
\]
This is a quadratic equation in $p$ with discriminant $D = s^4 + 2s^3 + 9s^2 - 152$.

For each $s$ we have $D < (s^2+s+5)^2$ as this is equivalent to $(2s+5)^2 + 329 > 0$.

For $s \geqslant 11$ and $s \leqslant -8$ we have $D > (s^2+s+3)^2$ as this is equivalent to $2s^2 - 6s - 161 > 0$, and thus also to $2(s+8)(s-11) > -15$.

We have the following cases:
\begin{itemize}
    \item If $s \geqslant 11$ or $s \leqslant -8$, then $D$ is a perfect square only when $D = (s^2+s+4)^2$, or equivalently, when $s = -21$. From (1) we get $p = \dfrac{23}{2}$ (which yields no solution) or $p = 20$, giving the solutions $(-1, -20)$ and $(-20, -1)$.
    \item If $-7 \leqslant s \leqslant 10$, then $D$ is directly checked to be a perfect square only for $s = 3$. Then $p = \pm 2$ and only $p = 2$ gives solutions, namely $(2, 1)$ and $(1, 2)$.
\end{itemize}

\medskip

\textbf{Remark by PSC.} In the second bullet point, one actually needs to check 18 possible values of $s$ which is actually quite time consuming. We did not see many possible shortcuts. For example, $D$ is always a perfect square modulo 2 and modulo 3, while modulo 5 we can only get rid of the four cases of the form $s \equiv 0 \pmod{5}$.",68,1313,Number Theory,23
93,shl_jbmo_2019_n5,shl_jbmo,2019,n,"Find all positive integers $x, y, z$ such that
\[
45^x - 6^y = 2019^z.
\]","\textbf{Solution.} We define $v_3(n)$ to be the non-negative integer $k$ such that $3^k \mid n$ but $3^{k+1} 
mid n$. The equation is equivalent to
\[
3^{2x} \cdot 5^x - 3^y \cdot 2^y = 3^z \cdot 673^z.
\]
We will consider the cases $y 
eq 2x$ and $y = 2x$ separately.

\textbf{Case 1.} Suppose $y 
eq 2x$. Since $45^x > 45^x - 6^y = 2019^z > 45^z$, then $x > z$ and so $2x > z$. We have
\[
z = v_3(3^z \cdot 673^z) = v_3(3^{2x} \cdot 5^x - 3^y \cdot 2^y) = \min\{2x, y\},
\]
as $y 
eq 2x$. Since $2x > z$, we get $z = y$. Hence the equation becomes $3^{2x} \cdot 5^x - 3^y \cdot 2^y = 3^y \cdot 673^y$, or equivalently,
\[
3^{2x-y} \cdot 5^x = 2^y + 673^y.
\]

\textbf{Case 1.1.} Suppose $y = 1$. Doing easy manipulations we have
\[
3^{2x-1} \cdot 5^x = 2 + 673 = 675 = 3^3 \cdot 5^2 \implies 45^{x-2} = 1 \implies x = 2.
\]
Hence one solution which satisfies the condition is $(x, y, z) = (2, 1, 1)$.

\textbf{Case 1.2.} Suppose $y \geqslant 2$. Using properties of congruences we have
\[
1 \equiv 2^y + 673^y \equiv 3^{2x-y} \cdot 5^y \equiv (-1)^{2x-y} \pmod{4}.
\]
Hence $2x - y$ is even, which implies that $y$ is even. Using this fact we have
\[
0 \equiv 3^{2x-y} \cdot 5^y \equiv 2^y + 673^y \equiv 1 + 1 \equiv 2 \pmod{3},
\]
which is a contradiction.

\textbf{Case 2.} Suppose $y = 2x$. The equation becomes $3^{2x} \cdot 5^x - 3^{2x} \cdot 2^{2x} = 3^z \cdot 673^z$, or equivalently,
\[
5^x - 4^x = 3^{z-2x} \cdot 673^z.
\]
Working modulo 3 we have
\[
(-1)^x - 1 \equiv 5^x - 4^x \equiv 3^{z-2x} \cdot 673^z \equiv 0 \pmod{3},
\]
hence $x$ is even, say $x = 2t$ for some positive integer $t$. The equation is now equivalent to
\[
(5^t - 4^t)(5^t + 4^t) = 3^{z-4t} \cdot 673^z.
\]
It can be checked by hand that $t = 1$ is not possible. For $t \geqslant 2$, since 3 and 673 are the only prime factors of the right hand side, and since, as it is easily checked $\gcd(5^t - 4^t, 5^t + 4^t) = 1$ and $5^t - 4^t > 1$, the only way for this to happen is when $5^t - 4^t = 3^{z-4t}$ and $5^t + 4^t = 673^z$ or $5^t - 4^t = 673^z$ and $5^t + 4^t = 3^{z-4t}$. Adding together we have
\[
2 \cdot 5^t = 3^{z-4t} + 673^z.
\]
Working modulo 5 we have
\[
0 \equiv 2 \cdot 5^t \equiv 3^{z-4t} + 673^z \equiv 3^{4t} \cdot 3^{z-4t} + 3^z \equiv 2 \cdot 3^z \pmod{5},
\]
which is a contradiction. Hence the only solution which satisfies the equation is $(x, y, z) = (2, 1, 1)$.

\medskip

\textbf{Alternative Solution by PSC.} Working modulo 5 we see that $-1 \equiv 4^z \pmod{5}$ and therefore $z$ is odd. Now working modulo 4 and using the fact that $z$ is odd we get that $1 - 2^y \equiv 3^z \equiv 3 \pmod{4}$. This gives $y = 1$. Now working modulo 9 we have $-6 \equiv 3^z \pmod{9}$ which gives $z = 1$. Now since $y = z = 1$ we get $x = 2$ and so $(2, 1, 1)$ is the unique solution.",73,2782,Number Theory,24
94,shl_jbmo_2019_n6,shl_jbmo,2019,n,"Find all triples $(a, b, c)$ of nonnegative integers that satisfy
\[
a! + 5^b = 7^c.
\]","\textbf{Solution.} We cannot have $c = 0$ as $a! + 5^b \geqslant 2 > 1 = 7^0$.

Assume first that $b = 0$. So we are solving $a! + 1 = 7^c$. If $a \geqslant 7$, then $7 \mid a!$ and so $7 
mid a! + 1$. So $7 
mid 7^c$ which is impossible as $c 
eq 0$. Checking $a < 7$ by hand, we find the solution $(a, b, c) = (3, 0, 1)$.

We now assume that $b > 0$. In this case, if $a \geqslant 5$, we have $5 \mid a!$, and since $5 \mid 5^b$, we have $5 \mid 7^c$, which obviously cannot be true. So we have $a \leqslant 4$. Now we consider the following cases:

\textbf{Case 1.} Suppose $a = 0$ or $a = 1$. In this case, we are solving the equation $5^b + 1 = 7^c$. However the Left Hand Side of the equation is always even, and the Right Hand Side is always odd, implying that this case has no solutions.

\textbf{Case 2.} Suppose $a = 2$. Now we are solving the equation $5^b + 2 = 7^c$. If $b = 1$, we have the solution $(a, b, c) = (2, 1, 1)$. Now assume $b \geqslant 2$. We have $5^b + 2 \equiv 2 \pmod{25}$ which implies that $7^c \equiv 2 \pmod{25}$. However, by observing that $7^4 \equiv 1 \pmod{25}$, we see that the only residues that $7^c$ can have when divided by 25 are $7, 24, 18, 1$. So this case has no more solutions.

\textbf{Case 3.} Suppose $a = 3$. Now we are solving the equation $5^b + 6 = 7^c$. We have $5^b + 6 \equiv 1 \pmod{5}$ which implies that $7^c \equiv 1 \pmod{5}$. As the residues of $7^c$ modulo 5 are $2, 4, 3, 1$, in that order, we obtain $4 \mid c$.

Viewing the equation modulo 4, we have $7^c \equiv 5^b + 6 \equiv 1 + 2 \equiv 3 \pmod{4}$. But as $4 \mid c$, we know that $7^c$ is a square, and the only residues that a square can have when divided by 4 are 0, 1. This means that this case has no solutions either.

\textbf{Case 4.} Suppose $a = 4$. Now we are solving the equation $5^b + 24 = 7^c$. We have $5^b \equiv 7^c - 24 \equiv 1 - 24 \equiv 1 \pmod{3}$. Since $5 \equiv 2 \pmod{3}$, we obtain $2 \mid b$. We also have $7^c \equiv 5^b + 24 \equiv 4 \pmod{5}$, and so we obtain $c \equiv 2 \pmod{4}$. Let $b = 2m$ and $c = 2n$. Observe that
\[
24 = 7^c - 5^b = (7^n - 5^m)(7^n + 5^m).
\]
Since $7^n + 5^m > 0$, we have $7^n - 5^m > 0$. There are only a few ways to express $24 = 24 \cdot 1 = 12 \cdot 2 = 8 \cdot 3 = 6 \cdot 4$ as a product of two positive integers. By checking these cases we find one by one, the only solution in this case is $(a, b, c) = (4, 2, 2)$.

Having exhausted all cases, we find that the required set of triples is
\[
(a,b,c) \in \{(3,0,1),(2,1,1),(4,2,2)\}.
\]",87,2528,Number Theory,25
95,shl_jbmo_2019_n7,shl_jbmo,2019,n,Find all perfect squares $n$ such that if the positive integer $a \leqslant 15$ is some divisor of $n$ then $a + 15$ is a prime power.,"\textbf{Solution.} We call a positive integer $a$ ``nice'' if $a + 15$ is a prime power.

From the definition, the numbers $n = 1, 4, 9$ satisfy the required property. Suppose that for some $t \in \mathbb{Z}^+$, the number $n = t^2 \geqslant 15$ also satisfies the required property. We have two cases:

\begin{enumerate}
    \item If $n$ is a power of 2, then $n \in \{16, 64\}$ since
    \[
    2^4 + 15 = 31,\quad 2^5 + 15 = 47,\quad\text{and}\quad 2^6 + 15 = 79
    \]
    are prime, and $2^7 + 15 = 143 = 11 \cdot 13$ is not a prime power. (Thus $2^7$ does not divide $n$ and therefore no higher power of 2 satisfies the required property.)
    
    \item Suppose $n$ has some odd prime divisor $p$. If $p > 3$ then $p^2 \mid n$ and $p^2 > 15$ which imply that $p^2$ must be a nice number. Hence
    \[
    p^2 + 15 = q^m
    \]
    for some prime $q$ and some $m \in \mathbb{Z}^+$. Since $p$ is odd, then $p^2 + 15$ is even, thus we can conclude that $q = 2$. I.e.\
    \[
    p^2 + 15 = 2^m.
    \]
    Considering the above modulo 3, we can see that $p^2 + 15 \equiv 0, 1 \pmod{3}$, so $2^m \equiv 1 \pmod{3}$, and so $m$ is even. Suppose $m = 2k$ for some $k \in \mathbb{Z}^+$. So we have
    \[
    (2^k - p)(2^k + p) = 15 \quad\text{and}\quad (2^k + p) - (2^k - p) = 2p \geqslant 10.
    \]
    Thus
    \[
    2^k - p = 1 \quad\text{and}\quad 2^k + p = 15,
    \]
    giving $p = 7$ and $k = 3$. Thus we can write $n = 4^x \cdot 9^y \cdot 49^z$ for some non-negative integers $x, y, z$.
    
    Note that 27 is not nice, so $27 
mid n$ and therefore $y \leqslant 1$. The numbers 18 and 21 are also not nice, so similarly, $x, y$ and $y, z$ cannot both be positive. Hence, we just need to consider $n = 4^x \cdot 49^z$ with $z \geqslant 1$.
    
    Note that $7^3$ is not nice, so $z = 1$. By checking directly, we can see that $7^2 + 15 = 2^6$, $2 \cdot 7^2 + 15 = 113$, $4 \cdot 7^2 + 15 = 211$ are nice, but $8 \cdot 7^2$ is not nice, so only $n = 49, 196$ satisfy the required property.
\end{enumerate}

Therefore, the numbers $n$ which satisfy the required property are $1, 4, 9, 16, 49, 64$ and $196$.

\medskip

\textbf{Remark by PSC.} One can get rid of the case $3 \mid n$ by noting that in that case, we have $9 \mid n$. But then $n^2 + 15$ is a multiple of 3 but not a multiple of 9 which is impossible. This simplifies a little bit the second case.",134,2373,Number Theory,26
424,tst_jbmo_ro_2019_1_p1,tst_jbmo,2019,n,Let $n$ be a given positive integer. Determine all positive divisors $d$ of $3n^2$ such that $n^2 + d$ is the square of an integer.,"
oindent\textbf{Solution 1.}
If $d$ divides $3n^2$, then there exist positive integers $k$ and $m$ such that $3n^2 = d \cdot k$ and $n^2 + d = m^2$. We substitute to get $n^2 + \frac{3n^2}{k} = m^2$, so $(mk)^2 = n^2(k^2 + 3k)$. We deduce that $k^2 + 3k$ is a perfect square. From the inequalities $k^2 < k^2 + 3k < (k+2)^2$ we deduce that $k^2 + 3k = (k+1)^2$, which implies $k = 1$ and $d = 3n^2$, which verifies the problem.

\bigskip

oindent\textbf{Solution 2.}
Let $d$ be a divisor of $3n^2$ such that $n^2 + d = m^2$. We have $m > n$ and $d = (m-n)(m+n)$. Denote $D = \gcd(m,n)$, $m = Da$ and $n = Db$. Since $(m-n)(m+n)$ divides $3n^2$, we have $(a-b)(a+b) \mid 3b^2$. The numbers $a$ and $b$ are coprime, therefore $(a-b)(a+b) \mid 3$. The case $a - b = a + b = 1$ implies $b = 0$, which is not possible. The only possibility is $a - b = 1$ and $a + b = 3$, meaning that $a = 2$ and $b = 1$. Finally we have $m = 2n$ and $d = (2n)^2 - n^2 = 3n^2$.",131,956,Number Theory,1
425,tst_jbmo_ro_2019_1_p2,tst_jbmo,2019,a,"Find the maximum value of the expression
\[
E(a,b) = \frac{a+b}{(4a^2+3)(4b^2+3)}
\]
when $a, b \in \mathbb{R}$.","
oindent\textbf{Solution 1.}
We will show that the maximum value is $\dfrac{1}{16}$, obtained when $a = b = \dfrac{1}{2}$.

The inequality $E(a,b) \leq \dfrac{1}{16}$ is equivalent to $16(a+b) \leq (4a^2+3)(4b^2+3)$, which we rewrite as
\[
(4ab - 1)^2 + 4(a+b-1)^2 + 2(2a-1)^2 + 2(2b-1)^2 \geq 0,
\]
obviously true.

\medskip

oindent\textit{Remark.} The inequality $16(a+b) \leq (4a^2+3)(4b^2+3)$ can be written as $f(a) \geq 0$, $\forall\, a, b \in \mathbb{R}$, where $f(a) = a^2(16b^2+12) - 16a + (12b^2 - 16b + 9)$ is a quadratic function in $a$.

\bigskip

oindent\textbf{Solution 2.}
We will show that the maximum value is $\dfrac{1}{16}$, obtained when $a = b = \dfrac{1}{2}$.

We combine the following inequalities:
\[
a + b \leq \frac{(1 + a + b)^2}{4}
\]
and (from CBS)
\[
(4a^2+3)(4b^2+3) = (4a^2+1+2)(1+4b^2+2) \geq (2a+2b+2)^2.
\]
We get:
\[
\frac{a+b}{(4a^2+3)(4b^2+3)} \leq \frac{(a+b+1)^2}{4(4a^2+3)(4b^2+3)} \leq \frac{(a+b+1)^2}{4(2a+2b+2)^2} \leq \frac{1}{16}.
\]",112,982,Algebra,2
426,tst_jbmo_ro_2019_1_p3,tst_jbmo,2019,g,"Let $ABC$ be a triangle, $I$ the incenter, $D$ the contact point of the incircle with the side $BC$, and $E$ the foot of the bisector of angle $A$. If $M$ is the midpoint of the arc $BC$ which contains the point $A$ of the circumcircle of triangle $ABC$, and $\{F\} = DI \cap AM$, prove that $MI$ passes through the midpoint of $[EF]$.","
oindent\textbf{Solution.}
The bisector of angle $\widehat{A}$ passes through the midpoint of the arc $BC$ which does not contain the point $A$. Denote this point by $S$. $MS$ is the perpendicular bisector of $[BC]$, so $MS \parallel ID$ (both lines are perpendicular on $BC$). Also, $\angle ASC = \angle ABC$ and $\angle SAC = \angle BAE$ proves that triangles $SAC$ and $BAE$ are similar, so $\dfrac{AB}{BE} = \dfrac{AS}{SC}$. Using the angle bisector theorem, we get $\dfrac{AB}{BE} = \dfrac{AI}{IE}$.

It is known that $SI = SC$ (one might compute the angles of triangle $SCI$). We have $\dfrac{AS}{SC} = \dfrac{AS}{SI} = \dfrac{AM}{MF}$. We proved $\dfrac{AI}{IE} = \dfrac{AM}{MF}$. Using Menelaus' theorem in the triangle $AEF$ and the transversal $I - X - M$ (where $\{X\} = IM \cap EF$) we have
\[
\frac{AI}{IE} \cdot \frac{EX}{XF} \cdot \frac{MF}{MA} = 1.
\]
Using the equality we proved before, we find $\dfrac{EX}{XF} = 1$, which means that $X$ is the midpoint of $[EF]$.",335,982,Geometry,3
427,tst_jbmo_ro_2019_1_p4,tst_jbmo,2019,c,"Ana and Bogdan play the following turn-based game: Ana starts with a pile of $n$ ($n \geq 3$) stones. At his turn each player has to split one pile. The winner is the player who can make at his turn all the piles to have at most two stones. Depending on $n$, determine which player has a winning strategy.","
oindent\textbf{Solution 1.}
If $n = 3$ or $n = 4$, Ana wins at her first move. If $n$ is odd, greater than 3, Bogdan will win. In this case, Ana has to start by making a pile with an even number of stones. Bogdan will split this pile into a pile with one stone and the rest into an odd pile. Ana has to make an even pile again and Bogdan continues his strategy unless he can win. The game may end in two different ways. If Ana leaves a pile of 2, one of 3, and the rest of 1, Bogdan will win by splitting the pile with 3 stones. If Ana leaves a pile of 4 and the rest of 1, Bogdan wins by splitting the 4-pile into two piles of 2 stones.

If $n \geq 6$ is even, then Ana will split the pile in 1 and $n-1$ and continue with the strategy described for Bogdan above. So, in this case, Ana will win.

\bigskip

oindent\textbf{Solution 2.}
For $n$ even, Ana splits the pile into two equal piles. Then, after Bogdan's move into one pile, she will make the same move in the other pile. This strategy will end up with Ana winning. For $n \geq 5$ odd, we prove by induction that Bogdan wins by always leaving an even number of piles with 2 stones. We verify the cases $n = 5, 7, 9$. For odd $n \geq 11$, we have two situations. If Ana splits $n = a + b$, with $a \geq 5$ odd, Bogdan will split the pile with $b$ stones into two equal piles. If Ana moves in one of the equal piles, Bogdan will make the same move in the other pile. If Ana moves in the odd pile, Bogdan will follow the strategy provided by the induction hypothesis to end up with piles of 1 stone and an even number of piles of 2 stones. Ana might still be able to move in a pile of 2, but whenever she splits a pile of 2, so will Bogdan, leaving an even number of piles of 2 stones. If $a = 3$, Bogdan will split $b = 3 + (b-3)$, where $b - 3 \geq 5$, and apply the same strategy as above.",305,1848,Combinatorics,4
428,tst_jbmo_ro_2019_2_p1,tst_jbmo,2019,n,"Let $n$ be a nonnegative integer and $M = \{n^3,\, n^3+1,\, n^3+2,\, \ldots,\, n^3+n\}$. Consider $A$ and $B$ two nonempty, disjoint subsets of $M$ such that the sum of elements of the set $A$ divides the sum of elements of the set $B$. Prove that the number of elements of the set $A$ divides the number of elements of the set $B$.","
oindent\textbf{Solution.}
Denote $A = \{n^3+n_1,\, n^3+n_2,\, \ldots,\, n^3+n_a\}$, $B = \{n^3+m_1,\, n^3+m_2,\, \ldots,\, n^3+m_b\}$ and $k \in \mathbb{N}$ such that
\[
n^3+m_1 + n^3+m_2 + \cdots + n^3+m_b = k(n^3+n_1 + n^3+n_2 + \cdots + n^3+n_a).
\]
Then $n^3(ka - b) = m_1 + m_2 + \cdots + m_b - k(n_1 + n_2 + \cdots + n_a)$.

The case $n = 1$ is obviously true since $M = \{1, 2\}$ and we can only have $A = \{1\}$ and $B = \{2\}$.

Now let us prove that $n > 1$ implies $k < n+1$ (so $k \leq n$). Indeed, supposing $k \geq n+1$, we would get
\[
n^3+1 + n^3+2 + \cdots + n^3+n \geq n^3+m_1 + \cdots + n^3+m_b \geq (n+1)(n^3+n_1 + \cdots + n^3+n_a) \geq (n+1)n^3,
\]
therefore $n^4 + \dfrac{n(n+1)}{2} \geq n^4 + n^3$. This would imply $n^2 + n \geq 2n^3$, which is false.

We are left with the case $k \leq n$. Now we have
\[
m_1 + m_2 + \cdots + m_b - k(n_1 + n_2 + \cdots + n_a) < 1 + 2 + \cdots + n = \frac{n(n+1)}{2} < n^3
\]
and
\[
m_1 + m_2 + \cdots + m_b - k(n_1 + n_2 + \cdots + n_a) \geq -n(n_1+n_2+\cdots+n_a) \geq -n(1+2+\cdots+n) = -\frac{n^2(n+1)}{2} > -n^3.
\]
To summarize, we have the inequalities $-n^3 < n^3(ka - b) < n^3$, therefore $ka - b = 0$, showing that $a$ divides $b$.",332,1201,Number Theory,5
429,tst_jbmo_ro_2019_2_p2,tst_jbmo,2019,a,"If $x, y$ and $z$ are real numbers such that $x^2 + y^2 + z^2 = 2$, prove that $x + y + z \leq xyz + 2$.","
oindent\textbf{Solution 1.}
Notice that $2xy \leq x^2 + y^2 \leq x^2 + y^2 + z^2 = 2$, therefore $xy \leq 1$. Similarly, $xz \leq 1$, and $yz \leq 1$. We also have $(x+y)^2 = x^2 + y^2 + 2xy \leq 4$, so $x + y \leq |x+y| \leq 2$. Also, $x + z \leq 2$, and $y + z \leq 2$. If one of the numbers is negative, let us say $z$, then the conclusion follows from the inequalities $x + y \leq 2$ and $z \leq xyz$.

In the case $x, y, z \in [0, 1]$, we have $z(1 - xy) \leq 1 - xy \leq 2 - x - y$ (the last inequality is equivalent to $(1-x)(1-y) \geq 0$).

We may have at most one number greater than 1, otherwise the product of two such numbers would be greater than 1. We are left with the case $x, y \in [0,1]$ and $z > 1$. In this case
\[
(1-x)(1-y)(z-1) \geq 0 \iff xyz + 2 \geq xy + yz + xz + 3 - (x+y+z) \geq x+y+z,
\]
the last inequality being equivalent to $2(xy + yz + zx) + 6 = (x+y+z)^2 + 4 \geq 4(x+y+z)$.

\bigskip

oindent\textbf{Solution 2.}
Using the Cauchy--Buniakowsky--Schwarz inequality we get
\begin{align*}
(x + y + z - xyz)^2 &= ((x+y) + z(1-xy))^2 \leq ((x+y)^2 + z^2)(1 + (1-xy)^2) \\
&= (2 + 2xy)(2 - 2xy + x^2y^2) = 4 + 2x^2y^2(xy - 1) \leq 4
\end{align*}
since $x^2 + y^2 \leq 2$ and $xy \leq 1$.",104,1218,Algebra,6
430,tst_jbmo_ro_2019_2_p3,tst_jbmo,2019,g,"Let $d$ be the tangent at $B$ to the circumcircle of the acute scalene triangle $ABC$. Let $K$ be the orthogonal projection of the orthocenter $H$ of triangle $ABC$ to the line $d$, and $L$ the midpoint of the side $AC$. Prove that the triangle $BKL$ is isosceles.","
oindent\textbf{Solution 1.}
Without loss of generality, we suppose $AB < BC$. Denote by $A'$ and $C'$ the feet of the altitudes from $A$ and $C$, respectively. Points $A'$, $C'$, $K$ belong to the circle of diameter $BH$. The quadrilateral $ACA'C'$ is cyclic, therefore $\angle BC'A' = \angle C = \angle KBA$, so the trapezoid $BKC'A'$ is cyclic, hence isosceles. On the other hand, $C'L$ and $A'L$ are medians in the right triangles $ACC'$ and $ACA'$, respectively, so $C'L = A'L = \dfrac{BC}{2}$ and the triangle $LA'C'$ is isosceles. It is easy to prove the congruence of the triangles $LC'K$ and $LA'B$ (SAS) and the conclusion follows.

\bigskip

oindent\textbf{Solution 2.}
Without loss of generality, we suppose $AB < BC$. Denote by $O$ the circumcenter of triangle $ABC$ and $T$ the midpoint of $[BH]$. $BTLO$ is a parallelogram (we have $LO \parallel BH$, $BH = 2R\cos B$ and $LO = R\cos B$). $LT$ is perpendicular to $d$ and it passes through the midpoint of $BK$, therefore $LT$ is the perpendicular bisector of $BK$ and we are done.",264,1045,Geometry,7
431,tst_jbmo_ro_2019_2_p4,tst_jbmo,2019,c,"In every unit square of an $n \times n$ table ($n \geq 11$) a real number is written, such that the sum of the numbers in any $10 \times 10$ square is positive and the sum of the numbers in any $11 \times 11$ square is negative. Determine all possible values for $n$.","
oindent\textbf{Solution.}
First we prove that $n = 19$ is possible. We write $100$ in the central square and $-1$ in the other ones. Any $10 \times 10$ square contains the central square and $99$ other squares, therefore the sum in any $10 \times 10$ square is $1$. Any $11 \times 11$ square contains the central square and $120$ other squares, therefore it has the sum $-20$. A similar example works for any $11 \leq n \leq 19$: write $100$ in the central square (or into one of the four central squares if $n$ is even) and $-1$ into the other ones. (Or simply take the squares situated in the first $n$ rows and first $n$ columns.)

$n = 20$ is not possible. Suppose the contrary. There are $11^2$ squares $10 \times 10$ and $10^2$ squares $11 \times 11$. We will show that the total sums of the $10 \times 10$ squares and of the $11 \times 11$ squares are the same, leading to a contradiction. Divide the table into four $10 \times 10$ disjoint regions. Let us label the rows and the columns starting from the top left. Consider a unit square in the top left region. Suppose the unit square we considered has coordinates $(i, j)$. This unit square is part of any $10 \times 10$ square which has the top left unit square of coordinates $(a, b)$ with $1 \leq a \leq i$ and $1 \leq b \leq j$. In the total sum of the $10 \times 10$ squares, it will contribute $i \cdot j$ times. The same unit square is part of any $11 \times 11$ square which has the top left unit square of coordinates $(a, b)$ with $1 \leq a \leq i$ and $1 \leq b \leq j$. In the total sum of the $11 \times 11$ squares it will have the same contribution, $i \cdot j$ times. We use the same argument in the other three regions to conclude that the two sums are equal and we get the desired contradiction.

$n > 20$ is also impossible since we can apply the previous reasoning to any $20 \times 20$ sub-square.",267,1879,Combinatorics,8
432,tst_jbmo_ro_2019_3_p1,tst_jbmo,2019,n,"Determine positive integers $a$ and $b$, coprime, such that $a^2 + b = (a - b)^3$.","
oindent\textbf{Solution 1.}
Denote $c = a - b$. Obviously $c \in \mathbb{Z}$, $c \geq 2$. Substituting $a = b + c$ in the equation we have $(b+c)^2 + b = c^3$, which implies $c \mid b^2 + b$ and, since $\gcd(b,c) = 1$, we find that $c \mid b + 1$. Similarly $b$ divides $c^3 - c^2 = c^2(c-1)$, so $b$ divides $c - 1 > 0$. We have $c \leq b + 1$ and $b \leq c - 1$, so $b = c - 1$. We plug into the initial equation to get $4c^2 - 3c = c^3$, which is equivalent to $c(c-1)(c-3) = 0$. Since $c \geq 2$, the only possibility is $c = 3$, $b = 2$, $a = 5$, which satisfy the conditions of the problem.

\bigskip

oindent\textbf{Solution 2.}
From the equation $a^2 + b = a^3 - 3a^2b + 3ab^2 - b^3$, we deduce that $a$ divides $b^3 + b = b(b^2+1)$ and $b$ divides $a^3 - a^2 = a^2(a-1)$. Since $\gcd(a,b) = 1$, we have $a \mid b^2 + 1$ and $b \mid a - 1$. Let $k \in \mathbb{Z}_+$ such that $a - 1 = kb$. Then $a = kb + 1$ divides $b^2 + 1$ and also divides $k(b^2+1) - b(kb+1) = k - b$. Obviously $kb + 1 > |k - b|$, therefore $k - b = 0$, which implies $a = b^2 + 1$. Plugging into the initial equation gives
\[
b(b^5 - 3b^4 + 5b^3 - 7b^2 + 4b - 4) = 0.
\]
The positive integer solutions of the equation $b^5 - 3b^4 + 5b^3 - 7b^2 + 4b - 4 = 0$ must be positive divisors of 4. We check $b = 1$, $b = 2$, $b = 4$ and we get the only solution $b = 2$ and $a = b^2 + 1 = 5$.",82,1366,Number Theory,9
433,tst_jbmo_ro_2019_3_p2,tst_jbmo,2019,n,"Let $n$ be a positive integer and $A$ a set containing $8n + 1$ positive integers coprime with $6$ and less than $30n$. Prove that there exist $a, b \in A$ two different numbers such that $a$ divides $b$.","
oindent\textbf{Solution.}
In the set $\{1, 2, \ldots, 30n\}$ there are $8n$ numbers coprime with $30$.

Indeed, let us denote
\[
B_1 = \{1 \leq x \leq 30n \mid 2 \text{ divides } x\},\quad B_2 = \{1 \leq x \leq 30n \mid 3 \text{ divides } x\},\quad B_3 = \{1 \leq x \leq 30n \mid 5 \text{ divides } x\}.
\]
We have $B = \{1, 2, \ldots, 30n\} \setminus (B_1 \cup B_2 \cup B_3)$ and since
\[
|B_1 \cup B_2 \cup B_3| = 15n + 10n + 6n - 5n - 3n - 2n + n = 22n,
\]
we have $|B| = 30n - 22n = 8n$. (Different approach: we have $n$ numbers for each of the remainders $1, 7, 11, 13, 17, 19, 23, 29$ modulo $30$.)

Denote $x_1, x_2, \ldots, x_{8n}$ the elements of $B$. Partition the set $A$ into $8n$ classes $C_i = \{x_i \cdot 5^k \in A\}$. We have $8n + 1$ numbers and $8n$ classes, so, by the Pigeonhole Principle, one of the classes $C_i$ must contain at least two elements $a = x_i \cdot 5^s$ and $b = x_i \cdot 5^t$. Then $a \mid b$ or $b \mid a$.",204,946,Number Theory,10
434,tst_jbmo_ro_2019_3_p3,tst_jbmo,2019,g,"A circle with center $O$ is internally tangent to two circles inside it at points $S$ and $T$. Suppose the two circles inside intersect at $M$ and $N$ with $N$ closer to $ST$. Show that $OM$ and $MN$ are perpendicular if and only if $S$, $N$, $T$ are collinear.","
oindent\textbf{Solution.}
The tangent lines at $S$ and $T$ to the big circle meet at a point $K$. The quadrilateral $KSOT$ is cyclic and $KS = KT$. $KS$ and $KT$ are also tangent to the small circles. The point $K$ has the same power with respect to the two small circles, therefore $K$ is on their radical axis, which is $MN$. From $KS^2 = KN \cdot KM$ we get the similarity of the triangles $KSN$ and $KMS$, so $\angle KNS = \angle KSM$. Similarly, we get $\angle KNT = \angle KTM$. Then we have:
\begin{align*}
S, N, T \text{ are collinear} &\iff \angle KNS + \angle KNT = 180^\circ \\
&\iff \angle KSM + \angle KTM = 180^\circ \\
&\iff SMTK \text{ is cyclic} \\
&\iff O, M, S, T, K \text{ are concyclic} \\
&\iff \angle OMK = 90^\circ \\
&\iff OM \perp MN.
\end{align*}",261,774,Geometry,11
435,tst_jbmo_ro_2019_3_p4,tst_jbmo,2019,a,Let $a$ and $b$ be positive real numbers such that $3(a^2 + b^2 - 1) = 4(a + b)$. Find the minimum value of the expression $\dfrac{16}{a} + \dfrac{1}{b}$.,"
oindent\textbf{Solution.}
From the Cauchy--Schwarz inequality, we have
\begin{equation}
\frac{16}{a} + \frac{1}{b} = \frac{8^2}{4a} + \frac{1}{b} \geq \frac{(8+1)^2}{4a + b} = \frac{81}{4a+b}. \tag{1}
\end{equation}

On the other side, from the condition in the hypothesis we get:
\[
9(a^2 + b^2 - 1) = 12(a+b),
\]
\[
(3a-2)^2 + (3b-2)^2 = 17.
\]
Now, using the Cauchy--Buniakowsky--Schwarz inequality, we have
\[
\bigl[(3a-2)^2 + (3b-2)^2\bigr](4^2 + 1^2) \geq [4(3a-2) + 3b-2]^2,
\]
or
\[
(12a + 3b - 10)^2 \leq 17^2,
\]
so
\[
4a + b \leq 9.
\]
Going back to (1), we find the minimum value:
\[
\frac{16}{a} + \frac{1}{b} \geq \frac{81}{4a+b} \geq 9.
\]
Equality is obtained for $a = 2$ and $b = 1$, which satisfy the condition of the problem. Thus, the minimum value is $\mathbf{9}$.

\medskip

oindent\textit{Remark.} The inequality $4a + b \leq 9$ can also be obtained as follows: from $(a-2)^2 \geq 0$ and $(b-1)^2 \geq 0$ we get $a^2 + b^2 \geq 4a + 2b - 5$, or $4a + 4b = 3(a^2+b^2-1) \geq 3(4a + 2b - 6)$, which means $8a + 2b \leq 9$.

Wait, this should read: $4a + 4b = 3(a^2+b^2-1) \geq 3(4a+2b-6)$, yielding $4a+4b \geq 12a+6b-18$, i.e.\ $18 \geq 8a+2b$, so $4a+b \leq 9$ after division by 2.",154,1205,Algebra,12
436,tst_jbmo_ro_2019_4_p1,tst_jbmo,2019,n,"Determine all positive integers $k$ for which there exist positive integers $n$ and $m$, $m \geq 2$, such that $3^k + 5^k = n^m$.","
oindent\textbf{Solution.}
Clearly $k = 1$ satisfies the desired property: $3^1 + 5^1 = 2^3$. We prove that the other positive integers do not satisfy it.

If $k$ is even, then $3^k \equiv 5^k \equiv 1 \pmod{4}$, hence $3^k + 5^k \equiv 2 \pmod{4}$. It follows that the exponent of 2 in the prime factorization of $3^k + 5^k$ is 1, while the exponent of 2 in the prime factorization of $n^m$ must be a multiple of $m$. Thus, $k$ cannot be even.

For $k > 1$ odd, we can write $3^k + 5^k = (3+5)(3^{k-1} - 3^{k-2} \cdot 5 + \cdots + 5^{k-1})$. The second factor is a sum consisting of an odd number of odd terms, thus it is odd. The exponent of 2 in $3^k + 5^k$ being 3, we must have $m = 3$.

Notice that $3^k \equiv 0 \pmod{9}$ and $n^3$, a perfect cube, can be congruent only to $-1$, $0$, or $1$ modulo 9. But $5^k \equiv 5 \pmod{9}$ if $k \equiv 1 \pmod{6}$, $5^k \equiv -1$ if $k \equiv 3 \pmod{6}$, and $5^k \equiv 3 \pmod{9}$ if $k \equiv 5 \pmod{6}$, which leaves the only possibility $k \equiv 3 \pmod{6}$, i.e.\ $k$ is a multiple of 3.

We have thus obtained an equation of the form $x^3 + y^3 = z^3$, $x, y, z \in \mathbb{N}$, which is known not to have positive solutions (Fermat's Last Theorem).

Alternatively, analyzing modulo 7 we get $3^k + 5^k \equiv 5 \pmod{7}$ if $k \equiv 3 \pmod{6}$, but it is easy to see a cube can only be congruent to $-1$, $0$, or $1$ modulo 7.",129,1388,Number Theory,13
437,tst_jbmo_ro_2019_4_p2,tst_jbmo,2019,g,"Let $O$ be the circumcenter of an acute triangle $ABC$ in which $\angle B < \angle C$. Line $AO$ intersects side $BC$ at $D$. Let $E$ and $F$ be the circumcenters of triangles $ABD$ and $ACD$, respectively. On $[AB]$ produced and $[AC]$ produced, beyond $A$, consider points $G$ and $H$, respectively, such that $AG = AC$ and $AH = AB$. Prove that the quadrilateral $EFGH$ is a rectangle if and only if $\angle ACB - \angle ABC = 60^\circ$.","
oindent\textbf{Solution.}
Obviously, $EF \perp AD$. Also, $\angle BAO = 90^\circ - \angle C = 90^\circ - \angle AGH$, so $AD \perp GH$. Thus, $EF \parallel GH$.

On the other hand, $\angle ADC = \angle B + \angle BAD = 90^\circ - \angle C + \angle B < 90^\circ$, which means that triangle $ADC$ is acute and, consequently, $F \in \mathrm{int}(\angle DAC)$. Angle $\angle ADB$ is obtuse, therefore $E \in \mathrm{int}(\angle EAD)$.

It is easy to see that $\angle AFC = 2 \cdot \angle ADC = \angle AEB$. Triangles $AFC$ and $AEB$ are isosceles and have equal angles at their apex, so they are similar. It follows that $\angle EAF = \angle A$ and $\dfrac{EA}{AB} = \dfrac{FA}{AC}$, which shows that triangles $AEF$ and $ABC$ are also similar.

It follows that $EFGH$ is a parallelogram if and only if $EF = GH$, i.e.\ if and only if triangles $AEF$ and $ABC$ are equal. This is equivalent to triangles $ABE$ and $ACF$ being equilateral, i.e.\ to $\angle ADC = 30^\circ$, which is equivalent to $\angle ACB - \angle ABC = 60^\circ$.

We still have to prove that, in the situation described above, $EFGH$ is a rectangle. But $AE = AB = AH$ and $AF = AC = AG$ show that in the parallelogram $EFGH$ the perpendicular bisectors of two opposite sides do meet, which only happens when the parallelogram is a rectangle.",440,1310,Geometry,14
438,tst_jbmo_ro_2019_4_p3,tst_jbmo,2019,a,"Real numbers $a, b, c, d$ satisfy conditions $|a|, |b|, |c|, |d| > 1$ and $abc + abd + acd + bcd + a + b + c + d = 0$. Prove that
\[
\frac{1}{a-1} + \frac{1}{b-1} + \frac{1}{c-1} + \frac{1}{d-1} > 0.
\]","
oindent\textbf{Solution.}
The second condition is equivalent to $(a-1)(b-1)(c-1)(d-1) = (a+1)(b+1)(c+1)(d+1)$, i.e.\
\[
\frac{a+1}{a-1} \cdot \frac{b+1}{b-1} \cdot \frac{c+1}{c-1} \cdot \frac{d+1}{d-1} = 1.
\]
From the AM--GM inequality it follows that
\begin{align*}
\frac{2}{a-1} + \frac{2}{b-1} + \frac{2}{c-1} + \frac{2}{d-1} &= \frac{a+1}{a-1} - 1 + \frac{b+1}{b-1} - 1 + \frac{c+1}{c-1} - 1 + \frac{d+1}{d-1} - 1 \\
&\geq 4\sqrt[4]{\frac{a+1}{a-1} \cdot \frac{b+1}{b-1} \cdot \frac{c+1}{c-1} \cdot \frac{d+1}{d-1}} - 4 = 0.
\end{align*}

Equality holds if and only if $\dfrac{a+1}{a-1} = \dfrac{b+1}{b-1} = \dfrac{c+1}{c-1} = \dfrac{d+1}{d-1}$ and $abc + abd + acd + bcd + a + b + c + d = 0$, i.e.\ for $a = b = c = d$ satisfying $4a^3 + 4a = 0$, which means $a = b = c = d = 0$, but which does not fulfill $|a|, |b|, |c|, |d| > 1$.

We conclude that the inequality is strict.",202,883,Algebra,15
439,tst_jbmo_ro_2019_4_p4,tst_jbmo,2019,c,"Let $n$ be a positive integer. $2n+1$ tokens are in a row, each being black or white. A token is said to be \emph{balanced} if the number of white tokens on its left plus the number of black tokens on its right is $n$. Determine whether the number of balanced tokens is even or odd.","
oindent\textbf{Solution.}
Define the \emph{score} of each token as being the sum between the number of white tokens on its left and the number of black tokens to its right. Thus, a token is balanced if its score is $n$. It is easy to see that two neighboring tokens have the same score if and only if they have different colors.

From this point, one may continue in at least two ways: either eliminate all pairs consisting of two neighboring tokens of different color, or swap two neighboring tokens if the one on the left is white and the one on the right is black, until all the black tokens get to the beginning of the row.

\medskip
\textbf{Elimination approach.} If we eliminate, the two eliminated tokens had the same score, so they were either both balanced or neither of them was balanced. The score of each of the remaining tokens decreases by 1. Those that were balanced remain balanced (their score decreases by 1, but so does $n$), and those that were unbalanced remain unbalanced. Thus, the elimination does not change the parity of the number of balanced tokens. After doing all possible eliminations, say after $k$ eliminations, we are left with $2n - 2k + 1$ tokens of the same color. The scores of these tokens are $0, 1, 2, \ldots, 2n - 2k$ (from left to right or from right to left, depending on the color of the remaining tokens). Among the scores of the $2(n-k)+1$ remaining tokens, $n-k$ appears only once, which means that at the end of the elimination process we have exactly one balanced token, so their number has always been \textbf{odd}.

\medskip
\textbf{Swap approach.} If we swap, the scores of the other tokens do not change, while the tokens we have swapped continue to have the same score, so the parity of the number of balanced tokens has not changed. We continue to swap until we get to the configuration in which all black tokens are at the left end of the row, while all the white ones are at their right-hand side. If we have $k$ black and $2n+1-k$ white tokens, their scores are, in order, $k-1, k-2, \ldots, 1, 0, 0, 1, \ldots, 2n-k$. In this list, the number $n$ appears exactly once, which means that the number of balanced tokens has always been \textbf{odd}.",282,2206,Combinatorics,16
440,tst_jbmo_ro_2019_5_p1,tst_jbmo,2019,n,"For a positive integer $m$ we denote by $\tau(m)$ the number of its positive divisors, and by $\sigma(m)$ their sum. Determine all positive integers $n$ for which
\[
n\sqrt{\tau(n)} \leq \sigma(n).
\]","
oindent\textbf{Solution.}
It is easy to check that the inequality holds for all $n \in \{1, 2, 4, 6\}$ and it does not hold for any $n \in \{3, 5\}$. We have equality for $n = 1$ and $n = 6$. We prove that the inequality does not hold for any larger $n$. It is easy to see that $\tau$ and $\sigma$ are multiplicative, i.e.\ that $\tau(a \cdot b) = \tau(a) \cdot \tau(b)$ and $\sigma(a \cdot b) = \sigma(a) \cdot \sigma(b)$ for all coprime positive integers $a, b$. From here it follows that if two coprime integers $a, b$ do not fulfill the equation, then neither does their product.

We prove that $2^k$ is no solution for any $k > 3$, that $p^k$, $2p^k$ and $4p^k$ are no solutions for any odd prime $p$ and positive exponent $k$. It will follow that the only solutions to the equation are the four ones mentioned at the beginning.

\begin{itemize}
\item $2^k\sqrt{\tau(2^k)} > \sigma(2^k) \iff 2^k\sqrt{k+1} > 2^{k+1} - 1$ follows from $\sqrt{k+1} \geq 2$, which holds for all $k \geq 3$.

\item $p^k\sqrt{\tau(p^k)} > \sigma(p^k) \iff p^k\sqrt{k+1}(p-1) > p^{k+1} - 1$.
But $p^k\sqrt{k+1}(p-1) \geq p^k(p-1)\sqrt{2} > p^{k+1} > p^{k+1} - 1$, the inequality in the middle being equivalent to $p > \sqrt{2} + 2$, and thus true for all $p > 3$. For $p = 3$, we have $k \geq 2$ and $p^k\sqrt{k+1}(p-1) \geq 3^k \cdot 2\sqrt{3} > 3^{k+1}$.

\item $2p^k\sqrt{\tau(2p^k)} > \sigma(2p^k) \iff 2p^k\sqrt{2(k+1)}(p-1) > 3(p^{k+1}-1)$.
But $2p^k\sqrt{2(k+1)}(p-1) \geq 4p^k(p-1) > 3p^{k+1} > 3(p^{k+1}-1)$, the inequality in the middle being equivalent to $p > 4$, and thus true for all $p \geq 5$. Also, if $p = 3$, then $k \geq 2$, so the inequality to be proven, $4 \cdot 3^k\sqrt{2(k+1)} > 3(3^{k+1}-1)$, follows from $4 \cdot 3^k\sqrt{2(k+1)} \geq 4\sqrt{6} \cdot 3^k > 3^{k+2} > 3(3^{k+1}-3)$.

\item $4p^k\sqrt{\tau(4p^k)} > \sigma(4p^k) \iff 4p^k\sqrt{3(k+1)}(p-1) > 7(p^{k+1}-1)$.
But $4p^k\sqrt{3(k+1)}(p-1) \geq 4\sqrt{6}\,p^k(p-1) > 7p^{k+1} > 7(p^{k+1}-1)$, the inequality in the middle following from $4\sqrt{6}(p-1) > 7p$, true for all $p \geq 5$. For $p = 3$, the inequality $8 \cdot 3^k\sqrt{3(k+1)} > 7(3^{k+1}-1)$ follows from $8 \cdot 3^k\sqrt{3(k+1)} > 7 \cdot 3^{k+1} \iff 8\sqrt{3(k+1)} > 21$, true for all $k \geq 2$. For $p = 3$, $k = 1$, a direct check shows that $n = 12$ is no solution.
\end{itemize}",200,2322,Number Theory,17
441,tst_jbmo_ro_2019_5_p2,tst_jbmo,2019,a,"Let $a, b, c, d \geq 0$ such that $a^2 + b^2 + c^2 + d^2 = 4$. Prove that
\[
\frac{a + b + c + d}{2} \geq 1 + \sqrt{abcd}.
\]
When does the equality hold?","
oindent\textbf{Solution 1.}
We have $ab + cd \geq 2\sqrt{abcd}$, $ac + bd \geq 2\sqrt{abcd}$, and $ad + bc \geq 2\sqrt{abcd}$, so
\[
ab + ac + ad + bc + bd + cd \geq 6\sqrt{abcd},
\]
and thus it is sufficient to show that
\[
\frac{a+b+c+d}{2} \geq 1 + \frac{ab+ac+ad+bc+bd+cd}{6}.
\]
But $ab + ac + ad + bc + bd + cd = \dfrac{(a+b+c+d)^2 - a^2 - b^2 - c^2 - d^2}{2} = 2p^2 - 2$, where $p = \dfrac{a+b+c+d}{2}$.

From this relation it is clear that $p \geq 1$, while from the inequality between the arithmetic and quadratic mean it follows that $p \leq 2$. It is sufficient to prove that
\[
p \geq 1 + \frac{p^2 - 1}{3},
\]
which is equivalent to $p^2 - 3p + 2 \leq 0$, i.e.\ to $(p-1)(p-2) \leq 0$, which is obvious.

Equality holds either when $p = 2$, i.e.\ in case of equality between the arithmetic and quadratic mean, thus for $a = b = c = d = 1$, or when $p = 1$, which means that $ab + ac + ad + bc + bd + cd = 0$, i.e.\ three of the variables are 0 and the fourth one is 2.

\bigskip

oindent\textbf{Solution 2.}
Squaring, the inequality becomes
\[
a^2 + b^2 + c^2 + d^2 + 2(ab+ac+ad+bc+bd+cd) \geq 4 + 8\sqrt{abcd} + 4abcd,
\]
i.e.\ $ab + ac + ad + bc + bd + cd \geq 4\sqrt{abcd} + 2abcd$. From the AM--GM inequality, $4 = a^2 + b^2 + c^2 + d^2 \geq 4\sqrt[4]{a^2b^2c^2d^2} = 4\sqrt{abcd}$, hence $abcd \leq 1$. Also from AM--GM,
\[
ab + ac + ad + bc + bd + cd \geq 6\sqrt{abcd} = 4\sqrt{abcd} + 2\sqrt{abcd} \geq 4\sqrt{abcd} + 2abcd.
\]
Equality holds when $ab = ac = ad = bc = bd = cd$, i.e.\ when either $a = b = c = d = 1$ or three of the variables are 0 (and then the fourth one must be 2).",154,1606,Algebra,18
442,tst_jbmo_ro_2019_5_p3,tst_jbmo,2019,g,"In the acute triangle $ABC$, point $I$ is the incenter, $O$ is the circumcenter, while $I_a$ is the excenter opposite the vertex $A$. Point $A'$ is the reflection of $A$ across the line $BC$. Prove that angles $\angle IOI_a$ and $\angle IA'I_a$ are equal.","
oindent\textbf{Solution.}
The quadrilateral $IBI_aC$ is cyclic, so $\angle AI_aC = \angle IBC = \angle ABI$ and from $\angle BAI = \angle I_aAC$ it follows that triangles $ABI$ and $AI_aC$ are similar (AA). We obtain that $AI \cdot AI_a = AB \cdot AC$. On the other hand,
\[
AB \cdot AC = \frac{2S}{\sin A} = \frac{a \cdot h_a}{\sin A} = 2R \cdot h_a = AA' \cdot AO.
\]
As $\angle A'AI = \angle I_aAO$ (because $AA'$ and $AO$ are isogonal in $\angle BAC$), it follows that triangles $AOI_a$ and $AIA'$ are similar.

Let $J$ be the point where the parallel through $I$ to $I_aA'$ intersects $AA'$. Then
\[
\frac{AJ}{AI} = \frac{AA'}{AI_a} = \frac{AI}{AO},
\]
therefore triangles $AJI$ and $AIO$ are similar. It follows that
\[
\angle IOI_a = \angle AOI_a - \angle AOI = \angle AIA' - \angle AIJ = \angle JIA' = \angle IA'I_a,
\]
and we obtain the conclusion.",255,858,Geometry,19
443,tst_jbmo_ro_2019_5_p4,tst_jbmo,2019,c,The numbers from 1 through 100 are written in some order on a circle. We call a pair of numbers on the circle \emph{good} if the two numbers are not neighbors on the circle and if at least one of the two arcs they determine on the circle only contains numbers smaller than both of them. What may be the total number of good pairs on the circle?,"
oindent\textbf{Solution 1.}
We prove that the number of good pairs is always 97, irrespective of the order of the numbers on the circle.

We will perform a succession of transforms on the order of the numbers on the circle. We will prove that none of these transforms changes the total number of good pairs. First, we will successively swap 1 with one of its neighbors until 1 gets immediately after 100 in clockwise order. Swapping 1 with one of its neighbors, $n$, does not change the number of good pairs. Indeed, all good pairs that do not contain $n$ remain good. There are never good pairs that contain the number 1. Also, all the good pairs that did contain $n$ remain good pairs with the exception of the pair consisting of $n$ and the other neighbor of 1. This pair was a good one but will cease to be so after the swap. The swap only produces one good pair: $n$ together with its former neighbor (the one different from 1). In conclusion, such a swap does not change the total number of good pairs. After moving 1 immediately after 100 by such swaps, we will perform swaps that move 2 immediately after 1. The proof of the fact that such swaps do not change the total number of good pairs is similar to the above one. We continue in the same manner until we arrange the numbers in increasing order (clockwise) on the circle. All this is done without changing the number of good pairs. In this final configuration, it is easy to count the good pairs: the only good pairs are $\{k, 100\}$, where $k \in \{2, 3, \ldots, 98\}$. In conclusion, there are $\mathbf{97}$ good pairs.

\bigskip

oindent\textbf{Solution 2.} \textit{(given in the contest by Iustinian Constantinescu)}

There are 97 good pairs. We prove by induction on $n$ that for a circle containing the numbers from 1 through $n$ there are $n - 3$ good pairs, irrespective of the order of the numbers on the circle.

For $n = 3$ the statement is obvious since all the numbers on the circle are neighbors. For a circle containing the numbers from 1 through $n+1$, we remove the number 1 and subtract 1 from all the other numbers. There were no good pairs containing the number 1, and all the pairs that were good before the removal of 1 remain good pairs for the circle containing the numbers from 1 through $n$, with the exception of the pair made of the two neighbors of 1. This pair was clearly good before the removal of 1 but not after (they have become neighbors). Similarly, a pair that was not a good pair before will not become a good one after the removal of 1 (because the relative order of the remaining numbers does not change). Thus, removing 1 diminishes the number of good pairs by 1.",344,2669,Combinatorics,20
444,tst_jbmo_ro_2019_6_p1,tst_jbmo,2019,a,"If $a, b, c$ are real numbers such that $ab + bc + ca = 0$, prove the inequality
\[
2(a^2+b^2+c^2)(a^2b^2 + b^2c^2 + c^2a^2) \geq 27a^2b^2c^2.
\]
When does the equality hold?","
oindent\textbf{Solution.}
The inequality can be written
\[
2(a^4b^2 + a^4c^2 + b^4a^2 + b^4c^2 + c^4a^2 + c^4b^2) \geq 21a^2b^2c^2.
\]
If $abc = 0$, there is nothing to prove. In this case, two of the variables must be 0, and the inequality is satisfied with equality.

If $abc 
eq 0$, we divide by $a^2b^2c^2$ and get
\[
2\left(\frac{a^2}{b^2} + \frac{a^2}{c^2}
ight) + 2\left(\frac{b^2}{c^2} + \frac{b^2}{a^2}
ight) + 2\left(\frac{c^2}{a^2} + \frac{c^2}{b^2}
ight) \geq 21,
\]
i.e.\
\[
\frac{a^2b^2 + a^2c^2}{b^2c^2} + \frac{a^2b^2 + b^2c^2}{a^2c^2} + \frac{a^2c^2 + b^2c^2}{a^2b^2} \geq \frac{21}{2}.
\]
Putting $ab = x$, $bc = y$, $ca = z$, we know that $x + y + z = 0$ and we want to prove that
\[
\frac{x^2+y^2}{z^2} + \frac{y^2+z^2}{x^2} + \frac{z^2+x^2}{y^2} \geq \frac{21}{2}.
\]
Two of the variables $x, y, z$ must have one sign, while the third one has the opposite sign. We may assume that $xy > 0$. Substituting $z = -x - y$, the inequality to be proven becomes
\[
\frac{x^2+y^2}{(x+y)^2} + \frac{x^2 + 2xy + 2y^2}{x^2} + \frac{y^2 + 2xy + 2x^2}{y^2} \geq \frac{21}{2},
\]
or
\[
\frac{x^2+y^2}{(x+y)^2} + 2\left(\frac{x^2}{y^2} + \frac{y^2}{x^2}
ight) + 2\left(\frac{x}{y} + \frac{y}{x}
ight) \geq \frac{17}{2}.
\]
But adding the obvious inequalities $\dfrac{x^2+y^2}{(x+y)^2} \geq \dfrac{1}{2}$, $\dfrac{x^2}{y^2} + \dfrac{y^2}{x^2} \geq 2$, and $\dfrac{x}{y} + \dfrac{y}{x} \geq 2$ (all satisfied with equality if and only if $x = y$) leads to the inequality above.

In this case, we have equality if $ab = ac$ and $ab + bc + ca = 0$, i.e.\ when $b = c = -2a$. In conclusion, equality holds for triples $(\alpha, -2\alpha, -2\alpha)$, $(\alpha, 0, 0)$, $\alpha \in \mathbb{R}$, and for their permutations.",174,1721,Algebra,21
445,tst_jbmo_ro_2019_6_p2,tst_jbmo,2019,n,Determine all positive integers $n$ such that $4k^2 + n$ is a prime number for all non-negative integers $k$ smaller than $n$.,"
oindent\textbf{Solution.}
For $k = 0$ we get that $n$ must be a prime. It is easy to check that $n = 3$ and $n = 7$ satisfy the given condition while $n = 2$ and $n = 5$ do not. We prove that there are no solutions $n > 10$ to the problem.

If $n = 4m+1$, choosing $k = m < n$, we obtain a composite number $(2m+1)^2$. Thus, there exists a positive integer $m$ such that $n = 4m - 1$. If $m$ has an odd divisor larger than 1, $d$, choosing $2k - 1 = d$ we obtain that $4k^2 + 4m - 1 = (2k-1)(2k+1) + 4m$ is a multiple of $d$, larger than $d$, i.e.\ a composite number.

We are left with the case when $n = 2^u - 1$ ($m$ is a power of 2), $u \geq 4$. If $u$ is composite, $n$ has a proper divisor $2^a - 1$ and we can take $k = 2^a - 1$.

If $u = 4j+1$, for $k = 1$ we get $2^{4j+1} + 3 \equiv 0 \pmod{5}$, $2^{4j+1}+3 > 5$, which is composite.

If $u = 4j+3$, for $k = 2^{2j-1}$ we get $4k^2 + n = 2^{4j} + 2^{4j+3} - 1 = (3 \cdot 4^j)^2 - 1 = (3 \cdot 4^j - 1)(3 \cdot 4^j + 1)$, which is composite.

\medskip

oindent\textit{Another ending.} As above, if $n = 4m - 1$ and $m$ has an odd divisor larger than 1, we are done. If not, write $n = 4s - 9$. Again, if $s$ has an odd divisor larger than 1, $d$, choosing $2k - 3 = d$ we obtain that $4k^2 + 4s - 9 = (2k-3)(2k+3) + 4s$ is a multiple of $d$, larger than $d$, i.e.\ a composite number. The only remaining case is when both $m$ and $s$ are powers of 2. Their difference being 8, they can only be 8 and 16, which makes $n = 7$.",126,1484,Number Theory,22
446,tst_jbmo_ro_2019_6_p3,tst_jbmo,2019,g,"Let $ABC$ be a triangle in which $AB < AC$, $D$ is the foot of the altitude from $A$, $H$ is the orthocenter, $O$ is the circumcenter, $M$ is the midpoint of the side $BC$, $A'$ is the reflection of $A$ across $O$, and $S$ is the intersection of the tangents at $B$ and $C$ to the circumcircle. The tangent at $A'$ to the circumcircle intersects $SC$ and $SB$ at $X$ and $Y$, respectively.

If $M$, $S$, $X$, $Y$ are concyclic, prove that lines $OD$ and $SA'$ are parallel.","
oindent\textbf{Solution.}
Let $N$ be the second intersection point of the circumcircle of triangle $SXY$ with the line $BC$. Writing the powers of points $B$ and $C$ with respect to the circumcircle of quadrilateral $MXSY$, we get $BM \cdot BN = BY \cdot BS$ and $CN \cdot CM = CX \cdot CS$. As $BM = CM$, $BS = CS$, $YB = YA'$, and $XC = XA'$, it follows that $\dfrac{BY}{CX} = \dfrac{BY}{CX} = \dfrac{BA'}{CA'}$.

Next we use the following lemma.

\begin{lemma}
Consider a triangle $ABC$ and points $D$ and $E$ on rays $(BA$ and $(CA$ respectively such that $\dfrac{BD}{CE} = k$. Let $M$ and $N$ be points on the line segments $BC$ and $DE$ respectively such that $\dfrac{BM}{MC} = \dfrac{DN}{NE} = k$. Then $MN$ is parallel to the bisector of angle $\angle BAC$.
\end{lemma}

\begin{proof}
We have $\overrightarrow{NM} = \dfrac{1}{k+1}\left(\overrightarrow{DB} + k \cdot \overrightarrow{EC}
ight)$. If $X$ and $Y$ are points on rays $(AB$ and $(AC$ such that $AX = DB$ and $AY = k \cdot EC$, then $\overrightarrow{NM}$ is parallel to the median from $A$ in triangle $AXY$. But $AX = AY$, which means that in triangle $AXY$ the median from $A$ has the direction of the bisector of angle $\angle BAC$.
\end{proof}

From the Lemma we obtain that $A'N \parallel SM$, which means that $N$ is the reflection of point $D$ across $M$. In triangle $SXY$, the circumcenter lies on $SN$, therefore the orthocenter lies on its isogonal, $SD$. It follows that $SD \perp XY$, i.e.\ $SD \parallel AA'$. We obtain that $ADSO$ is a parallelogram, hence $SD = R$. But then $SDOA'$ is also a parallelogram, hence the conclusion.

\medskip

oindent\textit{Remark.} The configuration of the problem has many other interesting properties: $M$ is the circumcenter of triangle $OXY$. Indeed, as $SM$ is the bisector of angle $\angle XSY$, we have $MX = MT$. From $\angle XMY = 180^\circ - \angle XSY = 2 \cdot \angle A = \angle BOC = 2 \cdot \angle XOY$ we obtain the statement.

$OSA'N$ is an isosceles trapezoid, hence $SN = OA' = R$. Thus, the circumcircle of $SXY$ passes through $M$ and $N$ and has the same radius as the Euler circle of triangle $ABC$, which shows that it is the reflection of the latter across point $M$.",473,2208,Geometry,23
447,tst_jbmo_ro_2019_6_p4,tst_jbmo,2019,c,"Consider two disjoint finite sets of positive integers, $A$ and $B$, with $n$ and $m$ elements, respectively. It is known that all $k$ belonging to $A \cup B$ satisfy at least one of the conditions $k + 17 \in A$ and $k - 31 \in B$. Prove that $17n = 31m$.","
oindent\textbf{Solution.}
We choose an arbitrary element $x_1 \in A \cup B$. If $x_1 + 17 \in A$, we choose $x_2 = x_1 + 17$, while if $x_1 + 17 
otin A$, we choose $x_2 = x_1 - 31 \in B$. We continue to choose $x_{j+1} = x_j + 17$ if $x_j + 17 \in A$ and $x_{j+1} = x_j - 31 \in B$ if not, until we obtain (for the first time) an element $x_{j+1}$ that has already been chosen (for which there exists $k < j$ such that $x_{j+1} = x_k$). Such an index $j$ must exist because $A \cup B$ is finite. Let us prove that $x_k = x_1$ (i.e.\ $k = 1$). Assume the contrary to be true. Then $x_j 
eq x_{k-1}$ because of the minimality of index $j+1$. But then we have either $x_{j+1} = x_j + 17 \in A$ and $x_k = x_{k-1} - 31 \in B$, or the other way around, $x_{j+1} = x_j - 31 \in B$ and $x_k = x_{k-1} + 17 \in A$. As $A$ and $B$ are disjoint, we cannot have $x_{j+1} = x_k$. Thus, $x_{j+1} = x_1$. If among the indices $2, 3, \ldots, j+1$ there exist $a$ indices for which $x_{m+1} = x_m + 17$ and $b = j - a$ indices for which $x_{m+1} = x_m - 31$, then adding these relations yields $x_1 = x_1 + a \cdot 17 - b \cdot 31$, i.e.\ $17a = 31b$. Moreover, those $a$ indices for which $x_{t+1} = x_t + 17$ are precisely those for which $x_{t+1} \in A$, while those $b$ indices for which $x_{t+1} = x_t - 31$ are precisely the ones for which $x_{t+1} \in B$. In other words, among the numbers $x_1, x_2, \ldots, x_j$ there are $a$ elements of $A$ and $b$ elements of $B$, with $17a = 31b$.

If by this procedure we have not covered $A \cup B$, we choose $y_1 \in A \cup B \setminus \{x_1, x_2, \ldots, x_j\}$ and define a new (finite) sequence $y_1, y_2, \ldots, y_s$ where $s$ is the first index such that $y_s \in \{x_1, x_2, \ldots, x_j, y_1, y_2, \ldots, y_{s-1}\}$. If $y_s \in A$, then $y_s = y_{s-1} + 17$. If $y_s = x_r$, with $r \in \{1, 2, \ldots, j\}$, then $x_r \in A$ means that $x_r = x_{r-1} + 17$ (indices are taken modulo $j$), hence $y_{s-1} = x_{r-1}$, which contradicts the choice of $s$. Also, if $y_s = y_r$ with $r < s$, then $r 
eq 1$ leads to $y_{s-1} = y_s - 17 = y_r - 17 = y_{r-1}$, contradicting the minimality of $s$. Similarly one gets a contradiction if $y_s \in B$. Thus, $y_s = y_1$. As we have done for the numbers $x_1, x_2, \ldots, x_j$, one can prove for the numbers $y_1, y_2, \ldots, y_s$ that the ratio between the number of elements of $A$ and the number of elements of $B$ within this sequence is $\dfrac{31}{17}$.

We continue this procedure until finishing up the set $A \cup B$. At the end of the procedure, $A \cup B$ is decomposed into disjoint cycles, and the ratio between the number of elements of $A$ and the number of elements of $B$ within each cycle is $\dfrac{31}{17}$. This ratio is preserved when reuniting the cycles, giving $17n = 31m$.",256,2786,Combinatorics,24
255,jbmo_2020_p1,jbmo,2020,a,"Find all triples $(a, b, c)$ of real numbers such that the following system holds:
\[
a + b + c = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}
\]
\[
a^2 + b^2 + c^2 = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}
\]","\textbf{Solution.} First of all if $(a, b, c)$ is a solution of the system then also $(-a, -b, -c)$ is a solution. Hence we can suppose that $abc > 0$. From the first condition we have
\begin{equation}
a + b + c = \frac{ab + bc + ca}{abc}. \label{eq1}
\end{equation}
Now, from the first condition and the second condition we get
\[
(a+b+c)^2 - (a^2+b^2+c^2) = \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}
ight)^2 - \left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}
ight).
\]
The last one simplifies to
\begin{equation}
ab + bc + ca = \frac{a+b+c}{abc}. \label{eq2}
\end{equation}

First we show that $a+b+c$ and $ab+bc+ca$ are different from $0$. Suppose on contrary then from relation \eqref{eq1} or \eqref{eq2} we have $a+b+c = ab+bc+ca = 0$. But then we would have
\[
a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ca) = 0,
\]
which means that $a = b = c = 0$. This is not possible since $a, b, c$ should be different from $0$.

Now multiplying \eqref{eq1} and \eqref{eq2} we have
\[
(a+b+c)(ab+bc+ca) = \frac{(a+b+c)(ab+bc+ca)}{(abc)^2}.
\]
Since $a+b+c$ and $ab+bc+ca$ are different from $0$, we get $(abc)^2 = 1$ and using the fact that $abc > 0$ we obtain that $abc = 1$. So relations \eqref{eq1} and \eqref{eq2} transform to
\[
a + b + c = ab + bc + ca.
\]
Therefore,
\[
(a-1)(b-1)(c-1) = abc - ab - bc - ca + a + b + c - 1 = 0.
\]
This means that at least one of the numbers $a, b, c$ is equal to $1$. Suppose that $c = 1$ then relations \eqref{eq1} and \eqref{eq2} transform to $a + b + 1 = ab + a + b \Rightarrow ab = 1$. Taking $a = t$ then we have $b = \frac{1}{t}$. We can now verify that any triple $(a,b,c) = \left(t, \frac{1}{t}, 1
ight)$ satisfies both conditions, $t \in \mathbb{R} \setminus \{0\}$. From the initial observation any triple $(a,b,c) = \left(t, \frac{1}{t}, -1
ight)$ satisfies both conditions, $t \in \mathbb{R} \setminus \{0\}$. So, all triples that satisfy both conditions are $(a,b,c) = \left(t, \frac{1}{t}, 1
ight)$, $\left(t, \frac{1}{t}, -1
ight)$ and all permutations for any $t \in \mathbb{R} \setminus \{0\}$.

\medskip
\textbf{Comment by PSC.} After finding that $abc = 1$ and
\[
a + b + c = ab + bc + ca,
\]
we can avoid the trick considering $(a-1)(b-1)(c-1)$ as follows. By the Vieta's relations we have that $a, b, c$ are roots of the polynomial
\[
P(x) = x^3 - sx^2 + sx - 1
\]
which has one root equal to $1$. Then, we can conclude as in the above solution.",210,2391,Algebra,1
256,jbmo_2020_p2,jbmo,2020,g,"Let $\triangle ABC$ be a right-angled triangle with $\angle BAC = 90^\circ$ and let $E$ be the foot of the perpendicular from $A$ on $BC$. Let $Z 
eq A$ be a point on the line $AB$ with $AB = BZ$. Let $(c)$ be the circumcircle of the triangle $\triangle AEZ$. Let $D$ be the second point of intersection of $(c)$ with $ZC$ and let $F$ be the antidiametric point of $D$ with respect to $(c)$. Let $P$ be the point of intersection of the lines $FE$ and $CZ$. If the tangent to $(c)$ at $Z$ meets $PA$ at $T$, prove that the points $T, E, B, Z$ are concyclic.","\textbf{Solution.} We will first show that $PA$ is tangent to $(c)$ at $A$.

Since $E, D, Z, A$ are concyclic, then $\angle EDC = \angle EAZ = \angle EAB$. Since also the triangles $\triangle ABC$ and $\triangle EBA$ are similar, then $\angle EAB = \angle BCA$, therefore $\angle EDC = \angle BCA$.

Since $\angle FED = 90^\circ$, then $\angle PED = 90^\circ$ and so
\[
\angle EPD = 90^\circ - \angle EDC = 90^\circ - \angle BCA = \angle EAC.
\]
Therefore the points $E, A, C, P$ are concyclic. It follows that $\angle CPA = 90^\circ$ and therefore the triangle $\angle PAZ$ is right-angled. Since also $B$ is the midpoint of $AZ$, then $PB = AB = BZ$ and so $\angle ZPB = \angle PZB$.

Furthermore, $\angle EPD = \angle EAC = \angle CBA = \angle EBA$ from which it follows that the points $P, E, B, Z$ are also concyclic.

Now observe that
\[
\angle PAE = \angle PCE = \angle ZPB - \angle PBE = \angle PZB - \angle PZE = \angle EZB.
\]
Therefore $PA$ is tangent to $(c)$ at $A$ as claimed.

It now follows that $TA = TZ$. Therefore
\begin{align*}
\angle PTZ &= 180^\circ - 2(\angle TAB) = 180^\circ - 2(\angle PAE + \angle EAB) \\
&= 180^\circ - 2(\angle ECP + \angle ACB) \\
&= 180^\circ - 2(90^\circ - \angle PZB) = 2(\angle PZB) \\
&= \angle PZB + \angle BPZ = \angle PBA.
\end{align*}
Thus $T, P, B, Z$ are concyclic, and since $P, E, B, Z$ are also concyclic then $T, E, B, Z$ are concyclic as required.",556,1409,Geometry,2
257,jbmo_2020_p3,jbmo,2020,c,"Alice and Bob play the following game: Alice picks a set $A = \{1, 2, \ldots, n\}$ for some natural number $n > 2$. Then starting with Bob, they alternatively choose one number from the set $A$, according to the following conditions: initially Bob chooses any number he wants, afterwards the number chosen at each step should be distinct from all the already chosen numbers, and should differ by $1$ from an already chosen number. The game ends when all numbers from the set $A$ are chosen. Alice wins if the sum of all of the numbers that she has chosen is composite. Otherwise Bob wins. Decide which player has a winning strategy.","\textbf{Solution.} To say that Alice has a winning strategy means that she can find a number $n$ to form the set $A$, so that she can respond appropriately to all choices of Bob and always get at the end a composite number for the sum of her choices. If such $n$ does not exist, this would mean that Bob has a winning strategy instead.

Alice can try first to check the small values of $n$. Indeed, this gives the following winning strategy for her: she initially picks $n = 8$ and responds to all possible choices made by Bob as in the list below (in each row the choices of Bob and Alice are given alternatively, starting with Bob):

\[
\begin{array}{cccccccc}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
2 & 3 & 1 & 4 & 5 & 6 & 7 & 8 \\
2 & 3 & 4 & 1 & 5 & 6 & 7 & 8 \\
3 & 2 & 1 & 4 & 5 & 6 & 7 & 8 \\
3 & 2 & 4 & 5 & 1 & 6 & 7 & 8 \\
3 & 2 & 4 & 5 & 6 & 1 & 7 & 8 \\
4 & 5 & 3 & 6 & 2 & 1 & 7 & 8 \\
4 & 5 & 3 & 6 & 7 & 8 & 2 & 1 \\
4 & 5 & 6 & 7 & 3 & 2 & 1 & 8 \\
4 & 5 & 6 & 7 & 3 & 2 & 8 & 1 \\
4 & 5 & 6 & 7 & 8 & 3 & 2 & 1 \\
5 & 4 & 3 & 2 & 1 & 6 & 7 & 8 \\
5 & 4 & 3 & 2 & 6 & 7 & 1 & 8 \\
5 & 4 & 3 & 2 & 6 & 7 & 8 & 1 \\
5 & 4 & 6 & 3 & 2 & 1 & 7 & 8 \\
5 & 4 & 6 & 3 & 7 & 8 & 2 & 1 \\
6 & 7 & 5 & 4 & 3 & 8 & 2 & 1 \\
6 & 7 & 5 & 4 & 8 & 3 & 2 & 1 \\
6 & 7 & 8 & 5 & 4 & 3 & 2 & 1 \\
7 & 6 & 8 & 5 & 4 & 3 & 2 & 1 \\
7 & 6 & 5 & 8 & 4 & 3 & 2 & 1 \\
8 & 7 & 6 & 5 & 4 & 3 & 2 & 1 \\
\end{array}
\]

In all cases, Alice's sum is either an even number greater than $2$, or else $15$ or $21$, thus Alice always wins.",632,1519,Combinatorics,3
258,jbmo_2020_p4,jbmo,2020,n,"Find all pairs $(p, q)$ of prime numbers such that
\[
1 + \frac{p^q - q^p}{p + q}
\]
is a prime number.","\textbf{Solution.} It is clear that $p 
eq q$. We set
\[
1 + \frac{p^q - q^p}{p + q} = r
\]
and we have that
\begin{equation}
p^q - q^p = (r-1)(p+q). \label{eq3}
\end{equation}
From Fermat's Little Theorem we have
\[
p^q - q^p \equiv -q \pmod{p}.
\]
Since we also have that
\[
(r-1)(p+q) \equiv -rq - q \pmod{p},
\]
from \eqref{eq3} we get that
\[
rq \equiv 0 \pmod{p} \Rightarrow p \mid qr,
\]
hence $p \mid r$, which means that $p = r$. Therefore, \eqref{eq3} takes the form
\begin{equation}
p^q - q^p = (p-1)(p+q). \label{eq4}
\end{equation}
We will prove that $p = 2$. Indeed, if $p$ is odd, then from Fermat's Little Theorem we have
\[
p^q - q^p \equiv p \pmod{q}
\]
and since
\[
(p-1)(p+q) \equiv p(p-1) \pmod{q},
\]
we have
\[
p(p-2) \equiv 0 \pmod{q} \Rightarrow q \mid p(p-2) \Rightarrow q \mid p-2 \Rightarrow q \leq p-2 < p.
\]
Now, from \eqref{eq4} we have
\[
p^q - q^p \equiv 0 \pmod{p-1} \Rightarrow 1 - q^p \equiv 0 \pmod{p-1} \Rightarrow q^p \equiv 1 \pmod{p-1}.
\]
Clearly $\gcd(q, p-1) = 1$ and if we set $k = \mathrm{ord}_{p-1}(q)$, it is well-known that $k \mid p$ and $k < p$, therefore $k = 1$. It follows that
\[
q \equiv 1 \pmod{p-1} \Rightarrow p-1 \mid q-1 \Rightarrow p-1 \leq q-1 \Rightarrow p \leq q
\]
a contradiction.

Therefore, $p = 2$ and \eqref{eq4} transforms to
\[
2^q = q^2 + q + 2.
\]
We can easily check by induction that for every positive integer $n \geq 6$ we have $2^n > n^2 + n + 2$. This means that $q \leq 5$ and the only solution is for $q = 5$. Hence the only pair which satisfies the condition is $(p, q) = (2, 5)$.

\medskip
\textbf{Comment by the PSC.} From the problem condition, we get that $p^q$ should be bigger than $q^p$, which gives
\[
q \ln p > p \ln q \iff \frac{\ln p}{p} > \frac{\ln q}{q}.
\]
The function $\frac{\ln x}{x}$ is decreasing for $x > e$, thus if $p$ and $q$ are odd primes, we obtain $q > p$.",103,1868,Number Theory,4
53,shl_jbmo_2020_a1,shl_jbmo,2020,a,"Find all triples $(a, b, c)$ of real numbers such that the following system holds:
\[
\begin{cases}
a + b + c = \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} \\[8pt]
a^2 + b^2 + c^2 = \dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2}
\end{cases}
\]","\begin{solution}
First of all if $(a, b, c)$ is a solution of the system then also $(-a, -b, -c)$ is a solution. Hence
we can suppose that $abc > 0$. From the first condition we have
\begin{equation}\label{eq:A1-1}
a + b + c = \frac{ab + bc + ca}{abc}.
\end{equation}
Now, from the first condition and the second condition we get
\[
(a + b + c)^2 - (a^2 + b^2 + c^2) = \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}
ight)^2 - \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}
ight).
\]
The last one simplifies to
\begin{equation}\label{eq:A1-2}
ab + bc + ca = \frac{a + b + c}{abc}.
\end{equation}
First we show that $a + b + c$ and $ab + bc + ca$ are different from $0$. Suppose on contrary then from
relation \eqref{eq:A1-1} or \eqref{eq:A1-2} we have $a + b + c = ab + bc + ca = 0$. But then we would have
\[
a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca) = 0,
\]
which means that $a = b = c = 0$. This is not possible since $a, b, c$ should be different from $0$.

Now multiplying \eqref{eq:A1-1} and \eqref{eq:A1-2} we have
\[
(a + b + c)(ab + bc + ca) = \frac{(a + b + c)(ab + bc + ca)}{(abc)^2}.
\]
Since $a + b + c$ and $ab + bc + ca$ are different from $0$, we get $(abc)^2 = 1$ and using the fact that $abc > 0$
we obtain that $abc = 1$. So relations \eqref{eq:A1-1} and \eqref{eq:A1-2} transform to
\[
a + b + c = ab + bc + ca.
\]
Therefore,
\[
(a - 1)(b - 1)(c - 1) = abc - ab - bc - ca + a + b + c - 1 = 0.
\]
This means that at least one of the numbers $a, b, c$ is equal to $1$. Suppose that $c = 1$ then relations
\eqref{eq:A1-1} and \eqref{eq:A1-2} transform to $a + b + 1 = ab + a + b \Rightarrow ab = 1$. Taking $a = t$ then we have $b = \frac{1}{t}$. We can
now verify that any triple $(a, b, c) = \left(t, \frac{1}{t}, 1
ight)$ satisfies both conditions. $t \in \mathbb{R} \setminus \{0\}$. From the initial
observation any triple $(a, b, c) = \left(t, \frac{1}{t}, -1
ight)$ satisfies both conditions. $t \in \mathbb{R} \setminus \{0\}$. So, all triples that
satisfy both conditions are $(a, b, c) = \left(t, \frac{1}{t}, 1
ight)$, $\left(t, \frac{1}{t}, -1
ight)$ and all permutations for any $t \in \mathbb{R} \setminus \{0\}$.

\medskip

\textit{Comment by PSC.} After finding that $abc = 1$ and
\[
a + b + c = ab + bc + ca,
\]
we can avoid the trick considering $(a - 1)(b - 1)(c - 1)$ as follows. By the Vieta's relations we have
that $a, b, c$ are roots of the polynomial
\[
P(x) = x^3 - sx^2 + sx - 1
\]
which has one root equal to $1$. Then, we can conclude as in the above solution.
\end{solution}",244,2532,Algebra,1
54,shl_jbmo_2020_a2,shl_jbmo,2020,a,"Consider the sequence $a_1, a_2, a_3, \ldots$ defined by $a_1 = 9$ and
\[
a_{n+1} = \frac{(n+5)a_n + 22}{n+3}
\]
for $n > 1$.
Find all natural numbers $n$ for which $a_n$ is a perfect square of an integer.","\begin{solution}
Define $b_n = a_n + 11$. Then
\[
22 = (n+3)a_{n+1} - (n+5)a_n = (n+3)b_{n+1} - 11n - 33 - (n+5)b_n + 11n + 55
\]
giving $(n+3)b_{n+1} = (n+5)b_n$. Then
\[
b_{n+1} = \frac{n+5}{n+3}\,b_n = \frac{(n+5)(n+4)}{(n+3)(n+2)}\,b_{n-1} = \frac{(n+5)(n+4)}{(n+2)(n+1)}\,b_{n-2} = \cdots = \frac{(n+5)(n+4)}{5 \cdot 4}\,b_1 = (n+5)(n+4).
\]
Therefore $b_n = (n+4)(n+3) = n^2 + 7n + 12$ and $a_n = n^2 + 7n + 1$.

Since $(n+1)^2 = n^2 + 2n + 1 < a_n < n^2 + 8n + 16 = (n+4)^2$, if $a_n$ is a perfect square, then
$a_n = (n+2)^2$ or $a_n = (n+3)^2$.

If $a_n = (n+2)^2$, then $n^2 + 4n + 4 = n^2 + 7n + 1$ giving $n = 1$. If $a_n = (n+3)^2$, then $n^2 + 6n + 9 =
n^2 + 7n + 1$ giving $n = 8$.

\medskip

\textit{Comment.} We provide some other methods to find $a_n$.

\textbf{Method 1:} Define $b_n = \dfrac{a_n + 11}{n+3}$. Then $b_1 = 5$ and $a_n = (n+3)b_n - 11$. So
\[
a_{n+1} = (n+4)b_{n+1} - 11 = \frac{(n+5)a_n + 22}{n+3} = a_n + \frac{2(a_n + 11)}{n+3} = (n+3)b_n - 11 + 2b_n
\]
giving $(n+4)b_{n+1} = (n+5)b_n$. Then
\[
b_{n+1} = \frac{n+5}{n+4}\,b_n = \frac{n+5}{n+3}\,b_{n-1} = \cdots = \frac{n+5}{5}\,b_1 = n+5.
\]
Then $b_n = n+4$, so $a_n = (n+3)(n+4) - 11 = n^2 + 7n + 1$.

\textbf{Method 2:} We have
\[
(n+3)a_{n+1} - (n+5)a_n = 22
\]
and therefore
\[
\frac{a_{n+1}}{(n+5)(n+4)} - \frac{a_n}{(n+4)(n+3)} = \frac{22}{(n+3)(n+4)(n+5)} = 11\left[\frac{1}{n+3} - \frac{2}{n+4} + \frac{1}{n+5}
ight].
\]
Now define $b_n = \dfrac{a_n}{(n+4)(n+3)}$ to get
\[
b_{n+1} = b_n + 11\left[\frac{1}{n+3} - \frac{2}{n+4} + \frac{1}{n+5}
ight]
\]
which telescopically gives
\[
b_{n+1} = b_1 + 11\left[\left(\frac{1}{4} - \frac{2}{5} + \frac{1}{6}
ight) + \left(\frac{1}{5} - \frac{2}{6} + \frac{1}{7}
ight) + \cdots + \left(\frac{1}{n+3} - \frac{2}{n+4} + \frac{1}{n+5}
ight)
ight]
= \frac{9}{20} + \frac{11}{20} - \frac{11}{(n+4)(n+5)}.
\]
We get $b_{n+1} = (n+4)(n+5) - 11$ from which it follows that $b_n = (n+3)(n+4) - 11 = n^2 + 7n + 1$.
\end{solution}",205,1961,Algebra,2
55,shl_jbmo_2020_a3,shl_jbmo,2020,a,"Find all triples of positive real numbers $(a, b, c)$ so that the expression
\[
M = \frac{(a+b)(b+c)(a+b+c)}{abc}
\]
gets its least value.","\begin{solution}
The expression $M$ is homogeneous, therefore we can assume that $abc = 1$. We set $s = a + c$
and $p = ac$ and using $b = \dfrac{1}{ac}$, we get
\[
M = \left(a + \frac{1}{ac}
ight)\left(\frac{1}{ac} + c
ight)\left(a + \frac{1}{ac} + c
ight) = (a + p^{-1})(c + p^{-1})(s + p^{-1}).
\]
Expanding the right-hand side we get
\[
M = ps + \frac{s^2}{p} + 1 + \frac{2s}{p^2} + \frac{1}{p^3}.
\]
Now by $s \geq 2\sqrt{p}$ and setting $x = p\sqrt{p} > 0$ we get
\[
M \geq 2x + 5 + \frac{4}{x} + \frac{1}{x^2}.
\]
We will now prove that
\[
2x + 5 + \frac{4}{x} + \frac{1}{x^2} \geq \frac{11 + 5\sqrt{5}}{2}.
\]
Indeed, the latter is equivalent to $4x^3 - (5\sqrt{5} + 1)x^2 + 8x + 2 \geq 0$, which can be rewritten as
\[
\left(x - \frac{1 + \sqrt{5}}{2}
ight)^2 (4x + 3 - \sqrt{5}) \geq 0,
\]
which is true.

\textit{Remark:} Notice that the equality holds for $a = c = \sqrt{p} = \sqrt[3]{\dfrac{1 + \sqrt{5}}{2}}$ and $b = \dfrac{1}{ac}$.
\end{solution}",138,962,Algebra,3
56,shl_jbmo_2020_c1,shl_jbmo,2020,c,"Alice and Bob play the following game: starting with the number $2$ written on a blackboard,
each player in turn changes the current number $n$ to a number $n + p$, where $p$ is a prime divisor of $n$.
Alice goes first and the players alternate in turn. The game is lost by the one who is forced to write
a number greater than $\underbrace{2\ldots 2}_{2020}$. Assuming perfect play, who will win the game.","\begin{solution}
We prove that Alice wins the game. For argument's sake, suppose that Bob can win by
proper play regardless of what Alice does on each of her moves. Note that Alice can force the line
\[
2 \to 4 \to 6 \to 8 \to 10 \to 12
\]
at the beginning stages of the game. (As each intermediate `position' from which Bob has to play is a prime power.) Thus the player on turn when the number $12$ is written on the blackboard must be in a `winning position', i.e., can win the game with skillful play. However,
Alice can place herself in that position through the following line that is once again forced for Bob:
\[
2 \to 4 \to 6 \to 9 \to 12.
\]
(This time she is in turn with $12$ written on the blackboard.) The obtained contradiction proves our point.

\medskip

\textit{Comment by the PSC.} Notice that this is a game of two players which always ends in a finite
number of moves with a winner. For games like this, a player whose turn is to make a move, may be
in position to force a win for him. If not, then the other player is in position to force a win for him.
\end{solution}",405,1090,Combinatorics,4
57,shl_jbmo_2020_c2,shl_jbmo,2020,c,"Viktor and Natalia bought $2020$ buckets of ice-cream and want to organize a degustation schedule
with $2020$ rounds such that:
\begin{itemize}
  \item In every round, each one of them tries $1$ ice-cream, and those $2$ ice-creams tried in a single round
are different from each other.
  \item At the end of the $2020$ rounds, each one of them has tried each ice-cream exactly once.
\end{itemize}
We will call a degustation schedule \textit{fair} if the number of ice-creams that were tried by Viktor before
Natalia is equal to the number of ice creams tried by Natalia before Viktor.
Prove that the number of fair schedules is strictly larger than $2020!(2^{1010} + (1010!)^2)$.","\begin{solution}
If we fix the order in which Natalia tries the ice-creams, we may consider $2$ types of fair
schedules:

1) Her last $1010$ ice-creams get assigned as Viktor's first $1010$ ice-creams, and vice versa: Viktor's first
$1010$ ice-creams are assigned as Natalia's last $1010$ ice-creams. This generates $(1010!)^2$ distinct fair
schedules by permuting the ice-creams within each group.

2) We divide all ice-creams into disjoint groups of $4$, and in each group we swap the first $2$ ice-creams
with the last $2$, which gives us $((2!)^2)^{504} = 2^{1010}$ distinct schedules.

Now, to make the inequality strict, we consider $1$ more schedule like $2)$, but with groups of $2$ ice-creams
instead of $4$.
\end{solution}",679,732,Combinatorics,5
58,shl_jbmo_2020_c3,shl_jbmo,2020,c,"Alice and Bob play the following game: Alice begins by picking a natural number $n > 2$. Then,
with Bob starting first, they alternately choose one number from the set $A = \{1, 2, \ldots, n\}$ according to
the following condition: The number chosen at each step should be distinct from all the already
chosen numbers, and should differ by $1$ from an already chosen number. (At the very first step Bob
can choose any number he wants.) The game ends when all numbers from the set $A$ are chosen.

For example, if Alice picks $n = 4$, then a valid game would be for Bob to choose $2$, then Alice to
choose $3$, then Bob to choose $1$, and then Alice to choose $4$.

Alice wins if the sum $S$ of all of the numbers that she has chosen is composite. Otherwise Bob
wins. (In the above example $S = 7$, so Bob wins.)

Decide which player has a winning strategy.","\begin{solution}
Alice has a winning strategy. She initially picks $n = 8$. We will give a strategy so that she
can end up with $S$ even, or $S = 15$, or $S = 21$, so she wins.

\textbf{Case 1:} If Bob chooses $1$, then the game ends with Alice choosing $2, 4, 6, 8$ so $S$ is even (larger
than $2$) and Alice wins.

\textbf{Case 2:} If Bob chooses $2$, then Alice chooses $3$. Bob can now choose either $1$ or $3$.

\textbf{Case 2A:} If Bob chooses $1$, then Alice's numbers are $3, 4, 6, 8$. So $S = 21$ and Alice wins.

\textbf{Case 2B:} If Bob chooses $4$, then Alice chooses $1$ and ends with the numbers $1, 3, 6, 8$. So $S$ is even
and Alice wins.

\textbf{Case 3:} If Bob chooses $3$, then Alice chooses $2$. Bob can now choose either $1$ or $4$.

\textbf{Case 3A:} If Bob chooses $1$, then Alice's numbers are $2, 4, 6, 8$. So $S$ is even and Alice wins.

\textbf{Case 3B:} If Bob chooses $4$, then Alice chooses $5$. Bob can now choose either $1$ or $6$.

\textbf{Case 3Bi:} If Bob chooses $1$, then Alice's numbers are $2, 5, 6, 8$. So $S = 21$ and Alice wins.

\textbf{Case 3Bii:} If Bob chooses $6$, then Alice chooses $1$. Then Alice's numbers are $2, 5, 1, 8$. So $S$ is
even and Alice wins.

\textbf{Case 4:} If Bob chooses $4$, then Alice chooses $5$. Bob can now choose either $3$ or $6$.

\textbf{Case 4A:} If Bob chooses $3$, then Alice chooses $6$. Bob can now choose either $2$ or $7$.

\textbf{Case 4Ai:} If Bob chooses $2$, then Alice chooses $1$ and ends up with $5, 6, 1, 8$. So $S$ is even and
Alice wins.

\textbf{Case 4Aii:} If Bob chooses $7$, then Alice chooses $8$ and ends up with $5, 6, 8, 1$. So $S$ is even and
Alice wins.

\textbf{Case 4B:} If Bob chooses $6$, then Alice chooses $7$. Bob can now choose either $3$ or $8$.

\textbf{Case 4Bi:} If Bob chooses $3$, then Alice chooses $2$ and ends up with $5, 7, 2$ and either $1$ or $8$. So
$S = 15$ or $S = 22$ and Alice wins.

\textbf{Case 4Bii:} If Bob chooses $8$, then Alice's numbers are $5, 7, 3, 1$. So $S$ is even and Alice wins.

\textbf{Cases 5--8:} If Bob chooses $k \in \{5, 6, 7, 8\}$ then Alice follows the strategy in case $9 - k$ but whenever
she had to choose $\ell$, she instead chooses $9 - \ell$. If at the end of that strategy she ended up with $S$, she
will now end up with $S' = 4 \cdot 9 - S = 36 - S$. Then $S'$
is even or $S' = 15$ or $S' = 21$ so again she wins.
\end{solution}",856,2390,Combinatorics,6
59,shl_jbmo_2020_g1,shl_jbmo,2020,g,"Let $\triangle ABC$ be an acute triangle. The line through $A$ perpendicular to $BC$ intersects $BC$ at $D$.
Let $E$ be the midpoint of $AD$ and $\omega$ the the circle with center $E$ and radius equal to $AE$. The line
$BE$ intersects $\omega$ at a point $X$ such that $X$ and $B$ are not on the same side of $AD$ and the line $CE$
intersects $\omega$ at a point $Y$ such that $C$ and $Y$ are not on the same side of $AD$. If both of the intersection
points of the circumcircles of $\triangle BDX$ and $\triangle CDY$ lie on the line $AD$, prove that $AB = AC$.","\begin{solution}
Denote by $s$ the line $AD$. Let $T$ be the second intersection point of the circumcircles of
$\triangle BDX$ and $\triangle CDY$. Then $T$ is on the line $s$. Note that $CDYT$ and $BDXT$ are cyclic. Using this
and the fact that $AD$ is perpendicular to $BC$ we obtain:
\[
\angle TYE = \angle TYC = \angle TDC = 90^\circ.
\]
This means that $EY$ is perpendicular to $TY$, so $TY$ must be tangent to $\omega$. We similarly show that $TX$
is tangent to $\omega$. Thus, $TX$ and $TY$ are tangents from $T$ to $\omega$ which implies that $s$ is the perpendicular
bisector of the segment $XY$. Now denote by $\sigma$ the reflection of the plane with respect to $s$. Then the
points $X$ and $Y$ are symmetric with respect to $s$, so $\sigma(X) = Y$. Also note that $\sigma(E) = E$, because $E$
is on $s$. Using the fact that $BC$ is perpendicular to $s$, we see that $BC$ is the reflection image of itself
with respect to $s$. Now note that $B$ is the intersection point of the lines $EX$ and $BC$. This means
that the image of $B$ is the intersection point of the lines $\sigma(EX) = EY$ and $\sigma(BC) = BC$, which is $C$.
From here we see that $\sigma(B) = C$, so $s$ is the perpendicular bisector of $BC$, which is what we needed
to prove.
\end{solution}

\medskip

\textbf{Alternative Solution.} Let the circle $\odot CDY$ intersects the line $AD$ at another point $Z$. Then
we have $\angle CED = \angle YEZ$. We also have $ED = EY$ because $E$ is the center of the circle $\omega$. Also note
that
\[
\angle DCE = \angle DCY = \angle DZY = \angle EZY.
\]
We conclude that $\triangle CDE$ and $\triangle ZYE$ are congruent. From here we have that $EZ = EC$. Now denote
by $Z'$ the other intersection point of $AD$ and $\odot BDX$. In the same way we prove that $EZ' = EB$. By
the assumption of the problem, we must have that $Z = Z'$. We now conclude that
\[
BE = CE = EZ = EZ'.
\]
Also, $\angle BDE = 90^\circ = \angle CDE$. Now we see that $\triangle BDE$ and $\triangle CDE$ are congruent (they share the
side $ED$), so $BD = CD$. But $D$ is both the midpoint of $BC$ and the foot of the altitude from $A$,
which means that $AB = AC$.

\medskip

\textbf{Alternative Solution.} Let $\alpha = \angle BXD$. Denote by $T$ the second intersection point of the
circumcircles of $\triangle BDX$ and $\triangle CDY$, which is on $AD$. We have
\[
ED = EA = EX
\]
because $E$ is the center of $\omega$. Now $EX = ED$ implies $\angle EDX = \alpha$. From here we have that
$\angle BED = 2\alpha$. Using that $BTXD$ is cyclic we obtain $\angle BTD = \angle BXD = \alpha$. We also have that
\[
\angle TBE = 180^\circ - \angle BET - \angle ETB = 180^\circ - (180^\circ - 2\alpha) - \alpha = \alpha = \angle BTE.
\]
This gives us $BE = TE$. We similarly show that $CE = TE$, and so $BE = CE$. In the same way as in
the second solution, this now gives us that $BD = CD$, so $D$ is also the midpoint of $BC$ and we must
have $AB = AC$.

\medskip

\textbf{Alternative Solution.} We can solve the problem using only calculations. Note that the condition
of the problem is that $E$ lies on the radical axis of the circumcircles of $\triangle BDX$ and $\triangle CDY$. This
gives us $EB \cdot EX = EC \cdot EY$. However, $EX = EY$ because $E$ is the center of $\omega$ and this means that
$BE = CE$. Now using Pythagoras' theorem we have the following:
\[
AB^2 = AD^2 + BD^2 = AD^2 + (BE^2 - DE^2) = AD^2 + (CE^2 - DE^2) = AD^2 + CD^2 = AC^2.
\]
From here we obtain $AB = AC$.",562,3473,Geometry,7
60,shl_jbmo_2020_g2,shl_jbmo,2020,g,"Let $\triangle ABC$ be a right-angled triangle with $\angle BAC = 90^\circ$, and let $E$ be the foot of
the perpendicular from $A$ on $BC$. Let $Z 
eq A$ be a point on the line $AB$ with $AB = BZ$. Let $(c)$,
$(c_1)$ be the circumcircles of the triangles $\triangle AEZ$ and $\triangle BEZ$, respectively. Let $(c_2)$ be an arbitrary
circle passing through the points $A$ and $E$. Suppose $(c_1)$ meets the line $CZ$ again at the point $F$, and
meets $(c_2)$ again at the point $N$. If $P$ is the other point of intersection of $(c_2)$ with $AF$, prove that the
points $N, B, P$ are collinear.","\begin{solution}
Since the triangles $\triangle AEB$ and $\triangle CAB$ are similar, then
\[
\frac{AB}{EB} = \frac{CB}{AB}.
\]
Since $AB = BZ$ we get
\[
\frac{BZ}{EB} = \frac{CB}{BZ}
\]
from which it follows that the triangles $\triangle ZBE$ and $\triangle CBZ$ are also similar. Since $FEBZ$ is cyclic,
then $\angle BEZ = \angle BFZ$. So by the similarity of triangles $\triangle ZBE$ and $\triangle CBZ$ we get
\[
\angle BFZ = \angle BEZ = \angle BZC = \angle BZF
\]
and therefore the triangle $\triangle BFZ$ is isosceles. Since $BF = BZ = AB$, then the triangle $\triangle AFZ$ is
right-angled with $\angle AFZ = 90^\circ$.

It now follows that the points $A, E, F, C$ are concyclic. Since $A, P, E, N$ are also concyclic, then
\[
\angle ENP = \angle EAP = \angle EAF = \angle ECF = \angle BCZ = \angle BZE,
\]
where in the last equality we used again the similarity of the triangles $\triangle ZBE$ and $\triangle CBZ$. Since
$N, B, E, Z$ are concyclic, then $\angle ENP = \angle BZE = \angle ENB$, from which it follows that the points
$N, B, P$ are collinear.
\end{solution}",593,1083,Geometry,8
61,shl_jbmo_2020_g3,shl_jbmo,2020,g,"Let $\triangle ABC$ be a right-angled triangle with $\angle BAC = 90^\circ$ and let $E$ be the foot of the perpendicular from $A$ on $BC$. Let $Z 
eq A$ be a point on the line $AB$ with $AB = BZ$. Let $(c)$ be the circumcircle
of the triangle $\triangle AEZ$. Let $D$ be the second point of intersection of $(c)$ with $ZC$ and let $F$ be the
antidiametric point of $D$ with respect to $(c)$. Let $P$ be the point of intersection of the lines $FE$ and
$CZ$. If the tangent to $(c)$ at $Z$ meets $PA$ at $T$, prove that the points $T, E, B, Z$ are concyclic.","\begin{solution}
We will first show that $PA$ is tangent to $(c)$ at $A$.

Since $E, D, Z, A$ are concyclic, then $\angle EDC = \angle EAZ = \angle EAB$. Since also the triangles $\triangle ABC$
and $\triangle EBA$ are similar, then $\angle EAB = \angle BCA$, therefore $\angle EDC = \angle BCA$.

Since $\angle FED = 90^\circ$, then $\angle PED = 90^\circ$ and so
\[
\angle EPD = 90^\circ - \angle EDC = 90^\circ - \angle BCA = \angle EAC.
\]
Therefore the points $E, A, C, P$ are concyclic. It follows that $\angle CPA = 90^\circ$ and therefore the triangle
$\angle PAZ$ is right-angled. Since also $B$ is the midpoint of $AZ$, then $PB = AB = BZ$ and so $\angle ZPB =
\angle PZB$.

Furthermore, $\angle EPD = \angle EAC = \angle CBA = \angle EBA$ from which it follows that the points $P, E, B, Z$
are also concyclic.

Now observe that
\[
\angle PAE = \angle PCE = \angle ZPB - \angle PBE = \angle PZB - \angle PZE = \angle EZB.
\]
Therefore $PA$ is tangent to $(c)$ at $A$ as claimed.

It now follows that $TA = TZ$. Therefore
\begin{align*}
\angle PTZ &= 180^\circ - 2(\angle TAB) = 180^\circ - 2(\angle PAE + \angle EAB) = 180^\circ - 2(\angle ECP + \angle ACB)\\
&= 180^\circ - 2(90^\circ - \angle PZB) = 2(\angle PZB) = \angle PZB + \angle BPZ = \angle PBA.
\end{align*}
Thus $T, P, B, Z$ are concyclic, and since $P, E, B, Z$ are also concyclic then $T, E, B, Z$ are concyclic as
required.
\end{solution}",556,1413,Geometry,9
62,shl_jbmo_2020_n1,shl_jbmo,2020,n,Determine whether there is a natural number $n$ for which $8^n + 47$ is prime.,"\begin{solution}
The number $m = 8^n + 47$ is never prime.

If $n$ is even, say $n = 2k$, then $m = 64^k + 47 \equiv 1 + 2 \equiv 0 \pmod{3}$. Since also $m > 3$, then $m$ is not prime.

If $n \equiv 1 \pmod{4}$, say $n = 4k+1$, then $m = 8 \cdot (8^k)^4 + 47 \equiv 3 + 2 \equiv 0 \pmod{5}$. Since also $m > 3$, then $m$ is not prime.

If $n \equiv 3 \pmod{4}$, say $n = 4k+3$, then $m = 8(64^{2k+1} + 1) \equiv 8((-1)^{2k+1} + 1) \equiv 0 \pmod{13}$. Since also $m > 13$, then $m$ is not prime.
\end{solution}",78,511,Number Theory,10
63,shl_jbmo_2020_n2,shl_jbmo,2020,n,"Find all positive integers $a, b, c$ and $p$, where $p$ is a prime number, such that
\[
73p^2 + 6 = 9a^2 + 17b^2 + 17c^2.
\]","\begin{solution}
Since the equation is symmetric with respect to the numbers $b$ and $c$, we assume that $b \geq c$.

If $p \geq 3$, then $p$ is an odd number. We consider the equation modulo $8$. Since,
\[
73p^2 + 6 \equiv 79 \equiv 7 \pmod{8},
\]
we get that
\[
a^2 + b^2 + c^2 \equiv 7 \pmod{8}.
\]
This cannot happen since for any integer $x$ we have that
\[
x^2 \equiv 0, 1, 4 \pmod{8}.
\]
Hence, $p$ must be an even prime number, which means that $p = 2$. In this case, we obtain the equation
\[
9a^2 + 17(b^2 + c^2) = 289.
\]
It follows that $b^2 + c^2 \leq 17$. This is possible only for
\[
(b, c) \in \{(4, 1), (3, 2), (3, 1), (2, 2), (2, 1), (1, 1)\}.
\]
It is easy to check that among these pairs only the $(4, 1)$ gives an integer solution for $a$, namely $a = 1$.
Therefore, the given equation has only two solutions,
\[
(a, b, c, p) \in \{(1, 1, 4, 2), (1, 4, 1, 2)\}.
\]
\end{solution}",124,900,Number Theory,11
64,shl_jbmo_2020_n3,shl_jbmo,2020,n,"Find the largest integer $k$ $(k \geq 2)$, for which there exists an integer $n$ $(n \geq k)$ such that from
any collection of $n$ consecutive positive integers one can always choose $k$ numbers, which verify the
following conditions:
\begin{enumerate}
  \item each chosen number is not divisible by $6$, by $7$ and by $8$;
  \item the positive difference of any two different chosen numbers is not divisible by at least one of the
numbers $6$, $7$ or $8$.
\end{enumerate}","\begin{solution}
An integer is divisible by $6$, $7$ and $8$ if and only if it is divisible by their Least Common Multiple, which equals $6 \times 7 \times 4 = 168$.

Let $n$ be a positive integer and let $A$ be an arbitrary set of $n$ consecutive positive integers. Replace
each number $a_i$ from $A$ with its remainder $r_i \pmod{168}$. The number $a_i$ is divisible by $6$ (7 or 8) if
and only if its remainder $r_i$ is divisible by $6$ (respectively $7$ or $8$). The difference $|a_i - a_j|$ is divisible
by $168$ if and only if their remainders $r_i = r_j$.

Choosing $k$ numbers from the initial set $A$, which verify the required conditions, is the same as
choosing $k$ their remainders $\pmod{168}$ such that:
\begin{enumerate}
  \item each chosen remainder is not divisible by $6$, $7$ and $8$;
  \item all chosen remainders are different.
\end{enumerate}

Suppose we have chosen $k$ numbers from $A$, which verify the conditions. Therefore, all remainders
are different and $k \leq 168$ (otherwise, there would be two equal remainders).

Denote by $B = \{0, 1, 2, 3, \ldots, 167\}$ the set of all possible remainders $\pmod{168}$ and by $B_m$ the
subset of all elements of $B$, which are divisible by $m$. Compute the number of elements of the following
subsets:
\[
|B_6| = 168 \div 6 = 28, \quad |B_7| = 168 \div 7 = 24, \quad |B_8| = 168 \div 8 = 21,
\]
\[
|B_6 \cap B_7| = |B_{42}| = 168 \div 42 = 4, \quad |B_6 \cap B_8| = |B_{24}| = 168 \div 24 = 7,
\]
\[
|B_7 \cap B_8| = |B_{56}| = 168 \div 56 = 3, \quad |B_6 \cap B_7 \cap B_8| = |B_{168}| = 1.
\]
Denote by $D = B_6 \cup B_7 \cup B_8$, the subset of all elements of $B$, which are divisible by at least one
of the numbers $6$, $7$ or $8$. By the Inclusion-Exclusion principle we got
\begin{align*}
|D| &= |B_6| + |B_7| + |B_8| - (|B_6 \cap B_7| + |B_6 \cap B_8| + |B_7 \cap B_8|) + |B_6 \cap B_7 \cap B_8| \\
&= 28 + 24 + 21 - (4 + 7 + 3) + 1 = 60.
\end{align*}
Each chosen remainder belongs to the subset $B \setminus D$, since it is not divisible by $6$, $7$ and $8$. Hence,
$k \leq |B \setminus D| = 168 - 60 = 108$.

Let us show that the greatest possible value is $k = 108$. Consider $n = 168$. Given any collection $A$
of $168$ consecutive positive integers, replace each number with its remainder $\pmod{168}$. Choose from
these remainders $108$ numbers, which constitute the set $B \setminus D$. Finally, take $108$ numbers from the
initial set $A$, having exactly these remainders. These $k = 108$ numbers verify the required conditions.
\end{solution}",472,2530,Number Theory,12
65,shl_jbmo_2020_n4,shl_jbmo,2020,n,"Find all prime numbers $p$ such that
\[
(x + y)^{19} - x^{19} - y^{19}
\]
is a multiple of $p$ for any positive integers $x, y$.","\begin{solution}
If $x = y = 1$ then $p$ divides
\[
2^{19} - 2 = 2(2^{18} - 1) = 2(2^9 - 1)(2^9 + 1) = 2 \cdot 511 \cdot 513 = 2 \cdot 3^3 \cdot 7 \cdot 19 \cdot 73.
\]
If $x = 2$, $y = 1$ then
\[
p \mid 3^{19} - 2^{19} - 1.
\]
We will show that $3^{19} - 2^{19} - 1$ is not a multiple of $73$. Indeed,
\[
3^{19} \equiv 3^3 \cdot (3^4)^4 \equiv 3^3 \cdot 8^4 \equiv 3^3 \cdot (-9)^2 \equiv 27 \cdot 81 \equiv 27 \cdot 8 \equiv 70 \pmod{73}
\]
and
\[
2^{19} \equiv 2 \cdot 64^3 \equiv 2 \cdot (-9)^3 \equiv -18 \cdot 81 \equiv -18 \cdot 8 \equiv -144 \equiv 2 \pmod{73}.
\]
Thus $p$ can be only among $2, 3, 7, 19$. We will prove all these work.

\begin{itemize}
  \item For $p = 19$ this follows by Fermat's Theorem as
  \[
  (x + y)^{19} \equiv x + y \pmod{19}, \quad x^{19} \equiv x \pmod{19}, \quad y^{19} \equiv y \pmod{19}.
  \]
  \item For $p = 7$, we have that
  \[
  a^{19} \equiv a \pmod{7},
  \]
  for every integer $a$. Indeed, if $7 \mid a$, it is trivial, while if $7 
mid a$, then by Fermat's Theorem we have
  \[
  7 \mid a^6 - 1 \mid a^{18} - 1,
  \]
  therefore $7 \mid a(a^{18} - 1)$.
  \item For $p = 3$, we will prove that
  \[
  b^{19} \equiv b \pmod{3}.
  \]
  Indeed, if $3 \mid b$, it is trivial, while if $3 
mid b$, then by Fermat's Theorem we have
  \[
  3 \mid b^2 - 1 \mid b^{18} - 1,
  \]
  therefore $3 \mid b(b^{18} - 1)$.
  \item For $p = 2$ it is true, since among $x + y$, $x$ and $y$ there are $0$ or $2$ odd numbers.
\end{itemize}
\end{solution}",128,1482,Number Theory,13
66,shl_jbmo_2020_n5,shl_jbmo,2020,n,"The positive integer $k$ and the set $A$ of different integers from $1$ to $3k$ inclusive are such that
there are no distinct $a, b, c$ in $A$ satisfying $2b = a + c$. The numbers from $A$ in the interval $[1, k]$ will
be called \textit{small}; those in $[k+1, 2k]$ -- \textit{medium} and those in $[2k+1, 3k]$ -- \textit{large}. Is it always true that
there are no positive integers $x$ and $d$ such that if $x$, $x + d$ and $x + 2d$ are divided by $3k$ then the
remainders belong to $A$ and those of $x$ and $x + d$ are different and are:
\begin{enumerate}[label=\alph*)]
  \item small?
  \item medium?
  \item large?
\end{enumerate}
\textit{(In this problem we assume that if a multiple of $3k$ is divided by $3k$ then the remainder is $3k$ rather
than $0$.)}","\begin{solution}
A counterexample for a) is $k = 3$, $A = \{1, 2, 9\}$, $x = 2$ and $d = 8$.

A counterexample for c) is $k = 3$, $A = \{1, 8, 9\}$, $x = 8$ and $d = 1$.

We will prove that b) is true.

Suppose the contrary and let $x, d$ have the above properties. We can assume $0 < d < 3k$, $0 < x \leq 3k$
(since for $d = 3k$ the remainders for $x$ and $x + d$ are equal). Hence $0 < x + d < 6k$ and there are two
cases:

\begin{itemize}
  \item If $x + d > 3k$, then since the remainder for $x + d$ is medium we have $4k < x + d \leq 5k$. This
means that the remainder of $x + d$ when it is divided by $3k$ is $x + d - 3k$.
Since $x$ is medium we have $x \leq 2k$ so $d = (x+d) - x > 2k$. Therefore $6k = 4k + 2k < (x+d)+d <
8k$. This means that the remainder of $x + 2d$ when it is divided by $3k$ is $x + 2d - 6k$.
Thus the remainders $(x + 2d - 6k)$, $(x + d - 3k)$ and $x$ are in $[1, 3k]$, they belong to $A$ and
\[
2(x + d - 3k) = (x + 2d - 6k) + x,
\]
a contradiction.

  \item If $x + d \leq 3k$ then as $x + d$ is medium we have $k < x + d \leq 2k$. From the limitations on $x$, we
have $x > k$ so $d = (x + d) - x < k$. Hence $0 \leq x + 2d = (x + d) + d < 3k$. Thus the remainders
$x$, $x + d$ and $x + 2d$ are in $A$ and
\[
2(x + d) = (x + 2d) + x,
\]
a contradiction.
\end{itemize}
\end{solution}",762,1314,Number Theory,14
67,shl_jbmo_2020_n6,shl_jbmo,2020,n,"Are there any positive integers $m$ and $n$ satisfying the equation
\[
m^3 = 9n^4 + 170n^2 + 289 \,?
\]","\begin{solution}
We will prove that the answer is no. Note that
\[
m^3 = 9n^4 + 170n^2 + 289 = (9n^2 + 17)(n^2 + 17).
\]
If $n$ is odd then $m$ is even, therefore $8 \mid m^3$. However,
\[
9n^4 + 170n^2 + 289 \equiv 9 + 170 + 289 \equiv 4 \pmod{8},
\]
which leads to a contradiction. If $n$ is a multiple of $17$ then so is $m$ and hence $289$ is a multiple of $17^3$, which is absurd. For $n$ even and not multiple of $17$, since
\[
\gcd(9n^2 + 17, n^2 + 17) \mid 9(n^2 + 17) - (9n^2 + 17) = 2^3 \cdot 17,
\]
this gcd must be $1$. Therefore $n^2 + 17 = a^3$ for an odd $a$, so
\[
n^2 + 25 = (a + 2)(a^2 - 2a + 4).
\]
For $a \equiv 1 \pmod{4}$ we have $a + 2 \equiv 3 \pmod{4}$, while for $a \equiv 3 \pmod{4}$ we have $a^2 - 2a + 4 \equiv 3
\pmod{4}$. Thus $(a + 2)(a^2 - 2a + 4)$ has a prime divisor of type $4\ell + 3$. As it divides $n^2 + 25$, it has to
divide $n$ and $5$, which is absurd.
\end{solution}",103,910,Number Theory,15
68,shl_jbmo_2020_n7,shl_jbmo,2020,n,"Prove that there doesn't exist any prime $p$ such that every power of $p$ is a palindrome (palindrome is a number that is read the same from the left as it is from the right; in particular, number
that ends in one or more zeros cannot be a palindrome).","\begin{solution}
Note that by criterion for divisibility by $11$ and the definition of a palindrome we have that
every palindrome that has even number of digits is divisible by $11$.

Since $11^5 = 161051$ is not a palindrome and since $11$ cannot divide $p^k$ for any prime other than $11$ we
are now left to prove that no prime whose all powers have odd number of digits exists.

Assume the contrary. It means that the difference between the numbers of digits of $p^m$ and $p^{m+1}$ is
even number. We will prove that for every natural $m$, the difference is the same even number.

If we assume not, that means that the difference for some $m_1$ has at least $2$ digits more than the
difference for some $m_2$. We will prove that this is impossible.

Let $p^{m_1} = 10^{t_1} \cdot a_1$, $p^{m_2} = 10^{t_2} \cdot a_2$ and $p = 10^h \cdot z$, where $1 < a_1, a_2, z < 10$.

This implies that $1 < a_1 \cdot z, a_2 \cdot z < 100$, which further implies that multiplying these powers of $p$ by $p$ can increase their number of digits by either $h$ or $h + 1$.

This is a contradiction. Call the difference between numbers of digits of consecutive powers $d$. Number
$p$ clearly cannot be equal to $10^d$ for $d \geq 1$ because $10$ is divisible by two primes, but for $d = 0$, we
would have that $1$ is a prime which is not true.

\textbf{Case 1.} $p > 10^d$. Let $p = 10^d \cdot a$, for some real number $a$ greater than $1$. \hfill (1)

From the definition of $d$ we also see that $a$ is smaller than $10$. \hfill (2)

From (1) we see that powering $a$ gives us arbitrarily large numbers and from (2) we conclude that
there is some natural power of $a$, call it $b$, greater than $1$, such that $10 < a^b < 100$.

It is clear that $p^b$ has exactly $(b - 1)d + 1$ digits more than $p$ has, which is an odd number, but sum
of even numbers is even.

\textbf{Case 2.} $p < 10^d$. Let $p = \dfrac{10^d}{a}$, for some real number $a$ greater than $1$. \hfill (1)

From the definition of $d$ we also see that $a$ is smaller than $10$. \hfill (2)

From (1) we see that powering $a$ gives us arbitrarily large numbers and from (2) we conclude that
there is some natural power of $a$, call it $b$, greater than $1$, such that $10 < a^b < 100$.

It is clear that $p^b$ has exactly $(b - 1)d - 1$ digits more than $p$ has, which is an odd number, but sum
of even numbers is even.

We have now arrived at the desired contradiction for both cases and have thus finished the proof.

\medskip

\textbf{Alternative solution.} Note that the sequence $\{p^n\}$ is periodic $\pmod{10}$. Let the period be $d$. Also,
let $p^d = g$.

Since all powers of $p$ are palindromes, all powers of $g$ are as well. Since $\{g^n\}$ is constant $\pmod{10}$,
the leftmost digit of each power of $g$ is equal to some $f$.

We will prove that the difference between numbers of digits of $g^m$ and $g^{m+1}$ is equal to some $r$ for
every natural number $m$.

This is true due to size reasons. Namely, to add exactly $k$ digits, and yet to have the same leftmost
digit, we need to multiply the number by at least $5 \cdot 10^{k-1}$ (if $k = 0$ then it's $1$) and by at most
$2 \cdot 10^k$ (values depend on the leftmost digit, it can easily be seen that leftmost digit being $1$ yields the
extremal values). Notice that
\[
2 \cdot 10^k < 5 \cdot 10^{k+1-1}.
\]
Since this inequality has clearly shown that the interval of multipliers which add exactly $k$ digits and
leave the leftmost digit the same is disjunct from the same kind of interval for $k + 1$ digits, which
implies that no number can belong to both intervals, we have successfully proven the claim.

Clearly, $g$ cannot be equal to $10^r$ for $r \geq 1$ because a palindrome cannot be divisible by $10$, but for
$r = 0$ we again cannot have the equality because $1$ is not a natural power of a prime.

\textbf{Case 1.} $g > 10^r$. Let $g = 10^r \cdot a$, where $a$ is a real number, $10 > a > 1$. (Here, $a$ is less
than $10$ because if it was not, multiplying by $g$ would add at least $r + 1$ digits, which is impossible.)
From this we see that powering $a$ gives us arbitrarily large numbers and that there is some natural
power of $a$, call it $b$, greater than $1$, such that $10 < a^b < 100$.
Pick smallest such $b$.
Now we easily see that the difference between numbers of digits of numbers $g^{b-1}$ and $g^b$ is exactly $r + 1$.

\textbf{Case 2.} $g < 10^r$. Let $g = \dfrac{10^r}{a}$, where $a$ is a real number, $10 > a > 1$. (Here, $a$ is less than
$10$ because if it was not, multiplying by $g$ would add at most $r - 1$ digits, which is impossible.)
From this we see that powering $a$ gives us arbitrarily large numbers and that there is some natural
power of $a$, call it $b$, greater than $1$, such that $10 < a^b < 100$.
Pick smallest such $b$.
Now we easily see that the difference between numbers of digits of numbers $g^{b-1}$ and $g^b$ is exactly $r - 1$.

We have arrived at the desired contradiction for both cases and have thus finished the proof.

\medskip

\textit{Comment.} In both solutions, after introducing $a$, there are multiple ways to finish the problem.
In particular, solution $1$ and solution $2$ could be finished in the same way, but distinct finishes were
purposely offered. Third, maybe even the most intuitive finishing argument, could be using non-exact
size arguments; namely, just the fact that powers of $a$ grow arbitrarily large is enough to reach the
contradiction.

\medskip

\textit{Alternative problem.} Find all positive integers $n$ such that every power of $n$ is a palindrome
(palindrome is a number that is read the same from the left as it is from the right; in particular,
number that ends in one or more zeros cannot be a palindrome).

Suggested difficulty for this problem is hard. Noting why solution $2$ doesn't work for $n = 1$ (because
$g$ now could be equal to $1$) and saying that $n = 1$ actually works are all necessary modifications to
solution $2$ to make it work for the alternative problem as well.
\end{solution}",252,6015,Number Theory,16
69,shl_jbmo_2020_n8,shl_jbmo,2020,n,"Find all pairs $(p, q)$ of prime numbers such that
\[
1 + \frac{p^q - q^p}{p + q}
\]
is a prime number.","\begin{solution}
It is clear that $p 
eq q$. We set
\[
1 + \frac{p^q - q^p}{p + q} = r
\]
and we have that
\begin{equation}\label{eq:N8}
p^q - q^p = (r - 1)(p + q).
\end{equation}
From Fermat's Little Theorem we have
\[
p^q - q^p \equiv -q \pmod{p}.
\]
Since we also have that
\[
(r - 1)(p + q) \equiv -rq - q \pmod{p},
\]
from \eqref{eq:N8} we get that
\[
rq \equiv 0 \pmod{p} \Rightarrow p \mid qr,
\]
hence $p \mid r$, which means that $p = r$. Therefore, \eqref{eq:N8} takes the form
\begin{equation}\label{eq:N8b}
p^q - q^p = (p - 1)(p + q).
\end{equation}
We will prove that $p = 2$. Indeed, if $p$ is odd, then from Fermat's Little Theorem we have
\[
p^q - q^p \equiv p \pmod{q}
\]
and since
\[
(p - 1)(p + q) \equiv p(p - 1) \pmod{q},
\]
we have
\[
p(p - 2) \equiv 0 \pmod{q} \Rightarrow q \mid p(p - 2) \Rightarrow q \mid p - 2 \Rightarrow q \leq p - 2 < p.
\]
Now, from \eqref{eq:N8b} we have
\[
p^q - q^p \equiv 0 \pmod{p - 1} \Rightarrow 1 - q^p \equiv 0 \pmod{p - 1} \Rightarrow q^p \equiv 1 \pmod{p - 1}.
\]
Clearly $\gcd(q, p - 1) = 1$ and if we set $k = \mathrm{ord}_{p-1}(q)$, it is well-known that $k \mid p$ and $k < p$, therefore
$k = 1$. It follows that
\[
q \equiv 1 \pmod{p - 1} \Rightarrow p - 1 \mid q - 1 \Rightarrow p - 1 \leq q - 1 \Rightarrow p \leq q,
\]
a contradiction.

Therefore, $p = 2$ and \eqref{eq:N8b} transforms to
\[
2^q = q^2 + q + 2.
\]
We can easily check by induction that for every positive integer $n \geq 6$ we have $2^n > n^2 + n + 2$. This
means that $q \leq 5$ and the only solution is for $q = 5$. Hence the only pair which satisfy the condition
is $(p, q) = (2, 5)$.

\medskip

\textit{Comment by the PSC.} From the problem condition, we get that $p^q$ should be bigger than $q^p$, which
gives
\[
q \ln p > p \ln q \iff \frac{\ln p}{p} > \frac{\ln q}{q}.
\]
The function $\dfrac{\ln x}{x}$ is decreasing for $x > e$, thus if $p$ and $q$ are odd primes, we obtain $q > p$.
\end{solution}",103,1939,Number Theory,17
259,jbmo_2021_p1,jbmo,2021,a,"Let $n$ ($n \geq 1$) be an integer. Consider the equation
\[
2 \cdot \left\lfloor \frac{1}{2x} 
ight
floor - n + 1 = (n+1)(1 - nx),
\]
where $x$ is the unknown real variable.
\begin{enumerate}[label=(\alph*)]
    \item Solve the equation for $n = 8$.
    \item Prove that there exists an integer $n$ for which the equation has at least 2021 solutions.
\end{enumerate}
(For any real number $y$ by $\lfloor y 
floor$ we denote the largest integer $m$ such that $m \leq y$.)","\subsection*{Solution}

Let $k = \left\lfloor \frac{1}{2x} 
ight
floor$, $k \in \mathbb{Z}$.

\begin{enumerate}[label=(\alph*)]
    \item For $n = 8$, the equation becomes
    \[
    k = \left\lfloor \frac{1}{2x} 
ight
floor = 8 - 36x \implies x 
eq 0 \text{ and } x = \frac{8-k}{36}.
    \]
    Since $x 
eq 0$, we have $k 
eq 8$, and the last relation implies
    \[
    k = \left\lfloor \frac{1}{2x} 
ight
floor = \left\lfloor \frac{18}{8-k} 
ight
floor.
    \]
    Checking signs, we see that $0 < k < 8$. By direct verification, we find the solutions $k = 3$ (hence $x = \frac{5}{36}$) and $k = 4$ (hence $x = \frac{1}{9}$).

    \item From the given equation we have $x 
eq 0$ and $x = \frac{2(n-k)}{n(n+1)}$. Therefore, $k 
eq n$ and
    \[
    k = \left\lfloor \frac{1}{2x} 
ight
floor = \left\lfloor \frac{n(n+1)}{4(n-k)} 
ight
floor.
    \]
    Again, checking signs we see that $0 \leq k < n$. The last equation implies
    \[
    k \leq \frac{n(n+1)}{4(n-k)} < k+1 \implies
    \begin{cases}
    (2k-n)^2 + n \geq 0 \\
    (2k+1-n)^2 < n+1
    \end{cases}
    \implies
    \]
    \begin{equation}
    \frac{n - 1 - \sqrt{n+1}}{2} < k < \frac{n - 1 + \sqrt{n+1}}{2} \tag{2}
    \end{equation}
    Conversely, if $k \in \mathbb{Z}$ satisfies (2) and $0 < k < n$, then $x = \frac{2(n-k)}{n(n+1)}$ is a solution to the given equation. It remains to note that choosing $n$ such that $\sqrt{n+1} > 2021$ ensures that there exist at least 2021 integer values of $k$ which satisfy (2).
\end{enumerate}",471,1505,Algebra,1
260,jbmo_2021_p2,jbmo,2021,a,"For any set $A = \{x_1, x_2, x_3, x_4, x_5\}$ of five distinct positive integers denote by $S_A$ the sum of its elements, and denote by $T_A$ the number of triples $(i,j,k)$ with $1 \leq i < j < k \leq 5$ for which $x_i + x_j + x_k$ divides $S_A$.

Find the largest possible value of $T_A$.","\subsection*{Solution}

We will prove that the maximum value that $T_A$ can attain is $4$. Let $A = \{x_1, x_2, x_3, x_4, x_5\}$ be a set of five positive integers such that $x_1 < x_2 < x_3 < x_4 < x_5$. Call a triple $(i,j,k)$ with $1 \leq i < j < k \leq 5$ \emph{good} if $x_i + x_j + x_k$ divides $S_A$. None of the triples $(3,4,5), (2,4,5), (1,4,5), (2,3,5), (1,3,5)$ is good, since, for example
\[
x_5 + x_3 + x_1 \mid S_A \implies x_5 + x_3 + x_1 \mid x_2 + x_4
\]
which is impossible since $x_5 > x_4$ and $x_3 > x_2$. Analogously we can show that any triple of form $(x, y, 5)$ where $y > 2$ isn't good.

By above, the number of good triples can be at most $5$ and only triples $(1,2,5)$, $(2,3,4)$, $(1,3,4)$, $(1,2,4)$, $(1,2,3)$ can be good. But if triples $(1,2,5)$ and $(2,3,4)$ are simultaneously good we have that:
\[
x_1 + x_2 + x_5 \mid x_3 + x_4 \implies x_5 < x_3 + x_4 \tag{1}
\]
and
\[
x_2 + x_3 + x_4 \mid x_1 + x_5 \implies x_2 + x_3 + x_4 \leq x_1 + x_5 \overset{(1)}{<} x_1 + x_3 + x_4 < x_2 + x_3 + x_4,
\]
which is impossible. Therefore, $T_A \leq 4$.

Alternatively, one can prove the statement above by adding up the two inequalities $x_1 + x_2 + x_4 < x_3 + x_4$ and $x_2 + x_3 + x_4 < x_1 + x_5$ that are derived from the divisibilities.

To show that $T_A = 4$ is possible, consider the numbers $1, 2, 3, 4, 494$. This works because $6 \mid 498$, $7 \mid 497$, $8 \mid 496$, and $9 \mid 495$.

\paragraph{Remark.} The motivation for construction is to realize that if we choose $x_1, x_2, x_3, x_4$ we can get all the conditions $x_5$ must satisfy. Let $S = x_1 + x_2 + x_3 + x_4$. Now we have to choose $x_5$ such that
\[
S - x_i \mid x_i + x_5, \text{ i.e. } x_5 \equiv -x_i \pmod{S - x_i} \quad \forall i \in \{1, 2, 3, 4\}.
\]
By the Chinese Remainder Theorem it is obvious that if $S - x_1, S - x_2, S - x_3, S - x_4$ are pairwise coprime, such $x_5$ must exist. To make all these numbers pairwise coprime it's natural to take $x_1, x_2, x_3, x_4$ to be all odd and then solve mod 3 issues. Fortunately it can be seen that $1, 5, 7, 11$ easily works because $13, 17, 19, 23$ are pairwise coprime.

However, even without the knowledge of this theorem it makes sense intuitively that this system must have a solution for some $x_1, x_2, x_3, x_4$. By taking $(x_1, x_2, x_3, x_4) = (1, 2, 3, 4)$ we get a pretty simple system which can be solved by hand rather easily.",290,2405,Algebra,2
261,jbmo_2021_p3,jbmo,2021,g,"Let $ABC$ be an acute scalene triangle with circumcenter $O$. Let $D$ be the foot of the altitude from $A$ to the side $BC$. The lines $BC$ and $AO$ intersect at $E$. Let $s$ be the line through $E$ perpendicular to $AO$. The line $s$ intersects $AB$ and $AC$ at $K$ and $L$, respectively. Denote by $\omega$ the circumcircle of triangle $AKL$. Line $AD$ intersects $\omega$ again at $X$.

Prove that $\omega$ and the circumcircles of triangles $ABC$ and $DEX$ have a common point.","\subsection*{Solution}

Let us denote angles of triangle $ABC$ with $\alpha, \beta, \gamma$ in a standard way. By basic angle-chasing we have
\[
\angle BAD = 90^\circ - \beta = \angle OAC \quad \text{and} \quad \angle CAD = \angle BAO = 90^\circ - \gamma.
\]

Using the fact that lines $AE$ and $AX$ are isogonal with respect to $\angle KAL$ we can conclude that $X$ is an $A$-antipode on $\omega$. (This fact can be purely angle-chased: we have
\[
\angle KAX + \angle AXK = \angle KAX + \angle ALK = 90^\circ - \beta + \beta = 90^\circ
\]
which implies $\angle AKX = 90^\circ$). Now let $F$ be the projection of $X$ on the line $AE$. Using that $AX$ is a diameter of $\omega$ and $\angle EDX = 90^\circ$ it's clear that $F$ is the intersection point of $\omega$ and the circumcircle of triangle $DEX$. Now it suffices to show that $ABFC$ is cyclic. We have $\angle KLF = \angle KAF = 90^\circ - \gamma$ and from $\angle FEL = 90^\circ$ we have that $\angle EFL = \gamma = \angle ECL$ so quadrilateral $EFCL$ is cyclic. Next, we have
\[
\angle AFC = \angle EFC = 180^\circ - \angle ELC = \angle ELA = \beta
\]
(where last equality holds because of $\angle AEL = 90^\circ$ and $\angle EAL = 90^\circ - \beta$). $\square$

\subsection*{Solution 2}

We have $\angle BAD = 90^\circ - \beta = \angle OAC$ and that $AX$ is the diameter of $\omega$. Also we note that
\[
\angle ALK = \beta, \quad \angle KLC = 180^\circ - \beta = \angle KBC
\]
so $BKCL$ is cyclic. Let $AO$ intersect circumcircle of $ABC$ again at $A'$. We will show that $A'$ is the desired concurrence point. Obviously $AA'$ is the diameter of circumcircle of triangle $ABC$ so $\angle A'CA = 90^\circ$ which implies that $A'CLE$ is cyclic. From power of point $E$ we have that $EK \cdot EL = EB \cdot EC = EA \cdot EA'$ so we can conclude that $A' \in \omega$. Now using the fact that $AX$ is a diameter of $\omega$ implies $\angle AXA' = 90^\circ$ we have that $DXA'E$ is cyclic because of $\angle EDX = 90^\circ$ which finishes the proof. $\square$",481,2013,Geometry,3
262,jbmo_2021_p4,jbmo,2021,c,"Let $M$ be a subset of the set of 2021 integers $\{1, 2, 3, \ldots, 2021\}$ such that for any three elements (not necessarily distinct) $a, b, c$ of $M$ we have $|a + b - c| > 10$. Determine the largest possible number of elements of $M$.","\subsection*{Solution}

The set $M = \{1016, 1017, \ldots, 2021\}$ has 1006 elements and satisfies the required property, since $a, b, c \in M$ implies that $a + b - c > 1016 + 1016 - 2021 = 11$. We will show that this is optimal.

Suppose $M$ satisfies the condition in the problem. Let $k$ be the minimal element of $M$. Then $k = |k + k - k| > 10 \implies k > 11$. Note also that for every $m$, the integers $m$, $m + k - 10$ cannot both belong to $M$, since $k + m - (m + k - 10) = 10$.

\begin{claim}
$M$ contains at most $k - 10$ out of any $2k - 20$ consecutive integers.
\end{claim}
\begin{proof}
We can partition the set $\{m, m+1, \ldots, m + 2k - 21\}$ into $k - 10$ pairs as follows:
\[
\{m, m+k-10\},\ \{m+1, m+k-9\},\ \ldots,\ \{m+k-11, m+2k-21\}.
\]
It remains to note that $M$ can contain at most one element of each pair.
\end{proof}

\begin{claim}
$M$ contains at most $\left\lfloor \frac{t + k - 10}{2} 
ight
floor$ out of any $t$ consecutive integers.
\end{claim}
\begin{proof}
Write $t = q(2k - 20) + r$ with $r \in \{0, 1, 2, \ldots, 2k - 21\}$. From the set of the first $q(2k - 20)$ integers, by Claim 1 at most $q(k - 10)$ can belong to $M$. Also by Claim 1, it follows that from the last $r$ integers, at most $\min\{r, k-10\}$ can belong to $M$.

Thus,
\begin{itemize}
    \item If $r \leq k - 10$, then at most
    \[
    q(k-10) + r = \frac{t + r}{2} \leq \frac{t + k - 10}{2}
    \]
    integers belong to $M$.
    \item If $r > k - 10$, then at most
    \[
    q(k-10) + k - 10 = \frac{t - r + 2(k-10)}{2} \leq \frac{t + k - 10}{2}
    \]
    integers belong to $M$.
\end{itemize}
\end{proof}

By Claim 2, the number of elements of $M$ amongst $k+1, k+2, \ldots, 2021$ is at most
\[
\left\lfloor \frac{(2021 - k) + (k - 10)}{2} 
ight
floor = 1005.
\]
Since amongst $\{1, 2, \ldots, k\}$ only $k$ belongs to $M$, we conclude that $M$ has at most $\mathbf{1006}$ elements as claimed.",238,1912,Combinatorics,4
302,shl_jbmo_2021_a1,shl_jbmo,2021,a,"Let $n$ ($n \ge 1$) be an integer. Consider the equation

$2\cdot \lfloor{\frac{1}{2x}}
floor - n + 1 = (n + 1)(1 - nx)$,

where $x$ is the unknown real variable.

(a) Solve the equation for $n = 8$.
(b) Prove that there exists an integer $n$ for which the equation has at least $2021$ solutions.
(For any real number $y$ by $\lfloor{y} 
floor$ we denote the largest integer $m$ such that $m \le y$.)","\textbf{Solution.} Let $A = \floor{\dfrac{1}{2x}}$ --- then the equation gives $x = \dfrac{2(n-A)}{n(n+1)}$ and now substituting in the definition of $A$ yields
\[
A = \floor{\frac{n(n+1)}{4(n-A)}}.
\]
The latter equality is a necessary and sufficient condition for the corresponding $x$ to be a solution to the equation. Let us also observe that $A$ is an integer and that $1 \leq A \leq n-1$ for $n \geq 3$ --- indeed, if $A = 0$, then $0 = \floor{\frac{n+1}{4}} \geq 1$; if $A = n$, the right-hand side is undefined; and if $A < 0$ or $A > n$, then the sides have different signs.

\medskip
\textbf{a) i)} For $n = 8$ we want $A = \floor{\dfrac{18}{8-A}}$. By the above, $A$ is an integer between $1$ and $7$ inclusive. A direct verification shows that only $A = 3$ and $A = 4$ are solutions, with the corresponding $x$ being $x = \dfrac{5}{36}$ and $x = \dfrac{1}{9}$.

\medskip
\textbf{a) ii)} For $n = 51$ we want $A = \floor{\dfrac{663}{51-A}}$ for integers $1 \leq A \leq 50$. This holds if and only if $A \leq \dfrac{663}{51-A} < A+1$. The left inequality is equivalent to $(2A-51)^2 + 51 \geq 0$ and holds for all $A$. The right one is equivalent to $(A-25)^2 < 13$ and hence has only $22 \leq A \leq 28$ as solutions. Hence all solutions are $x = \dfrac{51-A}{26 \cdot 51}$ for $22 \leq A \leq 28$.

\medskip
\textbf{b)} It suffices to have at least $2021$ integer solutions $1 \leq A \leq n-1$ to $A \leq \dfrac{n(n+1)}{4(n-A)} < A+1$ whenever $n \geq N$ for some suitable $N$. The left inequality is equivalent to $(2A-n)^2 + n \geq 0$ and holds for all $A$. The right inequality is equivalent to $(2A-n+1)^2 < n+1$ and hence holds precisely for $\dfrac{n-1-\sqrt{n+1}}{2} < A < \dfrac{n-1+\sqrt{n+1}}{2}$. Observe that this range for $A$ is tighter than $1 \leq A \leq n-1$ for $n \geq 6$, as $(n-3)^2 > n+1$ and $(n-1)^2 > (n+1)$ for these $n$. Finally, the difference between the endpoints of the interval $\left(\dfrac{n-1-\sqrt{n+1}}{2}, \dfrac{n-1+\sqrt{n+1}}{2}
ight)$ is $\sqrt{n+1}$ and hence for sufficiently large $n$ this interval must contain at least $2021$ integers. This completes the proof. \hfill$\square$",400,2136,Algebra,1
303,shl_jbmo_2021_a2,shl_jbmo,2021,a,"Let $n > 3$ be a positive integer. Find all integers $k$ such that $1 \le k \le n$ and for
which the following property holds:
If $x_1, . . . , x_n$ are $n$ real numbers such that $x_i + x_{i + 1} + ... + x_{i + k - 1} = 0$ for all integers $i > 1$ (indexes are taken modulo $n$), then $x_1 = . . . = x_n = 0$.","\textbf{Solution.} First, if some integer $d \geq 2$ divides both $k$ and $n$, the sequence
\[
x_1, x_2, \ldots, x_n = \underbrace{1, 0, \ldots, 0, -1}_{d \text{ numbers}},\; \underbrace{1, 0, \ldots, 0, -1}_{d \text{ numbers}},\; \ldots,\; \underbrace{1, 0, \ldots, 0, -1}_{d \text{ numbers}}
\]
is such that $x_i + x_{i+1} + \cdots + x_{i+k-1} = 0$ for all integers $i \geq 1$, but it contains non-zero terms. Thus, if $k$ is not coprime to $n$, it cannot be a solution of the problem.

Now, consider some integer $k \in \{1, 2, \ldots, n\}$ that is coprime with $n$, and let $x_1, x_2, \ldots, x_n$ be real numbers such that $x_i + x_{i+1} + \cdots + x_{i+k-1} = 0$ for all integers $i \geq 1$. Given any integer $i$, we have
\[
x_i = -x_{i+1} - \cdots - x_{i+k-1} = x_{i+k}.
\]
Thus, the sequence $(x_m)_{m \geq 1}$ is periodic, with period $k$. Since $k$ is coprime with $n$, there exists an integer $\ell$ such that $k\ell \equiv 1 \pmod{n}$. It follows that $x_i = x_{i+k\ell} = x_{i+1}$ for all $i \geq 1$, i.e., that the real numbers $x_i$ are all equal. Hence, $0 = x_1 + x_2 + \cdots + x_k = kx_1$ and $0 = x_1 = x_2 = \cdots = x_n$.

In conclusion, the solutions are the integers $k$ such that $1 \leq k \leq n$ and $k$ is coprime with $n$. \hfill$\square$",310,1268,Algebra,2
304,shl_jbmo_2021_a3,shl_jbmo,2021,a,"Let $n$ be a positive integer. A finite set of integers is called $n$-divided if there are exactly $n$ ways to partition this set into two subsets with equal sums. For example, the set $\{1, 3, 4, 5, 6, 7\}$ is $2$-divided because the only ways to partition it into two subsets with equal sums is by dividing it into $\{1, 3, 4, 5\}$ and $\{6, 7\}$, or $\{1, 5, 7\}$
and $\{3, 4, 6\}$. Find all the integers $n > 0$ for which there exists a $n$-divided set.","\textbf{Solution.} First, note that the set $\{1\}$ is $0$-divided and the set $\{1,2,3\}$ is $1$-divided. Now consider an integer $n \geq 2$ and let us show that the set
\[
E = \{k : 3n \leq k \leq 4n-1\} \cup \{k : 4n+1 \leq k \leq 5n\} \cup \{8n(n-2)\}
\]
is $n$-divided. Indeed, if $\Sigma_X$ denotes the sum of the elements of a set $X$, choosing a way to divide the set $E$ into two subsets with equal sums corresponds to selecting a subset $X$ of $E$ containing the number $8n(n-2)$, and for which
\[
\Sigma_X = \frac{\Sigma_E}{2} = \frac{1}{2}\left(8n(n-2) + \sum_{\ell=1}^{n}\bigl((4n-\ell)+(4n+\ell)\bigr)
ight) = \frac{1}{2}\bigl(8n(n-2)+8n^2\bigr) = 8n(n-1).
\]
In other words, it corresponds to selecting a subset $Y$ of the set $E' = E \setminus \{8n(n-2)\}$ for which $\Sigma_Y = 8n$.

Since $\max E' < 8n < 3\min E'$, such a set $Y$ contains exactly $2$ elements. So the corresponding sets $Y$ are the sets of the form $\{4n-\ell,\, 4n+\ell\}$ with $1 \leq \ell \leq n$. There exist $n$ such sets, so $E$ is $n$-divided as desired. \hfill$\square$",457,1063,Algebra,3
305,shl_jbmo_2021_g1,shl_jbmo,2021,g,"Let $ABC$ be an acute scalene triangle with circumcenter $O$. Let $D$ be the foot of the altitude from $A$ to the side $BC$. The lines $BC$ and $AO$ intersect at $E$. Let $s$ be the line through $E$ perpendicular to $AO$. The line $s$ intersects $AB$ and $AC$ at $K$ and $L$, respectively. Denote by $\omega$ the circumcircle of triangle $AKL$. Line $AD$ intersects $\omega$ again at $X$.
Prove that $\omega$ and the circumcircles of triangles $ABC$ and $DEX$ have a common point.","\textbf{Solution 1.}

Let us denote angles of triangle $ABC$ with $\alpha, \beta, \gamma$ in a standard way. We easily get that $\angle BAD = 90^\circ - \beta = \angle OAC$ and $\angle CAD = \angle BAO = 90^\circ - \gamma$.

Using the fact that lines $AE$ and $AX$ are isogonal with respect to $\angle KAL$ we can conclude that $X$ is an $A$-antipode on $\omega$. (This fact can be purely angle-chased, for example we have $\angle KAX + \angle AXK = \angle KAX + \angle ALK = 90^\circ - \beta + \beta = 90^\circ$ which implies $\angle AKX = 90^\circ$.) Now let us denote $F$ point on line $AE$ such that $XF \perp AE$. Using that $AX$ is a diameter of $\omega$ and $\angle EDX = 90^\circ$ it's clear that $F$ is the intersection point of $\omega$ and the circumcircle of $DEX$. Now it suffices to show that $ABFC$ is cyclic. Now we have that $\angle KLF = \angle KAF = 90^\circ - \gamma$ and from $\angle FEL = 90^\circ$ we have that $\angle EFL = \gamma = \angle ECL$ so quadrilateral $EFCL$ is cyclic. Now we have that $\angle AFC = \angle EFC = 180^\circ - \angle ELC = \angle ELA = \beta$ (where last equality holds because of $\angle AEL = 90^\circ$ and $\angle EAL = 90^\circ - \beta$). \hfill$\square$

\medskip
\textbf{Solution 2.}

As in the first solution we have that $\angle BAD = 90^\circ - \beta = \angle OAC$ and that $AX$ is the diameter of $\omega$. Also we note that $\angle ALK = \beta$, $\angle KLC = 180^\circ - \beta = \angle KBC$ so $BKCL$ is cyclic. Let $AO$ intersect circumcircle of $ABC$ again at $A'$. We will show that $A'$ is the desired concurrence point. Obviously $AA'$ is the diameter of circumcircle of triangle $ABC$ so $\angle A'CA = 90^\circ$ which implies that $A'CLE$ is cyclic. From power of point $E$ we have that $EK \cdot EL = EB \cdot EC = EA \cdot EA'$ so we can conclude that $A' \in \omega$. Now using the fact that $AX$ being the diameter of $\omega$ implies $\angle AXA' = 90^\circ$ we have that $DXA'E$ is cyclic because of $\angle EDX = 90^\circ$ which finishes the proof. \hfill$\square$",480,2040,Geometry,4
306,shl_jbmo_2021_g2,shl_jbmo,2021,g,"Let $P$ be an interior point of the isosceles triangle $ABC$ with $\hat{A} = 90^{\circ}$. If
$$\widehat{PAB} + \widehat{PBC} + \widehat{PCA} = 90^{\circ},$$prove that $AP \perp BC$.","\textbf{Solution.}

Let $D$ be the point on $BC$ with $AD \perp BC$. If $AP$ is not perpendicular to $BC$, without loss of generality, assume $P$ is inside the triangle $ABD$. Write $\angle PBC = x + y$ and $\angle PCB = y$. From the given angle equality, it is easy to see that $\angle PAD = x$. Let $F$ be the point on $AD$ with $|PA| = |PF|$. Firstly, we have $\angle PFD = x$. Then, when we consider the triangle $PBC$, we have
\[
|FB| = |FC| \quad \text{and} \quad \angle PFC - \angle PFB = 2x = 2 \cdot (\angle PBC - \angle PCB).
\]
This implies $F$ is the circumcenter of the triangle $PBC$, and hence we get $|FB| = |FP|$. On the other hand, let $K$ be the point on $AB$ such that $PK \perp AD$. Since $\angle AKF = 90^\circ$ and $KP \perp AF$, we get $|FB| > |FK| > |FP|$, which leads to a contradiction. As a result, we conclude that $AP \perp BC$. \hfill$\square$",181,874,Geometry,5
307,shl_jbmo_2021_g3,shl_jbmo,2021,g,"Let $ABC$ be an acute triangle with circumcircle $\omega$ and circumcenter $O$. The perpendicular from $A$ to $BC$ intersects $BC$ and $\omega$ at $D$ and $E$, respectively. Let $F$ be a point on the segment $AE$, such that $2 \cdot FD = AE$. Let $l$ be the perpendicular to $OF$ through $F$. Prove that $l$, the tangent to $\omega$ at $E$, and the line $BC$ are concurrent.","\textbf{Solution 1.} Let $\ell \cap BC = G$. We will prove that $GE$ is tangent to $\omega$. Let $H$ be the orthocenter of $ABC$. It is well-known that $HD = DE$. From $2FD = AE$ we get that $F$ is the midpoint of $AH$.

Let $M$ be the midpoint of $BC$. It is well known that $MH$ passes through $A'$ --- the antipode of $A$ in $\omega$. If $S$ is the second intersection of $MH$ and $\omega$, then $\angle HSA \equiv \angle A'SA = 90^\circ$, so $S$ lies on the circle with diameter $AH$, which is centered at $F$. Therefore, $FS = FH$ \ldots (1)

Since $\angle MSA \equiv \angle A'SA = 90^\circ$ we have $MH \perp AS$. But since $O$ and $F$ are centers of $ABC$ and $(ASH)$ and $AS$ is their common chord, we have $OF \perp AS$. Therefore,
\[
MH \parallel OF \quad \ldots\; (2)
\]
Since $FH$ and $OM$ are both perpendicular to $BC$, we get $FH \parallel OM$. Using (2), we get that $OFHM$ is a parallelogram. Therefore $FH = OM$. (This can alternatively be proven by the well-known fact that $AH = 2OM$). Using (1), we get that $FS = OM$. Using (2) again, we get that $OFSM$ is an isosceles trapezoid and therefore it's cyclic. Using that $OFGM$ is also cyclic ($\angle OFG = 90^\circ = \angle OMG$), we get that $OFSGM$ is cyclic and therefore $\angle OSG = \angle OFG = 90^\circ$. We have $HS \parallel OF$ and $OF \perp FG$, so $HS \perp FG$. Using (1), we get that $FG$ is the side bisector of $SH$, so $GS = GH$. Since $HD = DE$, we also get $GH = GE$. Therefore $GS = GH = GE$. Finally, we get that $\triangle OSG \cong \triangle OEG$ (by SSS), so $\angle OEG = \angle OSG = 90^\circ$, i.e.\ $GE$ is tangent to $\omega$. \hfill$\square$

\medskip
\textbf{Solution 2.} Let $H$ be the orthocenter of $\triangle ABC$ and $M$ the midpoint of $BC$ (thus $OM \perp BC$). Since $2FD = AE$, we get that $F$ is the midpoint of $AH$. It is known that $HD = DE$ and $AH = 2OM$. Since $FH = \tfrac{1}{2}AH = OM$ and $FH \parallel OM$, we obtain that $FHMO$ is a parallelogram, hence $FO = HM$. Since $H$ and $E$ are symmetric with respect to the line $BC$, we obtain that $EM = HM = FO$, which shows that $FOME$ is an isosceles trapezium, hence cyclic. Let $G$ be the intersection point of $BC$ with the tangent to $\omega$ at $E$. Then $\angle GEO = 90^\circ = \angle GMO$, hence $OG$ is the diameter of the circumcircle of $FOME$, and therefore we must have $\angle GFO = 90^\circ$. \hfill$\square$",374,2395,Geometry,6
308,shl_jbmo_2021_g4,shl_jbmo,2021,g,"Let $ABCD$ be a convex quadrilateral with $\angle B = \angle D = 90^{\circ}$. Let $E$ be the point of intersection of $BC$ with $AD$ and let $M$ be the midpoint of $AE$. On the extension of $CD$, beyond the point $D$, we pick a point $Z$ such that $MZ = \frac{AE}{2}$. Let $U$ and $V$ be the projections of $A$ and $E$ respectively on $BZ$. The circumcircle of the triangle $DUV$ meets again $AE$ at the point $L$. If $I$ is the point of intersection of $BZ$ with $AE$, prove that the lines $BL$ and $CI$ intersect on the line $AZ$.","\textbf{Solution.}

Since $MZ = \dfrac{AE}{2} = AM = ME$ then $\angle AZE = 90^\circ$ and $A, B, E, Z$ belong to a circle with center $M$.

Let $O$ be the projection of $M$ on $BZ$. Then $BO = OZ$. Since $AUEV$ is a trapezium, and $M$ is the midpoint of the diagonal $AE$, then $O$ is the midpoint of $UV$. Thus $UO = OV$. Since also $BU = ZV$, then $BO = OZ$.

Let $J$ be the point of intersection of $CD$ with the circumcircle of the triangle $DUV$. Since $\angle LDJ = 90^\circ$, then $L$ and $J$ are antidiametric points of the circle and so the points $L, O, J$ are collinear with $OL = OJ$. Since also $\angle ZOJ = \angle LOB$ then the triangles $ZOJ$ and $BOL$ are equal.

We deduce that $\angle OZJ = \angle LBO$ and therefore $ZD$ is parallel to $BL$. Since $ZD$ is perpendicular to $AE$, it follows that $BL$ is also perpendicular to $AE$.

Let $T$ be the point of intersection of $BL$ with $AZ$. Since the triangles $ABL$ and $CED$ are similar, then
\[
\frac{BL}{ED} = \frac{AL}{CD} \tag{1}
\]
Since the triangles $ALT$ and $ZDE$ are similar, then
\[
\frac{LT}{ED} = \frac{AL}{ZD} \tag{2}
\]
From (1) and (2) we deduce that
\[
\frac{BL}{LT} = \frac{ZD}{CD}
\]
from which it follows that $CI$ passes through $T$.

\medskip
\textbf{Alternative Approach:} We can also show that $BL$ is perpendicular to $AE$ as follows.

Since $\angle ZVE = 90^\circ = \angle ZDE$, then $D, E, Z, V$ are concyclic. Then $\angle IDV = \angle IZE$ and $\angle IVD = \angle IEZ$. So the triangles $ZIE$ and $IVD$ are similar.

Since $A, B, E, Z$ are concyclic, then the triangles $ZIE$ and $AIB$ are similar. Since $D, U, L, V$ are concyclic, then the triangles $IVD$ and $IUL$ are similar.

From the above, it follows that the triangles $AIB$ and $IUL$ are similar. Then $\angle ILU = \angle ABI$, and so $A, B, U, L$ are concyclic. Thus $\angle ALB = \angle AUB = 90^\circ$, i.e.\ $BL$ is perpendicular to $AE$. \hfill$\square$",532,1918,Geometry,7
309,shl_jbmo_2021_g5,shl_jbmo,2021,g,"Let $ABC$ be an acute scalene triangle with circumcircle $\omega$. Let $P$ and $Q$ be interior points of the sides $AB$ and $AC$, respectively, such that $PQ$ is parallel to $BC$. Let $L$ be a point on $\omega$ such that $AL$ is parallel to $BC$. The segments $BQ$ and $CP$ intersect at $S$. The line $LS$ intersects $\omega$ at $K$. Prove that $\angle BKP = \angle CKQ$.","\textbf{Solution 1.}

Denote the intersection of $SL$ with $PQ$ as $R$. We prove $BKRP$ and $CKRQ$ are cyclic (this is a direct consequence of Reim's theorem): $\angle ALR = \angle ALK = 180^\circ - \angle ABK = 180^\circ - \angle PBK$ but also $\angle ALR = \angle LRQ = \angle PRK$. Now obviously $\angle KRQ = 180^\circ - \angle KCQ$. Now notice that we need to prove $\angle BRP = \angle CRQ$ since $\angle BRP = \angle BKP$ and $\angle CKQ = \angle CRQ$. This is equivalent to proving $BR = CR$ ($\angle BRP = \angle RBC$ and $\angle CRQ = \angle RCB$). Notice that we would need $PR$ and $RM$ to be the interior and exterior angle bisectors of $\angle ARS$. This would mean $AR$ and $LR$ are symmetric w.r.t.\ bisector of $BC$, which is sufficient. From here we can proceed in multiple ways:

Let $U$ and $V$ be the intersections of circles $CKRQ$ and $BKRP$ with $CP$ and $BQ$ respectively. Since $S$ lies on the radical axis of these two circles, we must have $US \cdot SC = VS \cdot SB$ so quadrilateral $BUVC$ is cyclic and thus $PUVQ$ is cyclic ($\angle UVB = \angle UCB = \angle CPQ$). Since $PUVQ$ is cyclic we get $\angle QUC = \angle PVB = \angle PRB = \angle CRQ$, and we're done. \hfill$\square$

\medskip
\textbf{Solution 2.}

Let $M$ be the midpoint of $BC$ and $D$ the midpoint of $AL$. The intersection of $AS$ with $PQ$ is $T$. We have $A, S, M$ are collinear by Ceva or similar triangles (Thales). We need to prove $M, R, D$ are collinear. This is equivalent to proving $\dfrac{MT}{MA} = \dfrac{TR}{AD} = \dfrac{2TR}{AL}$ but $\dfrac{TR}{AL} = \dfrac{ST}{SA}$ so it suffices to prove $\dfrac{MT}{MA} = \dfrac{2ST}{SA}$ and here we are basically done since $A, T, S, M$ lie on one line:
\[
\frac{SA}{MA} = \frac{2ST}{MT} \iff 1 - \frac{SA}{MA} = 1 - \frac{2ST}{MT} \iff \frac{MS}{MA} = \frac{MS-ST}{MT} \iff \frac{MT}{MA} = \frac{MS-ST}{MS} = 1 - \frac{ST}{MS} \iff \frac{ST}{MS} = \frac{AT}{MA}
\]
which is true since $\dfrac{AQ}{AC} = \dfrac{PQ}{BC}$. \hfill$\square$

\medskip
\textbf{Solution 3.} Same notations as Solution 2: We need to prove that $M, R, D$ are collinear. By Menelaus it's enough to prove
\[
\frac{AD}{DL} \cdot \frac{LR}{RS} \cdot \frac{MS}{MA} = 1 \iff 1 \cdot \frac{AT}{TS} \cdot \frac{MS}{MA} = 1 \iff \frac{AT}{TS} = \frac{MA}{MS},
\]
which is true since $\dfrac{AQ}{AC} = \dfrac{PQ}{BC}$. \hfill$\square$

\medskip
\textbf{Solution 4.}

Let $R'$ be the point on $PQ$ such that $BR' = CR'$. We'll prove $S$, $R'$, $L$ are collinear. Let $M$ be the midpoint of $BC$. By Thales we know $A, S, M$ are collinear. Denote by $T$ the intersection of $AS$ with $PQ$. We prove that $R'T$ is the interior angle bisector of $\angle AR'S$ by

\medskip

oindent\textbf{Lemma.} For a given triangle $ABC$, let $P, Q \in BC$, such that $Q \in (BC)$ and $B \in (PQ)$. If $\dfrac{PB}{PC} = \dfrac{BQ}{QC}$ and $\angle PAQ = 90^\circ$, then $AP$, $AQ$ are the exterior and, respectively, interior angle bisectors of $\angle BAC$.

\medskip

oindent\textit{Proof.} Let $X$ and $Y$ be the intersections of the line through $Q$ perpendicular to $AQ$ with $AB$ and $AC$ respectively. By similar triangles ($\triangle BXQ \sim \triangle APB$ and $\triangle QYC \sim \triangle PAC$; we get these since $XY$ is parallel to $AP$) we have $QX = BQ \cdot \dfrac{AP}{PB}$ and $QY = QC \cdot \dfrac{AP}{PC}$. Now obviously $\dfrac{QX}{QY} = 1$, by the condition, hence by SAS congruence ($\triangle AXQ$ and $\triangle AYQ$) the conclusion follows.

Apply this lemma on triangle $AR'S$ and the segments $RT$ and $RM$. From here the collinearity is obvious since $\angle AR'P = \angle LR'Q = \angle PR'S$. (The condition $\dfrac{AT}{AM} = \dfrac{TS}{MS}$ can be checked to be true by simple similar triangles or Thales). \hfill$\square$",371,3754,Geometry,8
310,shl_jbmo_2021_n1,shl_jbmo,2021,n,"Find all positive integers $a, b, c$ such that $ab + 1$, $bc + 1$, and $ca + 1$ are all equal to
factorials of some positive integers.","\textbf{Solution.} Because of symmetry, we can assume that $a \geq b \geq c$. In particular, this means that $ab+1 \geq ac+1$. Now if $ac+1 = x!$ and $ab+1 = y!$, we have
\[
x! = ac+1 \leq ab+1 = y!
\]
So $x! \mid y!$ or $ac+1 \mid ab+1$. From here we obtain:
\[
ac+1 \mid (ab+1) - (ac+1) = a(b-c).
\]
Using that $ac+1$ and $a$ are relatively prime, this means that $ac+1 \mid b-c$. We conclude that either $b = c$ or $ac+1 \leq b-c$. However, we also see that $0 \leq b-c < b \leq a < ac+1$, which means that we must have $b = c$.

Let $bc+1 = z!$ for some positive integer $z$. Because $b = c$, this means that $b^2+1 = z!$, but this is only possible if $z = 2$ and $b = 1$. Indeed, if $z \geq 3$, then $3 \mid z!$, so $3 \mid b^2+1$. However, this would mean that $b^2 \equiv 2 \pmod{3}$, which is not possible. Checking $z = 1$ directly we see that this is not possible as well, so $b = c = 1$.

Finally, $a+1 = ac+1 = x!$, so $a = x!-1$ for some positive integer $x$. Because $a$ is also a positive integer, we must have $x \geq 2$. Taking symmetry into consideration we obtain that all solutions $(a, b, c)$ are of the form $(x!-1, 1, 1)$, $(1, x!-1, 1)$ or $(1, 1, x!-1)$ for some positive integer $x \geq 2$. We can easily see that all such triples indeed satisfy the condition of the problem. \hfill$\square$",134,1317,Number Theory,9
311,shl_jbmo_2021_n2,shl_jbmo,2021,n,"The real numbers $x, y$ and $z$ are such that $x^2 + y^2 + z^2 = 1$.
a) Determine the smallest and the largest possible values of $xy + yz - xz$.
b) Prove that there does not exist a triple $(x, y, z)$ of rational numbers, which attains any of the two values in a).","\textbf{Solution.}

\textbf{a)} We have $xy+yz-xz \geq -(x^2+y^2+z^2) = -1 \iff (x+y)^2+(y+z)^2+(x-z)^2 \geq 0$ and equality holds only for $\left(\pm\tfrac{1}{\sqrt{3}}, \mp\tfrac{1}{\sqrt{3}}, \pm\tfrac{1}{\sqrt{3}}
ight)$. On the other hand, $(x+z-y)^2 \geq 0 \iff xy+yz-xz \leq \tfrac{1}{2}$ and equality holds for example when $\left(0, \tfrac{1}{\sqrt{2}}, \tfrac{1}{\sqrt{2}}
ight)$.

\textbf{b)} The minimum case is ruled out since $\sqrt{3}$ is irrational. For the maximum case, it is enough to consider $x^2+y^2+z^2 = 1$ with $y = x+z$ --- that is, the equation $x^2+z^2+(x+z)^2 = 1$. Suppose the latter has a rational solution and write it as $x = \dfrac{p}{r}$, $z = \dfrac{q}{r}$, where $p, q, r 
eq 0$ are integers. Then $p^2+q^2+(p+q)^2 = r^2 \iff (2p+q)^2+3q^2 = 2r^2$. Now modulo 3 gives that $2p+q$ and $r$ are divisible by 3, whence $q$ (and thus $p$) is divisible by 3. Writing $p = 3p_1$, $q = 3q_1$, $r = 3r_1$, we reach $(2p_1+q_1)^2+3q_1^2 = 2r_1^2$, which is the same equation as the above one for $p, q$ and $r$. Finally, if the integer $s$ is such that $3^{s+1}$ does not divide $r$, then performing the above $s$ more times will yield an equation of the form $(2p'+q')^2+3q'^2 = 2r'^2$, with the right-hand side not divisible by 3, which is impossible. \hfill$\square$",265,1296,Number Theory,10
312,shl_jbmo_2021_n3,shl_jbmo,2021,n,"For any set $A = \{x_1, x_2, x_3, x_4, x_5\}$ of five distinct positive integers denote by $S_A$ the sum of its elements, and denote by $T_A$ the number of triples $(i, j, k)$ with $1 \le i < j < k \le 5$ for which $x_i + x_j + x_k$ divides $S_A$.
Find the largest possible value of $T_A$.","\textbf{Solution.} We will prove that maximum value $T_A$ can attain is $4$. Let $A = \{x_1, x_2, x_3, x_4, x_5\}$ be set of five distinct positive integers such that $x_1 < x_2 < x_3 < x_4 < x_5$. Call triple $(i,j,k)$ with $1 \leq i < j < k \leq 5$ \emph{good} if $x_i+x_j+x_k$ divides $S_A$. Obviously by size argument triplets $(3,4,5)$, $(2,4,5)$, $(1,4,5)$, $(2,3,5)$, $(1,3,5)$ aren't good because for example
\[
x_5+x_3+x_1 \mid S_A \iff x_5+x_3+x_1 \mid x_2+x_4
\]
which is impossible since $x_5 > x_4$ and $x_3 > x_2$. Analogously we can show that any triple of form $(x,y,5)$ where $y > 2$ isn't good.

Because of that number of good triplets is at most 5 and only triplets $(1,2,5)$, $(2,3,4)$, $(1,3,4)$, $(1,2,4)$, $(1,2,3)$ can be good. But if triplets $(1,2,5)$ and $(2,3,4)$ are simultaneously good we have that:
\[
x_1+x_2+x_5 \mid x_3+x_4 \implies x_5 < x_3+x_4 \tag{1}
\]
and
\[
x_2+x_3+x_4 \mid x_1+x_5 \implies x_2+x_3+x_4 \leq x_1+x_5 \overset{(1)}{<} x_1+x_3+x_4 < x_2+x_3+x_4,
\]
which is impossible. Therefore, $T_A \leq 4$.

To show that $T_A = 4$ is possible consider numbers $1, 2, 3, 4, 494$. This works because $6 \mid 498$, $7 \mid 497$, $8 \mid 496$, and $9 \mid 495$. \hfill$\square$",289,1217,Number Theory,11
313,shl_jbmo_2021_n4,shl_jbmo,2021,n,"Dragos, the early ruler of Moldavia, and Maria the Oracle play the following game. Firstly, Maria chooses a set $S$ of prime numbers. Then Dragos gives an infinite sequence $x_1, x_2, ...$ of distinct positive integers. Then Maria picks a positive integer $M$ and a prime number $p$ from her set $S$. Finally, Dragos picks a positive integer $N$ and the
game ends. Dragos wins if and only if for all integers $n \ge N$ the number $x_n$ is divisible by $p^M$; otherwise, Maria wins. Who has a winning strategy if the set S must be: $\hspace{5px}$a) finite; $\hspace{5px}$b) infinite?","\textbf{Solution.} We show that in both cases Drago\c{s} can win.

Suppose firstly that Maria chooses the finite set $S = \{p_1, p_2, \ldots, p_k\}$, where $p_1 < p_2 < \cdots < p_k$. Then Drago\c{s} can use the sequence $x_n = (p_1 p_2 \cdots p_k)^n$ (which is increasing and hence consists of distinct terms). Now, no matter what $M$ and $p$ Maria picks, Drago\c{s} can give $N = M$ in order to win --- indeed, $x_n$ is divisible by $p^n$ for each $p$ in $S$ and hence by $p^M$ for all $n \geq N$.

Now consider the case when Maria chooses the infinite set $S = \{p_1, p_2, \ldots\}$, where $p_1 < p_2 < \cdots$. Then Drago\c{s} can use the sequence $x_n = (p_1 p_2 \cdots p_n)^n$ (which is increasing and hence consists of distinct terms). Now, no matter what $M$ and $p_k$ Maria picks, Drago\c{s} can give $N = \max(M, k)$ in order to win --- indeed, $x_n$ is divisible by $p_k^n$ for each $p_k$ in $S$ whenever $n \geq k$ and hence by $p_k^M$ for all $n \geq \max(M, k)$. \hfill$\square$",582,992,Number Theory,12
314,shl_jbmo_2021_n5,shl_jbmo,2021,n,"Find all pairs of integers $(x, y)$ such that $x^2 + 5y^2 = 2021y$.","\textbf{Solution.} The cases $x = 0$ and $y = 0$ are immediate; we can without loss of generality treat $x > 0$ and hence $y > 0$. Clearly, $x^2 = y(2021-5y)$ and the greatest common divisor of the multipliers on the right divides $2021 = 43 \cdot 47$. Since $2021-5y > 0$, i.e.\ $y \leq 404$, we have the following possibilities: $43 \mid y$, $47 \mid y$ and $\gcd(y, 2021-5y) = 1$.

If $y = 43z$, then $x = 43t$ and we get $t^2 = z(47-5z)$, $z \leq 9$ and a direct verification shows there is no solution. If $y = 47z$, then $x = 47t$ and we get $t^2 = z(43-5z)$, $z \leq 8$ and a direct verification shows there is no solution. In the last case we necessarily have $y = m^2$ and $2021-5y = n^2$ for positive integers $m$ and $n$, i.e.\ $n^2+5m^2 = 2021$. Now we get $5m^2 < 2021$, i.e.\ $m \leq 20$; moreover $m$ is not divisible by $3$ (else $n^2 \equiv 2 \pmod{3}$) and does not give remainder $0$, $2$ or $5$ when divided by $7$ (else $n^2 \equiv 5, 6 \pmod{7}$). The remaining ones are $m = 1, 4, 8, 10, 11, 13, 17, 20$ and for $2021-5m^2$ we calculate $44^2 < 1941 < 2016 < 45^2$ (so $m = 1$ and $m = 4$ do not work), $41^2 < 1701 < 42^2$ (so $m = 8$ does not work), $1521 = 39^2$ (respectively $y = 100$), $37^2 < 1376 < 1416 < 38^2$ (so $m = 11$ and $m = 13$ do not work), $576 = 24^2$ (respectively $y = 289$) and $4^2 < 21 < 5^2$ (so $m = 20$ does not work). \hfill$\square$",67,1386,Number Theory,13
315,shl_jbmo_2021_n6,shl_jbmo,2021,n,"Given a positive integer $n \ge 2$, we define $f(n)$ to be the sum of all remainders obtained by dividing $n$ by all positive integers less than $n$. For example dividing $5$ with $1, 2, 3$ and $4$ we have remainders equal to $0, 1, 2$ and $1$ respectively. Therefore $f(5) = 0 + 1 + 2 + 1 = 4$. Find all positive integers $n \ge 3$ such that $f(n) = f(n - 1) + (n - 2)$.","\textbf{Solution.} Given any $d < n$ we write $a_d$ for the remainder when $n-1$ is divided by $d$ and $b_d$ for the remainder when $n$ is divided by $d$.

If $d \mid n$ then we have $a_d = d-1$ and $b_d = 0$. If $d 
mid n$ then we have $b_d - a_d = 1$. Thus
\begin{align*}
f(n) - f(n-1) &= \sum_{d 
mid n}(b_d - a_d) + \sum_{d \mid n}(b_d - a_d) \\
&= \sum_{d 
mid n} 1 - \sum_{d \mid n}(d-1) \\
&= \sum_d 1 - \sum_{d \mid n} d \\
&= (n-1) - [\sigma(n) - n] \\
&= 2n-1 - \sigma(n)
\end{align*}
Here, all sums are over all integers $d \in \{1, 2, \ldots, n-1\}$ satisfying the claimed properties, $\sigma(n)$ is the sum of all positive divisors of $n$ (including $n$) and $d(n)$ is the total number of positive divisors of $n$ (including $n$).

So $f(n) = f(n-1)+(n-2)$ if and only if $\sigma(n) = n+1$. But since $1, n$ are divisors of $n$, then $\sigma(n) \geq n+1$ with equality if and only if $n$ is a prime number. \hfill$\square$",371,935,Number Theory,14
316,shl_jbmo_2021_n7,shl_jbmo,2021,n,"Alice chooses a prime number $p > 2$ and then Bob chooses a positive integer $n_0$. Alice, in the first move, chooses an integer $n_1 > n_0$ and calculates the expression $s_1 = n_0^{n_1} + n_1^{n_0}$; then Bob, in the second move, chooses an integer $n_2 > n_1$ and calculates the expression $s_2 = n_1^{n_2} + n_2^{n_1}$; etc. one by one. (Each player knows the numbers chosen by the other in the previous moves.) The winner is the one who first chooses the number $n_k$ such that $p$ divides $s_k(s_1 + 2s_2 + · · · + ks_k)$. Who has a winning strategy?","\textbf{Solution.} We will prove that for any prime $p > 2$, Bob can win by choosing $n_0 = (p-1)^2$. Then
\[
s_1 = n_0^{n_1} + n_1^{n_0} \equiv \begin{cases} 1 & \text{if } p \mid n_1 \\ 2 & \text{if } p 
mid n_1 \end{cases} 
ot\equiv 0 \pmod{p} \quad \text{for any } n_1 > n_0.
\]
We will use the following two properties:
\begin{enumerate}
    \item If $p \mid n_k$ then it is obvious that for $n_{k+1} = pn_k > n_k$ we have $p \mid s_{k+1}$.
    \item If $p 
mid n_k$ and $n_k$ is odd then for $n_{k+1} = (p-1)(pn_k+1) > n_k$ we have $n_{k+1} \equiv -1 \pmod{p}$, and $s_{k+1} = n_k^{n_{k+1}} + n_{k+1}^{n_k} \equiv 1 + (-1)^{n_k} \equiv 0 \pmod{p}$.
\end{enumerate}
From 1) and 2) it follows that: If one of the players chooses $n_k$ to be odd or even number divisible by $p$ but does not win in his current move then the other can win in the next move. If the player cannot choose $n_k$ for which he wins, then it is clear that the only possibility not to lose is to choose $n_k$ to be an even and $p 
mid n_k$.

Let $n_{2j} = (p-1)^2 m_{2j}$ for $j = 1, \ldots, \dfrac{p-1}{2}$, where $m_{2j}$ is chosen so that $n_{2j} > n_{2j-1}$. Then
\[
s_j = n_{j-1}^{n_j} + n_j^{n_{j-1}} \equiv \begin{cases} 1 & \text{if } p 
mid n_{j-1}n_j \\ 2 & \text{if } p \mid n_{j-1}n_j \end{cases} 
ot\equiv 0 \pmod{p}
\]
and for $k < p-1$ we have
\[
s_1 + 2s_2 + \cdots + ks_k \equiv \begin{cases} k^2 & \text{if } p \mid n_k \\ k(k+1) & \text{if } p 
mid n_k \end{cases} 
ot\equiv 0 \pmod{p}
\]
and $s_1 + 2s_2 + \cdots + (p-1)s_{p-1} \equiv 2(1 + 2 + \cdots + (p-1)) \equiv 0 \pmod{p}$.

As $p-1$ is even, we get that the winning move belongs to Bob. \hfill$\square$",556,1657,Number Theory,15
317,shl_jbmo_2021_c1,shl_jbmo,2021,c,"In Mathcity, there are infinitely many buses and infinitely many stations. The stations are indexed by the powers of $2: 1, 2, 4, 8, 16, ...$ Each bus goes by finitely many stations, and the bus number is the sum of all the stations it goes by. For simplifications, the mayor of Mathcity wishes that the bus numbers form an arithmetic progression with common difference $r$ and whose first term is the favourite number of the mayor. For which positive integers $r$ is it always possible that, no matter the favourite number of the mayor, given any $m$ stations, there is a bus going by all of them?","\textbf{Solution.} If $r$ is even and the favourite number of the mayor is $2$, no bus will ever go by the station number $1$. Thus, even numbers $r$ do not satisfy the problem requirement.

If $r$ is odd, consider $m$ bus stations with numbers $2^{a_1}, \ldots, 2^{a_m}$, such that $a_1 < a_2 < \cdots < a_m$, and let $f$ be the favourite number of the mayor. Since $r$ is odd, it is coprime with $2^{a_m+1}$, and thus there exists a positive integer $q$ such that $rq \equiv 1 \pmod{2^{a_m+1}}$. Then, let $s$ be a positive integer such that $s \equiv -1-f \pmod{2^{a_m+1}}$. According to the problem statement, there is a bus with number $f + r(qs)$.

Since $f + r(qs) \equiv f + s \equiv -1 \equiv 2^0 + 2^1 + \cdots + 2^{a_m} \pmod{2^{a_m+1}}$, this bus will go by each of the stations $2^0, 2^1, \ldots, 2^{a_m}$. In particular, it goes by each of our initial $m$ stations, and thus $r$ satisfies the problem statement.

In conclusion, the solutions of the problem are the odd integers $r$.",598,996,Combinatorics,16
318,shl_jbmo_2021_c2,shl_jbmo,2021,c,"Let $n$ be a positive integer. We are given a $3n \times 3n$ board whose unit squares are colored in black and white in such way that starting with the top left square, every third diagonal is colored in black and the rest of the board is in white. In one move, one can take a $2 \times 2$ square and change the color of all its squares in such way that white squares become orange, orange ones become black and black ones become white. Find all $n$ for which, using a finite number of moves, we can make all the squares which were initially black white, and all squares which were initially white black.","\textbf{Solution 1.} (by proposers) Firstly, observe that if we change the color of one square $3$ times, it goes back to its original color. As the final configuration does not depend on the order of moves and placing a move on a square $3$ times does not change the configuration, we can suppose that a move is placed on any square exactly $0$, $1$ or $2$ times. This further implies that a white square must change its colour $2$ (modulo $3$) times and that a black square must be changed $1$ (modulo $3$) times.

Suppose we can reverse the colours of the squares, as required in the problem statement, for some $3n \times 3n$ board.

\medskip
\textbf{Claim:} $n$ cannot be odd.

\textit{Proof.} Let us associate each $2 \times 2$ square with its top left unit square. Let's take a look at the first column. Its top square is white and is included in only one $2 \times 2$ square, so that square has to be placed $2$ times. The second square in this column is also white and is included in its $2 \times 2$ square as well as in the one associated with the first unit square in the column. This square has already been turned black by taking the $2 \times 2$ square of the first unit square in this column twice, so its own $2 \times 2$ square has to be taken $0$ times. We use similar arguments to show that next four $2 \times 2$ squares have to be called $1, 1, 1$ and $0$ times. After that we have the white square included only in its own $2 \times 2$ square and in the one above it, which has been taken $0$ times, so we have the same situation we had in the beginning. That means that in the first column numbers of times successive $2 \times 2$ squares have to be taken make a cycle $(2, 0, 1, 1, 1, 0)$. Now take a look at the last square in this column. It is obviously black. Also, its $2 \times 2$ square is not included in the board, so this unit square is included only in the $2 \times 2$ square of a unit square above it, which has been taken $0$ times if $6 \mid 3n+3$ and $1$ time if $6 \mid 3n$. We know that this square must change its color once and that is only possible if $6 \mid 3n$, which means that $2 \mid n$.

For $n = 2k$, $k > 1$, we can partition the board into several disjoint $6 \times 6$ boards, which we can solve separately, which will solve the entire board (it is trivial that those $6 \times 6$ boards are colored the same way the board in case $n = 2$ is colored). \hfill$\square$

\medskip
\textbf{Solution 2.} (by Milica Vugdeli\'{c}, Serbia) As in the first solution, we can assume that a move is placed on every square $0$, $1$ or $2$ times. Also, use the same fact that white squares must change colour $2$ (modulo $3$) times and that black squares must change colour $1$ (modulo $3$) times. Use the same construction for even $n$. Suppose that we can obtain the required board for some $n$.

\medskip
\textbf{Claim 1:} The number of moves is divisible by $3$.

\textit{Proof.} Fix a sequence of moves and assign to each unit square the number of times it has changed colour in the sequence. Then the sum of all assigned values is equal to $2W + B$ modulo $3$, where $B$ and $W$ are the number of black and white squares on the initial board, respectively. It is easy to see that $3 \mid W$ and $3 \mid B$ so $3 \mid 2W+B$. On the other hand, the sum of all assigned values is just $4$ times the number of moves (every change of colours of a $2 \times 2$ square adds $4$ to our sum). So the number of moves is divisible by three. \hfill$\square$

\medskip
\textbf{Claim 2:} $n$ cannot be odd.

\textit{Proof.} Consider the set of every other square in every other row of the board (if we set the coordinate system with the top left square as $(1,1)$ and the bottom right as $(3n, 3n)$, we see that these squares are just the ones with both coordinates even). Observe that every move changes the colour of exactly one of these squares, so the sum of assigned values of these squares must be equal to the number of moves. If $2 
mid n$ ($n = 2k+1$, $k > 0$) then we have $\dfrac{3n^2}{2} = (3k+1)^2$ squares with both coordinates even, of which $b = \dfrac{2}{3}(3k)^2 + 3 = 6k+3$ are initially black and the rest are white. But the sum of assigned values of these squares will be $2((3k+1)^2 - b) + b$ modulo $3$ and since $3 \mid b$ we see that this is not divisible by three. But then the number of moves is not divisible by three, a contradiction. \hfill$\square$

\medskip
\textbf{Solution 3.} (by Mateja Vukeli\'{c}, Serbia) As in the first two solutions, we can assume that a move is placed on every square $0$, $1$ or $2$ times. Also, use the same fact that white squares must change color $2$ (modulo $3$) times and that black squares must change color $1$ (modulo $3$) times. Use the same construction for even $n$. Let power of each square be $0$ if it is white, $1$ if it is orange and $2$ if it is black. Now color the board as a chessboard in blue and red, and let the top left square be red. Define the $\mathit{PowerOfRed}$ as the sum of powers of all red squares. We define the $\mathit{PowerOfBlue}$ the same way. Let $S = \mathit{PowerOfRed} - \mathit{PowerOfBlue}$.

\medskip
\textbf{Claim 1:} After each move, $S$ stays the same modulo $3$.

\textit{Proof.} It is obvious that each move affects exactly $2$ red squares and $2$ blue ones. Notice that, when looking at powers of squares modulo $3$, in each move we add $1$ to the power of each square. That means that each move increases both the $\mathit{PowerOfBlue}$ and $\mathit{PowerOfRed}$ by $2$ each, so $S$ increases by $2-2 = 0$, which means it stays the same. \hfill$\square$

\medskip
\textbf{Claim 2:} $n$ cannot be odd.

\textit{Proof.} Let $n = 2k+1$. The number of red squares is $R = (3k+1)(6k+3)+3k+2$ and the number of blue squares is $B = (3k+1)(6k+3)+3k+1$. Notice that, at the beginning if some blue square is black, its entire diagonal is black and it contains $3m$ squares, so the number of all black blue squares is divisible by $3$, which means that the $\mathit{PowerOfBlue}$ is divisible by $3$. We use the same argument to show that, at the beginning, the $\mathit{PowerOfRed}$ is divisible by $3$ as well. This means that $S$ is divisible by $3$. Now let's take a look at our desired endboard. Now the number of black blue squares is the same as the number of white blue squares on the starting board, which is $B$ minus some number divisible by $3$. This equals $3k+1$ modulo $3$, so now $\mathit{PowerOfBlue} \equiv 2(3k+1) \pmod{3}$. We use the same argument to show that now $\mathit{PowerOfRed} \equiv 2(3k+2) \pmod{3}$, so $S \equiv 2(3k+2-3k-1) \equiv 2 \pmod{3}$. However, we already proved that after each move $S$ stays the same modulo $3$, so there is no way to apply some sequence of moves which will turn our starting board into desired endboard. \hfill$\square$",604,6821,Combinatorics,17
319,shl_jbmo_2021_c3,shl_jbmo,2021,c,"We have a set of $343$ closed jars, each containing blue, yellow and red marbles with the number of marbles from each color being at least $1$ and at most $7$. No two jars have exactly the same contents. Initially all jars are with the caps up. To flip a jar will mean to change its position from cap-up to cap-down or vice versa. It is allowed to choose a
triple of positive integers $(b; y; r) \in \{1; 2; ...; 7\}^3$ and flip all the jars whose number of blue, yellow and red marbles differ by not more than $1$ from $b, y, r$, respectively. After $n$ moves all the jars turned out to be with the caps down. Find the number of all possible values of $n$, if $n \le 2021$.","\textbf{Solution.} Call a jar \emph{important} if each of the quantities of blue, yellow and red marbles in it is $1$, $4$ or $7$. There are $3 \cdot 3 \cdot 3 = 27$ important jars. It is easy to check that any move flips exactly one important jar. Thus, after an even number of moves, the number of flipped jars is even. Therefore, the required outcome can appear only after an odd number of moves which is no less than $27$. In order to achieve any such number, make a move for each of the $27$ important jars, followed by any number of parasite pairs of moves with (say) $(b, y, r) = (1, 1, 1)$. Thus $n$ can be any odd number which is at least $27$.",674,653,Combinatorics,18
320,shl_jbmo_2021_c4,shl_jbmo,2021,c,"Alice and Bob play a game together as a team on a $100 \times 100$ board with all unit squares initially white. Alice sets up the game by coloring exactly $k$ of the unit squares red at the beginning. After that, a legal move for Bob is to choose a row or column with at least $10$ red squares and color all of the remaining squares in it red. What is the
smallest $k$ such that Alice can set up a game in such a way that Bob can color the entire board red after finitely many moves?","\textbf{Solution.} We will show by induction on $m+n$ the following:

Fix integers $m, n \geq 1$ and $0 \leq a < n$, $0 \leq b < m$. Assume that the game is being played on a board with $m$ rows and $n$ columns, but assume that a row is a legal move for Bob if it has at least $a$ red squares and a column is a legal move for Bob if it has at least $b$ red squares. Then if it is possible to win the game, there must be at least $ab$ squares initially red (colored by Alice at the beginning of the game).

The base case $m = n = 1$ and $a = b = 0$ is trivial.

Assume the contrary. This means that there are strictly less than $ab$ red squares initially. If $a = 0$ or $b = 0$ this is trivially not possible. Otherwise, provided the game can be won by the children, there is either a row with at least $a$ red squares or a column with at least $b$ red squares (if this is not the case then there are no legal moves). Assume without loss of generality that Bob's first turn is a row. Now make a new board by cutting out this row (which is now completely red).

We obtain a new board with $a_1 = a$ and $b_1 = b-1$, because all columns from the first board now have at least one red square (from the row we cut out). The total number of red squares on the new board is less than $ab - a = a(b-1) = a_1 b_1$, while the board is $(m-1) \times n$. We also have $0 \leq a_1 < n$ and $0 \leq b_1 < m-1$. But this board cannot be won by induction (we have $(m-1)+n < m+n$), so the original board cannot be won as well. We conclude that we must have at least $ab$ squares which are initially red.

Now taking $n = m = 100$ and $a = b = 10$ we see that we need at least $10 \times 10$ red squares, or in other words, $k \geq 100$. To show that this is indeed possible, notice that Alice can color the upper left corner $10 \times 10$ subsquare red. Now Bob first makes moves on the top $10$ rows, then on all the columns, in this order. This makes the entire board red, so $k = 100$. \hfill$\square$",483,1989,Combinatorics,19
321,shl_jbmo_2021_c5,shl_jbmo,2021,c,"Let $M$ be a subset of the set of $2021$ integers $\{1, 2, 3, ..., 2021\}$ such that for any three elements (not necessarily distinct) $a, b, c$ of $M$ we have $|a + b - c | > 10$.
Determine the largest possible number of elements of $M$.","\textbf{Solution.} The set $A = \{1016, 1017, \ldots, 2021\}$ has the necessary property as $a, b, c \in A$ implies that $a+b-c > 1016+1016-2021 = 11$. Notice that this set has $1006$ elements. We will show that this is optimal.

Let $k$ be the minimal element of $A$. Then $k = |k+k-k| > 10$. For every $m$, at least one of $m$, $m+k-10$ does not belong to $A$, since $k + m - (m+k-10) = 10$.

\medskip
\textbf{Claim 1:} $A$ contains at most $k-11$ out of any $2k-22$ consecutive integers.

\textit{Proof:} We can partition the set $\{m+1, m+2, \ldots, m+2k-22\}$ into $k-11$ pairs as follows:
\[
\{m+1, m+k-10\},\; \{m+2, m+k-9\},\; \ldots,\; \{m+k-11, m+k-22\}.
\]
It remains to note that $A$ can contain at most one element of each pair. \hfill$\square$

\medskip
\textbf{Claim 2:} $A$ contains at most $\dfrac{t+k-11}{2}$ out of any $t$ consecutive integers.

\textit{Proof:} Write $t = q(2k-22)+r$ with $r \in \{0, 1, 2, \ldots, 2k-21\}$. From the set of the first $q(2k-22)$ integers, by Claim 1 at most $q(k-11) = \dfrac{t-r}{2}$ can belong to $A$. From the last $r$ integers, at most $\min\{r, k-11\}$ can belong to $A$. Therefore, at most
\[
\frac{t-r}{2} + \min\{r, k-11\} \leq \frac{t-r}{2} + \frac{r+k-11}{2} = \frac{t+k-11}{2}
\]
of the $t$ consecutive integers can belong to $A$, as claimed. \hfill$\square$

By Claim 2, amongst $k+1, k+2, \ldots, 2021$ at most
\[
\frac{(2021-k)+(k-11)}{2} = 1005
\]
integers belong to $A$.

Since amongst $\{1, 2, \ldots, k\}$ only $k$ belongs to $A$, then $A$ has at most $1006$ elements as required. \hfill$\square$",238,1567,Combinatorics,20
322,shl_jbmo_2021_c6,shl_jbmo,2021,c,"Given an $m \times n$ table consisting of $mn$ unit cells. Alice and Bob play the following game: Alice goes first and the one who moves colors one of the empty cells with one of the given three colors. Alice wins if there is a figure, such as the ones below, having three different colors. Otherwise Bob is the winner. Determine the winner for all cases of $m$
and $n$ where $m, n \ge 3$.","\textbf{Solution.} For the sake of simplicity, we will label colors as $a$, $b$, $c$. Assume that $m \geq 5, n \geq 4$ or $m \geq 4, n \geq 5$. Without loss of generality, consider second case (for other case, we may rotate table). Since $m \geq 4, n \geq 5$, we can take the following subfigure from the table:

\medskip
Assume Alice colors $3$ as $a$. After Bob makes move, at least one of triples $\{1,2,4\}$ and $\{5,6,7\}$ will remain uncolored. Suppose it is $\{1,2,4\}$. Then, Alice colors $2$ as $b$. After next move of Bob, again at least one of $1$ and $4$ is still empty. Alice will choose uncolored one and color it with $c$. Thus, Alice wins.

Now, consider the case $m = n = 4$: Let's split the squares of the $4 \times 4$ table in pairs as follows:

\medskip
Whenever Alice chooses a cell, Bob will color its pair using same color as Alice. Thus, Bob wins (since for any diagonal of $3$ squares, $2$ of the squares will be paired).

Now, suppose $m = 3$, $n = 6$ (the symmetric case $m = 6, n = 3$ is treated similarly). Similar to above, we split the cells of the table into pairs. Whenever Alice chooses a cell, Bob will color its pair using same color as Alice. Thus, Bob wins.

Likewise, for $m = 3$, $n = 4$ (and $n = 3$, $m = 4$), a similar pairing shows that Bob has winning strategy.

Consider now the case $m = n = 3$. Look at the following pairs: $(4,2)$, $(8,6)$. Whenever Alice chooses a cell numbered $2, 4, 6,$ or $8$, Bob will color its pair with same color as Alice. If Alice colors $1$ ($3$, $7$, and $9$ cases are just rotation) with color $a$, Bob will color $5$ with the color $a$. Then Bob will color $3$ or $7$ (which one is empty) with color $a$ after Alice's move. If Alice colors $5$ with color $a$, then Bob will color at least one cell from pairs $(1,9)$ and $(3,7)$ with color $a$ in his next $2$ moves. So Bob wins.

Suppose $m = 3$, $n \geq 7$ (the case $m \geq 7, n = 3$ is similar). Consider the $3 \times 7$ subtable from the top left. Let Alice color $1$ as $a$.

\begin{enumerate}
    \item Assume Bob doesn't select $9, 17, 15, 3, 11, 5, 19$ in his next move. Then, Alice colors $9$ as $b$. Bob must color $17$ as one of $a, b$ (suppose it is $a$). Then, Alice chooses $11$ and colors it as $b$. Again, Bob must color $5$. Finally, Alice will choose $3$ and color it as $a$. Now, Bob needs to make move on both $15$ and $19$ simultaneously. Alice wins.
    \item Assume Bob's next move is either $11$ or $5$. At least one color is unused. Then, Alice colors $17$ with any color among remaining ones. Now, Bob should make move on $9$ and uncolored cell between $5$ and $11$. Thus, Alice wins.
    \item Assume Bob's next move is $3$. At least one color is unused. Then, Alice colors $9$ with any color among remaining ones. Now, Bob should make move on $15$ and $17$. Thus, Alice wins.
    \item Assume Bob's next move is $15$. At least one color is unused. Then, Alice colors $9$ with any color among remaining ones. Now, Bob should make move on $3$ and $17$. Thus, Alice wins.
    \item Assume Bob's next move is $19$. Alice colors $9$ as $b$. Then, Bob must color $17$ as one of $a, b$ (suppose it is $a$). Then, Alice colors $11$ as $b$. Again, Bob must choose $5$ and color it as one of $a, b$. Now, there is at least one unused color among $5$ and $19$. Alice colors $13$ with any color among remaining ones. Bob should make move on $7$ and $21$. Alice wins.
    \item Assume Bob's next move is $17$. Obviously, Bob should color it as $a$. Otherwise, Alice may choose remaining color for $9$ and win. So, suppose Alice colors $11$ as $b$. Then, Bob must color $5$ as one of $a, b$ (suppose it is $a$). Then, Alice colors $13$ as $b$. Bob's next move is $21$. Alice colors $19$ as any of unused color among $11$ and $13$ (this clearly exists). Now, Bob should make move on $3$ and $7$. Alice wins.
    \item Assume Bob's next move is $9$. Obviously he should color it as $a$. Alice colors $15$ as $b$. Then, Bob must color $3$ as one of $a, b$ (suppose it is $a$). Alice colors $11$ as $b$. Again, Bob must color $19$ as one of $a, b$. Alice colors $13$ as different color from $19$ so that Bob makes his next move on $7$. Since there is at least one unused color among $11$ and $13$, Alice may choose any of remaining ones, and force Bob to color $17$ and $21$ simultaneously. Alice wins.
\end{enumerate}

Finally, consider the case $m = 3$, $n = 5$ (the case $n = 3$, $m = 5$ is similar). Alice will color one of the even-numbered cells in the first move. Since there are $7$ (i.e.\ an odd number of) even-numbered cells in total, Alice can force Bob to start coloring odd-numbered cells.

\begin{enumerate}
    \item If Bob colors cell $1$ with $a$, Alice will color cell $7$ with $b$. Then, Bob will have to color cell $13$ with $a$ or $b$. Next, Alice will color cell $9$ with $c$. So, Bob will need to color cell $5$, and Alice will color $3$ with $a$. Bob's subsequent move will guarantee to have at least $11$ or $15$ to be uncolored. If $11$ is remained uncolored, then, Alice will color it with $c$, and in the end, $3, 7, 11$ will have different colors. Alice wins. Otherwise, if $15$ remained uncolored, Alice will color it with $b$, and $3, 9, 15$ will have different colors. Alice wins.
    \item If Bob colors $5$, $11$ or $15$, the case is similar (by symmetry) to 1). Alice wins.
    \item If Bob colors $7$ with $a$, Alice will color $1$ with $b$, and then the process will continue as in case 1). Alice wins.
    \item If Bob colors $9$, the case is similar (by symmetry) to 3). Alice wins.
    \item Assume Bob colors cell $3$ with $a$. Alice will then color $11$ with $b$. Since Bob will color $7$ either $a$ or $b$, Alice will color $1$ with $c$. Therefore, Bob will choose one of the colors of $1$ or $7$ to paint $13$. As there will be at most two different colors used for $3$ and $13$, Alice paints $9$ with one of the remaining colors. Since Bob will paint one of the cells $5$ or $15$ in the next move, Alice will paint with different color and win.
    \item If Bob colors $13$, the case is similar (by symmetry) to 5). Alice wins.
\end{enumerate}",389,6131,Combinatorics,21
411,tst_jbmo_ro_2021_1_p1,tst_jbmo,2021,a,"Let $n \in \mathbb{N}$, $n \geq 2$ and real numbers $a_1, a_2, \ldots, a_n \in [0,1]$. Find the maximum value of the minimum of the numbers:
\[
a_1 - a_1 a_2,\; a_2 - a_2 a_3,\; \ldots,\; a_n - a_n a_1.
\]","\subsection*{Solution}

We will show that the required maximum is $\dfrac{1}{4}$.

Agreeing to denote $a_{n+1} = a_1$, we choose an index $k \in \{1, 2, \ldots, n\}$ such that $a_k \leq a_{k+1}$. Such an index exists because we cannot have $a_1 > a_2 > \cdots > a_n > a_1$. (Alternatively, we may take $a_k = \min\{a_1, a_2, \ldots, a_n\}$.) Then
\[
a_k - a_k a_{k+1} \leq a_k - a_k^2 \leq \frac{1}{4},
\]
the last inequality being equivalent to $\left(a_k - \dfrac{1}{2}
ight)^2 \geq 0$. Therefore the required minimum is always at most $\dfrac{1}{4}$, so the required maximum is at most $\dfrac{1}{4}$.

Choosing $a_1 = a_2 = \cdots = a_n = \dfrac{1}{2}$, all numbers of the form $a_k - a_k a_{k+1}$ are equal to $\dfrac{1}{4}$, so the minimum is in this case $\dfrac{1}{4}$. Therefore the maximum value of the minimum of the numbers is $\dfrac{1}{4}$.",205,854,Algebra,1
412,tst_jbmo_ro_2021_1_p2,tst_jbmo,2021,n,"Find all nonzero natural numbers $x, y$ with the property that $x \leq y$ and
\[
\frac{(x+y)(xy-1)}{xy+1} = p,
\]
where $p$ is a prime number.","\subsection*{Solution}

Clearly, $x = y = 1$ does not work, so we may assume $xy - 1 > 0$. Let $d = \gcd(xy-1, xy+1)$. Then $d = 1$ or $d = 2$.

\textbf{Case $d = 1$:} From $xy + 1 \mid (x+y)(xy-1)$ it follows that $xy + 1 \mid x+y$. Since $x+y > 0$, we get $xy + 1 \leq x + y$, that is $(x-1)(y-1) \leq 0$. We deduce that $x = 1$.

Returning to the relation in the statement we obtain $y = p+1$. We get the solution $(1, p+1)$.

(The solution $x = 1$, $y = 3$ does not satisfy $d = 1$, so it should be found in the next case.)

\textbf{Case $d = 2$:} With $xy + 1 = 2k$, $k \in \mathbb{N}^*$, we have $k \mid x+y$, so $k \leq x+y$. Since $2k = xy+1 \geq x+y$, we can only have $x+y = k$ or $x+y = 2k$.

In the first case, $2x + 2y = xy + 1$ becomes $(x-2)(y-2) = 3$, with solution $x = 3$, $y = 5$.

In the second case, $x + y = xy + 1$ becomes $(x-1)(y-1) = 0$, hence $x = 1$. One obtains $y = p+1$. The condition $d = 2$ is satisfied only if $p$ is even, i.e.\ $p = 2$.

In conclusion, the solutions are $(1, p+1)$ and $(3, 5)$.",142,1031,Number Theory,2
413,tst_jbmo_ro_2021_1_p3,tst_jbmo,2021,g,"The circle with center $I$, inscribed in triangle $ABC$, is tangent to sides $AB$, $AC$, and $BC$ at points $M$, $N$, and $K$ respectively. The median $AD$ of triangle $ABC$ intersects $MN$ at point $L$. Show that points $K$, $I$, and $L$ are collinear.","\subsection*{Solution}

(Evan Chen --- \textit{Euclidean Geometry in Mathematical Olympiads}, p.\ 64)

Let $\{X\} = MN \cap KI$. We will show that $X \in AD$. Through point $X$ we draw a line parallel to $BC$. It intersects $AB$ and $AC$ at $B'$ and $C'$ respectively. We consider the case $B' \in [AM]$, $C' \in [CN]$, the other case being analogous. Since $KI \perp B'C'$, the quadrilaterals $B'MIX$ and $C'NXI$ are cyclic. Since $IM = IN$, we have $\angle XMI \equiv \angle XNI$, so $\angle XC'I \equiv \angle XNI \equiv \angle XMI \equiv \angle XB'I$. We deduce that triangle $IB'C'$ is isosceles, so the altitude $IX$ is also a median. From the fact that $X$ is the midpoint of $[B'C']$, it follows immediately that $X \in [AD]$.

\textbf{Remark:} We have $m(\angle AC'I) = 180^\circ - m(\angle NXI) = m(\angle MXI) = m(\angle MB'I) = 180^\circ - m(\angle AB'I)$, so the quadrilateral $AB'IC'$ is cyclic.

(One could invoke directly the converse of Simson's theorem: since the projections of $I$ onto the sides of triangle $AB'C'$ are collinear, $I$ lies on the circumscribed circle of this triangle.)",253,1106,Geometry,3
414,tst_jbmo_ro_2021_1_p4,tst_jbmo,2021,c,"Let $n \geq 2$ be a natural number. On an $n \times n$ board, $n$ rooks are placed such that no two attack each other. All rooks move simultaneously once and are only allowed to move to a square adjacent to the one they occupy.

Determine all values of $n$ for which there exists an arrangement of rooks such that, after one move, the rooks still do not attack each other.

\textit{Note: Two squares are adjacent if they share a common side.}","\subsection*{Solution}

We will show that $n$ can be any nonzero even number.

If $n$ is even, placing the rooks on the diagonal starting from the top-left corner of the board, no two rooks attack each other. We number the rows of the board from top to bottom, from 1 to $n$. We now move the rooks on odd-numbered rows one square to the right (we can do this; the rook on the last column is on an even-numbered row, so it is not affected) and the rooks on even-numbered rows one square to the left. Again, no two rooks attack each other, so for every even $n$ there exists an arrangement of rooks with the required property.

We now show that if $n$ is odd, no arrangement with the required property exists.

We number the rows and columns from 1 to $n$. We observe that, for no two rooks to attack each other, we must have exactly one rook on each row and each column, so the sum of the coordinates of the $n$ rooks must be
\[
1 + 2 + \cdots + n + 1 + 2 + \cdots + n = n(n+1),
\]
an even number. This is a necessary but not sufficient condition for a suitable arrangement to exist.

At each move, each rook changes one of its coordinates by 1 (a horizontal move leaves the row number unchanged and increases or decreases the column number by 1; similarly for a vertical move). After the move, each of the $n$ rooks has changed the sum of its coordinates by 1. Since there is an odd number of rooks, the sum of the coordinates of all rooks has changed by an odd number, so it has become odd. Therefore, it is no longer $n(n+1)$, which is a necessary condition for the rooks not to attack each other. We conclude that if $n$ is odd, no arrangement with the desired property exists.",442,1680,Combinatorics,4
415,tst_jbmo_ro_2021_1_p5,tst_jbmo,2021,g,"Let $I$ be the incenter of triangle $ABC$. The circle with center $A$ and radius $AI$ intersects the circumscribed circle of triangle $ABC$ at points $M$ and $N$.

Show that line $MN$ is tangent to the incircle of triangle $ABC$.","If $MN$ intersects sides $AB$ and $AC$ at $D$ and $E$ respectively, the conclusion of the problem reduces to showing that $I$ is the center of the $A$-excircle of triangle $ADE$.

Since points $A, M, B, C, N$ are concyclic and $AM = AN$, we have:
\[
\angle AMN \equiv \angle MBA \equiv \angle ABN \equiv \angle ACM \equiv \angle ACN \equiv \angle ANM.
\]
It follows that the quadrilateral $BCED$ is cyclic. Moreover, we deduce that triangles $AMB$ and $ADM$ are similar, so $AI^2 = AM^2 = AD \cdot AB$. We deduce that triangles $ADI$ and $AIB$ are also similar, from which
\[
m(\angle AID) = \tfrac{1}{2} m(\angle B).
\]
Analogously, $m(\angle AIE) = \tfrac{1}{2} m(\angle C)$. Therefore, point $I$ is that point on the bisector of angle $\angle DAE$, lying in the half-plane opposite to $A$ determined by $DE$, from which segment $[DE]$ is seen at angle
\[
\frac{m(\angle B) + m(\angle C)}{2} = \frac{m(\angle AED) + m(\angle ADE)}{2}.
\]
The only such point is the center of the $A$-excircle, which gives the conclusion.",229,1022,Geometry,5
416,tst_jbmo_ro_2021_2_p1,tst_jbmo,2021,c,"Alina and Bogdan take turns, starting with Alina, writing a $0$ or a $1$, until each has written 2021 digits (each appends one digit to the right of those already written).

Alina wins if the decimal representation of the number obtained (in base 2) can be written as the sum of two perfect squares; otherwise, Bogdan wins.

Determine which of the two has a winning strategy.","\textbf{(A)} Since the remainder obtained by dividing a perfect square by 4 can be 0 or 1, if $N \equiv 3 \pmod{4}$, then $N$ cannot be written as the sum of two perfect squares.

\textbf{(B)} If the binary representation of a natural number $N$ ends in two $1$s followed by an even number of $0$s, then $N$ cannot be written as a sum of two perfect squares.

Indeed, let $N = \overline{a_1 a_2 \ldots a_t 11 \underbrace{00\ldots0}_{2p \text{ times}}}_{(2)}$ and $m = \overline{a_1 a_2 \ldots a_t}_{(2)}$. Then:
\[
N = m \cdot 2^{2p+2} + 2^{2p+1} + 2^{2p} = 4^p \cdot (4m + 3).
\]
If $p = 0$, then $N \equiv 3 \pmod{4}$, which cannot be written as a sum of two perfect squares.

If $p \geq 1$, assuming there exist $x, y \in \mathbb{N}$ such that $N = x^2 + y^2$, then $x$ and $y$ would be even. Setting $x = 2x_1$ and $y = 2y_1$, it would follow that $x_1^2 + y_1^2 = 4^{p-1}(4m+3)$.

Similarly, if $p-1 \geq 1$, continuing the argument, we deduce there exist $x_p, y_p \in \mathbb{N}$ such that $x_p^2 + y_p^2 = 4m+3$, a contradiction, so $N$ cannot be written as a sum of two perfect squares.
\hfill \textbf{[3p]}

We show that, regardless of Alina's moves, Bogdan has a winning strategy.

\textbf{(C)} If Alina writes a $1$ at some point, then Bogdan immediately writes a $1$ as well, and then repeats Alina's digit at every step; the conclusion follows from (B).
\hfill \textbf{[2p]}

\textbf{(D)} If Alina writes only zeros, then Bogdan writes zeros everywhere, except for the last 3 digits, which he chooses to be equal to $1$. One thus obtains $10101_{(2)} = 21$, which cannot be written as a sum of two perfect squares.
\hfill \textbf{[2p]}",375,1649,Combinatorics,6
417,tst_jbmo_ro_2021_2_p2,tst_jbmo,2021,a,"For any nonempty subset $X$ of the set $M = \{1, 2, 3, \ldots, 2021\}$, denote by $a_X$ the sum of the largest and smallest elements of $X$.

Determine the arithmetic mean of all numbers $a_X$ obtained.","We will separately compute the sum of the minimum elements of all subsets $X \subset M$ and the sum of the maximum elements.

\textbf{(A)} The number $k \in \{1, 2, \ldots, 2020\}$ is the minimum element for all sets of the form $X = \{k\} \cup S$, where $S$ is an arbitrary subset of $\{k+1, k+2, \ldots, 2021\}$, which has $2021 - (k+1) + 1 = 2021 - k$ elements.

Therefore, $k$ is the minimum element for $2^{2021-k}$ subsets. Since $2021$ is the minimum element for exactly one subset, the sum of the minimum elements of all subsets $X \subset M$ is:
\[
S_{\min} = 1 \cdot 2^{2020} + 2 \cdot 2^{2019} + 3 \cdot 2^{2018} + \cdots + 2019 \cdot 2^2 + 2020 \cdot 2^1 + 2021.
\]
\hfill \textbf{[3p]}

\textbf{(B)} Analogously, $k \in \{2, 3, \ldots, 2021\}$ is the maximum element for every set $X = \{k\} \cup S$, where $S \subset \{1, 2, \ldots, k-1\}$, so it is the maximum element for $2^{k-1}$ subsets. Since $1$ is the maximum element for exactly one subset, it follows that the sum of the maximum elements of all subsets $X \subset M$ is:
\[
S_{\max} = 1 + 2 \cdot 2^1 + 3 \cdot 2^2 + \cdots + 2019 \cdot 2^{2018} + 2020 \cdot 2^{2019} + 2021 \cdot 2^{2020}.
\]
\hfill \textbf{[3p]}

As a result, the sum of all numbers $a_X$ is:
\begin{align*}
S_{\min} + S_{\max} &= (2021+1) + (2020+2) \cdot 2^1 + (2019+3) \cdot 2^2 + \cdots + (1+2021) \cdot 2^{2020}\\
&= 2022 \cdot \left(1 + 2 + 2^2 + \cdots + 2^{2020}
ight) = 2022 \cdot \left(2^{2021} - 1
ight).
\end{align*}
Since $M$ has $2^{2021} - 1$ nonempty subsets, the arithmetic mean of all numbers $a_X$ is $\boxed{2022}$.
\hfill \textbf{[1p]}",202,1599,Algebra,7
418,tst_jbmo_ro_2021_2_p3,tst_jbmo,2021,g,"Consider the convex quadrilateral $ABCD$ in which angles $A$ and $C$ are not acute. On sides $AB$, $BC$, $CD$, and $DA$, take points $K$, $L$, $M$, and $N$ respectively. Show that the perimeter of quadrilateral $KLMN$ is at least equal to twice the length of diagonal $AC$.","\textbf{(A)} We first prove an auxiliary result:

\begin{lemma}
In a triangle $ABC$ with $m(\angle A) \geq 90^\circ$, the length of the median from $A$ is at most equal to half the length of side $BC$.
\end{lemma}

\begin{proof}
Since $m(\angle BAC) \geq 90^\circ$, it follows that:
\[
m(\angle ABC) + m(\angle ACB) \leq m(\angle BAC) = m(\angle BAD) + m(\angle CAD).
\]
Consequently, at least one of the inequalities $m(\angle ABD) \leq m(\angle BAD)$ or $m(\angle ACD) \leq m(\angle CAD)$ holds. As a result, $AD \leq BD$ or $AD \leq CD$, hence $AD \leq \dfrac{BC}{2}$.
\end{proof}
\hfill \textbf{[2p]}

\textbf{(B)} Let $P$, $Q$, $T$ be the midpoints of segments $KN$, $KM$, $LM$ respectively.

$PQ$ is the midline in triangle $KNM$ and $TQ$ is the midline in triangle $MKL$. As a result:
\[
AC \leq AP + PQ + QT + TC \leq \frac{KN}{2} + \frac{MN}{2} + \frac{KL}{2} + \frac{ML}{2},
\]
from which it follows that $P_{KLMN} \geq 2AC$.
\hfill \textbf{[5p]}",273,956,Geometry,8
419,tst_jbmo_ro_2021_2_p4,tst_jbmo,2021,n,"Show that for any natural number $n \geq 2$, there exists a nonzero multiple $m$ of $n$ with the following properties:
\begin{enumerate}[label=\alph*)]
    \item $m < n^4$;
    \item in the decimal (base 10) representation of $m$, at most four distinct digits are used.
\end{enumerate}","For any natural number $n$, there exists $k \in \mathbb{N}^*$ such that $2^{k-1} \leq n < 2^k$.

\textbf{(A)} If $k \leq 6$, then $n < 2^6 = 64$ and the number $2n$ satisfies the conditions of the statement (it contains at most three digits).

\textbf{(B)} Let $k > 6$. Denote by $M$ the set of all natural numbers containing at most $k$ digits using only $0$ and $1$. Then the set $M$ has $2^k$ elements, and since $2^k > n$ it follows that there exist in $M$ two numbers $a$ and $b$, $a > b$, whose difference is divisible by $n$.
\hfill \textbf{[3p]}

\textbf{(C)} From the structure of numbers $a$ and $b$, it follows that the number $a - b$ can only contain the digits $0$, $1$, $8$, or $9$.
\hfill \textbf{[2p]}

\textbf{(D)} Furthermore,
\[
a - b < a < 10^k = 10 \cdot 10^{k-1} < 1.6^5 \cdot 10^{k-1} < 1.6^{k-1} \cdot 10^{k-1} = 16^{k-1} = (2^{k-1})^4 < n^4.
\]
Therefore, the number $a - b$ satisfies the conclusion of the problem.
\hfill \textbf{[2p]}

\subsection*{Observation.}
For treating exclusively section \textbf{(A)}, or for treating trivial cases (for example, $n \leq 9999$), 0 points are awarded.",285,1118,Number Theory,9
420,tst_jbmo_ro_2021_3_p1,tst_jbmo,2021,a,"Show that for any $a, b, c > 0$ with $a + b + c = 1$ the following inequality holds:
\[
\frac{1}{a + bc} + \frac{1}{b + ca} + \frac{1}{c + ab} \geq \frac{7}{1 + abc}.
\]","\textbf{(A)} We have $a + bc = a(a+b+c) + bc = (a+b)(a+c)$ and analogously for the other terms. The inequality to be proved can be rewritten as:
\[
\frac{1}{(a+b)(a+c)} + \frac{1}{(b+c)(b+a)} + \frac{1}{(c+a)(c+b)} \geq \frac{7}{1+abc},
\]
or, bringing to a common denominator:
\[
\frac{2(a+b+c)}{(a+b)(b+c)(c+a)} \geq \frac{7}{1+abc},
\]
that is
\[
2 + 2abc \geq 7(a+b)(b+c)(c+a). \tag{1}
\]
\hfill \textbf{[2p]}

\textbf{(B)} Taking into account the equalities:
\[
abc = (a+b+c)(ab+bc+ca) - (a+b)(b+c)(c+a) = ab+bc+ca-(a+b)(b+c)(c+a),
\]
and respectively
\[
2 + abc = 1 + (a+b+c) + abc = (1+a)(1+b)(1+c) - (ab+bc+ca),
\]
by addition we deduce that
\[
2 + 2abc = (1+a)(1+b)(1+c) - (a+b)(b+c)(c+a),
\]
so inequality (1) is equivalent to
\[
(1+a)(1+b)(1+c) \geq 8(a+b)(b+c)(c+a). \tag{2}
\]
\hfill \textbf{[3p]}

\textbf{(C)} Since $1 + a = (a+b)+(c+a) \geq 2\sqrt{(a+b)(c+a)}$ and analogously, multiplying these inequalities yields the conclusion.
\hfill \textbf{[2p]}

\bigskip
\subsection*{Alternative Solution and Marking Scheme}

\textbf{(A')} Using the Bergstr\""{o}m inequality, we obtain:
\[
\frac{1}{a+bc} + \frac{1}{b+ca} + \frac{1}{c+ab} \geq \frac{(1+1+1)^2}{a+b+c+ab+bc+ca} = \frac{9}{1+ab+bc+ca},
\]
so it is sufficient to show that $2 + 9abc \geq 7(ab+bc+ca)$. \hfill \textbf{(3)} \quad \textbf{[2p]}

\textbf{(B')} From Schur's inequality we have
\[
a^3+b^3+c^3+3abc \geq ab(a+b)+bc(c+b)+ca(c+a),
\]
which is equivalent to
\[
(a+b+c)^3 + 9abc \geq 4(a+b+c)(ab+bc+ca),
\]
from which it follows that
\[
1 + 9abc \geq 4(ab+bc+ca). \tag{4}
\]
\hfill \textbf{[4p]}

\textbf{(C')} Taking into account that $(a+b+c)^2 \geq 3(ab+bc+ca)$, we obtain $1 \geq 3(ab+bc+ca)$, an inequality which, added to (4), leads to (3), thereby completing the solution.
\hfill \textbf{[1p]}",169,1778,Algebra,10
421,tst_jbmo_ro_2021_3_p2,tst_jbmo,2021,g,"In the acute-angled triangle $ABC$, let $O$ be the circumcenter and $D$ the foot of the altitude from $A$. Let $M$, $N$, $P$, $Q$ be the midpoints of segments $AB$, $AC$, $BD$, and $CD$ respectively. Show that one of the intersection points of the circumscribed circles of triangles $AMN$ and $POQ$ lies on the altitude $AD$.","\textbf{(A)} Let $\omega_1$ and $\omega_2$ be the circumscribed circles of triangles $AMN$ and $POQ$ respectively.

Since $OM \perp AB$ and $ON \perp AC$, the quadrilateral $AMON$ is cyclic, and $AO$ is a diameter of circle $\omega_1$.
\hfill \textbf{[1p]}

\textbf{(B)} Furthermore, it follows that $O$ is one of the intersection points of circles $\omega_1$ and $\omega_2$.

If $O \in AD$ (which amounts to triangle $ABC$ being isosceles), the problem is solved.

If $O 
otin AD$, let $X$ be the foot of the perpendicular from $O$ to $AD$. We will show that $X \in \omega_1 \cap \omega_2$.

Since $OX \perp XA$, it follows that $X \in \omega_1$.
\hfill \textbf{[3p]}

\textbf{(C)} Denoting by $R$ the midpoint of $BC$, it follows that $OR \perp BC$, so the quadrilateral $ORDX$ is a rectangle. Since
\[
PR = BR - BP = \frac{1}{2}(BC - BD) = \frac{1}{2}CD = DQ,
\]
we deduce that $OXPQ$ is an isosceles trapezoid, so $X \in \omega_2$.
\hfill \textbf{[3p]}",325,956,Geometry,11
422,tst_jbmo_ro_2021_3_p3,tst_jbmo,2021,n,"Let $p, q$ be nonzero natural numbers. For each $a, b \in \mathbb{R}$ define the sets
\[
P(a) = \left\{ a_n = a + n \cdot \frac{1}{p} \;\middle|\; n \in \mathbb{N} 
ight\}, \qquad
Q(b) = \left\{ b_n = b + n \cdot \frac{1}{q} \;\middle|\; n \in \mathbb{N} 
ight\}.
\]
We define the \emph{distance} from $P(a)$ to $Q(b)$ as the minimum value of $|x - y|$, with $x \in P(a)$, $y \in Q(b)$.

Determine the maximum distance between the sets $P(a)$ and $Q(b)$, as $a$ and $b$ range over $\mathbb{R}$.","We will show that the maximum value of the distance is equal to $\dfrac{1}{2 \cdot [p,q]}$, where $[p,q]$ denotes the least common multiple of $p$ and $q$.

\textbf{(A)} If $m$ and $n$ are two natural numbers, then
\[
|a_m - b_n| = \left| a - b + \frac{m}{p} - \frac{n}{q} 
ight| = \left| (a-b) + \frac{mq_1 - np_1}{[p,q]} 
ight|,
\]
where $p = dp_1$, $q = dq_1$, $(p_1, q_1) = 1$.

Since $(p_1, q_1) = 1$, there exist $k, l \in \mathbb{N}$ such that $kq_1 - lp_1 = 1$. Therefore, for any integer $t$ there exist $m, n \in \mathbb{N}$ such that
\[
|a_m - b_n| = \left| a - b + \frac{t}{[p,q]} 
ight|.
\]
\hfill \textbf{[3p]}

\textbf{(B)} Let $s$ be the integer part of $x = [p,q] \cdot (a - b)$. Then:
\[
\frac{s}{[p,q]} \leq a - b < \frac{s+1}{[p,q]},
\]
so the distance between $P(a)$ and $Q(b)$ is the smaller of the two numbers
\[
a - b - \frac{s}{[p,q]} \quad \text{and} \quad \frac{s+1}{[p,q]} - (a-b).
\]
\hfill \textbf{[3p]}

\textbf{(C)} Consequently, the maximum of this distance is attained when $a - b = \dfrac{1}{2[p,q]}$ and is equal to $\dfrac{1}{2[p,q]}$.
\hfill \textbf{[1p]}",494,1093,Number Theory,12
423,tst_jbmo_ro_2021_3_p4,tst_jbmo,2021,c,"Let $M$ be a set consisting of 13 three-digit natural numbers.

Show that there exists a nonempty subset $S \subset M$ and a combination of elementary arithmetic operations (addition, subtraction, multiplication, division --- without using parentheses) between the elements of $S$, such that the value of the resulting expression is a rational number in the interval $(3, 4)$.","\textbf{(A)} We will partition the set of three-digit numbers into 8 disjoint subsets such that for any two numbers $a$ and $b$ from the same subset, with $a > b$, we have $\dfrac{a}{b} < \dfrac{4}{3}$.

The subsets are:
\begin{align*}
A_1 &= \{100, 101, \ldots, 133\}, & A_2 &= \{134, 135, \ldots, 178\}, & A_3 &= \{179, 180, \ldots, 238\},\\
A_4 &= \{239, 240, \ldots, 318\}, & A_5 &= \{319, 320, \ldots, 425\}, & A_6 &= \{426, 427, \ldots, 567\},\\
A_7 &= \{568, 569, \ldots, 757\}, & A_8 &= \{758, 759, \ldots, 999\}.
\end{align*}
\hfill \textbf{[3p]}

\textbf{(B)} Since $M$ has 13 elements, one of the sets $A_i$, with $i \in \{1, 2, \ldots, 8\}$, contains at least two elements $a, b$ from $M$, with $a > b$, from which $1 < \dfrac{a}{b} < \dfrac{4}{3}$.
\hfill \textbf{[2p]}

\textbf{(C)} Among the 11 remaining numbers in $M \setminus \{a, b\}$, two of them, say $c, d$, with $c > d$, belong to one of the sets $A_j$, $1 \leq j \leq 8$. As such, $1 < \dfrac{c}{d} < \dfrac{4}{3}$.

Similarly, there exist $e, f \in M \setminus \{a, b, c, d\}$, with $e > f$, and $k \in \{1, 2, \ldots, 8\}$ such that $e, f \in A_k$, so $1 < \dfrac{e}{f} < \dfrac{4}{3}$.

The conclusion follows by observing that
\[
3 < \frac{a}{b} + \frac{c}{d} + \frac{e}{f} < 4.
\]
\hfill \textbf{[2p]}",376,1280,Combinatorics,13
263,jbmo_2022_p1,jbmo,2022,a,"Find all pairs $(a, b)$ of positive integers such that
\[
11ab \leq a^3 - b^3 \leq 12ab.
\]","
oindent\textbf{Solution 1.} Let $a - b = t$. Due to $a^3 - b^3 \geq 11ab$ we conclude that $a > b$ so $t$ is a positive integer and the condition can be written as
\[
11b(b+t) \leq t\!\left(b^2 + b(b+t) + (b+t)^2
ight) \leq 12b(b+t).
\]
Since
\[
t\!\left(b^2 + b(b+t) + (b+t)^2
ight) = t\!\left(b^2 + b^2 + bt + b^2 + 2bt + t^2
ight) = 3tb(b+t) + t^3,
\]
the condition can be rewritten as
\[
(11 - 3t)b(b+t) \leq t^3 \leq (12 - 3t)b(b+t).
\]
We cannot have $t \geq 4$ since in that case $t^3 \leq (12 - 3t)b(b+t)$ is not satisfied as the right hand side is not positive. Therefore it remains to check the cases when $t \in \{1, 2, 3\}$.

If $t = 3$, the above condition becomes
\[
2b(b+3) \leq 27 \leq 3b(b+3).
\]
If $b \geq 3$, the left hand side is greater than $27$ and if $b = 1$ the right hand side is smaller than $27$ so there are no solutions in these cases. If $b = 2$, we get a solution $(a, b) = (5, 2)$.

If $t \leq 2$, we have
\[
(11 - 3t)b(b+t) \geq (11 - 6) \cdot 1 \cdot (1+1) = 10 > t^3,
\]
so there are no solutions in this case.

In summary, the only solution is $(a, b) = (5, 2)$.

\bigskip


oindent\textbf{Solution 2.} First, from $a^3 - b^3 \geq 11ab > 0$ it follows that $a > b$, implying that $a - b \geq 1$. Note that
\[
a^3 - b^3 = (a-b)(a^2 + ab + b^2) = (a-b)\!\left((a-b)^2 + 3ab
ight) \geq (a-b)(1 + 3ab) > 3ab(a-b).
\]
Therefore $12ab \geq a^3 - b^3 > 3ab(a-b)$, which implies that $a - b < 4$ so $a - b \in \{1, 2, 3\}$. We discuss three possible cases:

\begin{itemize}
    \item $a - b = 1$

    After replacing $a = b + 1$, the condition $a^3 - b^3 \geq 11ab$ reduces to $1 \geq 8b^2 + 8b$, which is not satisfied for any positive integer $b$.

    \item $a - b = 2$

    After replacing $a = b + 2$, the condition $a^3 - b^3 \geq 11ab$ reduces to $8 \geq 5b^2 + 10b$, which is also not satisfied for any positive integer $b$.

    \item $a - b = 3$

    After replacing $a = b + 3$, the condition $a^3 - b^3 \geq 11ab$ reduces to $27 \geq 2b^2 + 6b$. The last inequality holds true only for $b = 1$ and $b = 2$. For $b = 1$ we get $a = 4$ and for $b = 2$ we get $a = 5$. Direct verification shows that $a^3 - b^3 \leq 12ab$ is satisfied only for $(a, b) = (5, 2)$.
\end{itemize}

In summary, $(a, b) = (5, 2)$ is the only pair of positive integers satisfying all conditions of the problem.",91,2327,Algebra,1
264,jbmo_2022_p2,jbmo,2022,g,"Let $ABC$ be an acute triangle such that $AH = HD$, where $H$ is the orthocenter of $ABC$ and $D \in BC$ is the foot of the altitude from the vertex $A$. Let $\ell$ denote the line through $H$ which is tangent to the circumcircle of the triangle $BHC$. Let $S$ and $T$ be the intersection points of $\ell$ with $AB$ and $AC$, respectively. Denote the midpoints of $BH$ and $CH$ by $M$ and $N$, respectively. Prove that the lines $SM$ and $TN$ are parallel.","
oindent\textbf{Solution 1.} In order to prove that $SM$ and $TN$ are parallel, it suffices to prove that both of them are perpendicular to $ST$. Due to symmetry, we will provide a detailed proof of $SM \perp ST$, whereas the proof of $TN \perp ST$ is analogous. In this solution we will use the following notation: $\angle BAC = \alpha$, $\angle ABC = \beta$, $\angle ACB = \gamma$.

We first observe that, due to the tangency condition, we have
\[
\angle SHB = \angle HCB = 90^\circ - \beta.
\]
Combining the above with
\[
\angle SBH = \angle ABH = 90^\circ - \alpha
\]
we get
\[
\angle BSH = 180^\circ - (90^\circ - \beta) - (90^\circ - \alpha) = \alpha + \beta = 180^\circ - \gamma
\]
from which it follows that $\angle AST = \gamma$.

Since $AH = HD$, $H$ is the midpoint of $AD$. If $K$ denotes the midpoint of $AB$, we have that $KH$ and $BC$ are parallel. Since $M$ is the midpoint of $BH$, the lines $KM$ and $AD$ are parallel, from which it follows that $KM$ is perpendicular to $BC$. As $KH$ and $BC$ are parallel, we have that $KM$ is perpendicular to $KH$ so $\angle MKH = 90^\circ$. Using the parallel lines $KH$ and $BC$ we also have
\[
\angle KHM = \angle KHB = \angle HBC.
\]
Now,
\[
\angle HMK = 90^\circ - \angle KHM = 90^\circ - \angle HBC = 90^\circ - (90^\circ - \gamma) = \gamma = \angle AST = \angle KSH,
\]
so the quadrilateral $MSKH$ is cyclic, which implies that $\angle MSH = \angle MKH = 90^\circ$. In other words, the lines $SM$ and $ST$ are perpendicular, which completes our proof.

\bigskip


oindent\textbf{Solution 2.} We will refer to the same figure as in the first solution. Since $\ell H$ is tangent to the circumcircle of triangle $BHC$, we have
\[
\angle SHB = \angle HCB = 90^\circ - \angle ABC = \angle HAB.
\]
From the above it follows that triangles $AHB$ and $HSB$ are similar. If $K$ denotes the midpoint of $AB$, then triangles $AHK$ and $HSM$ are also similar. Now, observe that $H$ and $K$ are respectively the midpoints of $AD$ and $AB$, which implies that $HK \parallel DB$, so
\[
\angle AHK = \angle ADB = 90^\circ.
\]
Now, from the last observation and the similarity of triangles $AHK$ and $HSM$, it follows that
\[
\angle HSM = \angle AHK = 90^\circ.
\]
Due to symmetry, analogously as above, we can prove that $\angle HTN = 90^\circ$, implying that both $SM$ and $TN$ are perpendicular to $TS$, hence they are parallel.",456,2376,Geometry,2
265,jbmo_2022_p3,jbmo,2022,n,"Find all quadruples of positive integers $(p, q, a, b)$, where $p$ and $q$ are prime numbers and $a > 1$, such that
\[
p^a = 1 + 5q^b.
\]","
oindent\textbf{Solution 1.} First of all, observe that if $p$, $q$ are both odd, then the left hand side of the given equation is odd and the right hand side is even so there are no solutions in this case. In other words, one of these numbers has to be equal to $2$ so we can discuss the following two cases:

\begin{itemize}
    \item $p = 2$

    In this case the given equation becomes
    \[
    2^a = 1 + 5 \cdot q^b.
    \]
    Note that $q$ has to be odd. In addition, $2^a \equiv 1 \pmod{5}$. It can be easily shown that the last equation holds if and only if $a = 4c$, for some positive integer $c$. Now, our equation becomes $2^{4c} - 1 = 5 \cdot q^b$, which can be written into its equivalent form
    \[
    (4^c - 1)(4^c + 1) = 5 \cdot q^b.
    \]
    Since $q$ is odd, it cannot divide both $4^c - 1$ and $4^c + 1$. Namely, if it divides both of these numbers then it also divides their difference, which is equal to $2$, and this is clearly impossible. Therefore, we have that either $q^b \mid 4^c - 1$ or $q^b \mid 4^c + 1$, which implies that one of the numbers $4^c - 1$ and $4^c + 1$ divides $5$. Since for $c \geq 2$ both of these numbers are greater than $5$, we only need to discuss the case $c = 1$. But in this case $5 \cdot q^b = 15$, which is obviously satisfied only for $b = 1$ and $q = 3$. In summary, $(p, q, a, b) = (2, 3, 4, 1)$ is the only solution in this case.

    \item $q = 2$

    In this case obviously $p$ must be an odd number and the given equation becomes
    \[
    p^a = 1 + 5 \cdot 2^b.
    \]
    First, assume that $b$ is even. Then $2^b \equiv 1 \pmod{3}$, which implies that $1 + 5 \cdot 2^b$ is divisible by $3$, hence $3 \mid p^a$ so $p$ must be equal to $3$ and our equation becomes
    \[
    3^a = 1 + 5 \cdot 2^b.
    \]
    From here it follows that $3^a \equiv 1 \pmod{5}$, which implies that $a = 4c$, for some positive integer $c$. Then the equation $3^a = 1 + 5 \cdot 2^b$ can be written into its equivalent form
    \[
    \frac{3^{2c} - 1}{2} \cdot \frac{3^{2c} + 1}{2} = 5 \cdot 2^{b-2}.
    \]
    Observe now that $3^{2c} \equiv 1 \pmod{4}$ from where it follows that $\frac{3^{2c}+1}{2} \equiv 1 \pmod{2}$. From here we can conclude that the number $\frac{3^{2c}+1}{2}$ is relatively prime to $2^{b-2}$, so it has to divide $5$. Clearly, this is possible only for $c = 1$ since for $c > 1$ we have $\frac{3^{2c}+1}{2} > 5$. For $c = 1$, we easily find $b = 4$, which yields the solution $(p, q, a, b) = (3, 2, 4, 4)$.

    Next, we discuss the case when $b$ is odd. In this case,
    \[
    p^a = 1 + 5 \cdot 2^b \equiv 1 + 5 \cdot 2 \equiv 2 \pmod{3}.
    \]
    The last equation implies that $a$ must be odd. Namely, if $a$ is even then we cannot have $p^a \equiv 2 \pmod{3}$ regardless of the value of $p$. Combined with the condition $a > 1$, we conclude that $a \geq 3$. The equation $p^a = 1 + 5 \cdot 2^b$ can be written as
    \[
    5 \cdot 2^b = p^a - 1 = (p - 1)\!\left(p^{a-1} + p^{a-2} + \cdots + 1
ight).
    \]
    Observe that
    \[
    p^{a-1} + p^{a-2} + \cdots + 1 \equiv 1 + 1 + \cdots + 1 = a \equiv 1 \pmod{2},
    \]
    so this number is relatively prime to $2^b$, which means that it has to divide $5$. But this is impossible, since $a \geq 3$ and $p \geq 3$ imply that
    \[
    p^{a-1} + p^{a-2} + \cdots + 1 \geq p^2 + p + 1 \geq 3^2 + 3 + 1 = 13 > 5.
    \]
    In other words, there are no solutions when $q = 2$ and $b$ is an odd number.
\end{itemize}

In summary, $(a, b, p, q) = (4, 4, 3, 2)$ and $(a, b, p, q) = (4, 1, 2, 3)$ are the only solutions.

\bigskip


oindent\textbf{Solution 2.} Analogously as in the first solution we conclude that at least one of the numbers $p$ and $q$ has to be even. Since these numbers are prime, this implies that at least one of $p$ and $q$ must be equal to $2$. Therefore it is sufficient to discuss the following two cases:

\begin{itemize}
    \item $p = 2$

    In this case the given equation then becomes
    \[
    2^a = 1 + 5 \cdot q^b.
    \]
    From here, it follows that $q$ is an odd number. In addition, $2^a \equiv 1 \pmod{5}$, which implies that $a = 4c$, for some positive integer $c$. Then the above equation can be written in its equivalent form
    \[
    (2^c - 1)(2^c + 1)(2^{2c} + 1) = 5 \cdot q^b.
    \]
    Since $2^c - 1$, $2^c$ and $2^c + 1$ are three consecutive integers, one of them must be divisible by $3$. Clearly it is not $2^c$, implying that one of the numbers $2^c - 1$ and $2^c + 1$ is divisible by $3$. This implies that $3 \mid (2^c - 1)(2^c + 1)(2^{2c} + 1)$ so $3 \mid 5 \cdot q^b$, hence $q$ must be equal to $3$ and we are left with solving the equation
    \[
    (2^c - 1)(2^c + 1)(2^{2c} + 1) = 5 \cdot 3^b.
    \]
    Note that $2^{2c} + 1 \equiv 2 \pmod{3}$ so from the above equation it follows that $2^{2c} + 1$ must be equal to $5$, which implies that $c = 1$. For $c = 1$ we have $b = 1$, so we get $(a, b, p, q) = (4, 1, 2, 3)$ as the only solution in this case.

    \item $q = 2$

    In this case the given equation becomes
    \[
    p^a = 1 + 5 \cdot 2^b
    \]
    so clearly $p$ must be an odd number.

    If $a$ is odd then we have
    \[
    5 \cdot 2^b = (p - 1)\!\left(p^{a-1} + p^{a-2} + \cdots + p + 1
ight).
    \]
    The second bracket on the right hand side is a sum of $a$ odd numbers so it is an odd number. Due to the condition $a > 1$ we must have $a \geq 3$. But then
    \[
    p^{a-1} + p^{a-2} + \cdots + p + 1 \geq p^2 + p + 1 \geq 3^2 + 3 + 1 > 5
    \]
    so we do not have solutions in this case. Therefore it remains to discuss the case when $a$ is even.

    Let $a = 2c$ for some positive integer $c$. Then we have the following equation
    \[
    (p^c - 1)(p^c + 1) = 5 \cdot 2^b.
    \]
    Note that $p^c - 1$ and $p^c + 1$ are two consecutive even numbers so one of them is divisible by $2$ but not by $4$. Looking into the right hand side of the above equation, we conclude that this number must be equal to either $2$ or $5 \cdot 2 = 10$. In other words, either $p^c - 1 \in \{2, 10\}$ or $p^c + 1 \in \{2, 10\}$ yielding the following possible values for $p^c$: $1, 3, 9, 11$. Clearly $p^c = 1$ is impossible, whereas $p^c = 3$ implies that $(p^c - 1)(p^c + 1)$ is not divisible by $5$ so there are no solutions if $p^c = 3$. Similarly, for $p^c = 11$ we have that $(p^c - 1)(p^c + 1)$ is divisible by $3$ so it cannot be equal to $5 \cdot 2^b$ for any positive integer $b$. Finally, if $p^c = 9$, we have solution $(a, b, p, q) = (4, 4, 3, 2)$.
\end{itemize}

In summary, $(a, b, p, q) = (4, 4, 3, 2)$ and $(a, b, p, q) = (4, 1, 2, 3)$ are the only solutions.",137,6599,Number Theory,3
266,jbmo_2022_p4,jbmo,2022,c,"We call an even positive integer $n$ \emph{nice} if the set $\{1, 2, \ldots, n\}$ can be partitioned into $\frac{n}{2}$ two-element subsets, such that the sum of the elements in each subset is a power of $3$. For example, $6$ is nice, because the set $\{1, 2, 3, 4, 5, 6\}$ can be partitioned into subsets $\{1, 2\}$, $\{3, 6\}$, $\{4, 5\}$. Find the number of nice positive integers which are smaller than $3^{2022}$.","
oindent\textbf{Solution.} For a nice number $n$ and a given partition of the set $\{1, 2, \ldots, n\}$ into two-element subsets such that the sum of the elements in each subset is a power of $3$, we say that $a, b \in \{1, 2, \ldots, n\}$ are \emph{paired} if both of them belong to the same subset.

Let $x$ be a nice number and $k$ be a (unique) non-negative integer such that $3^k \leq x < 3^{k+1}$. Suppose that $x$ is paired with $y < x$. Then, $x + y = 3^s$, for some positive integer $s$. Since
\[
3^s = x + y < 2x < 2 \cdot 3^{k+1} < 3^{k+2},
\]
we must have $s < k + 2$. On the other hand, the inequality
\[
x + y \geq 3^k + 1 > 3^k
\]
implies that $s > k$. From these we conclude that $s$ must be equal to $k + 1$, so $x + y = 3^{k+1}$. The last equation, combined with $x > y$, implies that $x > \frac{3^{k+1}}{2}$.

Similarly as above, we can conclude that each number $z$ from the closed interval $\left[3^{k+1} - x,\, x
ight]$ is paired with $3^{k+1} - z$. Namely, for any such $z$, the larger of the numbers $z$ and $3^{k+1} - z$ is greater than $\frac{3^{k+1}}{2}$ which is greater than $3^k$, so the numbers $z$ and $3^{k+1} - z$ must necessarily be in the same subset. In other words, each number from the interval $\left[3^{k+1} - x,\, x
ight]$ is paired with another number from this interval. Note that this implies that all numbers smaller than $3^{k+1} - x$ are paired among themselves, so the number $3^{k+1} - x - 1$ is either nice or equals zero. Also, the number $3^k$ must be paired with $2 \cdot 3^k$, so $x \geq 2 \cdot 3^k$.

Finally, we prove by induction that $a_n = 2^n - 1$, where $a_n$ is the number of nice positive integers smaller than $3^n$. For $n = 1$, the claim is obviously true, because $2$ is the only nice positive integer smaller than $3$. Now, assume that $a_n = 2^n - 1$ for some positive integer $n$. We will prove that $a_{n+1} = 2^{n+1} - 1$. To prove this, first observe that the number of nice positive integers between $2 \cdot 3^n$ and $3^{n+1}$ is exactly $a_{n+1} - a_n$. Next, observe that $3^{n+1} - 1$ is nice. For every nice number $2 \cdot 3^n \leq x < 3^{n+1} - 1$, the number $3^{n+1} - x - 1$ is also nice and is strictly smaller than $3^n$. Also, for every positive integer $y < 3^n$, obviously there is a unique number $x$ such that $2 \cdot 3^n \leq x < 3^{n+1} - 1$ and $3^{n+1} - x - 1 = y$. Thus,
\[
a_{n+1} - a_n = a_n + 1 \iff a_{n+1} = 2a_n + 1 = 2(2^n - 1) + 1 = 2^{n+1} - 1
\]
completing the proof.

In summary, there are $2^{2022} - 1$ nice positive integers smaller than $3^{2022}$.",418,2563,Combinatorics,4
279,shl_jbmo_2022_a1,shl_jbmo,2022,a,"Find all pairs of positive integers $(a, b)$ such that$$11ab \le a^3 - b^3 \le 12ab.$$","Clearly, $a>b$.

If $a\ge b+4$, then $a^3-b^3-12ab=3ab(a-b-4)+(a-b)^3>0$, contradiction.

If $a\leq b+2$, then $a^3-b^3-11ab=(a-b)^3-ab(11-3(a-b))\leq 8-5ab\leq -2<0$, contradiction.

Hence, $a=b+3$. This gives us$$11b(b+3)\leq (b+3)^3-b^3\leq 12b(b+3)\Leftrightarrow 9\leq b(b+3)\leq \frac{27}2\Rightarrow b=2$$
So the only solution is $(a,b)=(5,2)$.",86,351,Algebra,1
280,shl_jbmo_2022_a2,shl_jbmo,2022,a,"Let $x, y,$ and $z$ be positive real numbers such that $xy + yz + zx = 3$. Prove that
$$\frac{x + 3}{y + z} + \frac{y + 3}{z + x} + \frac{z + 3}{x + y} + 3 \ge 27 \cdot \frac{(\sqrt{x} + \sqrt{y} + \sqrt{z})^2}{(x + y + z)^3}.$$","Let $x+y+z=s$. By Cauchy-Schwarz,

$\frac{x + 3}{y + z} + \frac{y + 3}{z + x} + \frac{z + 3}{x + y} \geq \frac{(x+y+z+9)^2}{\displaystyle \sum(x+3)(y+z)}=\dfrac{(s+9)^2}{6s+6},$

and by Cauchy-Schwarz again

$27 \cdot \frac{(\sqrt{x} + \sqrt{y} + \sqrt{z})^2}{(x + y + z)^3} \leq \dfrac{27 \cdot 3s}{s^3}=\dfrac{81}{s^2},$

hence we are left to prove that $\dfrac{(s+9)^2}{6s+6}+3 \geq \dfrac{81}{s^2},$ which rewrites as $(s-3)(s^3+39s^2+216s+162) \geq 0,$ which is true as $s^2 \geq 3(xy+yz+zx)=9,$ that is $s \geq 3$.",228,520,Algebra,2
281,shl_jbmo_2022_a3,shl_jbmo,2022,a,"Let $a, b,$ and $c$ be positive real numbers such that $a + b + c = 1$. Prove the following inequality
$$a \sqrt[3]{\frac{b}{a}} + b \sqrt[3]{\frac{c}{b}} + c \sqrt[3]{\frac{a}{c}} \le ab + bc + ca + \frac{2}{3}.$$","By Holder inequality, we have
$$(ab+bc+ca)(a+b+c)(1+1+1) \ge \left( a \sqrt[3]{\frac{b}{a}} + b \sqrt[3]{\frac{c}{b}} + c \sqrt[3]{\frac{a}{c}} 
ight)^3$$Let $ab+bc+ca=t$, we have
$$\sqrt[3]{3t} \ge a \sqrt[3]{\frac{b}{a}} + b \sqrt[3]{\frac{c}{b}} + c \sqrt[3]{\frac{a}{c}} $$It suffices to prove $t+\frac{2}{3}\ge \sqrt[3]{t}$. Which is equal to,
$$(3t+2)^3 \ge 81t$$$$27t^3+54t^2-45t+8\ge 0$$$$(3t-1)^2(3t+8)\ge0$$With the equality holds when $a=b=c=\frac{1}{3}$.",214,466,Algebra,3
282,shl_jbmo_2022_a4,shl_jbmo,2022,a,"Suppose that $a, b,$ and $c$ are positive real numbers such that
$$a + b + c \ge \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.$$Find the largest possible value of the expression
$$\frac{a + b - c}{a^3 + b^3 + abc} + \frac{b + c - a}{b^3 + c^3 + abc} + \frac{c + a - b}{c^3 + a^3 + abc}.$$","Note that if $a+b-c,b+c-a,c+a-b$ are all non-negative, then in light of $x^3+y^3 \geq xy(x+y)$ we obtain

$\displaystyle \sum \dfrac{a+b-c}{a^3+b^3+abc} \leq \dfrac{1}{a+b+c} \cdot \dfrac{\displaystyle 2 \sum ab-\sum a^2}{abc} \leq \dfrac{\displaystyle \sum ab}{abc(a+b+c)}=\dfrac{\displaystyle \sum 1/a}{\displaystyle \sum a} \leq 1,$

as desired.

Now, if one of $a+b-c,b+c-a,c+a-b$ is negative, WLOG $a+b-c<0$, the other two must be positive as $(a+b-c)+(b+c-a)=2b>0$ and $(a+b-c)+(c+a-b)=2a>0$. Hence,

$\displaystyle \sum \dfrac{a+b-c}{a^3+b^3+abc} \leq \dfrac{b+c-a}{b^3+c^3+abc}+\dfrac{c+a-b}{c^3+a^3+abc} \leq \dfrac{b+c-a}{bc(a+b+c)}+\dfrac{c+a-b}{ca(a+b+c)}=$

$=\dfrac{2ab+ac+bc-a^2-b^2}{abc(a+b+c)} \leq \dfrac{ac+bc}{abc(a+b+c)} < \dfrac{ab+bc+ca}{abc(a+b+c)}=\dfrac{\displaystyle \sum 1/a}{\displaystyle \sum a} \leq 1,$

as desired.

Equality holds, for example, if $a=b=c=1$.",283,891,Algebra,4
283,shl_jbmo_2022_a5,shl_jbmo,2022,a,"The numbers $2, 2, ..., 2$ are written on a blackboard (the number $2$ is repeated $n$ times). One step consists of choosing two numbers from the blackboard, denoting them as $a$ and $b$, and replacing them with $\sqrt{\frac{ab + 1}{2}}$.
$(a)$ If $x$ is the number left on the blackboard after $n - 1$ applications of the above operation, prove that $x \ge \sqrt{\frac{n + 3}{n}}$.
$(b)$ Prove that there are infinitely many numbers for which the equality holds and infinitely many for which the inequality is strict.","So, we delete $a$ and $b$ and write $c=\sqrt{\frac{ab + 1}{2}}$. We have
$$c^2=\frac{ab + 1}{2} \le \frac{1}{4} a^2 + \frac{1}{4}b^2 + \frac{1}{2}.$$So, if in each operation instead of $c$ we write the number $c_1$ satisfying $c_1^2=\frac{1}{4} a^2 + \frac{1}{4}b^2 + \frac{1}{2}$, we will finally get a number $x_1$ greater than or equal to number obtained in the initial process. Thus, going to the squares of the numbers we write, it is enough to prove the following claim.

Claim. The numbers $4, 4, \ldots, 4$ are written on a blackboard (the number $4$ is repeated $n$ times). One step consists of choosing two numbers $a',b'$ from the blackboard and replacing them with $c' = \frac{1}{4}a' + \frac{1}{4}b'+ \frac{1}{4}$. Prove for the number $x'$ left on the blackboard holds $x'\ge 1+\frac{3}{n}$.

To prove this, we represent the operations as a binary graph - leaves (the initially written numbers) are at the top. The root $r$ (the last number) has two children $r_1$ and $r_2$. Assume there are $n_1$ leaves above $r_1$, and there are $n_2$ leaves above $r_2$ ($n_1+n_2=n$). By the inductive assumption (we do induction) $r_1 \ge 1+\frac{3}{n_1}$ and $r_2 \ge 1+\frac{3}{n_2}$. We have to check that $r\ge 1+\frac{3}{n}$. That is, it's enough to prove:
$$\frac{1}{4}\left(1+\frac{3}{n_1} 
ight)+\frac{1}{4}\left(1+\frac{3}{n_2} 
ight)\ge 1+\frac{3}{n_1+n_2},$$which boils down to prove
$$\frac{1}{n_1}+\frac{1}{n_2}\ge \frac{4}{n_1+n_2},$$which is Cauchy-Schwartz (Engel's form).

Now, the equality is only possible if we are dealing with equal numbers at each operation, which is only possible if the mentioned binary tree is perfectly balanced, that is, only if $n=2^k$.",518,1684,Algebra,5
284,shl_jbmo_2022_a6,shl_jbmo,2022,a,"Let $a, b,$ and $c$ be positive real numbers such that $a^2 + b^2 + c^2 = 3$. Prove that
$$\frac{a^2 + b^2}{2ab} + \frac{b^2 + c^2}{2bc} + \frac{c^2 + a^2}{2ca} + \frac{2(ab + bc + ca)}{3} \ge 5 + |(a - b)(b - c)(c - a)|.$$","The inequality rewrites as

$\displaystyle \sum \dfrac{(a-b)^2}{2ab} \geq 2-\dfrac{\displaystyle 2\sum ab}{3}+\prod |a-b|,$

with all sums and products being cyclic. Note that

$2-\dfrac{\displaystyle 2 \sum ab}{3}=\dfrac{\displaystyle 2(3-\sum ab)}{3}=\dfrac{\displaystyle 2(\sum a^2-\sum ab)}{3}=\dfrac{\displaystyle \sum (a-b)^2}{3},$

hence we are left to prove that

$\displaystyle \sum (a-b)^2(\dfrac{1}{2ab}-\dfrac{1}{3}) \geq \prod |a-b|$

However,

$\dfrac{1}{2ab}-\dfrac{1}{3}=\dfrac{3-2ab}{6ab}=\dfrac{(a-b)^2+c^2}{6ab} \geq \dfrac{c|a-b|}{3ab},$

hence the desired inequality boils down to

$\sum \dfrac{c|a-b|^3}{3ab} \geq \prod |a-b|,$

which readily follows due to the AM-GM inequality.",223,701,Algebra,6
285,shl_jbmo_2022_c1,shl_jbmo,2022,c,"Anna and Bob, with Anna starting first, alternately color the integers of the set $S = \{1, 2, ..., 2022 \}$ red or blue. At their turn each one can color any uncolored number of $S$ they wish with any color they wish. The game ends when all numbers of $S$ get colored. Let $N$ be the number of pairs $(a, b)$, where $a$ and $b$ are elements of $S$, such that $a$, $b$ have the same color, and $b - a = 3$.
Anna wishes to maximize $N$. What is the maximum value of $N$ that she can achieve regardless of how Bob plays?","Replace $2022$ with $6m$, for $m=337$. We then claim the answer is $3m-3$. For $i \in \{1,2,3 \}$, let $A_i=\{s: \, 1 \leq s \leq 6m : s \equiv i \pmod 3 \}$. Then, Anna wishes to maximize the number of unordered pairs $(a,b)$ such that $a,b \in A_i$ for some $i$ and $a,b$ are of the same colour. For each $i$ and each $x \in A_i$, call $x-3$ and $x+3$ its neighbors. The problem splits into two parts.

Part 1: Anna's strategy.

Anna's strategy is to simply pick a coloured integer and colour one of its uncoloured neighbors with the same colour. This cannot be done only if for all $i$, all or no elements of $A_i$ are coloured, which will happen at most $3$ times during the process. Hence, Anna will increase $N$ by at least $1$ at least $3m-3$ times (she moves $3m$ times in total). Hence, this strategy guarantees that $N \geq 3m-3$.

Part 2: Bob's strategy.

Bob's strategy is to simply pick a coloured integer and colour one of its uncoloured neighbors with the other colour. This cannot be done only if for all $i$, all or no elements of $A_i$ are coloured. However, whenever Bob moves, an odd number of integers have been coloured, hence the above cannot happen.

Therefore, Bob can always apply the above strategy. This strategy increases by at least one the number of pairs $(a,b)$ such that $a,b \in A_i$ and $a,b$ are of different colours. Thus, the number of these pairs $X$ in the end is $\geq 3m$, which implies that $N=(6m-3)-X \leq 3m-3$.

To sum up, the answer to the problem is $\max N=3m-3$.",518,1514,Combinatorics,7
286,shl_jbmo_2022_c2,shl_jbmo,2022,c,"Let $n \ge 2$ be an integer. Alex writes the numbers $1, 2, ..., n$ in some order on a circle such that any two neighbours are coprime. Then, for any two numbers that are not comprime, Alex draws a line segment between them. For each such segment $s$ we denote by $d_s$ the difference of the numbers written in its extremities and by $p_s$ the number of all other drawn segments which intersect $s$ in its interior.
Find the greatest $n$ for which Alex can write the numbers on the circle such that $p_s \le |d_s|$, for each drawn segment $s$.","Answer is $11$.
Let's show that $12$ or more are impossible.
There should be $6$ even number at least. And $2$ even numbers can't be neighbour. Also there should be edge between any $2$ evens.
There is no even numbers between any $2$ consecutive even numbers because $(s:2k-2k+2)$, $d_s=2 \implies p_s \leq 2$.
$\implies 2 - 4 - 6 - 8 - 10 - 12-$
Let's look at $3$ and $9$.
$3$ and $9$ have at least $3$ edges. So $3$ can't be between any consecutive even numbers.
$\implies 3$ is between $n$ and $2$.
Similarly $9$ is between $n$ and $2$. Also there is no number between $3$ and $9$. Contradiction.",543,599,Combinatorics,8
287,shl_jbmo_2022_c3,shl_jbmo,2022,c,"There are $200$ boxes on the table. In the beginning, each of the boxes contains a positive integer (the integers are not necessarily distinct). Every minute, Alice makes one move. A move consists of the following. First, she picks a box $X$ which contains a number $c$ such that $c = a + b$ for some numbers $a$ and $b$ which are contained in some other boxes. Then she picks a positive integer $k > 1$. Finally, she removes $c$ from $X$ and replaces it with $kc$. If she cannot make any mobes, she stops. Prove that no matter how Alice makes her moves, she won't be able to make infinitely many moves.","Assume otherwise. Note that there is a nonzero number of boxes whose number changes infinitely many times. We may ignore any boxes whose number changes a finite number of times, as after a finite number of moves we may assume that all those boxes retain their number, and since the number in each of the other boxes increases, we may assume that the process happens between among these particular boxes.

Hence, the number in each of the boxes changes infinitely may times. Consider the smallest number among the boxes. Note that this number cannot change, as it is smaller than all the remaining numbers, and the numbers on the other boxes keep increasing during the process, a contradiction.

Hence, the process will eventually stop.",603,735,Combinatorics,9
288,shl_jbmo_2022_c4,shl_jbmo,2022,c,"We call an even positive integer $n$ nice if the set $\{1, 2, \dots, n\}$ can be partitioned into $\frac{n}{2}$ two-element subsets, such that the sum of the elements in each subset is a power of $3$. For example, $6$ is nice, because the set $\{1, 2, 3, 4, 5, 6\}$ can be partitioned into subsets $\{1, 2\}$, $\{3, 6\}$, $\{4, 5\}$. Find the number of nice positive integers which are smaller than $3^{2022}$.","Let $P(k)$ be the number of nice $n$ smaller than $3^k$ and $T(k)$ be the set including the nice $n$ smaller than $3^k$.

Easily we get $P(1)=1, P(2)=3, P(3)=7$.

Lemma 1: $n=3^k - 1$ is always nice.
Proof: Pair the numbers as following: $(3^k - 1,1), (3^k - 2, 2) ....$, done.

Lemma 2: $n=3^{k+1} - m - 1, m \in T(k)$ is always nice.
Proof: Pair the numbers ${m+1, m+2 .... , 3^{k+1} - m - 1}$ as we did above. Now the numbers left are ${1,2...,m}$ and $m$ is nice, done.

Induction $Q(k)$: The only nice $n$ for each $k \geq 2$ are $l$:
1)$l=m, m \in T(k-1)$
2)$l=3^k -1$
3)$l=3^k - m - 1, m \in T(k-1)$

For $k=2$ just do the handwork.
Let $Q(k)$ be true for $k=j$
For $k=j+1$:
Obviously these numbers work due to the lemmas. If there exists another number $l$ so that $l \in T(j+1)$, then $l > 2 \cdot 3^j$ (why?), and also $3^{j+1} - l - 1$ is good.
But $3^{j+1} - l - 1 < 3^j$ and therefore due to $Q(j)$ it must be $3^{j+1} - l -1\in T(j)$, which is clearly absurd. Done.

Now by counting the number of these 3 cases of nice numbers we get that $P(1)=1$ and by straightforward induction $P(n)=2P(n-1) + 1, \forall n \geq 2$.
Solving the recurrence relation we get that $P(n)=2^n - 1 \implies P(2022) = 2^{2022} - 1$.",410,1224,Combinatorics,10
289,shl_jbmo_2022_c5,shl_jbmo,2022,c,"Let $S$ be a finite set of points in the plane, such that for each $2$ points $A$ and $B$ in $S$, the segment $AB$ is a side of a regular polygon all of whose vertices are contained in $S$. Find all possible values for the number of elements of $S$.","$|S| = 3$ only.
Consider $m$, the longest segment in $S$. It must be a side of some regular polygon in $S$. If that polygon is not a triangle, then one of its diagonals is in $S$ and is longer than $m$, contradicting the fact that $m$ is the longest segment.
Therefore the largest shape is $S$ must be an equilateral triangle (which has no interior diagonals). Let the big triangle be $ABC$, and WLOG let $AB = AC = BC = 3$.

Then you can show that the second-largest shape must also be a triangle $XYZ$, with a side length $XY = XZ = YZ = r$ where $2 \leq r \leq 3$.
If $r \leq 3$, you can get a contradiction by showing one of $\{AX, AY, AZ, BX, BY, BZ, CX, CY, CZ\} \geq r$. It follows that $r = 3$ and some more geo shows that $XYZ = ABC$ so there can only be a single triangle in $S$.",249,789,Combinatorics,11
290,shl_jbmo_2022_g1,shl_jbmo,2022,g,"Let $ABCDE$ be a cyclic pentagon such that $BC = DE$ and $AB$ is parallel to $DE$. Let $X, Y,$ and $Z$ be the midpoints of $BD, CE,$ and $AE$ respectively. Show that $AE$ is tangent to the circumcircle of the triangle $XYZ$.","Note that $XY \parallel BE$, as $BCDE$ is an isosceles trapezoid. Hence,

$\angle YZE=\angle CAE=180^\circ-\angle CDE=\angle BED=\angle(BE,ED)=\angle(XY,XZ)=\angle ZXY,$

as desired.",224,182,Geometry,12
291,shl_jbmo_2022_g2,shl_jbmo,2022,g,"Let $ABC$ be a triangle with circumcircle $k$. The points $A_1, B_1,$ and $C_1$ on $k$ are the midpoints of arcs $\widehat{BC}$ (not containing $A$), $\widehat{AC}$ (not containing $B$), and $\widehat{AB}$ (not containing $C$), respectively. The pairwise distinct points $A_2, B_2,$ and $C_2$ are chosen such that the quadrilaterals $AB_1A_2C_1, BA_1B_2C_1,$ and $CA_1C_2B_1$ are parallelograms. Prove that $k$ and the circumcircle of triangle $A_2B_2C_2$ have a common center.
Comment. Point $A_2$ can also be defined as the reflection of $A$ with respect to the midpoint of $B_1C_1$, and analogous definitions can be used for $B_2$ and $C_2$.","Let $O$ be the circumcenter of triangle $ABC$. Then,

$\angle OA_1B_2=\angle OA_1B-\angle B_2A_1B=90^\circ-\dfrac{\angle A}{2}-(180^\circ-\angle C_1BA_1)=90^\circ-\dfrac{\angle A}{2}-\angle C_1AA_1=$

$=90^\circ-\dfrac{\angle A}{2}-(\dfrac{\angle A+\angle C}{2})=\dfrac{\angle B-\angle A}{2},$

and in a similar manner we obtain $\angle OB_1A_2=\dfrac{\angle B-\angle A}{2},$ hence $\angle OA_1B_2=\angle OB_1A_2$. Thus, triangles $OA_1B_2$ and $OA_2B_1$ are equal, as $OA_1=OB_1$ and $A_1B_2=BC_1=AC_1=A_2B_1$ and $\angle OB_1A_2=\angle OA_1B_2$. Hence, $OA_2=OB_2$.

Similarly, $OB_2=OC_2$, and so $O$ is the circumcenter of triangle $A_2B_2C_2,$ as desired.",644,660,Geometry,13
292,shl_jbmo_2022_g3,shl_jbmo,2022,g,"Let $ABC$ be an acute triangle such that $AH = HD$, where $H$ is the orthocenter of $ABC$ and $D \in BC$ is the foot of the altitude from the vertex $A$. Let $\ell$ denote the line through $H$ which is tangent to the circumcircle of the triangle $BHC$. Let $S$ and $T$ be the intersection points of $\ell$ with $AB$ and $AC$, respectively. Denote the midpoints of $BH$ and $CH$ by $M$ and $N$, respectively. Prove that the lines $SM$ and $TN$ are parallel.","Let $O,P$ be the circumcenter of $\triangle ABC$ and the midpoint of $BC$, respectively.

Let $AO$ meet $ST$ at $Q$. Then $\displaystyle OP=\frac{1}{2}AH=\frac{1}{2}HD$.

This implies $M,O,N$ are collinear and $PO\perp MN$.

It’s easy to show that $\square BSTC$ is cyclic by angles chasing. Hence, $\triangle ATS \sim \triangle ABC$ and, consequently, $AO\perp ST$.

Moreover $\triangle MPN\sim \triangle CHB$.
Hence, $\displaystyle \frac{SQ}{QT}=\frac{CD}{DB}=\frac{MO}{ON}$, i.e. $SM\parallel TN$ as desired.",456,511,Geometry,14
293,shl_jbmo_2022_g4,shl_jbmo,2022,g,"Given is an equilateral triangle $ABC$ and an arbitrary point, denoted by $E$, on the line segment $BC$. Let $l$ be the line through $A$ parallel to $BC$ and let $K$ be the point on $l$ such that $KE$ is perpendicular to $BC$. The circle with centre $K$ and radius $KE$ intersects the sides $AB$ and $AC$ at $M$ and $N$, respectively. The line perpendicular to $AB$ at $M$ intersects $l$ at $D$, and the line perpendicular to $AC$ at $N$ intersects $l$ at $F$. Show that the point of intersection of the angle bisectors of angles $MDA$ and $NFA$ belongs to the line $KE$.","We have the following main Claim.

Claim: Triangle $KMN$ is equilateral.
Proof: Note that

$2R_{AKM}=\dfrac{KM}{\sin \angle KAM}=\dfrac{KN}{\sin \angle KAN}=2R_{AKN},$

hence triangles $AKM$ and $AKN$ have the same circumradious, therefore $K \in (AMN),$ which implies that $\angle MKN=\angle MAN=60^\circ,$ as desired $\blacksquare$

Now, let $DM,FN$ intersect at point $S$. Then, $\angle AMS=\angle ANS=90^\circ,$ hence $S \in (AMN)$. Thus,

$\angle AKS=90^\circ=\angle AKE,$

implying that points $K,E,S$ are collinear. Now the finish is trivial, as triangle $DFS$ is isosceles ($\angle KSD=\angle KSM=\angle KNM=\angle KMN=\angle KSN=\angle KSF$), and $SK$ is its $S-$ altitude.",571,682,Geometry,15
294,shl_jbmo_2022_g5,shl_jbmo,2022,g,"Given is an acute angled triangle $ABC$ with orthocenter $H$ and circumcircle $k$. Let $\omega$ be the circle with diameter $AH$ and $P$ be the point of intersection of $\omega$ and $k$ other than $A$. Assume that $BP$ and $CP$ intersect $\omega$ for the second time at points $Q$ and $R$, respectively. If $D$ is the foot of the altitude from $A$ to $BC$ and $S$ is the point of the intersection of $\omega$ and $QD$, prove that $HR = HS$.","We have the following main Claim.

Claim: $\angle ADQ=90^\circ-\angle A$.
Proof: Let $\omega$ meet $AB$ again at $X$. Then,

$BP \cdot BQ=BX \cdot BA=BD \cdot BC,$

hence $QPDC$ is cyclic. Therefore,

$\angle ADQ=\angle QDC-90^\circ=\angle QPC-90^\circ=180^\circ-\angle a-90^\circ=90^\circ-\angle A,$

as desired $\blacksquare$

Back to the problem, if $PH$ meets $(ABC)$ again at $A'$, then $A'$ is the antipode of $A$ in $(ABC)$. Hence, we have

$\angle HAR=\angle HPR=\angle A'PC=\angle A'AC=90^\circ-\angle B$

and, by Claim 1,

$\angle HAS=\angle HQS=\angle HQS=\angle QHA-\angle ADQ=\angle QPA-\angle ADQ=\angle C-(90^\circ-\angle A)=90^\circ-\angle B,$

and so $\angle HAS=\angle HAR,$ which implies that $HS=HR$, as desired.",440,732,Geometry,16
295,shl_jbmo_2022_g6,shl_jbmo,2022,g,"Let $ABC$ be a right triangle with hypotenuse $BC$. The tangent to the circumcircle of triangle $ABC$ at $A$ intersects the line $BC$ at $T$. The points $D$ and $E$ are chosen so that $AD = BD, AE = CE,$ and $\angle CBD = \angle BCE < 90^{\circ}$. Prove that $D, E,$ and $T$ are collinear.","Let $\angle DBC=\angle ECB=x$. Note that

$\angle TBD=180^\circ-x$ and $\angle TAD=\angle TAB+\angle BAD=\angle C+\angle ABD=\angle B+\angle C-x=90^\circ-x$

and

$\angle TCE=x$ and $TAE=\angle TAC-\angle EAC=90^\circ+\angle C-\angle ECA=90^\circ+x,$

and so by applying LoS in triangles $TBD,TAD$ and $TEC,TEA$, we may infer that

$\dfrac{\sin \angle DTB}{\sin \angle DTA}=\dfrac{\sin x}{\sin(90^\circ-x)}=\dfrac{\sin x}{\sin(90^\circ+x)}=\dfrac{\sin \angle ETC}{\sin \angle ETA},$

hence if we denote $\angle DTB=a$, $\angle ETC=b$ and $\angle ATB=s$, we obtain

$\dfrac{\sin a}{\sin(s-a)}=\dfrac{\sin b}{\sin(s-b)},$

which, due to the monotonicity of the function $\dfrac{\sin x}{\sin(s-x)}$ implies that $a=b$, i.e. $T,D,E$ are collinear, as desired.",289,755,Geometry,17
296,shl_jbmo_2022_n1,shl_jbmo,2022,n,"Determine all pairs $(k, n)$ of positive integers that satisfy
$$1! + 2! + ... + k! = 1 + 2 + ... + n.$$","We take mod 7.
Claim 1: If $k \ge 7$, then, $1! + 2! + ... + k! \equiv 5\pmod{7}.$
Proof: $1! + 2! + ... + k! \equiv 1!+2!+3!+4!+5!+6! \equiv 5\pmod{7}.$
Also, we can see that, $\frac{n(n+1)}{2}\equiv 0, 1, 3, 6 \pmod 7$, is always true, for $k \ge 7$, now, we have contradiction, hence, $k \le 7.$
Now, if $k=1$, then $n=1$, if $k=2$, then $n=2$, if $k=3,4$, no solutions, if $k=5,$ then $n=17$, so $(k,n)=\boxed{(1,1), (2,2), (5,17)}.$",104,437,Number Theory,18
297,shl_jbmo_2022_n2,shl_jbmo,2022,n,"Let $a < b < c < d < e$ be positive integers. Prove that
$$\frac{1}{[a, b]} + \frac{1}{[b, c]} + \frac{1}{[c, d]} + \frac{2}{[d, e]} \le 1$$where $[x, y]$ is the least common multiple of $x$ and $y$ (e.g., $[6, 10] = 30$). When does equality hold?","Alternatively, we may note that $[x,y]=\dfrac{xy}{(x,y)} \leq \dfrac{xy}{x-y}$ and $[x,y] \geq 2y$ for all $x>y$, hence

$\dfrac{1}{[a,b]}+\dfrac{1}{[b,c]}+\dfrac{1}{[c,d]}+\dfrac{2}{[d,e]} \leq (\dfrac{1}{a}-\dfrac{1}{b})+(\dfrac{1}{b}-\dfrac{1}{c})+(\dfrac{1}{c}-\dfrac{1}{d})+\dfrac{2}{2d}=\dfrac{1}{a} \leq 1,$

as desired. Eqality holds if $a=1$ and $(a,b)=b-a, (b,c)=c-b,(c,d)=d-c$ and $[d,e]=2d$. The last equality can only hold for $e=2d.$ Moreover, $b-1=(b,a)=(a,b)=(1,b)=1$, i.e. $b=2$. Now, if $c$ is odd then $c=3$, while if $c$ is even then $c=4$. If $c=3$ then $(3,d)=d-3$ and so $d \in \{4,6 \},$ while if $c=4$ then $(4,d)=d-4$ and so $d \in \{5,6, 8 \}$.

Hence, there is a total of $5$ triples attaining equality:

$(a,b,c,d,e)=(1,2,3,4,8),(1,2,3,6,12),(1,2,4,5,10),(1,2,4,6,12),(1,2,4,8,16)$",247,810,Number Theory,19
298,shl_jbmo_2022_n3,shl_jbmo,2022,n,"Find all quadruples of positive integers $(p, q, a, b)$, where $p$ and $q$ are prime numbers and $a > 1$, such that$$p^a = 1 + 5q^b.$$","$\pmod 2$ we get that either $p=2$ or $q=2$
1) $p=2$
$2^a\equiv 1+5q^b\equiv 1\pmod 5 \implies a\equiv 0\pmod 4$
Now we have $(2^{\frac a2}-1)(2^{\frac a2}+1)=5q^b$.
$GCD(2^{\frac a2}-1,2^{\frac a2}+1)=1$, thus 5 divides exactly one of the two multiples and one of the multiples is divisible by $q$.
One of the multiples is 5 and the other one is $q^b$ because there can't be a multiple equal to one since $2^a-1>1$.
If $b=1\implies 2^{\frac a2}+1=5$ and $2^{\frac a2}-1 = q^b\implies (a,b,p,q)=(4,1,2,3)$
If $b\geq2\implies q^b\geq9>5+2$ contradiction because the difference of the multiples is not two.



2) $q=2$
$p^a-1=5\cdot 2^b$
$(p-1)(p^{a-1}+p^{a-2}+p^{a-3}+\cdot \cdot \cdot+p+1)=5\cdot 2^b$

2.1) $a$ is odd $\implies a\geq3$ and $(p^{a-1}+p^{a-2}+p^{a-3}+\cdot \cdot \cdot+p+1)$ is odd $\implies 5\geq(p^{a-1}+p^{a-2}+p^{a-3}+\cdot \cdot \cdot+p+1)\geq(p^2+p+1)=13$ contradiction


2.2) $a$ is even
$(p^{\frac a2}-1)(p^{\frac a2}+1)=5\cdot 2^b$
$GCD(p^{\frac a2}-1,p^{\frac a2}+1)=2$, thus 5 divides exactly one of the two multiples and one of the multiples is divisible by 2 but not 4.
One of the multiples is 10 and the other one 2^{b-1} because $p^a-1>2$

2.2.1)$p^{\frac a2}-1=10$ contradiction $p
otin \mathbb{N}$

2.2.2)$p^{\frac a2}+1=10 \implies p=3$ and $a=4$
$3^4=1+5\cdot 2^b \implies b=4$

$(a,b,p,q)=(4,4,3,2)$",134,1335,Number Theory,20
299,shl_jbmo_2022_n4,shl_jbmo,2022,n,"Consider the sequence $u_0, u_1, u_2, ...$ defined by $u_0 = 0, u_1 = 1,$ and $u_n = 6u_{n - 1} + 7u_{n - 2}$ for $n \ge 2$. Show that there are no non-negative integers $a, b, c, n$ such that
$$ab(a + b)(a^2 + ab + b^2) = c^{2022} + 42 = u_n.$$","We find that the characteristic equation of the recursion is $\lambda^2-6\lambda - 7 =0$ with roots $\lambda = -7$ and $\lambda=1$. This together with $u_0=0$ and $u_1=1$ gives
\[
u_n = \frac{7^n-(-1)^n}{8}.
\]Assume the contrary that such $a,b,c,n$ exists. If $n$ is odd then $7^n-(-1)^n\equiv 8\pmod{16}$, so $u_n$ is odd. This yields $a,b,a+b$ to be all odd, a contradiction. Hence, $n$ is even and we have
\[
u_{2k} = \frac{7^{2k}-1}{8} = c^{2022}+42.
\]Taking both sides modulo $7$, we get $u_{2k}\equiv -1\pmod{7}$, so $c^{2022}\equiv -1\pmod{7}$. But it is clear $-1$ is not a quadratic residue modulo 7, yielding a contradiction.",245,637,Number Theory,21
300,shl_jbmo_2022_n5,shl_jbmo,2022,n,"Find all pairs $(a, p)$ of positive integers, where $p$ is a prime, such that for any pair of positive integers $m$ and $n$ the remainder obtained when $a^{2^n}$ is divided by $p^n$ is non-zero and equals the remainder obtained when $a^{2^m}$ is divided by $p^m$.","Note that the given condition implies that $p^n \mid a^{2^m-2^n}-1$ for all $m>n$. We have the following 2 Claims, which solve the problem.

Claim 1: No $p>2$ works.
Proof: Take $m=2$ and $n=1$ to obtain that $p \mid a^2-1$. Hence, since $p$ is odd we obtain, by LTE,

$v_p(a^{2^m-2^n}-1)=v_p((a^2)^{2^{m-1}-2^{n-1}}-1)=v_p(a^2-1)+v_p(2^{m-1}-2^{n-1})=v_p(a^2-1)+v_p(2^{m-n}-1)$

Therefore, $v_p(a^2-1)+v_p(2^{m-n}-1) \geq n$ for all $m>n$, which is a contradiction by taking $m=n+1$ and $n$ sufficiently large $\blacksquare$

Claim 2: $p=2$ works if and only if $a$ is odd.
Proof Note that if $a$ is even, then $a^{2^2} \equiv \pmod 2^2$ is equal to $0$, a contradiction. Hence, $a$ is odd. We claim that, in this case, $a^{2^n} \equiv 1 \pmod 2^n$ for all $n$. Indeed, by LTE,

$v_2(a^{2^n}-1)=v_2(a-1)+v_2(a+1)+v_2(2^n)-1 \geq 1+1+n-1=n+1>n,$

as desired $\blacksquare$

Hence, all solutions are $(a,p)=(2k+1,2)$ for some $k \geq 0$.",263,936,Number Theory,22
301,shl_jbmo_2022_n6,shl_jbmo,2022,n,Find all positive integers $n$ for which there exists an integer multiple of $2022$ such that the sum of the squares of its digits is equal to $n$.,"Solution. We will show that $n
e 1, 2, 4, 7$. For any other positive integer $n$, there exists an integer multiple of $2022$ such that the sum of the squares of its digits is equal to $n$.

We first note that the prime factorization of $2022$ is $2022=2\cdot 3\cdot 337$. We have

$n
e1$. It is $1=1^2$. The sum of the squares of the digits of a positive integer $m$ is equal to $1$ if and only if $m$ is equal to a power of $10$. Then the sum of the digits of $m$ equals $1$, which means that $m$ is not a multiple of $3$. Hence $m$ cannot be a multiple of $2022$ either.

* $n
e 2$. It is $2=1^2+1^2$. The sum of the squares of the digits of a positive integer $m$ is equal to $2$ if and only if $m$ is equal to the sum of two distinct powers of $10$. Then the sum of the digits of $m$ equals $2$, which means that $m$ is not a multiple of $3$. Hence $m$ cannot be a multiple of $2022$ either.

* $n=3$. This value is feasible since, for example, it is $5\cdot 2022=10110$.

* $n
e 4$. It is $4=1^2+1^2+1^2+1^2$ and $4=2^2$. The sum of the squares of the digits of a positive integer $m$ is equal to $4$ if and only if $m$ is equal to either the sum of four distinct powers of $10$ or twice a power of $10$. Then the sum of the digits of $m$ equals $4$ or $2$, respectively, which means that $m$ is not a multiple of $3$. Hence $m$ cannot be a multiple of $2022$ either.

* $n=5$. This is the most interesting case. The only ways to express $5$ as a sum of the squares of positive integers (that are smaller than $10$) are $5=1^2+1^2+1^2+1^2+1^2$ and $5=2^2+1^2$. The sum of the squares of the digits of a positive integer $m$ is equal to $5$ if and only if $m$ is equal to either the sum of five distinct powers of $10$ or the sum of twice a power of $10$ with a distinct power of $10$.

In the first case, as above, we cannot obtain a multiple of $3$. So, we examine the second case. For instance, we wish to obtain an integer multiple of $2022$ which is of the form $200\dots 01..0$ or of the form $100\dots 02..0$.

Let's consider the latter one. Every integer of the form $100\dots 02..0$ is even and also a multiple of $3$. Thus it suffices to obtain a positive integer $k$ such that $10^k\equiv -2 \pmod{337}$.
We find that $10^{8}\equiv -32\equiv-2^5 \pmod{337}$ and we observe that
\[
2^{21}=\left(2^{10}
ight)^2\cdot 2 \equiv 13^2\cdot 2\equiv 1 \pmod{337}.
\]We may proceed in two ways:

(1st way) It is $10^{136}\equiv -2\pmod{337}$. Indeed, we have
\[
10^{136}=\left(10^8
ight)^{17}\equiv \left(-2^{5}
ight)^{17}\equiv -2^{85}\equiv -\left(2^{21}
ight)^4\cdot 2\equiv -2\pmod{337}.
\](2nd way) We have
\begin{align}
otag
10^{48}=\left(10^8
ight)^6&\equiv \left(-2^{5}
ight)^6\equiv 2^{30}\equiv 2^9\equiv 175\pmod{337},\\
otag 
10^{112}=\left(10^8
ight)^{14}&\equiv \left(-2^5
ight)^{14}\equiv 2^{70}\equiv 2^7 \equiv 128 \pmod{337}, \text{ and}\\
otag
10^{168}=\left(10^8
ight)^{21}&\equiv \left(-2^5
ight)^{21}\equiv -\left(2^{21}
ight)^5\equiv -1 \pmod{337}.
otag
\end{align}
Since $\varphi (337)=336=2^4\cdot 3\cdot 7$, $10^{\varphi (337)/7}=10^{48}
ot\equiv 1 \pmod{337}$, $10^{\varphi (337)/3}=10^{112}
ot\equiv 1 \pmod{337}$ and $10^{\varphi (337)/2}=10^{168}
ot\equiv 1 \pmod{337}$, it follows that $10$ is a primitive root of $337$. Hence there exists a positive integer $k$ such that $10^k\equiv -2 \pmod{337}$.

Note that $10^{136}+2$ (or $10^k+2$ in our 2nd way) is a multiple of $2$, $3$, and $337$, and so, it is a multiple of $2022$. The sum of the squares of its digits is equal to $1^2+2^2=5$, as desired.

* $n=6$. This value is feasible since, for example, we could start with $10110$ and append one more copy of it. It is ${\color{red}{10110}}{\color{blue}{10110}}=500005\cdot 2022$.

* $n
e 7$. The only ways to express $7$ as a sum of the squares of positive integers (that are smaller than $10$) are $7=1^2+1^2+1^2+1^2+1^2+1^2+1^2$ and $7=2^2+1^2+1^2+1^2$. The sum of the squares of the digits of a positive integer $m$ is equal to $7$ if and only if $m$ is equal to either the sum of seven distinct powers of $10$ or the sum of twice a power of $10$ with three distinct powers of $10$. Then the sum of the digits of $m$ equals $7$ or $5$, respectively, which means that $m$ is not a multiple of $3$. Hence $m$ cannot be a multiple of $2022$ either.

* $n\geq 8$. We will imitate what we did with $n=6$. Every positive integer $n\geq 8$ can be written in the form $3x+5y$ for some non-negative integer $x$, and $y=0$ or $1$ or $2$, depending on whether $n\equiv 0\pmod{3}$ or $n\equiv 2\pmod{3}$ or $n\equiv 1\pmod{3}$, respectively.

If we write the number $10110$ $x$ times next to each other, followed by $y$ times the number $10^{136}+2=1000...02$, then we obtain an integer multiple of $2002$ such that the sum of the squares of its digits equals $3x+5y=n$, as desired.

The proof is complete.

Comments:
(a) For the case where $n=5$, one could show that there exists a positive integer $\mu$ such that $10^\mu\equiv -5\pmod{337}$.
Indeed, it is
\[
10^{201}=\left(10^8
ight)^{25}\cdot 10 \equiv \left(-2^{5}
ight)^{25}\cdot 10 \equiv -2^{125}\cdot 10\equiv -\left(2^{21}
ight)^6\cdot 5\equiv -5\pmod{337}.
\]Otherwise, since $10$ is a primitive root of $337$, there exists a positive integer $\mu$ such that $10^\mu\equiv -5 \pmod{337}$.

Since the sum of the digits of $10^\mu+5$ is $6$, it follows that $\dfrac{10^\mu+5}{3}$ is an integer multiple of $ 337$. Hence
\[
2\left(10^\mu+5
ight)=6\frac{10^\mu+5}{3}
\]is a multiple of $2022$, which is of the form $2000...010$. Moreover, the sum of the squares of its digits is equal to $2^2+1^2=5$.

(b) The smallest multiple of $2022$ such that the sum of the squares of its digits is equal to $6$ is equal to
$50005\cdot 2022=101110110$.",147,5729,Number Theory,23
377,tst_jbmo_ro_2022_1_p1,tst_jbmo,2022,n,"A natural number $n \geq 2$ is called \emph{square-free} if it is not divisible by any perfect square greater than 1.

Determine all square-free natural numbers $n \geq 2$ with the property that the number
\[
\frac{1}{d_1} + \frac{1}{d_2} + \cdots + \frac{1}{d_k}
\]
is a natural number, where $\{d_1, d_2, \ldots, d_k\}$ is the set of natural divisors of $n$.","Let $n = p_1 p_2 \cdots p_j$ be the prime factorization of $n$. If
\[
S = \frac{1}{d_1} + \frac{1}{d_2} + \cdots + \frac{1}{d_k},
\]
then
\[
S = \frac{1}{n}\left(1 + \sum p_1 + \sum p_1 p_2 + \cdots + \sum p_1 p_2 \cdots p_{j-1} + p_1 p_2 \cdots p_j
ight) = \frac{(1+p_1)(1+p_2)\cdots(1+p_j)}{p_1 p_2 \cdots p_j}.
\]
\pts{3}

Since $\gcd(p_j,\, 1+p_j) = 1$, it follows that $p_j$ divides the product $(1+p_1)(1+p_2)\cdots(1+p_{j-1})$. Being prime, $p_j$ divides one of the factors of this product. \pts{1}

Assume $p_1 < p_2 < \cdots < p_j$. We deduce that $p_j$ divides $1 + p_{j-1}$, so $p_j = 1 + p_{j-1}$. Therefore $j = 2$, $p_1 = 2$, and $p_2 = 3$. \pts{2}

The only number with the stated property is $n = 6$. \pts{1}",360,724,Number Theory,1
378,tst_jbmo_ro_2022_1_p2,tst_jbmo,2022,g,"Consider triangle $ABC$ with $\angle A = 30^\circ$ and $\angle B = 80^\circ$. On sides $AC$ and $BC$, consider points $D$ and $E$, respectively, such that $\angle ABD \equiv \angle DBC$ and $DE \parallel AB$. Determine the measure of angle $\angle EAC$.","Let $AM \perp BC$ and $AF$ be the angle bisector of angle $\angle MAC$, with $M$ and $F$ on side $BC$. Clearly $\angle BAM = \angle MAF = 10^\circ$, so $AM$ is both an altitude and an angle bisector in triangle $ABF$. We deduce $BM = MF$. \pts{1}

Applying the angle bisector theorem in triangle $ABC$: $\dfrac{CD}{DA} = \dfrac{BC}{AB}$. \pts{1}

Applying the angle bisector theorem in triangle $AMC$: $\dfrac{CF}{MF} = \dfrac{AC}{AM}$. It follows that $\dfrac{CF}{FB} = \dfrac{1}{2} \cdot \dfrac{CF}{MF} = \dfrac{1}{2} \cdot \dfrac{AC}{AM}$. \pts{1}

But
\[
\frac{1}{2} \cdot \frac{AC}{AM} = \frac{1}{2} \cdot \frac{AC \cdot AB}{AM \cdot AB} = \frac{AC \cdot AB \cdot \sin 30^\circ}{AM \cdot AB} = \frac{2S_{ABC}}{AM \cdot AB} = \frac{AM \cdot BC}{AM \cdot AB} = \frac{BC}{AB}.
\]
\pts{3}

Thus $\dfrac{CD}{DA} = \dfrac{CF}{FB}$, so lines $DF$ and $AB$ are parallel.

This means $E$ and $F$ coincide, hence $\angle EAC = 10^\circ$. \pts{1}",253,940,Geometry,2
379,tst_jbmo_ro_2022_1_p3,tst_jbmo,2022,c,"Find how many natural numbers $k \in \{1, 2, 3, \ldots, 2022\}$ have the property that, if 2022 real numbers are written on a circle such that the sum of any $k$ consecutive numbers is equal to 2022, then all 2022 numbers are equal.","Let $k \in \{1, 2, 3, \ldots, 2022\}$ have the stated property. Given $x_1, x_2, \ldots, x_{2022}$ on the circle, with the sum of any $k$ consecutive numbers equal to 2022, for any $n \in \mathbb{N}$, define $x_n = x_r$ where $r \in \{1, 2, \ldots, 2022\}$ and $n \equiv r \pmod{2022}$.

\textbf{(A)} For any $i \in \mathbb{N}$:
\[
x_i + x_{i+1} + \cdots + x_{i+k-1} = x_{i+1} + x_{i+2} + \cdots + x_{i+k} = 2022,
\]
so $x_i = x_{i+k}$. \pts{2}

\textbf{(B)} If $\gcd(2022, k) = d \geq 2$, set $x = \dfrac{2022d}{k}$ and write on the circle, in order, the numbers
\[
\underbrace{x, 0, 0, \ldots, 0}_{d \text{ terms}},\; \underbrace{x, 0, 0, \ldots, 0}_{d \text{ terms}},\; \ldots,\; \underbrace{x, 0, 0, \ldots, 0}_{d \text{ terms}}.
\]
The sum of any $k$ consecutive numbers is 2022, yet the numbers are not all equal.

Therefore, values of $k$ for which $\gcd(2022, k) 
eq 1$ are not solutions. \pts{2}

\textbf{(C)} We show that all $k \in \{1, 2, \ldots, 2022\}$ with $\gcd(2022, k) = 1$ are solutions. Since $\gcd(k, 2022) = 1$, there exist $u, v \in \mathbb{N}$ such that $k \cdot u = 2022v + 1$. Then for any $m \in \mathbb{N}$:
\[
x_m = x_{m+ku} = x_{m+2022v+1} = x_{m+1},
\]
so all 2022 numbers are equal. \pts{2}

\textbf{(D)} In total there are $\varphi(2022) = 672$ solutions. \pts{1}",232,1296,Combinatorics,3
380,tst_jbmo_ro_2022_1_p4,tst_jbmo,2022,g,"Consider the acute triangle $ABC$ ($AB < AC$) with altitudes $AD$, $BE$, $CF$, where $D \in BC$, $E \in CA$, $F \in AB$. Denote by $M$ the midpoint of side $BC$ and by $H$ the orthocenter of triangle $ABC$.

Let $X$ be the point where the circle with diameter $MH$ meets line $AM$ for the second time, and let $T$ be the intersection of lines $HX$ and $BC$.

Prove that the circumscribed circles of triangles $TFD$ and $AEF$ are tangent.","Let $\omega_1$ be the circle with diameter $HM$.

Then $\angle HXM = 90^\circ$, so $\angle AXH = 90^\circ = \angle AFH = \angle AEH$. It follows that $E$, $F$, and $X$ lie on the circle $\omega_2$ with diameter $AH$. \pts{1}

Since $MC = MF$ (as $M$ is the midpoint of $BC$ and $F$ is the foot of the altitude, so actually $MC = MB$), we have $\angle MFC = \angle MCF = 90^\circ - \angle B = \angle BAD$. Therefore line $MF$ is tangent to circle $\omega_2$. \pts{2}

Let $\omega_3$ be the circle with diameter $BC$. The radical axis of $\omega_1$ and $\omega_2$ is line $HX$; the radical axis of $\omega_1$ and $\omega_3$ is line $BC$; and the radical axis of $\omega_2$ and $\omega_3$ is line $EF$. These three radical axes are either parallel or concurrent. Since $HX \cap BC = \{T\}$, it follows that $T \in EF$. \pts{2}

Then:
\[
\angle FTD = 180^\circ - \angle ABT - \angle BFT = 180^\circ - (180^\circ - \angle B) - \angle AFE = \angle B - \angle C = \angle MFB - \angle DFB = \angle MFD.
\]
We deduce that line $MF$ is tangent to the circumscribed circle of triangle $TFD$, and from this the conclusion follows. \pts{2}",437,1126,Geometry,4
381,tst_jbmo_ro_2022_1_p5,tst_jbmo,2022,a,"We say that a set $A \subset \mathbb{R}$ with at least three elements is \emph{free of arithmetic progressions} if for any distinct $a, b, c \in A$, we have $a + b 
eq 2c$.

Show that the set $\{0, 1, 2, \ldots, 3^8 - 1\}$ contains a subset $A$ with at least 256 elements that is free of arithmetic progressions.","We have $6560 = 3^8 - 1$.

Every number in the set $\{0, 1, 2, \ldots, 6560\}$ can be written in base 3 using 8 digits (0, 1, or 2). Define the set $A$ as the set of numbers whose base-3 representation uses only the digits 0 and 1. There are $2^8 = 256$ such numbers, and we show that $A$ is free of arithmetic progressions. \pts{2}

If, for contradiction, there existed $a, b, c \in A$ such that $a + b = 2c$, write
\[
a = \overline{a_7 a_6 a_5 \cdots a_1 a_0}_{(3)}, \quad
b = \overline{b_7 b_6 b_5 \cdots b_1 b_0}_{(3)}, \quad
c = \overline{c_7 c_6 c_5 \cdots c_1 c_0}_{(3)}
\]
with $a_i, b_i, c_i \in \{0,1\}$ for $i = 0, \ldots, 7$. The digits of $a + b$ are $a_7+b_7, a_6+b_6, \ldots, a_0+b_0$ and those of $2c$ are $2c_7, 2c_6, \ldots, 2c_0$. Equality holds if and only if
\[
a_i + b_i = 2c_i \quad (\text{even}), \quad i = 0, \ldots, 7,
\]
which forces $a_i = b_i = 1$ or $a_i = b_i = 0$, i.e.\ $a = b$ --- a contradiction. \pts{5}

\bigskip
\textit{Alternative solution.} We prove by mathematical induction on $n \geq 2$: the set $\{0, 1, 2, \ldots, 3^n - 1\}$ contains a subset with $2^n$ elements that is free of arithmetic progressions.

\textit{Base case.} For $n = 2$, the subset $A = \{1, 2, 4, 5\}$ of $\{0, 1, \ldots, 8\}$ is free of arithmetic progressions. \pts{1}

\textit{Inductive step.} Assume the statement holds for $n$; we prove it for $n+1$.

Let $X = \{a_1, \ldots, a_{2^n}\} \subset \{0, 1, \ldots, 3^n - 1\}$ be a subset free of arithmetic progressions, with $a_1 < \cdots < a_{2^n}$.

We show that
\[
A = \{a_1, \ldots, a_{2^n},\; a_1 + 2 \cdot 3^n,\; \ldots,\; a_{2^n} + 2 \cdot 3^n\}
\]
is a subset of $\{0, 1, \ldots, 3^{n+1}-1\}$ free of arithmetic progressions, of size $2^{n+1}$.

That $A \subset \{0,\ldots,3^{n+1}-1\}$ follows from $a_i + 2 \cdot 3^n < 3^n - 1 + 2\cdot 3^n = 3^{n+1}-1$. It has $2^{n+1}$ elements since for $i < j$: $a_i < a_j < a_i + 2\cdot 3^n < a_j + 2\cdot 3^n$. Moreover, $A \cap \{3^n, \ldots, 2\cdot 3^n - 1\} = \emptyset$ $(*)$ and $A \setminus X$ is free of arithmetic progressions.

The fact that $A$ is free of arithmetic progressions follows from:
\begin{enumerate}
  \item The arithmetic mean of $a_i$ and $a_j$ ($i 
eq j$) is not in $A$, since $X$ is free of arithmetic progressions.
  \item For any $i, j$ (possibly equal):
  \[
  3^n = \frac{0 + 2\cdot 3^n}{2} \leq \frac{a_i + (a_j + 2\cdot 3^n)}{2} \leq \frac{(3^n-1) + (3^n-1) + 2\cdot 3^n}{2} = 2\cdot 3^n - 1,
  \]
  so by $(*)$, the arithmetic mean of $a_i$ and $a_j + 2\cdot 3^n$ does not belong to $A$.
  \item For $i 
eq j$, the arithmetic mean of $a_i + 2\cdot 3^n$ and $a_j + 2\cdot 3^n$ satisfies
  \[
  \frac{(a_i + 2\cdot 3^n) + (a_j + 2\cdot 3^n)}{2} \geq 2 \cdot 3^n,
  \]
  so it does not lie in $X$; since $A \setminus X$ is free of arithmetic progressions, the conclusion follows.
\end{enumerate}
The induction is complete, and in particular for $n = 8$ we obtain the solution. \pts{6}",312,2926,Algebra,5
382,tst_jbmo_ro_2022_2_p1,tst_jbmo,2022,g,"Let $M$, $N$ and $P$ be the midpoints of sides $BC$, $CA$, and $AB$, respectively, of the acute triangle $ABC$. Denote by $A'$, $B'$, and $C'$ the antipodal points of vertices $A$, $B$, and $C$ in the circumscribed circle of triangle $ABC$. On the open segments $MA'$, $NB'$, and $PC'$, consider points $X$, $Y$, and $Z$, respectively, such that
\[
\frac{MX}{XA'} = \frac{NY}{YB'} = \frac{PZ}{ZC'}.
\]
\begin{enumerate}[label=\alph*)]
  \item Prove that lines $AX$, $BY$, and $CZ$ are concurrent at a point $S$.
  \item Show that $OS < OG$, where $O$ is the circumcenter and $G$ is the centroid of triangle $ABC$.
\end{enumerate}","a) Let $H$ be the orthocenter of triangle $ABC$. The quadrilateral $BHCA'$ is a parallelogram ($BH$ and $A'C$ are both perpendicular to line $AC$, and $CH$ and $A'B$ are both perpendicular to line $AB$), so $M$ is the midpoint of segment $HA'$. \pts{1}

If $\dfrac{MX}{XA'} = k$, then
\[
\frac{A'X}{XM} = \frac{1}{k} \Rightarrow \frac{A'X}{A'M} = \frac{1}{k+1} \Rightarrow \frac{A'X}{A'H} = \frac{1}{2k+2} \Rightarrow \frac{A'X}{XH} = \frac{1}{2k+1}.
\]
\pts{1}

Let $O$ be the circumcenter and $S$ the point where line $AX$ meets the Euler line $OH$. Applying Menelaus' theorem in triangle $HOA'$ with transversal $X - S - A$:
\[
\frac{HS}{SO} \cdot \frac{OA}{AA'} \cdot \frac{A'X}{XH} = 1 \Rightarrow \frac{HS}{SO} \cdot \frac{1}{2} \cdot \frac{1}{2k+1} = 1 \Rightarrow \frac{HS}{SO} = 4k+2.
\]
\pts{1}

Considering the intersection points of the Euler line with lines $BY$ and $CZ$ respectively, we find that they divide segment $HO$ in the same ratio $4k + 2$, so they coincide with $S$. Therefore, lines $AX$, $BY$ and $CZ$ are concurrent (and their common point lies on the Euler line). \pts{2}

b) The ratio $k$ ranges over $(0, \infty)$, so the ratio $\dfrac{HS}{SO} = 4k+2$ ranges over $(2, \infty)$. \pts{1}

We obtain $HS > 2\,OS$, hence $OS < \dfrac{OH}{3} = OG$. \pts{1}

\bigskip
\textit{Alternative solution to part a.}

$ABA'B'$ is a rectangle, so $A'B' = AB \overset{\text{def}}{=} c$. $MN$ is a midline of triangle $ABC$, so $MN \parallel AB \parallel A'B'$; hence $MNB'A'$ is a trapezoid. Since $\dfrac{MX}{XA'} = \dfrac{NY}{YB'}$, it follows that $XY \parallel MN \parallel A'B'$. \pts{1}

Let $E \in A'B'$ such that $ME \parallel NB'$ and $\{F\} = ME \cap XY$. Since $MNB'E$ is a parallelogram and $FY \parallel MN$, we have $EB' = FY = MN = \dfrac{c}{2}$. From the similarity of triangles $MXF$ and $MA'E$ we get $\dfrac{XF}{A'E} = \dfrac{MX}{MA'} \overset{\text{def}}{=} t$, so $XF = t \cdot \dfrac{c}{2}$ and $XY = XF + FY = \dfrac{(t+1)c}{2}$. \pts{2}

Since $XY \parallel A'B' \parallel AB$, the quadrilateral $ABYX$ is a trapezoid. Let $\{S\} = AX \cap BY$. From the similarity of triangles $ASB$ and $XSY$:
\[
\frac{SX}{SA} = \frac{SY}{SB} = \frac{XY}{AB} = \frac{t+1}{2}.
\]
\pts{1}

Analogously, the quadrilateral $BCYZ$ is a trapezoid. Let $\{S'\} = BY \cap CZ$. As before, $\dfrac{S'Y}{S'B} = \dfrac{t+1}{2}$, so $S = S'$, and therefore lines $AX$, $BY$ and $CZ$ are concurrent. \pts{1}",629,2418,Geometry,6
383,tst_jbmo_ro_2022_2_p2,tst_jbmo,2022,n,"Determine the largest natural number $n$ for which the following statement is true:

\textit{There exist $n$ distinct positive natural numbers $x_1, x_2, \ldots, x_n$ with the property that for any $a_1, a_2, \ldots, a_n \in \{-1, 0, 1\}$, not all zero, the number $n^3$ does not divide $a_1 x_1 + a_2 x_2 + \cdots + a_n x_n$.}","For $n = 9$, choose $x_1 = 2^0, x_2 = 2^1, \ldots, x_9 = 2^8$. For any $a_1, \ldots, a_9 \in \{-1, 0, 1\}$:
\[
|a_1 x_1 + \cdots + a_n x_n| \leq 1 + 2 + \cdots + 2^8 = 2^9 - 1 < 9^3.
\]
If $9^3$ divides $a_1 \cdot 1 + a_2 \cdot 2 + \cdots + a_9 \cdot 2^8 = 0$. By parity, $a_1 = 0$; dividing by 2 and repeating the parity argument gives $a_2 = 0$, etc. We obtain that all $a_i$ are zero, a contradiction. So $n = 9$ satisfies the statement. \pts{3}

If $n \geq 10$, then $2^n > n^3$. Let $A = \{x_1, \ldots, x_n\}$ be a set of $n$ distinct positive natural numbers, and $\mathcal{P}(A)$ its power set. Since $|\mathcal{P}(A)| = 2^n > n^3$, by the Pigeonhole Principle there exist two distinct subsets $B$ and $C$ of $A$ such that
\[
\sum_{x \in B} x \equiv \sum_{x \in C} x \pmod{n^3}.
\]
Setting $a_i = 1$ for elements of $B \setminus C$, $a_j = -1$ for elements of $C \setminus B$, and $a_k = 0$ otherwise, we find a combination $a_1 x_1 + \cdots + a_n x_n$ divisible by $n^3$. So numbers $n \geq 10$ do not satisfy the statement, and the answer is $n = 9$. \pts{4}

\textit{Observations.}
\begin{enumerate}
  \item[(O1)] Simply writing a correct example for $n = 9$ without justification earns no points.
  \item[(O2)] Failing to prove $2^n > n^3$ for $n \geq 10$ results in a deduction of one point from the 4 available.
\end{enumerate}",327,1340,Number Theory,7
384,tst_jbmo_ro_2022_2_p3,tst_jbmo,2022,c,"Consider a grid of 49 points, which are the vertices of the 36 unit squares into which a $6 \times 6$ square is decomposed.

We say that a square with vertices at the grid points is \emph{good} if its sides and diagonals do not lie on the sides of the grid squares.
\begin{enumerate}[label=\alph*)]
  \item Find the number of good squares that can be formed with the grid vertices.
  \item Show that there exist two disjoint, non-congruent good squares such that the minimum distance between their points is $\dfrac{\sqrt{5}}{5}$.
\end{enumerate}","a) We call a square \emph{normal} if its vertices are at grid points and its sides lie on grid lines parallel to the sides of the $6 \times 6$ square (or on the sides of the $6 \times 6$ square). Every good square has its vertices on the sides of a normal square, whose side length can be 3, 4, 5, or 6. Let a normal square have side length $x \in \{3, 4, 5, 6\}$. The side length of a good square inscribed in it has the form $\sqrt{a^2 + b^2}$, where $a, b \in \{1, 2, \ldots, x-1\}$, $a 
eq b$, and $a + b = x$.

Indeed, if $a = 0$ or $b = x$, then the sides of the good square would lie on grid lines; if $a = b$, the sides of the good square are parallel to the diagonals of the large square, so its diagonals would lie on the grid lines --- contradiction. \pts{1}

If $a, b \in \{1, 2, \ldots, x-1\}$ with $a 
eq b$ and $a + b = x$, each normal square of side $x$ contains exactly two good squares generated by the pair $(a, b)$, having side $\sqrt{a^2+b^2}$, with vertices on its sides. \pts{1}

For $x = 3 = 1 + 2$: in each normal $3 \times 3$ square there are exactly two good squares, generated by $(a,b) \in \{(1,2),(2,1)\}$. Thus in each $6 \times 3$ horizontal strip there are $2 \cdot 4 = 8$ good squares of side $\sqrt{5}$. Since there are 4 such strips of width 3, there are $8 \cdot 4 = 32$ good squares of side $\sqrt{5}$ in total.

Analogously:
\begin{itemize}
  \item 18 good squares of side $\sqrt{10}$, generated by $(a,b) \in \{(1,3),(3,1)\}$,
  \item 8 good squares of side $\sqrt{17}$, generated by $(a,b) \in \{(1,4),(4,1)\}$,
  \item 2 good squares of side $\sqrt{26}$, generated by $(a,b) \in \{(1,5),(5,1)\}$,
  \item 8 good squares of side $\sqrt{13}$, generated by $(a,b) \in \{(2,3),(3,2)\}$,
  \item 2 good squares of side $\sqrt{20}$, generated by $(a,b) \in \{(2,4),(4,2)\}$.
\end{itemize}
The total number of good squares is: $32 + 18 + 8 + 2 + 8 + 2 = \mathbf{70}$. \pts{3}

b) Choose good squares $ABCD$ and $MNPQ$ as in the figure, with $AB = \sqrt{5}$ and $MN = \sqrt{10}$, such that $A$, $D$, $M$ and $N$ lie on the sides of the large square.

Let $d$ be the line through $Q$ parallel to $BC$. Since $MNPQ$ is contained in the half-plane $[dM$, the sought distance is the distance from $Q$ to $ABCD$. Choose points $R$ and $F$ such that $Q$ is the midpoint of $BR$, triangle $BFQ$ is right-angled at $B$ with $BF = 2$, and $F$ and $M$ are on opposite sides of line $d$.

Since triangles $BQF$ and $RCB$ are congruent, lines $BC$ and $FQ$ are perpendicular, and the sought distance equals the length of segment $QE$, where $E$ is the intersection of lines $BC$ and $FQ$. By the leg theorem (geometric mean relation):
\[
EQ = \frac{\sqrt{5}}{5}.
\]
\pts{2}",546,2695,Combinatorics,8
385,tst_jbmo_ro_2022_2_p4,tst_jbmo,2022,a,"Let $a$, $b$, and $c$ be three positive real numbers with sum 3. Show that:
\[
\frac{ab}{ab+a+b} + \frac{bc}{bc+b+c} + \frac{ca}{ca+c+a} + \frac{1}{9}\left(\frac{(a-b)^2}{ab+a+b} + \frac{(b-c)^2}{bc+b+c} + \frac{(c-a)^2}{ca+c+a}
ight) \leq 1.
\]","\textit{Solution 1.} We have
\[
\frac{ab}{ab+a+b} + \frac{(a-b)^2}{9(ab+a+b)} = \frac{a^2+7ab+b^2}{9(ab+a+b)}.
\]
\pts{1}

We will show that
\[
\frac{a^2+7ab+b^2}{ab+a+b} \leq a + b + 1. \quad (1)
\]
This inequality is equivalent to
\[
a^2+7ab+b^2 \leq a^2+3ab+b^2+a^2b+ab^2+a+b \iff a^2b + ab^2 + a + b \geq 4ab.
\]
By AM-GM: $a^2b + b \geq 2ab$ and $ab^2 + a \geq 2ab$, so inequality (1) holds. \pts{5}

Writing the analogous inequalities and summing:
\[
\sum \frac{ab}{ab+a+b} + \frac{1}{9}\sum\frac{(a-b)^2}{ab+a+b} \leq \frac{1}{9}\sum(a+b+1) = 1.
\]
\pts{1}

\textit{Solution 2.} We have $ab + a + b = ab + (a+b) \cdot \dfrac{a+b+c}{3} = \dfrac{a^2+b^2+5ab+ac+bc}{3}$. \pts{1}

Therefore,
\begin{align*}
\frac{ab}{ab+a+b} + \frac{(a-b)^2}{9(ab+a+b)} - \frac{1}{3}
&= \frac{9ab + (a^2+b^2-2ab) - (a^2+b^2+5ab+ac+bc)}{9(ab+a+b)} \\
&= \frac{b(a-c)}{9(ab+a+b)} + \frac{a(b-c)}{9(ab+a+b)}.
\end{align*}
\pts{2}

Then
\begin{align*}
\sum \frac{ab}{ab+a+b} + \frac{1}{9}\sum \frac{(a-b)^2}{ab+a+b} - 1
&= \sum \left(\frac{b(a-c)}{9(ab+a+b)} + \frac{a(b-c)}{9(ab+a+b)}
ight) \\
&= \sum \left(\frac{b(a-c)}{9(ab+a+b)} + \frac{b(c-a)}{9(bc+b+c)}
ight) \\
&= \sum \frac{-b(a-c)^2(b+c)}{9(ab+a+b)(bc+b+c)} \leq 0.
\end{align*}
\pts{4}

\textit{Observation.} Equality holds for $a = b = c = 1$.",245,1288,Algebra,9
386,tst_jbmo_ro_2022_3_p1,tst_jbmo,2022,a,"Let $a \geq b \geq c \geq d$ be real numbers with the property that
\[
(a-b)(b-c)(c-d)(d-a) = -3.
\]
\begin{enumerate}[label=\alph*)]
  \item If $a + b + c + d = 6$, prove that $d < 0.36$.
  \item If $a^2 + b^2 + c^2 + d^2 = 14$, show that $(a+c)(b+d) \leq 8$. State the equality cases.
\end{enumerate}","\textit{Solution.} a) Let $a - b = x$, $b - c = y$, $c - d = z$. The relation from the hypothesis becomes $xyz(x+y+z) = 3$. Since $c = d+z$, $b = d+y+z$, $a = d+x+y+z$, we have $4d + 3z + 2y + x = 6$.

By the AM-GM inequality,
\[
x + 2y + 3z = \tfrac{1}{7}(x+y+z) + \tfrac{6}{7}x + \tfrac{13}{7}y + \tfrac{20}{7}z \geq \tfrac{4}{7}\sqrt[4]{1560\, xyz(x+y+z)} = \tfrac{4}{7}\sqrt[4]{4680}.
\]
\pts{3}

Therefore $4d \leq 6 - \dfrac{4}{7}\sqrt[4]{4680}$, from which
\[
d \leq \frac{3}{2} - \frac{1}{7}\sqrt[4]{4680} < \frac{3}{2} - \frac{1}{7}\sqrt[4]{4096} = \frac{3}{2} - \frac{8}{7} = \frac{5}{14} < \frac{9}{25} = 0.36.
\]
\pts{1}

b) We observe that
\begin{align*}
(a+c)(b+d) &= \left(a^2+b^2+c^2+d^2
ight) - \frac{1}{2}\left((a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2
ight) \\
&= 14 - \frac{1}{2}\left(x^2+y^2+z^2+(x+y+z)^2
ight).
\end{align*}

The inequality to be proved reduces to $x^2 + y^2 + z^2 + (x+y+z)^2 \geq 12$, \quad (1).

Since $x, y, z \geq 0$, we have $(xy + yz + zx)^2 \geq 3xyz(x+y+z)$ (Newton's inequality), from which $xy + yz + zx \geq 3$. Therefore,
\[
x^2+y^2+z^2+(x+y+z)^2 = 2(x^2+y^2+z^2) + 2(xy+yz+zx) \geq 4(xy+yz+zx) \geq 12,
\]
i.e.\ inequality (1). \pts{2}

Equality holds if and only if $x = y = z = 1$, so $a = 3+d$, $b = 2+d$, $c = 1+d$. We have
\[
a^2+b^2+c^2+d^2 = 14 \iff 14 + 12d + 4d^2 = 14 \iff d^2 + 3d = 0 \iff d \in \{0, -3\}.
\]
The resulting 4-tuples are: $(a,b,c,d) \in \{(3,2,1,0),\, (0,-1,-2,-3)\}$. \pts{1}",302,1446,Algebra,10
387,tst_jbmo_ro_2022_3_p2,tst_jbmo,2022,g,"Consider two circles $\mathcal{C}_1$ and $\mathcal{C}_2$ internally tangent at the point $P$ (circle $\mathcal{C}_2$ is interior to circle $\mathcal{C}_1$).

A chord $AB$ of circle $\mathcal{C}_1$ is tangent to circle $\mathcal{C}_2$ at point $C$. Let $D$ be the second intersection of line $CP$ with circle $\mathcal{C}_1$. A tangent drawn from $D$ to $\mathcal{C}_2$ intersects circle $\mathcal{C}_1$ again at $E$ and circle $\mathcal{C}_2$ at $F$.

Show that $F$ is the center of the circle inscribed in triangle $ABE$.","\textit{Solution.}
Let $O_1$ and $O_2$ be the centers of circles $\mathcal{C}_1$ and $\mathcal{C}_2$$. Triangles $O_2PC$ and $O_1PD$ are isosceles (with bases $PC$ and $PD$) and share the angle $\angle O_2PC$. Therefore $\angle O_2PC \equiv \angle O_1DC$, so $O_2C \parallel O_1D$ and, since $O_2C \perp AB$, we get $O_1D \perp AB$.

We deduce that $D$ is the midpoint of arc $AB$ not containing $P$. \pts{2}

Since $\angle APD = \tfrac{1}{2}m(\widehat{AD}) = \tfrac{1}{2}m(\widehat{DB}) = \angle DAC$, it follows that $\triangle APD \sim \triangle CAD$, from which $DA^2 = DC \cdot DP$. \pts{2}

But $DC \cdot DP = DF^2$ (the power of point $D$ with respect to circle $\mathcal{C}_2$), so $DA = DF$. Since $ED$ is the angle bisector of angle $AEB$, we deduce that $F$ is the center of the inscribed circle of triangle $ABE$. \pts{3}",522,833,Geometry,11
388,tst_jbmo_ro_2022_3_p3,tst_jbmo,2022,n,"Let $p_i$ ($i \in \mathbb{N}^*$) be the $i$-th prime number. For each positive natural number $k$, denote by $a_k$ the number of positive natural numbers $i$ with the property that the product $p_i p_{i+1}$ divides $k$.

If $n$ is a positive natural number, show that
\[
a_1 + a_2 + \cdots + a_n < \frac{n}{3}.
\]","\textit{Solution.}
Let $l$ be maximal with the property that $p_l \leq n$ (so $n < p_{l+1}$).

We observe that $a_1 + \cdots + a_n$ represents the number of pairs of positive natural numbers $(k, i)$ with $1 \leq k \leq n$, $1 \leq i \leq l-1$, such that $p_i p_{i+1} \mid k$.

Let us count these pairs differently: for a fixed $i$, the number of pairs $(k, i)$ represents the number of multiples of $p_i p_{i+1}$ that are at most $n$. This number equals $\left\lfloor \dfrac{n}{p_i p_{i+1}} 
ight
floor$.

It follows that
\[
a_1 + \cdots + a_n = \left\lfloor \frac{n}{p_1 p_2} 
ight
floor + \left\lfloor \frac{n}{p_2 p_3} 
ight
floor + \cdots + \left\lfloor \frac{n}{p_{l-1} p_l} 
ight
floor,
\]
and since $\lfloor x 
floor \leq x$ for any real $x$,
\[
a_1 + \cdots + a_n \leq \frac{n}{p_1 p_2} + \frac{n}{p_2 p_3} + \cdots + \frac{n}{p_{l-1} p_l}.
\]
\pts{3}

For $i \geq 2$ we have $p_{i+1} \geq p_i + 2$, so $p_i \geq p_2 + 2(i-2) = 2i - 1$ for all $i \geq 2$. Therefore,
\begin{align*}
\frac{n}{p_1 p_2} + \cdots + \frac{n}{p_{l-1}p_l}
&\leq \frac{n}{2 \cdot 3} + \frac{n}{3 \cdot 5} + \frac{n}{5 \cdot 7} + \cdots + \frac{n}{(2l-3)(2l-1)} \\
&= n\left(\frac{1}{2} - \frac{1}{3} + \frac{1}{2}\left(\frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \cdots + \frac{1}{2l-3} - \frac{1}{2l-1}
ight)
ight) \\
&= n\left(\frac{1}{2} - \frac{1}{3} + \frac{1}{6} - \frac{1}{4l-2}
ight) \\
&< n\left(\frac{1}{2} - \frac{1}{3} + \frac{1}{6}
ight) = \frac{n}{3},
\end{align*}
which completes the proof. \pts{4}",313,1509,Number Theory,12
389,tst_jbmo_ro_2022_3_p4,tst_jbmo,2022,c,"Let $n \geq 2$ be a natural number and $M = \mathbb{N} \times \mathbb{N} \times \cdots \times \mathbb{N}$ ($n$ copies of $\mathbb{N}$) be the set of $n$-tuples of natural numbers. For each $a = (a_1, a_2, \ldots, a_n) \in M$, denote by $d_a$ the number of pairs $(i, j)$, $i, j \in \{1, 2, \ldots, n\}$, for which $a_i - a_j = 1$.

Determine the maximum value of $d_a$ as $a$ ranges over $M$.","\textit{Solution.}
Let $a = (a_1, a_2, \ldots, a_n) \in M$. If we sort $a_1, a_2, \ldots, a_n$ in increasing order, or subtract the same natural number from all of them, the value of $d_a$ does not change. We may therefore assume $0 = a_1 \leq a_2 \leq \cdots \leq a_n$.

If there exists an index $i$ for which $a_{i+1} - a_i \geq 2$, decreasing $a_{i+1}, a_{i+2}, \ldots, a_n$ by the same amount so that the new values at positions $i$ and $i+1$ differ by 1, increases the value of $d_a$. \pts{1}

Thus, $d_a$ is maximal for an $n$-tuple of the form
\[
\alpha = (\underbrace{0,\ldots,0}_{x_0},\, \underbrace{1,\ldots,1}_{x_1},\, \underbrace{2,\ldots,2}_{x_2},\, \ldots,\, \underbrace{k-1,\ldots,k-1}_{x_{k-1}},\, \underbrace{k,\ldots,k}_{x_k}),
\]
where $k, x_0, x_1, \ldots, x_k \in \mathbb{N}^*$ and $x_0 + x_1 + \cdots + x_k = n$. For such an $\alpha$:
\[
d_\alpha = x_0 x_1 + x_1 x_2 + \cdots + x_{k-1} x_k.
\]
\pts{1}

If $k \geq 3$, consider the $n$-tuple $\beta$ with $x_0$ zeros, $x_1$ ones, \ldots, $x_{k-3}$ copies of $k-3$, $(x_{k-2} + x_k)$ copies of $k-2$, and $x_{k-1}$ copies of $k-1$ (i.e.\ we replace all components equal to $k$ with $k-2$). We have $d_\beta = d_\alpha + x_{k-3} x_k > d_\alpha$.

Thus, the maximum is achieved for $k \leq 2$ and $n$-tuples of the form $(0,\ldots,0,1,\ldots,1)$ or $(0,\ldots,0,1,\ldots,1,2,\ldots,2)$. \pts{1}

If $\gamma = (\underbrace{0,\ldots,0}_{x_0}, \underbrace{1,\ldots,1}_{x_1})$ with $x_0 + x_1 = n$, then $d_\gamma = x_0(n - x_0)$. We have:
\begin{itemize}
  \item $d_\gamma \leq \dfrac{n^2}{4}$ if $n$ is even, with equality for $x_0 = x_1 = \dfrac{n}{2}$;
  \item $d_\gamma \leq \dfrac{n^2-1}{4}$ if $n$ is odd, with equality for $x_0 = \dfrac{n+1}{2}$, $x_1 = \dfrac{n-1}{2}$ (or vice versa).
\end{itemize}
\pts{2}

If $\delta = (\underbrace{0,\ldots,0}_{x_0}, \underbrace{1,\ldots,1}_{x_1}, \underbrace{2,\ldots,2}_{x_2})$ with $x_0+x_1+x_2 = n$, then $d_\delta = x_0 x_1 + x_1 x_2 = x_1(n - x_1)$. We have:
\begin{itemize}
  \item $d_\delta \leq \dfrac{n^2}{4}$ if $n$ is even, with equality when $x_1 = \dfrac{n}{2}$ and $x_0 + x_2 = \dfrac{n}{2}$;
  \item $d_\delta \leq \dfrac{n^2-1}{4}$ if $n$ is odd, with equality when $x_1 = \dfrac{n-1}{2}$ and $x_0 + x_2 = \dfrac{n+1}{2}$, or $x_1 = \dfrac{n+1}{2}$ and $x_0 + x_2 = \dfrac{n-1}{2}$.
\end{itemize}

In conclusion, the maximum value of $d_a$ is $\dfrac{n^2}{4}$ if $n$ is even, and $\dfrac{n^2-1}{4}$ if $n$ is odd. \pts{2}",392,2448,Combinatorics,13
390,tst_jbmo_ro_2022_4_p1,tst_jbmo,2022,n,"Show that, for any prime natural number $p$, there exist natural numbers $x, y, z$ and $t$ such that $t < p$ and
\[
x^2 + y^2 + z^2 = tp.
\]","\textit{Solution.}
If $p = 2$, take $x = y = t = 1$, $z = 0$. \pts{1}

If $p \geq 3$, then
\[
1 + 0^2,\; 1 + 1^2,\; 1 + 2^2,\; \ldots,\; 1 + \left(\frac{p-1}{2}
ight)^2
\]
are $\dfrac{p+1}{2}$ integers that give distinct remainders upon division by $p$.

Indeed, if $a$ and $b$ are two distinct natural numbers, $0 \leq a, b \leq \dfrac{p-1}{2}$, then $p$ divides $(1+a^2) - (1+b^2)$ if and only if $p$ divides $a + b$ or $a - b$, which is impossible. \pts{2}

Likewise,
\[
-0^2,\; -1^2,\; -2^2,\; \ldots,\; -\left(\frac{p-1}{2}
ight)^2
\]
are $\dfrac{p+1}{2}$ integers that give distinct remainders upon division by $p$. \pts{2}

By the Pigeonhole Principle, there exist two numbers $x$ and $y$, $0 \leq x, y \leq \dfrac{p-1}{2}$, such that
\[
1 + x^2 \equiv -y^2 \pmod{p}.
\]
It follows that $x^2 + y^2 + 1^2 = tp$ for some natural number $t$. We have
\[
x^2 + y^2 + 1 \leq 2\left(\frac{p-1}{2}
ight)^2 + 1 < p^2,
\]
therefore $t < p$. \pts{2}",140,945,Number Theory,14
391,tst_jbmo_ro_2022_4_p2,tst_jbmo,2022,g,"In the scalene acute triangle $ABC$, let $D$ be the foot of the angle bisector from $A$ and $E$ the foot of the altitude from $A$. The perpendicular bisector of segment $AD$ intersects the semicircles with diameters $AB$ and $AC$, constructed in the exterior of triangle $ABC$, at $X$ and $Y$ respectively. Prove that the points $X$, $Y$, $D$ and $E$ are concyclic.","\textit{Solution.}
Let $O_1$ be the midpoint of $[AB]$ and $O_2$ the midpoint of $[AC]$.

Denote $\angle AXY = \angle DXY = a$, $\angle BAX = b$, $\angle CAY = x$, $\angle AYX = \angle DYX = y$.

The quadrilaterals $AXBE$ and $AYCE$ are cyclic, so $\angle XBE = \angle XAB = b$ and $\angle CEY = \angle YAC = x$. Then $\angle XEY = 180^\circ - (b + x)$, \quad (1). By symmetry, $\angle XDY = \angle XAY = 180^\circ - (b + x)$.

From triangle $AXY$ we get $\angle XAY = 180^\circ - (a + y)$, \quad (2). \pts{2}

Since $O_1$, $M$, $O_2$ are collinear and $O_1O_2 \parallel BC$, we get
\[
\frac{MO_1}{MO_2} = \frac{DB}{DC}.
\]
But $\dfrac{DB}{DC} = \dfrac{AB}{AC} = \dfrac{2XO_1}{2YO_2} = \dfrac{XO_1}{YO_2}$, so $\dfrac{MO_1}{MO_2} = \dfrac{XO_1}{YO_2}$.

Since the angles $\angle MO_1X$ and $\angle MO_2Y$ are both obtuse, by the SAS similarity criterion we get $\triangle MO_1X \sim \triangle MO_2Y$, hence $\angle MO_1X \equiv \angle MO_2Y$. Consequently, $XO_1 \parallel YO_2$, so $\angle MXO_1 \equiv \angle MYO_2$. \pts{2}

Since $\angle MXO_1 = a - \angle AXO_1 = a - \angle XAO_1 = a - b$ and $\angle MY O_2 = \angle AYO_2 - \angle AYM = \angle YAO_2 - y = x - y$, we deduce $a - b = x - y$.

Consequently, $b + x = a + y$, from which $\angle XEY \equiv \angle XDY$, i.e.\ the quadrilateral $XEDY$ is cyclic. \pts{3}

\bigskip
\textit{Comment.} The problem admits the following alternative solution.

Let $Z$ be a point on line $XY$ such that the quadrilateral $ZEDY$ is cyclic.

Since $XY$ is the perpendicular bisector of segment $AD$, we obtain $\angle AZY \equiv \angle DZY$, (1), and $\angle AYZ \equiv \angle DYZ$, (2).

The quadrilateral $ZEDY$ is cyclic, so $\angle DZY \equiv \angle DEY$, (3), and the quadrilateral $AECY$ is also cyclic, so $\angle DEY \equiv \angle CAY$, (4).

Let $\{M\} = XY \cap AB$ and $\{N\} = XY \cap AC$. Since $AD$ is the angle bisector of angle $BAC$, we deduce that triangle $AMN$ is isosceles and $\angle AMN \equiv \angle ANM$.

From (1), (3) and (4) we obtain $\angle AZY \equiv \angle CAY$, (5). We deduce that
\[
\angle ZAB = \angle AMN - \angle AZM = \angle ANM - \angle YAN = \angle AYZ = \angle DYZ = \angle ZEB.
\]
Therefore $\angle ZAB \equiv \angle ZEB$, so the quadrilateral $AZBE$ is cyclic. Consequently, $\angle AZB = 90^\circ$, from which it follows that $Z$ and $X$ coincide. Hence, the points $X$, $Y$, $D$ and $E$ are concyclic.",365,2391,Geometry,15
392,tst_jbmo_ro_2022_4_p3,tst_jbmo,2022,a,"Find all pairs of natural numbers $(a, b)$ for which the number
\[
\frac{(a+b)^2}{4 + 4a(a-b)^2}
\]
is an integer.","\textit{Solution.}
Denote $c = c(a,b) = \dfrac{(a+b)^2}{4 + 4a(a-b)^2}$. We observe directly that:
\begin{itemize}
  \item if $a = b = x$, $x \in \mathbb{N}$, then $c(a,b) = c(x,x) = x^2 \in \mathbb{Z}$;
  \item if $a = 0$, then $c(a,b) \in \mathbb{Z}$ if and only if $b = 2x$, $x \in \mathbb{N}$.
\end{itemize}
\pts{1}

In what follows, let $a, b \in \mathbb{N}$, $a \geq 1$, $b 
eq a$, such that $c(a,b) \in \mathbb{Z}$. In fact, we observe that $c(a,b) \in \mathbb{N}^*$. Since the denominator of $c(a,b)$ is an even number, $a$ and $b$ have the same parity. Thus, we may consider $b = a \pm 2x$, $x \in \mathbb{N}^*$, for which we obtain
\[
c(a,b) = \frac{(a \pm x)^2}{4ax^2 + 1}.
\]
Since $c(a,b) \geq 1$, it follows that $4ax^2 + 1 \leq (a-x)^2$ or $4ax^2 + 1 \leq (a+x)^2$. Therefore,
\[
4ax^2 < 4ax^2 + 1 \leq \max\{(a-x)^2, (a+x)^2\} = (a+x)^2,
\]
so $2\sqrt{a}\,x < a + x$, from which we get
\[
x < \frac{a}{2\sqrt{a}-1} \leq \sqrt{a}, \quad (1).
\]
\pts{2}

Since $c(a,b) \in \mathbb{Z}$, we deduce that
\[
4x^2 \cdot \frac{(a \pm x)^2}{4ax^2+1} \in \mathbb{Z} \iff a \pm 2x + \frac{4x^4 \mp 2x - a}{4ax^2+1} \in \mathbb{Z} \iff \frac{4x^4 \mp 2x - a}{4ax^2+1} \in \mathbb{Z}.
\]
One verifies immediately that $d(a,x) \overset{\text{def}}{=} \dfrac{4x^4 \mp 2x - a}{4ax^2+1} > -1$ for all $a, x \in \mathbb{N}^*$. \pts{1}

Suppose, for contradiction, that $d(a,x) \geq 1$. Then $4ax^2 + 1 \leq 4x^4 \pm 2x - a$, from which $a(4x^2+1) \leq 4x^4 \pm 2x - 1$. We deduce that
\[
a \leq \frac{4x^4 \pm 2x - 1}{4x^2+1} = \frac{x^2(4x^2+1) - (x \pm 1)^2}{4x^2+1} \leq x^2,
\]
a contradiction with (1). \pts{2}

Therefore $d(a,x) = 0$, so $a = 4x^4 \pm 2x$, from which we obtain $b = 4x^4$.

In conclusion, the problem admits the solutions $(0, 2x)$, $x \in \mathbb{N}$; $(x, x)$, $x \in \mathbb{N}^*$; and $(4x^4 \pm 2x,\; 4x^4)$, $x \in \mathbb{N}^*$. \pts{1}",114,1864,Algebra,16
393,tst_jbmo_ro_2022_4_p4,tst_jbmo,2022,c,"We say that a natural number is \emph{round} if its number of divisors is a perfect square. For a round natural number $n$ with $d^2$ divisors, we construct a $d \times d$ table and fill its cells with the divisors of $n$. At each step, we may choose a row of the table and move the divisor in column 1 to column 2, the one in column 2 to column 3, \ldots, and the one in column $d$ to column 1. Any configuration of the table we can reach is called \emph{feasible} if there exists a column whose elements $a_1, a_2, \ldots, a_d$ satisfy $a_1 \mid a_2 \mid \cdots \mid a_d$ or $a_d \mid a_{d-1} \mid \cdots \mid a_1$.

Determine all round numbers $n \geq 2$ for which, no matter how we initially fill the divisor table, there exists a feasible configuration reachable in a finite number of steps.","\textit{Solution.}
Let $n \geq 2$ be a round number with $d^2$ divisors. From $n \geq 2$ we deduce $d \geq 2$. If $d = 2$, then $n = p^3$ or $n = pq$, where $p$ and $q$ are distinct prime numbers. We observe that these work, since however we fill the $2 \times 2$ divisor table, the divisor $1$ will ensure the feasibility of the configuration. \pts{2}

Assume further that $d \geq 3$. Then in the $d \times d$ table, the second row is not the last. We aim to fill this row with divisors that do not divide other divisors, or that are not divisible by other divisors. \pts{1}

Let $n = d_1 > d_2 > \cdots > d_{d^2} = 1$ be the divisors of $n$. Fill the second row with the divisors $d_1, d_2, \ldots, d_d$ and fill the other cells arbitrarily. Then, however we rotate the divisors in any row, the smallest divisor in the second row is always $d_d$, which is greater than every divisor appearing in any row different from the second. \pts{2}

Then, if we assume we can reach a feasible configuration, we have $d_p \mid d_q \mid d_r$ or $d_r \mid d_q \mid d_p$, where $q \in \{1, 2, \ldots, d\}$ and $p, r \geq d + 1$. This implies $d_d 
eq \max\{d_p, d_q, d_r\}$, which contradicts the ordering of the divisors. Therefore, the only numbers that work are those of the form $pq$ or $p^3$, where $p$ and $q$ are any distinct prime numbers. \pts{2}",796,1343,Combinatorics,17
267,jbmo_2023_p1,jbmo,2023,n,"Find all pairs $(a, b)$ of positive integers such that $a! + b$ and $b! + a$ are both powers of 5.","
oindent\textbf{Solution.} The condition is symmetric so we can assume that $b \leq a$.

The first case is when $a = b$. In this case, $a! + a = 5^m$ for some positive integer $m$. We can rewrite this as $a \cdot ((a-1)! + 1) = 5^m$. This means that $a = 5^k$ for some integer $k \geq 0$. It is clear that $k$ cannot be 0. If $k \geq 2$, then $(a-1)! + 1 = 5^l$ for some $l \geq 1$, but $a - 1 = 5^k - 1 > 5$, so $5 \mid (a-1)!$, which is not possible because $5 \mid (a-1)! + 1$. This means that $k = 1$ and $a = 5$. In this case, $5! + 5 = 125$, which gives us the solution $(5, 5)$.

Let us now assume that $1 \leq b < a$. Let us first assume that $b = 1$. Then $a + 1 = 5^x$ and $a! + 1 = 5^y$ for integers $x, y \geq 1$. If $x \geq 2$, then $a = 5^x - 1 \geq 5^2 - 1 > 5$, so $5 \mid a!$. However, $5 \mid 5^y = a! + 1$, which leads to a contradiction. We conclude that $x = 1$ and $a = 4$. From here $a! + b = 25$ and $b! + a = 5$, so we get two more solutions: $(1, 4)$ and $(4, 1)$.

Now we focus on the case $1 < b < a$. Then we have $a! + b = 5^x$ for $x \geq 2$, so $b \cdot \left(\frac{a!}{b} + 1
ight) = 5^x$, where $b \mid a!$ because $a > b$. Because $b \mid 5^x$ and $b > 1$, we have $b = 5^z$ for $z \geq 1$. If $z \geq 2$, then $5 < b < a$, so $5 \mid a!$, which means that $\frac{a!}{b} + 1$ cannot be a power of 5. We conclude that $z = 1$ and $b = 5$. From here $5! + a$ is a power of 5, so $5 \mid a$, but $a > b = 5$, which gives us $a \geq 10$. However, this would mean that $25 \mid a!$, $5 \mid b$ and $25 
mid b$, which is not possible, because $a! + b = 5^x$ and $25 \mid 5^x$.

We conclude that the only solutions are $(1, 4)$, $(4, 1)$ and $(5, 5)$. \hfill$\square$",98,1695,Number Theory,1
268,jbmo_2023_p2,jbmo,2023,a,"Prove that for all non-negative real numbers $x, y, z$, not all equal to 0, the following inequality holds
\[
\frac{2x^2 - x + y + z}{x + y^2 + z^2} + \frac{2y^2 + x - y + z}{x^2 + y + z^2} + \frac{2z^2 + x + y - z}{x^2 + y^2 + z} \geqslant 3.
\]
Determine all the triples $(x, y, z)$ for which the equality holds.","
oindent\textbf{Solution.} Let us first write the expression $L$ on the left hand side in the following way
\begin{align*}
L &= \left(\frac{2x^2 - x + y + z}{x + y^2 + z^2} + 2
ight) + \left(\frac{2y^2 + x - y + z}{x^2 + y + z^2} + 2
ight) + \left(\frac{2z^2 + x + y - z}{x^2 + y^2 + z} + 2
ight) - 6 \\
&= (2x^2 + 2y^2 + 2z^2 + x + y + z)\left(\frac{1}{x + y^2 + z^2} + \frac{1}{x^2 + y + z^2} + \frac{1}{x^2 + y^2 + z}
ight) - 6.
\end{align*}

If we introduce the notation $A = x + y^2 + z^2$, $B = x^2 + y + z^2$, $C = x^2 + y^2 + z$, then the previous relation becomes
\[
L = (A + B + C)\left(\frac{1}{A} + \frac{1}{B} + \frac{1}{C}
ight) - 6.
\]
Using the arithmetic-harmonic mean inequality or Cauchy-Schwartz inequality for positive real numbers $A, B, C$, we easily obtain
\[
(A + B + C)\left(\frac{1}{A} + \frac{1}{B} + \frac{1}{C}
ight) \geqslant 9,
\]
so it holds $L \geqslant 3$.

The equality occurs if and only if $A = B = C$, which is equivalent to the system of equations
\[
x^2 - y^2 = x - y, \quad y^2 - z^2 = y - z, \quad x^2 - z^2 = x - z.
\]
It follows easily that the only solutions of this system are
\[
(x, y, z) \in \{(t, t, t) \mid t > 0\} \cup \{(t, t, 1-t) \mid t \in [0,1]\} \cup \{(t, 1-t, t) \mid t \in [0,1]\} \cup \{(1-t, t, t) \mid t \in [0,1]\}.
\]
\hfill$\square$

\medskip


oindent\textbf{PSC Remark} We feel the equality case needs more explanations in order to have a complete solution, our suggestion follows:

Clearly if $x, y, z$ are all equal and not 0 satisfy the condition. Now suppose that not all of them are equal it means we can't simultaneously have $x + y = y + z = z + x = 1$ otherwise we would have all $x, y, z$ equal to $\frac{1}{2}$ which we already discussed. We can suppose now that $x = y$ and $y + z = z + x = 1$ where we get $z = 1 - x$. So, all triples which satisfy the equality are $(x, y, z) = (a, a, a)$, $(b, b, 1-b)$ and all permutations for any $a > 0$ and $b \in [0, 1]$.",314,1942,Algebra,2
269,jbmo_2023_p3,jbmo,2023,c,"Alice and Bob play the following game on a $100 \times 100$ grid, taking turns, with Alice starting first. Initially the grid is empty. At their turn, they choose an integer from 1 to $100^2$ that is not written yet in any of the cells and choose an empty cell, and place it in the chosen cell. When there is no empty cell left, Alice computes the sum of the numbers in each row, and her score is the maximum of these 100 sums. Bob computes the sum of the numbers in each column, and his score is the maximum of these 100 sums. Alice wins if her score is greater than Bob's score, Bob wins if his score is greater than Alice's score, otherwise no one wins.

Find if one of the players has a winning strategy, and if so which player has a winning strategy.","
oindent\textbf{Solution.} We denote by $(i, j)$ the cell in the $i$-th line and in the $j$-th column for every $1 \leq i, j \leq n$. Bob associates the following pair of cells: $(i, 2k+1)$, $(i, 2k+2)$ for $1 \leq i \leq 100$ and $0 \leq k \leq 49$ except for $(i, k) = (100, 0)$ and $(100, 1)$, and the pairs $(100, 1)$, $(100, 3)$ and $(100, 2)$, $(100, 4)$.

Each time Alice writes the number $j$ in one of the cells, Bob writes the number $100^2 + 1 - j$ in the other cell of the pair.

One can prove by induction that after each of Bob's turn, for each pair of cells, either there is a number written in each of the cells of the pair, or in neither of them. And that if a number $j$ is written, $100^2 + 1 - j$ is also written. Thus Bob can always apply the previous strategy (since $j = 100^2 + 1 - j$ is impossible).

At the end, every line has sum $(100^2 + 1) \times 50$.

Assume by contradiction that Alice can stop Bob from winning if he applies this strategy. Let $c_j$ be the sum of the numbers in the $j$-th column for $1 \leq j \leq 100$: then $c_j \leq 50(100^2 + 1)$. Note that:
\[
100 \times 50(100^2 + 1) \geq c_1 + \cdots + c_{100} = 1 + \cdots + 100^2 = \frac{100^2(100^2 + 1)}{2} = 100 \times 50(100^2 + 1)
\]
Thus we have equality in the previous inequality: $c_1 = \cdots = c_{100} = 50(100^2 + 1)$. But if $a$ is the number written in the cell $(100, 1)$ and $b$ the number written in the cell $(100, 2)$, then $c_1 - b + c_2 - c = 99(100^2 + 1)$. Thus $b + c = 100(100^2 + 1) - 99(100^2 + 1) = 100^2 + 1$: by hypothesis $c$ is also written in the cell $(100, 3)$ which is a contradiction.

Thus Bob has a winning strategy. \hfill$\square$",755,1665,Combinatorics,3
270,jbmo_2023_p4,jbmo,2023,g,"Let $ABC$ be an acute triangle with circumcenter $O$. Let $D$ be the foot of the altitude from $A$ to $BC$ and let $M$ be the midpoint of $OD$. The points $O_b$ and $O_c$ are the circumcenters of triangles $AOC$ and $AOB$, respectively. If $AO = AD$, prove that the points $A$, $O_b$, $M$ and $O_c$ are concyclic.","
oindent\textbf{Solution.}

Note that $AB = AC$ cannot hold since $AO = AD$ would imply that $O$ is the midpoint of $BC$, which is not possible for an acute triangle. So we may assume without loss of generality that $AB < AC$.

Let $M_b$ and $M_c$ be the midpoints of $AC$ and $AB$, respectively. Since $\angle AM_bO = \angle AM_cO = 90^\circ = \angle AMO$ (the latter since $AO = AD$), the pentagon $AM_bOMM_c$ is cyclic.

Next, notice that $AM$ is the perpendicular bisector of $OD$, $O_bO_c$ is the perpendicular bisector of $AO$ and $M_bM_c$ is the perpendicular bisector of $AD$. Hence these three lines are concurrent -- denote their common point by $T$.

The quadrilateral $AO_bMO_c$ is cyclic if and only if $AT \cdot TM = O_bT \cdot O_cT$. From the cyclic $AM_bMM_c$ we have $AT \cdot TM = M_bT \cdot M_cT$. Hence it now suffices to argue $M_bT \cdot M_cT = O_bT \cdot O_cT$ -- or equivalently, that $M_b$, $M_c$, $O_b$ and $O_c$ are concyclic.

We assume that $\angle AOB < 90^\circ$ and $\angle AOC > 90^\circ$ so that $O_c$ is in the interior of triangle $AOB$ and $O_b$ is external to the triangle $AOC$ (the other cases are analogous and if $\angle AOB = 90^\circ$ or $\angle AOC = 90^\circ$, then $M_b \equiv O_b$ or $M_c \equiv O_c$ and we are automatically done). We have
\[
\angle M_cM_bO_b = 90^\circ + \angle AM_bM_c = 90^\circ + \angle ACB
\]
as well as (since $O_cO_b$ is a perpendicular bisector of $AO$ and hence bisects $\angle AO_CO$)
\[
\angle M_cO_cO_b = 180^\circ - \angle OO_cO_b = 90^\circ + \frac{\angle AO_cM_c}{2} = 90^\circ + \frac{\angle AO_cB}{4} = 90^\circ + \frac{\angle AOB}{2} = 90^\circ + \angle ACB
\]
and therefore $O_bM_bO_cM_c$ is cyclic, as desired. \hfill$\square$",313,1712,Geometry,4
28,shl_jbmo_2023_a1,shl_jbmo,2023,a,"Prove that for all positive real numbers $a, b, c, d$,
\[
\frac{2}{(a+b)(c+d)+(b+c)(a+d)} \leq \frac{1}{(a+c)(b+d)+4ac} + \frac{1}{(a+c)(b+d)+4bd}
\]
and determine when equality occurs.","Let
\begin{align*}
X &= (a+b)(c+d)+(b+c)(a+d),\\
Y &= (a+c)(b+d)+4ac,\\
Z &= (a+c)(b+d)+4bd.
\end{align*}
Next, note that
\begin{align*}
X - Y &= (ab+2ac+ad+bc+2bd+cd)-(ab+4ac+ad+bc+cd)\\
      &= 2(bd-ac)
\end{align*}
and
\begin{align*}
X - Z &= (ab+2ac+ad+bc+2bd+cd)-(ab+ad+bc+4bd+cd)\\
      &= 2(ac-bd)\\
      &= Y - X.
\end{align*}
Hence
\[
X = \frac{Y+Z}{2}.
\]
Therefore, the desired inequality is equivalent to
\[
\frac{4}{Y+Z} \leq \frac{1}{Y} + \frac{1}{Z}.
\]
The last inequality is equivalent to $4YZ \leq (Y+Z)^2$, which is true because
\[
(Y+Z)^2 - 4YZ = (Y-Z)^2 \geq 0.
\]
The equality holds if and only if $Y = Z$, that is, if and only if $ac = bd$.

\medskip

oindent\textbf{Proposer's Remark.} Note that
\[
Y - Z = 4(ac - bd)
\]
and
\[
X^2 = \left(\frac{Y+Z}{2}
ight)^2 = 4\left(\frac{Y-Z}{4}
ight)^2 + YZ = 4(ac-bd)^2 + YZ.
\]
Therefore, we obtain the stronger inequality
\[
\bigl((a+b)(c+d)+(b+c)(a+d)\bigr)^2 \geq \bigl((a+c)(b+d)+4ac\bigr)\bigl((a+c)(b+d)+4bd\bigr).
\]
The equality holds if and only if $ac = bd$.

\medskip

oindent\textbf{PSC Remark.} We suggest the following alternative easier way to solve this problem.

Taking $(a{+}c)(b{+}d)+4ac = x$ and $(a{+}c)(b{+}d)+4bd = y$ we can easily find that $(a{+}b)(c{+}d)+(b{+}c)(a{+}d) = \frac{x+y}{2}$. So, that is enough to prove,
\[
\frac{4}{x+y} \leq \frac{1}{x} + \frac{1}{y},
\]
which is very well known inequality which can be done multiple ways, Cauchy-Schwarz, AM-HM, etc. With equality iff $x = y$ so iff $ab = cd$.",185,1504,Algebra,1
29,shl_jbmo_2023_a2,shl_jbmo,2023,a,"For positive real numbers $x, y, z$ with $xy + yz + zx = 1$, prove that
\[
\frac{2}{xyz} + 9xyz \geq 7(x+y+z).
\]","\begin{solution}[Solution 1]
It is well known that $(x+y)(y+z)(z+x) = (x+y+z)(xy+yz+zx) - xyz$, which implies $(x+y)(y+z)(z+x) = x+y+z - xyz$. \hfill(1)

Now, consider the expression
\[
S = \frac{x}{x^2+1} + \frac{y}{y^2+1} + \frac{z}{z^2+1}.
\]
On the one hand,
\begin{align*}
S &= \sum_{\mathrm{cyc}} \frac{x}{x^2 + xy+yz+zx} = \sum_{\mathrm{cyc}} \frac{x}{(x+y)(x+z)}\\
  &= \frac{2(xy+yz+zx)}{(x+y)(y+z)(z+x)} \stackrel{(1)}{=} \frac{2}{x+y+z-xyz}.
\end{align*}
On the other hand,
\[
S = \sum_{\mathrm{cyc}} \frac{1}{x + \frac{1}{x}} \geq \frac{9}{x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}} = \frac{9}{x+y+z + \frac{1}{xyz}}
\]
by using Cauchy-Schwarz inequality. Then, by comparing these expressions, we get the following inequality:
\[
2(x+y+z) + \frac{2}{xyz} \geq 9(x+y+z) - 9xyz
\]
which gives the desired inequality. Equality holds when $x = y = z = \frac{1}{\sqrt{3}}$.
\end{solution}

\begin{solution}[Solution 2]
Let us make a substitution of $a = xy$, $b = yz$, $c = zx$, then the given condition would be $a+b+c = 1$ and we need to show that $2 + 9abc \geq 7(ab+bc+ca)$. However, we will prove the stronger inequality which is equivalent to the desired inequality with the given property.
\[
2(a+b+c)^3 + 9abc \geq 7(ab+bc+ca)(a+b+c).
\]
After some algebraic manipulations, we will prove the following inequality:
\[
2\sum_{\mathrm{cyc}} a^3 + 6\sum_{\mathrm{cyc}} a^2(b+c) + 21abc \geq 7\sum_{\mathrm{cycy}} a^2(b+c) + 21abc
\]
which is equivalent with the following:
\[
2\sum_{\mathrm{cyc}} a^3 \geq \sum_{\mathrm{cycy}} a^2(b+c).
\]
But, it is known that for all positive real numbers $m, n$, we have $(m-n)^2(m+n) \geq 0 \implies m^3 + n^3 \geq m^2 n + mn^2$. After applying this well-known inequality to the pairs $(a,b)$, $(b,c)$, $(c,a)$, and adding the results side-by-side, we get:
\begin{align*}
2(a^3+b^3+c^3) &= (a^3+b^3)+(b^3+c^3)+(c^3+a^3) \geq ab(a+b)+bc(b+c)+ca(c+a)\\
&= \sum_{\mathrm{cyc}} a^2(b+c)
\end{align*}
as wanted.
\end{solution}

\begin{solution}[Solution 3 -- PSC suggested solution]
Taking substitution $x = \frac{1}{a}$, $y = \frac{1}{b}$, $z = \frac{1}{c}$ condition transform to $a+b+c = abc$ and it is equivalent to prove,
\[
2abc + \frac{9}{abc} \geq 7\left(\frac{ab+bc+ca}{abc}
ight) \Leftrightarrow 2(abc)^2 + 9 \geq 7(ab+bc+ca)
\]
\[
\Leftrightarrow 2(a+b+c)^2 + \frac{9abc}{a+b+c} \geq 7(ab+bc+ca)
\]
\[
\Leftrightarrow 2(a+b+c)^3 + 9abc \geq 7(ab+bc+ca)(a+b+c).
\]
After transformation last inequality is equivalent to,
\[
2(a^3+b^3+c^3) \geq ab(a+b)+bc(b+c)+ca(c+a) \Leftrightarrow \sum_{\mathrm{cyc}}(a+b)(a-b)^2 \geq 0.
\]
Hence true.


oindent\textbf{Comment PSC} Last inequality can also be done using $AM - GM$ this way,
\[
\sum_{\mathrm{cyc}}(2a^3+b^3) + \sum_{\mathrm{cyc}}(a^3+2b^3) \geq 3\sum_{\mathrm{cyc}} a^2 b + 3\sum_{\mathrm{cyc}} ab^2.
\]
\end{solution}",113,2824,Algebra,2
30,shl_jbmo_2023_a3,shl_jbmo,2023,a,"Prove that for all non-negative real numbers $x, y, z$, such that at least one of them is not equal to 0, the following inequality holds
\[
\frac{2x^2-x+y+z}{x+y^2+z^2} + \frac{2y^2+x-y+z}{x^2+y+z^2} + \frac{2z^2+x+y-z}{x^2+y^2+z} \geqslant 3.
\]
Determine all the triples $(x,y,z)$ for which the equality holds.","\begin{solution}
Let us first write the expression $L$ on the left hand side in the following way
\begin{align*}
L &= \left(\frac{2x^2-x+y+z}{x+y^2+z^2}+2
ight) + \left(\frac{2y^2+x-y+z}{x^2+y+z^2}+2
ight) + \left(\frac{2z^2+x+y-z}{x^2+y^2+z}+2
ight) - 6\\
&= (2x^2+2y^2+2z^2+x+y+z)\left(\frac{1}{x+y^2+z^2}+\frac{1}{x^2+y+z^2}+\frac{1}{x^2+y^2+z}
ight) - 6.
\end{align*}
If we introduce the notation $A = x+y^2+z^2$, $B = x^2+y+z^2$, $C = x^2+y^2+z$, then the previous relation becomes
\[
L = (A+B+C)\left(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}
ight) - 6.
\]
Using the arithmetic-harmonic mean inequality or Cauchy-Schwartz inequality for positive real numbers $A, B, C$, we easily obtain
\[
(A+B+C)\left(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}
ight) \geqslant 9,
\]
so it holds $L \geqslant 3$.

The equality occurs if and only if $A = B = C$, which is equivalent to the system of equations
\[
x^2-y^2 = x-y, \quad y^2-z^2 = y-z, \quad x^2-z^2 = x-z.
\]
It follows easily that the only solutions of this system are
\[
(x,y,z) \in \{(t,t,t)\mid t>0\} \cup \{(t,t,1-t)\mid t\in[0,1]\} \cup \{(t,1-t,t)\mid t\in[0,1]\} \cup \{(1-t,t,t)\mid t\in[0,1]\}.
\]

\medskip

oindent\textbf{PSC Remark.} We feel the equality case needs more explanations in order to have a complete solution, our suggestion follows:

Clearly if $x,y,z$ are all equal and not 0 satisfy the condition. Now suppose that not all of them are equal. If $x+y = y+z = z+x = 1$ otherwise we would have all $x,y,z$ equal, it means we can't simultaneously have $x+y = y+z = z+x = 1$ otherwise we would have all $x,y,z$ equal to $\frac{1}{2}$ which we already discussed. We can suppose now that $x = y$ and $y+z = z+x = 1$ where we get $z = 1-x$. So, all triples which satisfy the equality are $(x,y,z) = (a,a,a)$, $(b,b,1-b)$ and all permutations for any $a > 0$ and $b \in [0,1]$.
\end{solution}",312,1851,Algebra,3
31,shl_jbmo_2023_a4,shl_jbmo,2023,a,"Let $a, b, c, d$ be positive real numbers with $abcd = 1$. Prove that
\[
\sqrt{\frac{a}{b+c+d^2+a^3}} + \sqrt{\frac{b}{c+d+a^2+b^3}} + \sqrt{\frac{c}{d+a+b^2+c^3}} + \sqrt{\frac{d}{a+b+c^2+d^3}} \leq 2.
\]","\begin{solution}
From Cauchy-Schwarz inequality we have
\[
(b+c+d^2+a^3)(b+c+1+1/a) \geq (a+b+c+d)^2
\]
therefore
\[
\frac{a}{b+c+d^2+a^3} \leq \frac{ab+ac+a+1}{(a+b+c+d)^2}.
\]
Using this in the original inequality with quadratic-arithmetic mean inequality, it suffices to prove
\[
\frac{2\sqrt{4+\sum a + \sum_{\mathrm{sym}} ab + (ac+bd)}}{a+b+c+d} \leq 2
\]
dividing both sides by 2 and taking squares, this inequality is equivalent to
\[
\sum a^2 + \sum_{\mathrm{cyc}} ab \geq \sum a + 4
\]
and this is true since $ab+bc+cd+da \geq 4$ by AM-GM, and $a+b+c+d \geq 4$ by AM-GM with $4(\sum a^2) \geq (\sum a)^2$ by Cauchy-Schwarz inequality.
\end{solution}",205,658,Algebra,4
32,shl_jbmo_2023_a5,shl_jbmo,2023,a,"Let $a \geq b \geq 1 \geq c \geq 0$ be real numbers such that $a+b+c = 3$. Show that
\[
3\left(\frac{a}{b}+\frac{b}{a}
ight) \geq 4c^2 + \frac{a^2}{b} + \frac{b^2}{a}.
\]","\begin{solution}[Solution 1]
We have
\[
a + 2bc \leq a + \frac{(b+c)^2}{2} = a + \frac{(3-a)^2}{2} = \frac{a^2-4a+9}{2} = 3 + \frac{(a-1)(a-3)}{2} \leq 3.
\]
because $a-1 \geq 0$ and $a-3 \leq 0$. Analogously, we obtain $b + 2ac \leq 3$.

Multiplying $a+2bc \leq 3$ by $\frac{a}{b}$, it follows that $\frac{a^2}{b} + 2ac \leq 3\frac{a}{b}$. In the same way, $\frac{b^2}{a} + 2bc \leq 3\frac{b}{a}$.

Adding the last two inequalities, we get that
\[
3\frac{a}{b}+3\frac{b}{a} \geq 2c(a+b) + \frac{a^2}{b}+\frac{b^2}{a} \stackrel{a+b\geq 2c}{\geq} 4c^2 + \frac{a^2}{b}+\frac{b^2}{a},
\]
and the conclusion follows.
\end{solution}

\begin{solution}[Solution 2 -- PSC suggested solution]
Inequality is equivalent to,
\[
\frac{(3-a)a}{b} + \frac{(3-b)b}{a} \geq 4c^2.
\]
From Cauchy-Schwarz inequality we have,
\[
\left(\frac{(3-a)a}{b}+\frac{(3-b)b}{a}
ight)\left(\frac{(3-a)b}{a}+\frac{(3-b)a}{b}
ight) \geq (3-a+3-b)^2 = (c+3)^2.
\]
On other hand using condition and that $a \geq 1$, and $b \geq 1$ we have,
\[
\frac{(3-a)b}{a}+\frac{(3-b)a}{b} = 3\left(\frac{a}{b}+\frac{b}{a}
ight) - a - b \leq 3(a+b)-a-b = 2(a+b) = 6-2c.
\]
So we have,
\[
\frac{(3-a)a}{b}+\frac{(3-b)b}{a} \geq \frac{(c+3)^2}{6-2c}.
\]
So it is enough to prove,
\[
\frac{(c+3)^2}{6-2c} \geq 4c^2 \Leftrightarrow (c+3)^2 \geq 4c^2(6-2c) \Leftrightarrow (1-c)(9+15c-8c^2) \geq 0.
\]
Last inequality it is true because from $1 \geq c$ we have $1-c \geq 0$ and also we have that $9+15c-8c^2 \geq 9+7c \geq 9$, multiplying the last inequalities we get the desired result.
\end{solution}",170,1550,Algebra,5
33,shl_jbmo_2023_a6,shl_jbmo,2023,a,"Find the maximum constant $C$ such that, whenever $\{a_n\}_{n=1}^{\infty}$ is a sequence of positive real numbers satisfying $a_{n+1} - a_n = a_n(a_n+1)(a_n+2)$ we have
\[
\frac{a_{2023}-a_{2020}}{a_{2022}-a_{2021}} > C.
\]","\begin{solution}
The answer is $\dfrac{13}{3}$.

The given condition is equivalent to $a_{n+1}+1 = (a_n+1)^3$. Let $b_n = a_n+1$, then all terms of the sequence $\{b_n\}_{n=1}^{\infty}$ is greater than 1 and satisfies $b_n = b_1^{3^{n-1}}$. Let $b_{2020} = t$, we have
\[
\frac{a_{2023}-a_{2020}}{a_{2022}-a_{2021}} = \frac{b_{2023}-b_{2020}}{b_{2022}-b_{2021}} = \frac{t^{27}-t}{t^9-t^3}
\]
as the function we want to minimise. Since
\[
\frac{t^{27}-t}{t^9-t^3} = \frac{t^{26}-1}{t^8-t^2} = \frac{1+t^2+t^4+\ldots t^{24}}{t^6+t^4+t^2} = 1 + \frac{1+t^8+t^{10}+\ldots t^{24}}{t^6+t^4+t^2}
\]
with $t > 1$, we have $t^{24}+1 > 2t^{12}$ and therefore substituting this into the final expression, we get a fraction with every term in the numerator is greater than every term in the denominator, so this value is always larger than $13/3$. Since $t$ can be arbitrarily close to 1 with a suitable choice of $b_1$, we see that $C = 13/3$ is the desired value.
\end{solution}",223,968,Algebra,6
34,shl_jbmo_2023_a7,shl_jbmo,2023,a,"Let $a_1, a_2, a_3, \ldots, a_{250}$ be real numbers such that $a_1 = 2$ and
\[
a_{n+1} = a_n + \frac{1}{a_n^2}
\]
for every $n = 1, 2, \ldots, 249$. Let $x$ be the greatest integer which is less than
\[
\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_{250}}.
\]
How many digits does $x$ have?","\begin{solution}[Solution 1]
Answer: 2.

Denote $S = \frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_{250}}$, then it is sufficient to show $10 < S < 100$.

We compute $a_1 = 2$, $a_2 = \frac{9}{4}$, $a_3 = \frac{793}{324} \in \left(\frac{12}{5}, \frac{5}{2}
ight)$ and hence $a_4 = a_3 + \frac{1}{a_3^2} > \frac{12}{5} + \frac{4}{25} = \frac{64}{25}$. In general all numbers are positive and $a_{n+1} > a_n$ for $n \geq 1$; in particular $a_n \geq a_4 > \frac{64}{25}$ for $n = 4,5,\ldots,250$. This implies
\[
S < \frac{1}{2}+\frac{4}{9}+\frac{2}{5}+247\cdot\frac{25}{64} < 2+97 < 100.
\]

On the other hand, since $a_n > a_2 = \frac{9}{4}$ for $n \geq 2$, we have $a_{n+1} < a_n + \frac{16}{81} < a_n + \frac{1}{5}$ for $n \geq 2$. Hence (by induction) $a_n < \frac{n+10}{5}$ for all $n \geq 2$. This implies
\[
S > \frac{1}{2} + 5\cdot\left(\frac{1}{12}+\frac{1}{13}+\frac{1}{12}+\cdots+\frac{1}{260}
ight).
\]
The latter expression contains 9 fractions, greater than or equal to $\frac{1}{20}$, 30 fractions, greater than or equal to $\frac{1}{50}$, 50 fractions, greater than or equal to $\frac{1}{100}$, as well as 100 fractions, greater than or equal to $\frac{1}{200}$. Hence it exceeds $5\left(\frac{9}{20}+\frac{3}{5}+\frac{1}{2}+\frac{1}{2}
ight) > 10$. Therefore $S > 10$.
\end{solution}

\begin{solution}[Solution 2 -- PSC suggested solution]
Answer: 2.

If we find a positive integer $k$ such that $10^{k-1} \leq x < 10^k$ then we can say that $x$ has $k$ digits. It is clear from condition that the sequence is increasing.

\medskip

oindent\textit{Lemma 1.} $f(z) = z + \frac{1}{z^2}$ is increasing for all $z \geq 2$.


oindent\textit{Proof.} Let $x > y \geq 2$ then we have $(x-1)(y-1) > 1 \Rightarrow xy > x+y$. We also have $(xy)^2 > 4xy > xy$, so,
\[
f(x)-f(y) = x + \frac{1}{x^2} - y - \frac{1}{y^2} = \frac{(x-y)((xy)^2-x-y)}{(xy)^2} > 0. \quad \square
\]


oindent\textit{Lemma 2.} $a_n \leq \sqrt{2n+2}$ for all positive integers $n = 1, 2, \ldots, 250$.


oindent\textit{Proof.} We use induction to prove the statement. For $n=1$ we get $2 = a_1 \leq \sqrt{2\cdot 1+2} = 2$ which is true. Suppose now that it is true for $n=k$ we prove for $n=k+1$. Since $a_k \leq \sqrt{2k+2}$ and using the condition and first lemma we get, $a_{k+1} = a_k + \frac{1}{a_k^2} \leq \sqrt{2k+2} + \frac{1}{2k+2}$, so it is enough to prove,
\begin{align*}
&\sqrt{2k+2} + \frac{1}{2k+2} \leq \sqrt{2k+4} \Leftrightarrow \frac{1}{2k+2} \leq \sqrt{2k+4}-\sqrt{2k+2} = \frac{2}{\sqrt{2k+4}+\sqrt{2k+2}}\\
&\Leftrightarrow \sqrt{2k+4}+\sqrt{2k+2} \leq 4k+4.
\end{align*}
Since,
\[
\sqrt{2k+4}+\sqrt{2k+2} < 2\sqrt{2k+4}.
\]
Hence that is enough to prove,
\[
\sqrt{2k+4} \leq 2k+2 \Leftrightarrow 2k+4 < 4k^2+8k+4 \Leftrightarrow 4k^2+6k > 0. \quad \square
\]
From last lemma we get that $a_{250} \leq \sqrt{2\cdot 251} = \sqrt{502} < \sqrt{625} = 25$.

Now since we have that the sequence is increasing and from knowing that $a_{250} < 25$ we have,
\[
\sum_{i=1}^{250}\frac{1}{a_i} > \frac{250}{a_{250}} > \frac{250}{25} = 10.
\]

On other hand summing up $\frac{1}{a_n^2} = a_{n+1}-a_n$ for $n=1,2,\ldots,249$ then adding both sides $\frac{1}{a_{250}^2}$ and using that $a_{250} < 25$ we have,
\[
\sum_{i=1}^{250}\frac{1}{a_i^2} = a_{250} + \frac{1}{a_{250}^2} - a_1 < a_{250} < 25.
\]
Multiplying both sides with 400 and using Cauchy-Schwarz inequality we have,
\[
10000 > 400\sum_{i=1}^{250}\frac{1}{a_i^2} \geq 250\sum_{i=1}^{250}\frac{1}{a_i^2} \geq \left(\sum_{i=1}^{250}\frac{1}{a_i}
ight)^2 \Rightarrow \sum_{i=1}^{250}\frac{1}{a_i} < 100.
\]

Since,
\[
10 < \sum_{i=1}^{250}\frac{1}{a_i} < 100 \Rightarrow 10 \leq x < 100
\]
we have that $x$ has two digits.
\end{solution}


oindent\textbf{PSC Remark.} We were able to get a solution of the analogous problem with 2023 numbers as suggested from the proposer, and the solution follows:

If we find a positive integer $k$ such that $10^{k-1} \leq x < 10^k$ then we can say that $x$ has $k$ digits. It is clear from condition that the sequence is increasing.

\textit{Lemma 1.} $f(z) = z + \frac{1}{z^2}$ is increasing for all $z \geq 2$.

\textit{Proof.} The proof is the same as in Solution 2.

\textit{Lemma 2*.} $a_n \leq \sqrt[3]{4n+4}$ for all positive integers $n = 1, 2, \ldots, 2023$.

\textit{Proof.} We use induction to prove the statement. For $n=1$ we get $2 = a_1 \leq \sqrt[3]{4\cdot 1+4} = \sqrt[3]{8} = 2$ which is true. Suppose now that it is true for $n=k$ we prove for $n=k+1$. Since $a_k \leq \sqrt[3]{4k+4}$ and using the condition and first lemma we get, $a_{k+1} = a_k + \frac{1}{a_k^2} \leq \sqrt[3]{4k+4} + \frac{1}{\sqrt[3]{(4k+4)^2}}$, so it is enough to prove,
\begin{align*}
\sqrt[3]{4k+4} + \frac{1}{\sqrt[3]{(4k+4)^2}} &\leq \sqrt[3]{4k+8}\\
\Leftrightarrow \frac{1}{\sqrt[3]{(4k+4)^2}} &\leq \sqrt[3]{k+2} - \sqrt[3]{k+1} = \frac{1}{\sqrt[3]{(k+2)^2}+\sqrt[3]{(k+2)(k+1)}+\sqrt[3]{(k+1)^2}}\\
\Leftrightarrow \sqrt[3]{(k+2)^2}&+\sqrt[3]{(k+2)(k+1)} \leq 3\sqrt[3]{(k+1)^2}
\end{align*}
Since we have, $\sqrt[3]{(k+2)^2}+\sqrt[3]{(k+2)(k+1)} < 2\sqrt[3]{(k+2)^2}$ so that is enough to prove,
\[
2\sqrt[3]{(k+2)^2} \leq 3\sqrt[3]{(k+1)^2} \Leftrightarrow \sqrt[3]{8}(k+2) \leq \sqrt[3]{27}(k+1).
\]
Which is true since,
\[
\sqrt{8}(k+2) \leq 3(k+2) \leq 5(k+1) \leq \sqrt{27}(k+1).
\]
Now since we have that the sequence is increasing and from knowing that $a_{2023} \leq \sqrt[3]{4\cdot 2024}$ we have,
\[
\sum_{i=1}^{2023}\frac{1}{a_i} > \frac{2023}{a_{2023}} \geq \frac{2023}{\sqrt[3]{4\cdot 2024}} > 100.
\]
Last inequality is true since noting with $t = 2000$ we have,
\begin{align*}
\frac{2023}{\sqrt[3]{4\cdot 2024}} > 100 &\Leftrightarrow \frac{2023^3}{4\cdot 2024} > 100^3 \Leftrightarrow 2023^3 > 2024\cdot 2000^2\\
&\Leftrightarrow (t+23)^3 > (t+24)t^2\\
&\Leftrightarrow 45t^2 + 3\cdot 23^2 t + 23^3 > 0.
\end{align*}
On other hand summing up $\frac{1}{a_n^2} = a_{n+1}-a_n$ for $n=1,2,\ldots,2022$ then adding both sides $\frac{1}{a_{2023}^2}$ we have,
\[
\sum_{i=1}^{2023}\frac{1}{a_i^2} = a_{2023} + \frac{1}{a_{2023}^2} - a_1 < a_{2023} \leq \sqrt[3]{4\cdot 2024} < \sqrt[3]{45\cdot 2025} = 45 < 49.
\]
Multiplying both sides with $49^2$ and using Cauchy-Schwarz inequality we have,
\[
49^3 > 49^2 \sum_{i=1}^{2023}\frac{1}{a_i^2} > 2023\sum_{i=1}^{2023}\frac{1}{a_i^2} \geq \left(\sum_{i=1}^{2023}\frac{1}{a_i}
ight)^2 \Rightarrow \sum_{i=1}^{2023}\frac{1}{a_i} < 7^3 = 343 < 1000.
\]
Since,
\[
100 < \sum_{i=1}^{2023}\frac{1}{a_i} < 1000 \Rightarrow 100 \leq x < 1000
\]
we have that $x$ has three digits.",297,6510,Algebra,7
35,shl_jbmo_2023_c1,shl_jbmo,2023,c,"Given is a square board with dimensions $2023 \times 2023$, in which each unit cell is colored blue or red. There are exactly 1012 rows in which the majority of cells are blue, and exactly 1012 columns in which the majority of cells are red. What's the maximal possible side length of the largest monochromatic square?","\begin{solution}[Solution 1]
Answer: 1011.

Experimenting with small dimensions like 5, 7, 9 etc. gives us the idea that for a board of size $2n+1$, the answer is $n$.

First we show that there can't be a monochromatic square with side 1012.
If there is such red square, then at least 1012 rows are majority red, thus at most 1011 can be majority blue, contradiction.
If there is such blue square, then at least 1012 columns are majority blue, so at most 1011 can be majority red, contradiction.

Furthermore, we will show that it's possible to have a monochromatic square of side 1011. In the following construction, we have \textbf{exactly} 1012 rows which are majority blue, and 1012 columns which are majority red.

\begin{enumerate}
  \item In the lower right region we color a blue square of side 1011 (this is the desired maximal monochromatic square).
  \item In the first row we color the first 1012 squares blue, in the second row the squares 2--1013 are blue, in the third row 3--1014, and so forth, and in the 1011th row the squares 1011--2022 are blue.
  \item In the 1012th row we color blue the first (leftmost) square, and the last (rightmost) 1011 squares.
\end{enumerate}

Now we can observe that:
\begin{itemize}
  \item[$I)$] The first (topmost) 1012 rows are majority blue (with exactly 1012 blue squares each), and the next 1011 rows are majority red (with exactly 1011 blue squares each), thus the ``blue'' condition is satisfied.
  \item[$II)$] The first (leftmost) 1012 columns are majority red, and the next 1011 columns have at least 1012 blue squares each, so they are majority blue, thus we have exactly 1012 majority red columns, so the ``red'' condition is satisfied as well.
\end{itemize}
\end{solution}

\begin{solution}[Solution 2 -- PSC suggested solution]
An alternative construction of a $(2n+1)\times(2n+1)$ board with an $n\times n$ monochromatic square would be as follows: Assign all the cell coordinates of the form $(a,b)$, with $1\leq a,b\leq 2n+1$, and use the red color only in the following squares:

\begin{enumerate}
  \item $(a,b)$ with $1\leq a,b\leq n$;
  \item $(n+1,n+1)$;
  \item $(a,b)$ with $n+2\leq a\leq 2n+1$, $1\leq b\leq 2n+1$.
\end{enumerate}

Checking the rows in which the majority of cells are blue:
\begin{enumerate}
  \item Rows 1 through $n$ have exactly $n+1$ blue cells (with $b$-coordinates $n+1$ through $2n+1$), hence, a majority of blue cells;
  \item Row $n+1$ has exactly $2n$ blue cells (only $(n+1,n+1)$ is red), so a majority of blue cells;
  \item Rows $n+2$ through $2n+1$ have no blue cells at all.
\end{enumerate}

So from (1) and (2), there are $n+1$ rows with a majority of blue cells.

Checking the columns in which the majority of cells are red:
\begin{enumerate}
  \item Columns 1 through $n$ have exactly $2n$ red cells (with $a$-coordinates 1 through $n$ from (1) and $n+2$ through $2n+1$ from (3)), hence, a majority of red cells;
  \item Column $n+1$ has exactly $n+1$ red cells (with $a$-coordinates $n+1$ through $2n+1$ from (2) and (3)), so a majority of red cells;
  \item Columns $n+2$ through $2n+1$ have exactly $n$ red cells (with $a$-coordinates $n+2$ through $2n+1$ from (3)), so a minority of red cells.
\end{enumerate}

So from (1) and (2), there are $n+1$ columns with a majority of blue cells.
\end{solution}",318,3315,Combinatorics,8
36,shl_jbmo_2023_c2,shl_jbmo,2023,c,"There are $n$ blocks are placed on the unit squares of $n \times n$ chessboard such that there are exactly one block in each row and each column. Find the maximum value $k$, in terms of $n$, such that however the blocks are arranged, we can place $k$ rooks on the board without any two of them threatening each other. (Two rooks are not threatening each other if there is a block lying between them.)","\begin{solution}
The answer is $2n-2$. There are at most 1 rook in the rows containing the block in the leftmost or rightmost unit square, and at most 2 rooks in the other rows. Therefore, we can place at most $2n-2$ rooks.

Now we prove that we can place $2n-2$ rooks no matter how the blocks are arranged. We will use induction with $n=1,2$ cases are trivial. Assume that it holds for $n-1$. For an $n\times n$ chessboard, consider the block at the first row, let that square be $C$. First, let's delete the row and column containing $C$ and place $2n-4$ rooks on the remaining $(n-1)\times(n-1)$ board. Then we add the row and column containing $C$. If $C$ is at one of the corners, then we can add the new rooks to just above the block in the row next to $C$ and just next to the block in the column next to $C$. If $C$ is not at the corner, consider the block in the row just below $C$. Wlog, say that this block lies on the left hand side of $C$. Now, consider the column next to the column containing $C$, lying on the right hand side. Considering the block in that column, there should be a rook above that block. We remove that rook, put one rook in the same row as that rook and same column under $C$. Finally, we can place another rook in the square next to $C$ (lying on the RHS), therefore we can place $2n-2$ rooks in the $n\times n$ board satisfying the conditions.
\end{solution}",400,1395,Combinatorics,9
37,shl_jbmo_2023_c3,shl_jbmo,2023,c,"There is a grid of size $100 \times 100$. Alice and Bob play the following game. Alice and Bob, with Anna starting first, chooses one integer between 1 and $100^2$ that is not yet written in any of the cell, and writes it in one cell without any number. At the end, Anna computes the sum of the numbers in every line, and her score is the maximum of these 100 numbers. Bob computes the sum of the numbers in every column, and his score is the maximum of these 100 numbers. Alice wins if her score is greater than Bob's score, Bob wins if his score is greater than Alice's score, otherwise no one wins.

Find if one of the player has a winning strategy, and if so which player has a winning strategy.","\begin{solution}
We denote by $(i,j)$ the cell in the $i$-th line and in the $j$-th column for every $1 \leq i,j \leq n$. Bob associates the following pair of cells: $(i, 2k+1), (i, 2k+2)$ for $1 \leq i \leq 100$ and $0 \leq k \leq 49$ except for $(i,k) = (100,0)$ and $(100,1)$, and the pairs $(100,1),(100,3)$ and $(100,2),(100,4)$.

Each time Alice writes the number $j$ in one of the cell, Bob writes the number $100^2+1-j$ in the other cell of the pair.

One can prove by induction that after each of Bob's turn, for each pair of cell, either there is a number written in each of the cell of the pair, or in neither of them. And that if a number $j$ is written, $100^2+1-j$ is also written. Thus Bob can always apply the previous strategy (since $j = 100^2+1-j$ is impossible).

At the end, every line has sum $(100^2+1)\times 50$.

Assume by contradiction that Alice can stop Bob from winning if he applies this strategy. Let $c_j$ be the sum of the number in the $j$-th column for $1 \leq j \leq 100$; then $c_j \leq 50(100^2+1)$. Note that:
\[
100 \times 50(100^2+1) \geq c_1 + \cdots + c_{100} = 1 + \cdots + 100^2 = \frac{100^2(100^2+1)}{2} = 100\times 50(100^2+1)
\]
Thus we have equality in the previous inequality: $c_1 = \cdots = c_{100} = 50(100^2+1)$. But if $a$ is the number written in the case $(100,1)$ and $b$ the number written in the case $(100,2)$, then $c_1 - b + c_2 - c = 99(100^2+1)$. Thus $b+c = 100(100^2+1) - 99(100^2+1) = 100^2+1$; by hypothesis $c$ is also written in the cell $(100,3)$ which is a contradiction.

Thus Bob has a winning strategy.
\end{solution}",699,1594,Combinatorics,10
38,shl_jbmo_2023_c4,shl_jbmo,2023,c,"Anna and Bob playing the following game: The number 2 is initially written on the blackboard. With Anna playing first, they alternately double the number currently written on the blackboard or square it. The person who first writes on the blackboard a number greater than $2023^{10}$ is the winner. Determine which player has a winning strategy.","\begin{solution}
\textbf{Claim.} $2^{109} < 2023^{10} < 2^{110}$.


oindent\textbf{Proof of Claim.} We have
\[
2023^{10} < 2048^{10} = (2^{11})^{10} = 2^{110}.
\]
Furthermore, by Bernoulli's inequality, we also have
\[
2023^{10} = 2048^{10}\left(1-\frac{25}{2048}
ight)^{10} > 2^{110}\cdot\left(1-\frac{10\cdot 25}{2048}
ight) = 2^{109}. \quad \square
\]

Observe now that the number written on the blackboard is always of the form $2^a$ where at each step $a$ is either doubled, or increased by 1. So from the above claim the game is equivalent to the game $G(110)$, where $G(n)$ is the following:

The number 1 is initially written on the blackboard. With Anna playing first, they alternately increase the number by 1 or double it. The person who first writes on the blackboard a number greater than or equal to $n$ is the winner.

We now show that Bob has a winning strategy in $G(110)$. We say that a number $m \in \{2,3,\ldots,109\}$ is winning if the person who writes it down at any step in the game has a winning strategy and losing otherwise.

Let $W = \{3,5,\ldots,13\}\cup\{28,30,\ldots,54\}$. We claim that a number is winning if and only if it belongs to $W$. To show this first observe that given any $w \in W$ both $w+1$ and $2w$ do not belong to $w$ and are smaller than 110. Furthermore, given any $x 
otin W$, either $x+1 \in W$ or $2x \in W$ or $2x \geq 110$. Indeed, for $x \in \{2,4,\ldots,12\}$ and $x \in \{27,29,\ldots,53\}$ we have $x+1 \in W$. For $x \in \{14,15,\ldots,26\}$ we have $2x \in W$, while for $x \in \{55,56,\ldots,109\}$ we have $2x \geq 110$.

So our claim is true. Since on her first move on $G(110)$ Anna writes the number 2, then Bob has a winning strategy.

\medskip

oindent\textbf{Note.} One can avoid the use of Bernoulli inequality and use the immediate but weaker inequality $2^{100} < 2023^{10} < 2^{110}$. This can be done by checking (with similar methods) that in each of $G(101), G(102), \ldots, G(110)$ Bob has a winning strategy.
\end{solution}",345,2001,Combinatorics,11
39,shl_jbmo_2023_c5,shl_jbmo,2023,c,"Consider an increasing sequence of real numbers $a_1 < a_2 < \ldots < a_{2023}$ such that all pairwise sums of the elements in the sequence are different. For such a sequence, denote by $M$ the number of pairs $(a_i, a_j)$ such that $a_i < a_j$ and such that $a_i + a_j < a_2 + a_{2022}$. Find the minimal and the maximal possible value of $M$.","\begin{solution}
We have an increasing sequences of pairwise sums
\begin{align*}
a_2+a_{2022} &< a_3+a_{2022} < a_4+a_{2022} < \ldots < a_{2021}+a_{2022},\\
a_2+a_{2022} &< a_2+a_{2023} < a_3+a_{2023} < \ldots < a_{2022}+a_{2023}.
\end{align*}
Thus, $a_2+a_{2022}$ is less than at least $2019+2021 = 4040$ other pairwise sums. This proves
\[
M \leq \frac{2023\cdot 2022}{2} - 4040 - 1 = 2041212.
\]

Now we will prove that for the sequence
\[
a_k = \begin{cases} 2^k, & 1 \leq k \leq 2022,\\ 2^{2022}+\sqrt{2}, & k = 2023, \end{cases}
\]
the desired bound can be achieved. First, this sequence satisfies properties given in the statement (the pairwise sums without $a_{2023}$ are different as they correspond to different numbers in binary representation with two 1's, and the ones containing $a_{2023}$ are the only ones that are irrational). If $1 \leq i,j \leq 2021$, we have
\begin{align*}
a_i+a_j &= 2^i+2^j \leq 2\cdot 2^{2021} = 2^{2022} < 2^{2022}+2^2,\\
a_1+a_{2022} &= 2^1+2^{2022} < 2^{2022}+2^2,\\
a_1+a_{2023} &= 2^1+2^{2022}+\sqrt{2} < 2^{2022}+2^2.
\end{align*}
All the other pairwise sums are greater, as it is shown in the proof. Therefore, for this sequence it holds
\[
M = \frac{2021\cdot 2020}{2}+2 = 2041212.
\]

For the lower bound, the proof is completely analogous. We prove that there are always 4040 pairwise sums that are smaller than $a_2+a_{2022}$, which proves $M \geq 4040$.

The sequence for which $M = 4040$ is, for example, mirrored sequence from the first example, given by
\[
a_k = \begin{cases} -2^{2022}-\sqrt{2}, & k = 1,\\ -2^{2024-k}, & 2 \leq k \leq 2023. \end{cases}
\]
\end{solution}",344,1627,Combinatorics,12
40,shl_jbmo_2023_g1,shl_jbmo,2023,g,"Let $ABC$ be triangle with circumcentre $O$ and circumcircle $\Omega$. $\Gamma$ is the circle passing through $O, B$ and tangent to $AB$ at $B$. Let $\Gamma$ intersects $\Omega$ second time at $P 
eq B$. The circle passing through $P, C$ and tangent to $AC$ at $C$ intersects with $\Gamma$ at $M$. Prove that $|MP| = |MC|$.","\begin{solution}[Solution 1]
First, we show that $M$ lies on $BC$. Let $M'$ be the second intersection of $\Gamma$ and $BC$. Then, $\angle BPM' = \angle B$ and since $\angle BPC = 180 - \angle A = \angle B + \angle C$ we have $\angle M'PC = \angle C$.

Hence we see that the circumcircle of $M'PC$ is tangent to $AC$ and $M \equiv M'$.

Then, $\angle OPM = \angle OBM = \angle OCM$ and since $O$ lies on the perpendicular bisector of $PC$, so is $M$.
\end{solution}

\begin{solution}[Solution 2 -- PSC suggested solution]
Since the $\triangle AOB$ is isosceles, let $\angle OAB = \angle OBA$, and since the circle $\Gamma$ is tangent to $AB$ at $B$ and the $\triangle BOP$ is isosceles we have:
\[
\angle OAB = \angle OBA = \angle OPB = \angle OBP
\]
Since $OA = OB = OC$ we have $\triangle AOB = \triangle BOP$, in other words $AP = BP$.

Doing some angle chasing we get
\[
\angle PMB = \angle POB = \angle ACP
\]
Using the tangency of the other circle at point $C$ with $AC$ we have
\[
\angle PMB + \angle PMC = \angle ACP + \angle PCR = 180^\circ
\]
This implies that point $M$ lies in the segment $BC$.

Doing some further angle chasing we have
\[
\angle MCO = \angle BCO = \angle OBC = \angle OPM
\]
Using the fact that $\triangle OPC$ is isosceles ($OP = OC$) we have $\angle OPC = \angle OCP$.

Finally, the above results imply
\[
\angle MPC = \angle OPC - \angle OPM = \angle OCP - \angle OCM = \angle MCP
\]
As result, the $\triangle MPC$ is isosceles, which implies $MP = MC$.
\end{solution}",323,1501,Geometry,13
41,shl_jbmo_2023_g2,shl_jbmo,2023,g,"Let $ABC$ be triangle with $AB < AC$ and $\omega$ be its circumcircle. Tangent line to $\omega$ at $A$ intersects with line $BC$ at $D$ and let $E$ be a point on circumcircle of $ABC$ such that $BE$ is parallel to $AD$. $DE$ intersects with segment $AB$ and $\omega$ at $F$ and $G$, respectively. Circumcircle of $BGF$ intersects with $BE$ at $N$. Line $NF$ intersects with $AD$ and $EA$ at $S$ and $T$, respectively. Prove that $DGST$ is cyclic.","\begin{solution}[Solution 1]
\textbf{Claim 1:} $ANBT$ is cyclic.


oindent\textbf{Proof:} As $ABCE$ and $BNFG$ are cyclic, we have $\angle TAB = \angle BCE = 180 - \angle BGE = 180 - \angle BGF = \angle BNF = \angle BNT$ which gives the desired result.

\medskip
\textbf{Claim 2:} Circumcircle of triangle $TSA$ touches line $AB$ at $A$.


oindent\textbf{Proof:} Using the facts $ANBT$ is cyclic and $AD \parallel BE$ we get $\angle ATS = \angle ABN = \angle BAD = \angle FAS$ which means $(TSA)$ touches $AB$ at $A$.

\medskip
\textbf{Claim 3:} $FA^2 = FG\cdot FD$.


oindent\textbf{Proof:} By parallel lines $\triangle FAD \sim \triangle FBE \implies \frac{FA}{FB} = \frac{FD}{FE}$ (1). By Power of Point we have $FA\cdot FB = FG\cdot GE$ (2). Multiplying (1) and (2) side-by-side gives $FA^2 = FG\cdot FD$.

\medskip
\textbf{Finishing:} By \textbf{Claim 2} and \textbf{Claim 3} we get $FS\cdot FT = FA^2 = FG\cdot FD$. So, $DGST$ is cyclic.
\end{solution}

\begin{solution}[Solution 2 -- PSC suggested solution]
Doing some angle chasing
\[
\angle GAF = \angle GAB = \angle GEB = \angle GBA
\]
This implies that $AF$ is tangent to the circumcircle of $\triangle DAG$ and as result $AF^2 = FG\cdot FD$.

Doing some more angle chasing
\[
\angle BAE = \angle EGB = \angle FNE = \angle TNE
\]
This implies that $\triangle TNE \sim \triangle BAE$ and as result
\[
\angle ETN = \angle EBA = \angle BAD
\]
Since from the conditions of the problem $BE \parallel AD$, we have
\[
\angle SAF = \angle BAD = \angle EAB = \angle STA
\]
This implies that $AF$ is tangent to the circumcircle of $\triangle TSA$ and as result we have $AF^2 = SF\cdot TF$. So finally we have
\[
FG\cdot FD = AF^2 = SF\cdot TF
\]
which implies $DGST$ is cyclic as requested.
\end{solution}",446,1756,Geometry,14
42,shl_jbmo_2023_g3,shl_jbmo,2023,g,"Let $A, B, C, D$ and $E$ be five points lying in this order on a circle, such that $AD = BC$. The lines $AD$ and $BC$ meet at the point $F$. The circumcircles of the triangles $CEF$ and $ABF$ meet again at the point $P$.

Prove that the circumcircles of the triangles $BDF$ and $BEP$ are tangent to each other.","\begin{solution}[Solution 1 -- PSC suggested solution]
Since $AB \parallel DC$, we have that $ABCD$ is isosceles trapezium. Lines $PF$, $AB$ and $EC$ are collinear since they are radical axes for the circles $(APFB)$, $(AECB)$ and $(FPEC)$, so let $K = PF \cap AB \cap EC$.

Let $\angle FDB = x$, $\angle APE = y$ and $\angle APB = z$ and let $l$ be the line that passes through point $B$ and is tangent to the circumcircle of $\triangle FDB$ and let $S, R$ be point on the line $l$ as showed in the figure.

It is easy to see that
\[
\angle FBS = x = \angle FDB = \angle RBC
\]
Using Miquel's theorem on the $\triangle KFC$ we have that $KPAE$ is cyclic.

Doing some angle chasing we have
\begin{align*}
\angle AFB &= \angle APB = z\\
\angle APE &= \angle AKE = \angle AKC = \angle KCD = \angle ECD = \angle EBD = y\\
\angle DBR &= \angle DBC - \angle RBC = (\angle DFB + \angle FDB) - x = z - x - x = z
\end{align*}
This implies that
\[
\angle EBR = \angle EBD + \angle DBR = y + z = \angle APE + \angle APB = \angle EPB
\]
which implies the desired result.
\end{solution}

\begin{solution}[Solution 2]
Since $AD = BC$, the quadrilateral $ABCD$ is an isosceles trapezium, and the lines $AB$ and $CD$ are parallel. Consequently, denoting by $\ell$ the tangent line to the circumcircle of triangle $BEP$ at point $B$, we find
\begin{align*}
(BD,\ell) &= (BD,BE)+(BE,\ell) = (BD,BE)+(PE,PB) = (BD,BE)+(PE,PF)+(PF,PB)\\
&= (CD,CE)+(CE,CF)+(AF,AB) = (CD,CE)+(CE,CF)+(AF,AB)\\
&= (CD,CE)+(CE,CF)+(DF,DC) = (CD,CF)+(DF,DC) = (FD,FC) = (FD,FB),
\end{align*}
which yields the desired equality.
\end{solution}",310,1601,Geometry,15
43,shl_jbmo_2023_g4,shl_jbmo,2023,g,"Let $ABCD$ be a cyclic quadrilateral, for which $B$ and $C$ are acute-angles. $M$ and $N$ are the projections of the vertex $B$ on the lines $AC$ and $AD$, respectively. $P$ and $T$ are the projections of the vertex $D$ on the lines $AB$, and $AC$, respectively. $Q$ and $S$ are the intersections of the pairs of lines $MN$ and $CD$, respectively $PT$ and $BC$. Prove the following statements:
\begin{itemize}
  \item[a)] $NS \parallel PQ \parallel AC$;
  \item[b)] $NP = SQ$;
  \item[c)] $NPQS$ is a rectangle iff $AC$ is a diameter of the circumscribed circle of the quadrilateral $ABCD$.
\end{itemize}","\begin{solution}
a) $BM \perp AC$ and $BN \perp AD$, therefore $MN$ is Simpson's line of the point $B$, by respect to the triangle $ACD$, whence we obtain $BQ \perp CD$. From $DP \perp AB$ and $DT \perp AC$, we deduce that $PT$ is Simpson's line of the point $D$ by respect to the triangle $ABC$, therefore $DS \perp BC$.

$BPDQ$ is a cyclic quadrilateral, $(\angle BPD = \angle BQD = 90^\circ)$, thus $\angle BPQ = \angle BDQ = \angle BDC = \angle BAC$, therefore $PQ \parallel AC$.

$BNDS$ is a cyclic quadrilateral, $(\angle BND = \angle BSD = 90^\circ)$, thus $\angle SND = \angle DBS = \angle DBC = \angle CAD$, so $NS \parallel AC$.

Consequently, $NS \parallel PQ \parallel AC$.

b) $BDPN$ is cyclic $(\angle BND = \angle BPD = 90^\circ)$, thus $\angle DNP = \angle PBD = \angle ABD$. As $\angle SND = \angle CAD = \angle CBD$, we obtain:
\[
\angle SNP = \angle SND + \angle DNP = \angle CBD + \angle ABD = \angle ABC. \tag{1}
\]
$BDQS$ is cyclic $(\angle BSD = \angle BQD = 90^\circ)$, therefore $\angle DSQ = \angle DBQ$. $BNDS$ is cyclic, so $\angle NSD = \angle NBD$. Consequently, we deduce:
\[
\angle NSQ = \angle NSD + \angle DSQ = \angle NBD + \angle DBQ = \angle NBQ \tag{2}
\]
As $\angle BAN = 180^\circ - \angle BAD = \angle BCD = \angle BCQ$, the triangles $ABN$ and $CBQ$ are similar, whence $\angle ABN = \angle CBQ$, and we deduce:
\[
\angle NBQ = \angle ABQ + \angle NAB = \angle ABQ + \angle CBQ = \angle ABC,
\]
From (1) and (2) follows: $\angle NSQ = \angle NBQ = \angle ABC = \angle SNP$, therefore $NPQS$ is a trapezoid or a rectangle, hence $NP = SQ$.

c) $NPQS$ is a rectangle iff $\angle SNP = 90^\circ$. As $\angle SNP = \angle ABC$, we deduce that $NPQS$ is a rectangle iff $\angle ABC = 90^\circ$, therefore iff $AC$ is a diameter of the circumscribed circle of $ABCD$.
\end{solution}",604,1818,Geometry,16
44,shl_jbmo_2023_g5,shl_jbmo,2023,g,"Let $D, E, F$ be the points of tangency of incircle of given triangle $ABC$ with sides $BC, CA, AB$, respectively. Denote by $I$ the incenter of $ABC$, by $M$ the midpoint of $BC$ and by $G$ the foot of perpendicular from $M$ to the line $EF$. Prove that the line $ID$ is tangent to the circumcircle of the triangle $MGI$.","\begin{solution}[Solution 1]
Let the intersection points of the line $EF$ with $BI$ and $CI$ be $P$ and $Q$, respectively, while $K$ is the intersection point of the lines $BQ$ and $CP$. It is a rather well-known fact that $\angle BPC = \angle BQC = 90^\circ$, since it holds
\[
\angle IPQ = \frac{\angle ACB}{2} = \angle ICE, \qquad \angle IQP = \frac{\angle ABC}{2} = \angle IBF,
\]
so the pentagons $CPIDE$ and $BDIFQ$ are both cyclic. This implies that $I$ is the orthocenter of triangle $BCK$, so $K$ lies on the line $ID$. Denote by $N$ the midpoint of $KI$. Then it holds
\[
MB = MC = MP = MQ, \qquad NK = NI = NP = NQ,
\]
so $MN$ is a line of symmetry of the segment $PQ$, i.e. $M, N, G$ are collinear. Moreover, a simple angle chasing shows that $\angle MPN = \angle MQN = 90^\circ$, so it follows easily that
\[
NP^2 = NQ^2 = NG\cdot NM.
\]
But then it also holds that $NI^2 = NG\cdot NM$, which implies that the triangles $NIG$ and $NMI$ are similar, so it holds $\angle NIG = \angle IMG$ and the proof is finished.
\end{solution}

\begin{solution}[Solution 2 -- PSC suggested solution]
Let $P = AI \cap FE$ and let $L$ be the intersection point of lines $EF$ and $BI$.

It is known that $AI \perp FE$ and $ML \parallel AB$ and $ML = MC = BM = \frac{BC}{2}$ or in other words $BL \perp LC$.

$IELC$ is cyclic since $\angle IEC = \angle ILC = 90^\circ$. This implies that
\[
\angle ILG = \angle ECI = \angle ICM = \frac{\angle ACB}{2}
\]

Doing some angle chasing we have
\[
\angle PIL = 90^\circ - \angle PLI = 90^\circ - \angle ICD = \angle DIC = 90^\circ - \frac{\angle ACB}{2}
\]

In $\triangle GLM$ we have $GL = ML\sin\angle GML = \frac{BC}{2}\sin\frac{\angle BAC}{2}$

In $\triangle ILC$ we have $IL = IC\cos\angle LIC = IC\sin\frac{\angle BAC}{2}$

Dividing both sides implies
\[
\frac{GL}{IL} = \frac{\frac{BC}{2}\sin\frac{\angle BAC}{2}}{IC\sin\frac{\angle BAC}{2}} = \frac{BC}{2IC} = \frac{MC}{IC}
\]
This results that $\triangle GIL \sim \triangle MIC$. Doing some further angle chasing we have
\begin{align*}
\angle GIL &= \angle MIC\\
\Rightarrow \angle PIL - \angle PIG &= \angle DIC - \angle DIM\\
\Rightarrow 90^\circ - \frac{\angle ACB}{2} - \angle PIG &= 90^\circ - \frac{\angle ACB}{2} - \angle DIM\\
\Rightarrow \angle PIG &= \angle DIM
\end{align*}
Since $IP \perp FL$ and $MG \perp FL$, we have that $IP \parallel MG$.

Doing some further angle chasing
\[
\angle PIG = \angle DIM \implies \angle IGM = \angle DIM
\]
This implies that $DI$ is tangent to the circumcircle of the $\triangle IGM$ as required.
\end{solution}",322,2553,Geometry,17
45,shl_jbmo_2023_g6,shl_jbmo,2023,g,"Let $ABC$ be an acute triangle with circumcenter $O$, let $D$ be the foot of the altitude from $A$ to $BC$ and denote by $M$ the midpoint of $OD$. Let $O_b$ and $O_c$ be the circumcenters of triangles $AOC$ and $AOB$, respectively. If $AO = AD$, prove that the points $A$, $O_b$, $M$ and $O_c$ are concyclic.","\begin{solution}
Note that $AB = AC$ cannot hold since $AO = AD$ would imply that $O$ is the midpoint of $BC$, which is not possible for an acute triangle. So we may assume without loss of generality that $AB < AC$.

Let $M_b$ and $M_c$ be the midpoints of $AC$ and $AB$, respectively. Since $\angle AM_bO = \angle AM_cO = 90^\circ = \angle AMO$ (the latter since $AO = AD$), the pentagon $AM_bOMM_c$ is cyclic.

Next, notice that $AM$ is the perpendicular bisector of $OD$, $O_bO_c$ is the perpendicular bisector of $AO$ and $M_bM_c$ is the perpendicular bisector of $AD$. Hence these three lines are concurrent -- denote their common point by $T$.

The quadrilateral $AO_bMO_c$ is cyclic if and only if $AT\cdot TM = O_bT\cdot O_cT$. From the cyclic $AM_bMM_c$ we have $AT\cdot TM = M_bT\cdot M_cT$. Hence it now suffices to argue $M_bT\cdot M_cT = O_bT\cdot O_cT$ -- or equivalently, that $M_b$, $M_c$, $O_b$ and $O_c$ are concyclic.

We assume that $\angle AOB < 90^\circ$ and $\angle AOC > 90^\circ$ so that $O_c$ is in the interior of triangle $AOB$ and $O_b$ in external to the triangles $AOC$ (the other cases are analogous and if $\angle AOB = 90^\circ$ or $\angle AOC = 90^\circ$, then $M_b \equiv O_b$ or $M_c \equiv O_c$ and we are automatically done). We have
\[
\angle M_cM_bO_b = 90^\circ + \angle AM_bM_c = 90^\circ + \angle ACB
\]
as well as (since $O_cO_b$ is a perpendicular bisector of $AO$ and hence bisects $\angle AO_cO$)
\begin{align*}
\angle M_cO_cO_b &= 180^\circ - \angle OO_cO_b = 90^\circ + \frac{\angle AO_cM_c}{2}\\
&= 90^\circ + \frac{\angle AO_cB}{4} = 90^\circ + \frac{\angle AOB}{2} = 90^\circ + \angle ACB
\end{align*}
and therefore $O_bM_bO_cM_c$ is cyclic, as desired.

\medskip

oindent\textbf{Proposer's Remark.} Another approach is to introduce $B' = OO_c \cap BC$ and $C' = OO_b \cap BC$, then it is evident that $AOC'B$ and $AOB'C$ are cyclic. It can be shown that all the six points $A$, $O_b$, $M$, $B'$, $C'$, $O_c$ are concyclic.
\end{solution}",308,1991,Geometry,18
46,shl_jbmo_2023_g7,shl_jbmo,2023,g,"Let $D$ and $E$ be arbitrary points on the sides $BC$ and $AC$ of triangle $\triangle ABC$, respectively. The circumcircle of $\triangle ADC$ meets for the second time the circumcircle of $\triangle BCE$ at the point $F$. The line $FE$ meets the line $AD$ at the point $G$, while the line $FD$ meets the line $BE$ at the point $H$. Prove that the lines $CF$, $AH$ and $BG$ pass through the same point.","\begin{solution}
Let the line $AH$ meet the circumcircle of $\triangle ADC$ at the point $I$, and the line $BG$ meet the circumcircle $\triangle BCF$ at the point $J$. Let $M$ be the common point of the lines $AD$ and $BE$.

Because the pentagons $AFIDC$ and $BCEJF$ are cyclic, we obtain the following angle equalities:
\[
\angle FAG \equiv \angle FAD = \angle FCD \equiv \angle FCB = \angle FJB \equiv \angle FJG,
\]
from where we conclude that $AFGJ$ is cyclic.

Using the above together with the fact that $BCEJF$ is cyclic we get:
\[
\angle MAJ \equiv \angle GAJ = \angle GFJ \equiv \angle EFJ = \angle EBJ \equiv \angle MBJ.
\]
This means that $ABMJ$ is cyclic.

In the same way we also prove that $ABIM$ is cyclic, from where we conclude that the point $I$ lies on the circumcircle of the quadrilateral $ABMJ$.

Because $BJ$ is the radical axis of the circles $(BCEJF)$ and $(ABIMJ)$, $AI$ is the radical axis of $(AIDC)$ and $(ABIMJ)$, and $(BCEJF)$, we get that the lines $BJ$, $CF$ and $AI$ meet at a common point, which concludes the proof.
\end{solution}",401,1066,Geometry,19
47,shl_jbmo_2023_n1,shl_jbmo,2023,n,Find all positive integers $a$ and $b$ such that $a! + b$ and $b! + a$ are powers of 5.,"\begin{solution}
The condition is symmetric so we can assume that $b \leq a$.

The first case is when $a = b$. In this case, $a! + a = 5^m$ for some positive integer $m$. We can rewrite this as $a\cdot((a-1)!+1) = 5^m$. This means that $a = 5^k$ for some integer $k \geq 0$. It is clear that $k$ cannot be 0. If $k \geq 2$, then $(a-1)!+1 = 5^l$ for some $l \geq 1$, but $a-1 = 5^k - 1 > 5$, so $5|(a-1)!$, which is not possible because $5|(a-1)!+1$. This means that $k = 1$ and $a = 5$. In this case, $5!+5 = 125$, which gives us the solution $(5,5)$.

Let us now assume that $1 \leq b < a$. Let us first assume that $b = 1$. Then $a+1 = 5^x$ and $a!+1 = 5^y$ for integers $x, y \geq 1$. If $x \geq 2$, then $a = 5^x - 1 \geq 5^2 - 1 > 5$, so $5|a!$. However, $5|5^y = a!+1$, which leads to a contradiction. We conclude that $x = 1$ and $a = 4$. From here $a!+b = 25$ and $b!+a = 5$, so we get two more solutions: $(1,4)$ and $(4,1)$.

Now we focus on the case $1 < b < a$. Then we have $a!+b = 5^x$ for $x \geq 2$, so $b\cdot\left(\frac{a!}{b}+1
ight) = 5^x$, where $b|a!$ because $a > b$. Because $b|5^x$ and $b > 1$, we have $b = 5^z$ for $z \geq 1$. If $z \geq 2$, then $5 < b < a$, so $5|a!$, which means that $\frac{a!}{b}+1$ cannot be a power of 5. We conclude that $z = 1$ and $b = 5$. From here $5!+a$ is a power of 5, so $5|a$, but $a > b = 5$, which gives us $a \geq 10$. However, this would mean that $25|a!$, $5|b$ and $25
mid b$, which is not possible, because $a!+b = 5^x$ and $25|5^x$.

We conclude that the only solutions are $(1,4)$, $(4,1)$ and $(5,5)$.
\end{solution}",87,1588,Number Theory,20
48,shl_jbmo_2023_n2,shl_jbmo,2023,n,"A positive integer is called \textit{Tiranian} if it can be written as $x^2 + 6xy + y^2$, where $x$ and $y$ are (not necessarily distinct) positive integers. The integer $36^{2023}$ is written as sum of $k$ Tiranian integers. What is the smallest possible value of $k$?","\begin{solution}
Answer: 2.

We firstly show that $k = 1$ is not possible, i.e. $x^2+6xy+y^2 = 2^{4046}\cdot 3^{4046}$ has no positive integer solutions.

Firstly, modulo 3 implies $x^2+y^2$ is divisible by 3 and hence $x$ and $y$ are both divisible by 3. Now writing $x = 3x_1$, $y = 3y_1$ and dividing by $3^2$ leads to $x_1^2+6x_1y_1+y_1^2 = 2^{4046}\cdot 3^{4044}$; repeating this 2022 times leads to an equation of the form
\[
u^2 + 6uv + v^2 = 2^{4046}
\]
where $u$ and $v$ are positive integers.

If $v$ is even, then $u$ must also be even and writing $u = 2u_1$, $v = 2v_1$ and dividing by 4 leads to $u_1^2+6u_1v_1+v_1^2 = 2^{4044}$; repeating several times leads to an equation of the form
\[
s^2 + 6st + t^2 = 2^{2A}
\]
where $s, t, A$ are positive integers with $t$ odd and $A \geq 2$. The latter can be rewritten as $(s+3t)^2 - 8t^2 = 2^{2A}$. Now by modulo 8 we see that $(s+3t)^2$ is divisible by 8, thus $s+3t$ is divisible by 4, thus $(s+3t)^2$ is divisible by 16. But then it must be the case that $8t^2$ is divisible by 16, contradicting that $t$ is odd. Therefore, $k = 1$ is not possible.

Regarding an example with $k = 2$, note that
\[
36 = [1^2+6\cdot 1\cdot 1+1^2] + [3^2+6\cdot 3\cdot 1+1^2]
\]
and hence multiplication by $((36)^{1011})^2$ leads to
\begin{align*}
36^{2023} &= [(36^{1011})^2 + 6\cdot 36^{1011}\cdot 36^{1011} + (36^{1011})^2]\\
&\quad + [(3\cdot 36^{1011})^2 + 6\cdot(3\cdot 36^{1011})\cdot 36^{1011} + (36^{1011})^2]
\end{align*}
that is, it is the sum of the integers $a_1^2+6a_1b_1+b_1^2$ and $a_2^2+6a_2b_2+b_2^2$, where $a_1 = a_2 = b_2 = 36^{1011}$ and $b_1 = 3\cdot 36^{1011}$.

\medskip

oindent\textbf{Proposer's Remark.} The modulo 3 step in the $k=1$ case is indeed essential; otherwise, if we do the modulo 4 only, we would reach $s^2+6st+t^2 = 2^{2A}\cdot 3^{4046}$. The problem here is that $A$ could actually be 0 and then in the form $(s+3t)^2-8t^2 = 3^{4046}$ one cannot probably get a contradiction without using modulo 3 repeatedly.
\end{solution}",269,2010,Number Theory,21
49,shl_jbmo_2023_n3,shl_jbmo,2023,n,"Let $A$ be a subset of $\{2, 3, \ldots, 28\}$ such that if $a \in A$, then the residue obtained when we divide $a^2$ by 29 also belongs to $A$.

Find the minimum possible value of $|A|$.","\begin{solution}[Solution 1]
Denote $p = 29$, which is prime. For 2 positive integers $x, y$, we will denote by $r(x,y)$ the residue obtained when we divide $x$ by $y$.

Then $a^2 \equiv a \pmod{p} \Leftrightarrow p|a(a-1) \Leftrightarrow a \equiv 0 \pmod{p}$ or $a \equiv 1 \pmod{p}$. Hence, if $a \in A$, then $r(a^2, p) 
eq a$, so $|A|$ cannot be 1.

Assume that $|A| = 2$. Then $A = \{a,b\}$ and $r(a^2, p) 
eq a$ and $r(b^2, p) 
eq b$, so $a^2 \equiv b \pmod{p}$ and $b^2 \equiv a \pmod{p}$.

Hence $a^4 \equiv b^2 \equiv a \pmod{p}$. As $2 \leq a \leq 28 = p-1$, we have $(a,p)=1$, so $a^3 \equiv 1 \pmod{p}$.

By Little Fermat's Theorem, we have $a^{p-1} \equiv 1 \pmod{p} \Leftrightarrow a^{28} \equiv 1 \pmod{p}$. But $a^3 \equiv 1 \pmod{p} \Rightarrow (a^3)^9 \equiv 1 \pmod{p} \Leftrightarrow a^{27} \equiv 1 \pmod{p}$. And as $a^{28} \equiv 1 \pmod{p}$, it means that $a \equiv 1 \pmod{p}$, which is impossible, as $2 \leq a \leq 28 = p-1$.

This means that $|A| \geq 3$.

We will give an example for $|A| = 3$.

We know that $2^{28} \equiv 1 \pmod{29}$, by Little Fermat's Theorem, so $(2^4)^7 \equiv 1 \pmod{29}$.

Let $a = 16$, for which we have $a^2 = 256 \equiv 24 \pmod{29}$, and let $b = 24$, which leads to $b^2 = 576 \equiv 25 \pmod{29}$. Also, for $c = 25$, we have $c^2 = 625 \equiv 16 \pmod{29} \equiv a \pmod{29}$.

So, we may choose $A = \{a,b,c\} = \{16, 24, 25\}$. Indeed $r(a^2, 29) = b \in A$, $r(b^2, 29) = c \in A$, $r(c^2, 29) = a \in A$.

Consequently, $|A|_{\min} = 3$.

\medskip

oindent\textbf{Proposer's Remarks:} The problem has the same answer for any prime $p$ with $p \equiv 8 \pmod{29}$. Moreover, if $p \equiv 1 \pmod{3}$, then $|A|_{\min} = 2$.

\textit{Proof:} Indeed, the equation $a^4 \equiv a \pmod{p}$ is equivalent to $a^3 \equiv 1 \pmod{p}$ and as $a 
eq 1$, it means that $a^2+a+1 \equiv 0 \pmod{p} \Leftrightarrow (2a+1)^2 \equiv -3 \pmod{p}$.

And as
\[
\left(\frac{-3}{p}
ight) = 1,
\]
it means that there is such $a$. We choose $b \equiv -a-1$, so $b = 28-a$, and the conclusion will hold.
\end{solution}

\begin{solution}[Solution 2 -- PSC suggested solution]
If we take the set $\{16, 24, 25\}$ we have,
\[
16^2 \equiv 24 \pmod{29}, \quad 24^2 \equiv 25 \pmod{29}, \quad 25^2 \equiv 16 \pmod{29}.
\]
So, $|\mathbb{A}| = 3$ works. We gonna prove that this is the minimum. In order to do that we gonna show that $|A| = 1$ and $|A| = 2$ are not possible.

\textit{Case 1.} $|\mathbb{A}| = 1$

Then we would have $a^2 \equiv a \pmod{29} \Rightarrow a(a-1) \equiv 0 \pmod{29} \Rightarrow a \equiv 0,1 \pmod{29}$, which is not possible.

\textit{Case 2.} $|\mathbb{A}| = 2$

Then we would have $a^2 \equiv b \pmod{29}$ and $b^2 \equiv a \pmod{29}$. So,
\[
a^2 - b^2 \equiv b - a \pmod{29} \Rightarrow (a-b)(a+b+1) \equiv 0 \pmod{29} \Rightarrow a+b \equiv -1 \pmod{29}.
\]
So, we would get,
\begin{align*}
b^2 &\equiv a \pmod{29} \equiv -b-1 \pmod{29} \Rightarrow b^2+b+1 \equiv 0 \pmod{29}\\
&\Rightarrow (2b+1)^2 \equiv -3 \pmod{29} \equiv 26 \pmod{29}.
\end{align*}
But we can easily prove that for all integers $x$ we have,
\[
x^2 \equiv 0, 1, 4, 5, 6, 7, 9, 13, 16, 20, 22, 23, 24, 25 \pmod{29},
\]
which is not possible.

So, minimal possible value of $|A|$ is 3.
\end{solution}",186,3237,Number Theory,22
50,shl_jbmo_2023_n4,shl_jbmo,2023,n,"The triangle $ABC$ is sectioned by $AD$, $BE$ and $CF$ (where $D \in (BC)$, $E \in (CA)$ and $F \in (AB)$) in seven disjoint polygons named \textit{regions}. In each one of the nine vertices of these regions we write a digit, such that each nonzero digit appears exactly once. We assign to each side of a region the lowest common multiple of the digits at its ends, and to each region the highest common divisor of the numbers assigned to its sides.

Find the highest possible value of the product of the numbers assigned to the regions.","\begin{solution}
Let $M$ be the required maximum. We shall show that $M = 2^5\cdot 3^3$. In the left hand side there is an example.

For any prime $p$, if $p$ divides the number associated to a region (we say that $p$ divides the region), then:
\begin{enumerate}
  \item If the region is a triangle, we have $p|[a,b]$, $p|[a,c]$ and $p|[c,a]$, where $a, b, c$, are numbers in the vertices. Then, $p$ divides at least two of the numbers in the vertices.
  \item If the region is a quadrilateral, then $p|[a,b]$, $p|[c,d]$ and $p|[a,d]$, $p|[b,c]$ where $a, b, c, d$ are the numbers in the vertices. Then, $p$ divides at least two of the numbers in the vertices, \textbf{situated in opposite vertices}.
\end{enumerate}

Then, in both cases we can associate to each region, with respect to $p$, an unique pair of two vertices (two triangles have no common side, two quadrilaterals have no common diagonal or one of the diagonals cannot be another's side). As for $p \geq 5$ there is at most one digit divisible with $p$, we have $M = 2^n\cdot 3^m$.

\textbf{For} $p = 3$, we have 3 digits multiple of 3, with which we can form maximum 3 distinct pairs. If 9 divides one of the regions, then $9|[a,b]$, $9|[b,c]$ and $9|[c,a]$ or $9|[a,b]$, $9|[b,c]$, $9|[c,d]$ and $9|[d,a]$ for some distinct digits $a, b, c, d$. In both cases we need two distinct digits multiple of 9, impossible. So all regions divisible by 3 are not divisible by 9. Therefore $m \leq 3$.

\textbf{For} $p = 2$, we have 4 even digits. Assume we have at least 5 even regions. If two of the regions 1, 3, 5 are even, let's say 1 and 3, as they have no common vertices, each has to have one pair of even numbers in their vertices. Then $B, M, D$ are all odd, so regions 4, 5, 6 cannot be even, so we have maximum 4 even regions, false. As we have 5 even regions, all regions are even apart from two among 1, 3, 5 -- for example 1, 2, 4, 6, 7.

Regions 1 and 4 have no common vertices, so the 4 even digits will be in their vertices. Then $B$ and $E$ are odd and as 2 and 6 are even, $M, F, A, N$ must be even, so region 4 is not even ($C$ and $D$ are odd). We have a contradiction, so at most 4 of the regions are even.

If an even region is divisible by 4, then $4|[a,b]$, $4|[b,c]$ and $4|[c,a]$ or $4|[a,b]$, $4|[b,c]$, $4|[c,d]$ and $4|[d,a]$ for distinct digits $a, b, c, d$. Then $\max\{\exp_2 a, \exp_2 b\} \geq 2$ and similarly, so we need two distinct digits divisible by 4, which can only be 4 and 8, so an unique pair, therefore such region is unique. As 8 cannot divide this region (a similar argument would imply two distinct digits divisible to 8), we conclude $n \leq 2+1+1+1 = 5$.
\end{solution}",537,2675,Number Theory,23
51,shl_jbmo_2023_n5,shl_jbmo,2023,n,"Find the greatest positive integer $k$ such that we can find $A \subseteq \{1, 2, \ldots, 100\}$ with $k$ elements such that, for any $a, b \in A$, $a$ divides $b$ if and only if $s(a)$ divides $s(b)$, where $s(k)$ denotes the sum of $k$'s digits.","\begin{solution}
Let $a, b \in A$ with $s(a) = s(b)$. As $s(a)|s(b)$, we have $a|b$, and as $s(b)|s(a)$, we get $b|a$. So $a = b$. As $s(a) \leq 18$, for any $a \in \{1, 2, \ldots, 100\}$, we get $k \leq 18$. Assume we can build an example for $k = 18$. So there is an $a \in A$ with $s(a) = p$, for every $p \in \{1, 2, \ldots, 18\}$.

The sole $a \in \{1, 2, \ldots, 100\}$ with $s(a) = 18$ is $a = 99$, so $99 \in A$. Let $b \in A$ such that $s(b) = 2$. Then $b \in \{2, 11, 20\}$. As $s(b) = 2|18 = s(a)$, we get $b|a$. So $b = 11$.

Let $c \in A$ such that $s(c) = 10$. We have $s(11)|s(c)$, so $11|c$ and so $c = 55$. Let $d \in A$ such that $s(d) = 5$. We have $s(d) = 5|s(55)$, so $d|55$. As the only numbers with sum of digits 5 from $\{1, 2, \ldots, 100\}$ are $\{5, 14, 23, 32, 41, 50\}$, we have $d = 5$. Let $e \in A$ for which $s(e) = 15$. Then $e \in \{69, 78, 87, 96\}$. But $s(5)|s(e)$ so $5|e$, false, so $k \leq 17$. The following example

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
$a$ & 1 & 11 & 3 & 22 & 5 & 33 & 7 & 44 & 9 & 55 & 92 & 66 & 94 & 77 & 88 & 89 & 99 \\
\hline
$s(a)$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 16 & 17 & 18 \\
\hline
\end{tabular}
\end{center}

shows that 17 is the required maximum.
\end{solution}",247,1304,Number Theory,24
52,shl_jbmo_2023_n6,shl_jbmo,2023,n,"\textbf{Problem (v1):} Find all primes $p$ satisfying the following conditions:
\begin{enumerate}[label=(
oman*)]
  \item $\dfrac{p+1}{2}$ is a prime number.
  \item There are at least three distinct positive integers $n$ in which $\dfrac{p^2+n}{p+n^2}$ is an integer.
\end{enumerate}

\textbf{Problem (v2):} Let $p 
eq 5$ be a prime number such that $\dfrac{p+1}{2}$ is also prime. Suppose there exist positive integers $a < b$ such that $\dfrac{p^2+a}{p+a^2}$ and $\dfrac{p^2+b}{p+b^2}$ are integers. Show that $b = (a-1)^2+1$.","\textbf{Problem (v2):} Let $p 
eq 5$ be a prime number such that $\dfrac{p+1}{2}$ is also prime. Suppose there exist positive integers $a < b$ such that $\dfrac{p^2+a}{p+a^2}$ and $\dfrac{p^2+b}{p+b^2}$ are integers. Show that $b = (a-1)^2+1$.

\begin{solution}[Solution 1 -- (v1)--(v2)]
Let $p$ be a prime number such that $r := \dfrac{p+1}{2}$ is also prime. Let $n 
eq p$ be a positive integer such that $\dfrac{p^2+n}{p+n^2}$ is an integer. First, it is clear that $n < p$. Note that $\gcd(p+n^2, n) = \gcd(p,n) = 1$ and
\[
\frac{p^2+n}{p+n^2} = p - n^2 + \frac{n(n+1)(n^2-n+1)}{p+n^2},
\]
so we have
\[
p + n^2 \mid (n+1)(n^2-n+1). \tag{$*$}
\]
Also, since $r$ is prime and $2r = p+1 > n+1$, we have
\[
\gcd(p+n^2, n+1) = \gcd(p+1, n+1) = \gcd(2r, n+1) \in \{1, 2, r\}.
\]
Now, if $\gcd(p+n^2, n+1) = r$, then $2r > n+1$ implies that $r = n+1$. Thus, plugging $n = r-1$ and $p = 2r-1$, $(*)$ gives that
\[
r^2 | r(r^2-3r+3) \implies r|(r^2-3r+3) \implies r|3,
\]
so $r = 3$, i.e. $n = 2$ and $p = 5$. If $\gcd(p+n^2, n+1) 
eq r$, by $(*)$, we obtain $p+n^2|2(n^2-n+1)$, which implies that $p+n^2 = 2(n^2-n+1)$ as $p+n^2 > n^2-n+1$.

As a result, if $\dfrac{p^2+n}{p+n^2}$ is an integer for some positive integer $n$ where $p$ and $\dfrac{p+1}{2}$ are prime numbers, then one of the following holds:
\begin{enumerate}[label=(
oman*)]
  \item $p = 5$ and $n = 2$,
  \item $p = n^2-2n+2$,
  \item $p = n$.
\end{enumerate}
\end{solution}

\begin{solution}[Solution 2 -- (v2): PSC suggested solution]
We gonna show that if $p+n^2 \mid p^2+n$ has at least two different values of $n$ which satisfy the condition then there are exactly two values of $n$ which satisfy the condition. First $n = p$ clearly satisfies the condition. Now if $n 
eq p$ from $p+n^2 \mid p^2+n$ we have that,
\[
p+n^2 \leq p^2+n \Rightarrow n^2-p^2+p-n \leq 0 \Rightarrow (n-p)(n+p-1) \leq 0 \Rightarrow n \leq p.
\]
Now since $n < p$ we can suppose $a = n$ and $b = p$ and in order to prove the statement we need to prove $p = (n-1)^2+1 = n^2-2n+2$.
\begin{align*}
p+n^2 &\mid p^2+n = p(p+n^2)+n-pn^2 \Rightarrow p+n^2 \mid pn^2-n = n(pn-1).
\end{align*}
Because $\gcd(p,n) = 1$ then $\gcd(p+n^2, n) = \gcd(p+n^2-n\cdot n, n) = \gcd(p,n) = 1$ we have that $p+n^2 \mid pn-1$ so,
\[
p+n^2 \mid pn-1 = p(n+p^2)-p^3-1 \Rightarrow p+n^2 \mid p^3+1 = \left(\frac{p+1}{2}
ight)\cdot 2(p^2-p+1).
\]
We gonna show that $\gcd\left(p+n^2, \frac{p+1}{2}
ight) = 1$. Suppose otherwise then since $\frac{p+1}{2}$ is a prime number we have that,
\[
\frac{p+1}{2} \mid p+n^2 \mid p^2+n = p^2-1+n+1 = (p-1)(p+1)+n+1 \Rightarrow \frac{p+1}{2} \mid n+1.
\]
Because $n+1 < p+1$ we have that $n+1 = \frac{p+1}{2} \Rightarrow n = \frac{p-1}{2}$. So we have,
\[
p + \left(\frac{p-1}{2}
ight)^2 \mid p^2 + \frac{p-1}{2} \Rightarrow (p+1)^2 \mid 4p^2+2p-2 = 4(p+1)^2-6p-6
\]
\[
\Rightarrow (p+1)^2 \mid 6(p+1) \Rightarrow p+1 \mid 6 \Rightarrow p+1 = 2, 3, 6 \Rightarrow p = 1, 2, 5.
\]
Which clearly is not possible from condition. Hence, $\gcd\left(p+n^2, \frac{p+1}{2}
ight) = 1$ so we have,
\[
p+n^2 \mid 2(p^2-p+1) = 2(p+n^2)+2(n-p)(p+n^2) \mid 2(p+n-1).
\]
Since clearly $2(p+n^2) > 2(p+n-1) \Rightarrow \frac{2(p+n-1)}{p+n^2} < 2$ so from last 2 relations we have,
\[
2(p+n-1) = p+n^2 \Rightarrow p = n^2-2n+2,
\]
as desired.

\medskip

oindent\textbf{PSC Remark.} The solution 2 is presented for v2. If we want to adapt it for v1 as well, we just have to discuss the case $p = 5$ separately.
\end{solution}",529,3464,Number Theory,25
394,tst_jbmo_ro_2023_1_p1,tst_jbmo,2023,a,"Determine the smallest natural number $n$ for which there exist distinct nonzero natural numbers $a, b, c$, with the property that $n = a + b + c$ and $(a+b)(b+c)(c+a)$ is a perfect cube.","Let $n$ be the minimum sought, and $a, b, c \in \mathbb{N}^*$ distinct, such that $n = a + b + c$ and $(a+b)(b+c)(c+a) = k^3$ with $k \in \mathbb{N}$.

\textbf{Case I:} If all of $a, b, c$ are even, then for distinct nonzero naturals $a' = \tfrac{a}{2}$, $b' = \tfrac{b}{2}$, $c' = \tfrac{c}{2}$, we have $\tfrac{n}{2} = a' + b' + c' \in \mathbb{N}$ and $(a' + b')(b' + c')(c' + a') = \left(\tfrac{k}{8}
ight)^3 \in \mathbb{N}$, contradicting the minimality of $n$. \hfill \textbf{[2p]}

\textbf{Case II:} If all of $a, b, c$ are odd, then $n \geq 1 + 3 + 5 = 9$. Since $(1, 3, 5)$ does not satisfy the condition that $(a+b)(b+c)(c+a)$ is a perfect cube, $n \geq 9 + 2 = 11$. \hfill \textbf{[1p]}

\textbf{Case III:} If $a, b, c$ do not all have the same parity, without loss of generality suppose only $a$ and $b$ have the same parity. Then $a+b$ is even, while $c+a$ and $b+c$ are odd. Then $2 \mid (a+b)(b+c)(c+a) = k^3$, so $2 \mid k$, hence $8 \mid (a+b)(b+c)(c+a)$. Since $b+c$ and $c+a$ are odd, $8 \mid a+b$, so $a + b \geq 8$. \hfill \textbf{[1p]}

If $n = 9$, then $a + b = 8$ and $c = 1$, and $(b+c)(c+a) = ab + 9$ must be a perfect cube, so $ab \geq 18$. But $4 = \dfrac{a+b}{2} \geq \sqrt{ab}$, so $ab \leq 16$, contradiction. Therefore $n \geq 10$. \hfill \textbf{[2p]}

We observe that for $n = 10$ and $a = 1, b = 2, c = 7$, the conditions are satisfied. Combining Cases I, II, and III, we deduce the minimum value of $n$ is $\boxed{10}$. \hfill \textbf{[1p]}",187,1475,Algebra,1
395,tst_jbmo_ro_2023_1_p2,tst_jbmo,2023,g,"Consider a triangle $ABC$. Let $P$ and $Q$ be points on sides $AB$ and $AC$ respectively, such that $AP = AQ$, and line $PQ$ passes through the incenter $I$ of triangle $ABC$. Denote by $M$ the second intersection point of the circumscribed circles of triangles $BPI$ and $CQI$, by $D$ the intersection of lines $PM$ and $BI$, and by $E$ the intersection of lines $QM$ and $CI$. Prove that line $MI$ passes through the midpoint of segment $DE$.","Triangle $APQ$ is isosceles, so $\angle APQ = \angle AQP$. Quadrilaterals $BMIP$ and $CMIQ$ are cyclic, so $\angle APQ = \angle BMI$ and $\angle AQP = \angle CMI$, consequently $\angle BMI = \angle CMI$. \hfill (1) \hfill \textbf{[1p]}

Let $K$ be the intersection of lines $IM$ and $BC$.

Since $BI$ bisects $\angle ABC$ and quadrilateral $BMIP$ is cyclic, $\angle IMP = \angle IBP = \angle IBK$. From $\angle KMD = \angle KBD$, quadrilateral $BDKM$ is cyclic, so $\angle IDK = \angle BMK$. Analogously, $CEKM$ is cyclic and $\angle IEK = \angle CMK$.

Using (1), $\angle IDK = \angle IEK$. \hfill \textbf{[3p]}

Quadrilaterals $BDKM$ and $BMIP$ are cyclic, so $\angle BKD = \angle BMD = \angle BIP$. \hfill (2)

Quadrilaterals $CEKM$ and $CMIQ$ are cyclic, so $\angle CKE = \angle CME = \angle CIQ$. \hfill (3)

From (2) and (3): $\angle BKD + \angle CKE = \angle BIP + \angle CIQ$. Consequently:
\[
\angle DKE = 180^\circ - (\angle BKD + \angle CKE) = 180^\circ - (\angle BIP + \angle CIQ) = \angle DIE.
\]
\hfill \textbf{[2p]}

Since $\angle IDK = \angle IEK$ and $\angle DKE = \angle DIE$, $DIEK$ is a parallelogram, so line $KI$ passes through the midpoint of segment $DE$. Since $I, K, M$ are collinear, the conclusion follows. \hfill \textbf{[1p]}",444,1255,Geometry,2
396,tst_jbmo_ro_2023_1_p3,tst_jbmo,2023,n,"On a blackboard, the numbers $2^3 - 2,\; 3^3 - 3,\; 4^3 - 4,\; \ldots,\; (2n+1)^3 - (2n+1)$ are written, where $n$ is a natural number with $n \geq 2$. An \emph{operation} consists of erasing three arbitrary numbers $a, b, c$ from the board and replacing them with $\dfrac{abc}{ab + bc + ca}$. Several operations are performed until two numbers remain. Show that the sum of the last two numbers remaining on the board is greater than $16$.","Since $\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = \dfrac{ab + bc + ca}{abc} = \dfrac{1}{\frac{abc}{ab+bc+ca}}$ for any positive $a, b, c$, after each operation the sum of the reciprocals of the numbers remaining on the board equals that of the original numbers. Therefore, the sum of the reciprocals of the last two numbers equals the sum $S$ of the reciprocals of the initial numbers. \hfill \textbf{[2p]}

Since $\dfrac{1}{k^3 - k} = \dfrac{1}{k(k-1)(k+1)} = \dfrac{1}{2}\left(\dfrac{1}{(k-1)k} - \dfrac{1}{k(k+1)}
ight)$ for any $k > 1$, we get:
\begin{align*}
S &= \frac{1}{2^3 - 2} + \frac{1}{3^3 - 3} + \cdots + \frac{1}{(2n+1)^3 - (2n+1)} \\
&= \frac{1}{2}\left(\frac{1}{1 \cdot 2} - \frac{1}{2 \cdot 3} + \frac{1}{2 \cdot 3} - \frac{1}{3 \cdot 4} + \cdots + \frac{1}{2n(2n+1)} - \frac{1}{(2n+1)(2n+2)}
ight) \\
&= \frac{1}{2}\left(\frac{1}{2} - \frac{1}{(2n+1)(2n+2)}
ight) = \frac{2n^2 + 3n}{8n^2 + 12n + 4}.
\end{align*}
\hfill \textbf{[2p]}

Let $x$ and $y$ be the last two numbers. We have $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{x+y}{xy} = \dfrac{2n^2 + 3n}{8n^2 + 12n + 4}$.

Since $\dfrac{4}{x+y} \leq \dfrac{x+y}{xy}$ for all $x, y > 0$, we get
\[
\frac{4}{x+y} \leq \frac{2n^2 + 3n}{8n^2 + 12n + 4} < \frac{1}{4},
\]
from which $x + y > 16$. \hfill \textbf{[3p]}",439,1279,Number Theory,3
397,tst_jbmo_ro_2023_1_p4,tst_jbmo,2023,c,"A cube of side length $3$ is divided into $27$ unit cubes, and inside each is written a nonzero natural number. A \emph{strip} is any $1 \times 1 \times 3$ rectangular parallelepiped formed by three unit cubes sharing three coplanar faces. Any number greater than $1$ in a unit cube is the sum of three numbers from other cubes, one from each strip it belongs to. Prove that, regardless of the choice of numbers, at least $16$ numbers are less than or equal to $60$.","If all numbers are equal to $1$, there is nothing to prove. If the cube contains numbers greater than $1$, suppose it also contains even numbers and let $n$ be the smallest of them. Then $n$ must be the sum of $3$ odd numbers, which is impossible since the sum of three odd numbers is odd. So all numbers are odd. \hfill \textbf{[1p]}

Suppose there is a strip with no number equal to $1$. Let $a$ be the smallest number on that strip. Since $a > 1$, it is the sum of three numbers smaller than $a$ and greater than $1$, contradicting the minimality of $a$. So every strip has at least one $1$, hence the cube contains at least $9$ ones. \hfill \textbf{[1p]}

Let $a_1 \leq a_2 \leq \cdots \leq a_{18}$ be the other numbers. If $a_1 > 3$, then one of the three numbers summing to $a_1$ must be greater than $1$ and less than $a_1$, contradicting minimality. So $a_1 \in \{1, 3\}$. \hfill \textbf{[1p]}

From this, $a_2 \in \{1, 3, 5\}$. If $a_2 \leq 3$, then $a_3 \leq 1 + a_1 + a_2 \leq 7$. If $a_2 = 5$, then $a_1$ and $a_2$ are on a common strip and cannot both appear in the sum for $a_3$, so $a_3 \leq 1 + 1 + a_2 = 7$. Consequently $a_3 \in \{1, 3, 5, 7\}$. \hfill \textbf{[1p]}

If $a_3 = 7$, then $a_3$ is on a common strip with $a_2$, so $a_3$ and $a_2$ cannot both appear in the sum for $a_4$, giving $a_4 \leq a_3 + a_1 + 1 \leq 7 + 3 + 1 = 11$. If $a_3 \leq 5$: if $a_2 = 5$, then $a_1$ and $a_2$ are on the same strip, so $a_4 \leq a_3 + a_2 + a_1 \leq 5 + 5 + 1 = 11$; if $a_2 \leq 3$, $a_4 \leq a_3 + a_2 + a_1 \leq 5 + 3 + 3 = 11$. In all cases $a_4 \leq 11$. \hfill \textbf{[1p]}

In general, $a_{k+3} \leq a_{k+2} + a_{k+1} + a_k$.

If $a_{k+1}$ and $a_{k+2}$ are on the same strip, they cannot both appear in the sum for $a_{k+3}$, and since $a_{k+2} \leq a_{k+1} + a_k + a_{k-1}$, we get $a_{k+3} \leq a_{k+2} + a_k + a_{k-1} \leq a_{k+1} + 2a_k + 2a_{k-1}$.

If $a_{k+1}$ and $a_{k+2}$ are on different strips, since $a_{k+2} \leq a_k + a_{k-1} + a_{k-2}$, $a_{k+3} \leq a_{k+2} + a_{k+1} + a_k \leq a_{k+1} + 2a_k + a_{k-1} + a_{k-2} \leq a_{k+1} + 2a_k + 2a_{k-1}$.

Therefore, $a_{k+3} \leq a_{k+1} + 2a_k + 2a_{k-1}$, for all $k \geq 2$. \hfill (1) \hfill \textbf{[1p]}

From (1) we successively deduce $a_5 \leq 23$, $a_6 \leq 35$, $a_7 \leq 59$, so at least $16$ numbers are less than or equal to $60$. \hfill \textbf{[1p]}",466,2350,Combinatorics,4
398,tst_jbmo_ro_2023_1_p5,tst_jbmo,2023,g,"On the exterior of trapezoid $ABCD$ with small base $AB$, construct squares $ADEF$ and $BCGH$. Prove that the perpendicular bisector of side $AB$ passes through the midpoint of segment $FH$.","Let $I$ be the midpoint of base $AB$. Choose points $M$ and $Q$ on base $CD$ with $DM = QC = AI$.

Clearly $AIMN$ and $BIQC$ are parallelograms, so $IM = AD$, $IM \parallel AD$, $IQ = BC$, $IQ \parallel BC$.

On the exterior of triangle $IMQ$, construct squares $IMNP$ and $IQRS$. Consider points $U$ and $V$ such that $IPUS$ is a parallelogram and $V$ is the projection of $I$ onto $CD$. \hfill \textbf{[1p]}

We have $\angle PIS + \angle PIM + \angle MIQ + \angle QIS = 360^\circ$, and since $IPUS$ is a parallelogram: $\angle PIS + \angle MIQ = 180^\circ = \angle IPU + \angle PIS$, so $\angle IPU = \angle MIQ$. Since $IP = IM$ and $PU = IS = IQ$, triangles $IPU$ and $MIQ$ are congruent (SAS), so $\angle PIU = \angle IMQ$. \hfill \textbf{[3p]}

Triangle $IMV$ is right-angled at $V$, so $\angle MIV = 90^\circ - \angle IMV = 90^\circ - \angle PIU$. We get $\angle MIV + \angle PIU + \angle MIP = 180^\circ$, so points $U$, $I$, $V$ are collinear. Therefore $UI$ is the perpendicular bisector of $AB$. Since $IPUS$ is a parallelogram, line $UI$ passes through the midpoint $J$ of diagonal $PS$. \hfill \textbf{[1p]}

From $\angle DAF = \angle MIP = 90^\circ$ and $AD \parallel IM$, we get $AF \parallel IP$ and $AF = IP$, so quadrilateral $AIPF$ is a parallelogram. Hence $FP \parallel AI$ and $FP = AI$. Analogously $SH \parallel IB$ and $SH = IB$. Since $I$ is the midpoint of $AB$, $FP \parallel SH$ and $FP = SH$, so $FPHS$ is a parallelogram. Consequently, diagonal $FH$ passes through the midpoint $J$ of segment $PS$, so lines $UI$ and $FH$ are concurrent. \hfill \textbf{[2p]}",190,1589,Geometry,5
399,tst_jbmo_ro_2023_2_p1,tst_jbmo,2023,a,"Determine the positive real numbers $x, y, z > 0$ for which
\[
xyz \leq \min\left\{4\!\left(x - \frac{1}{y}
ight),\; 4\!\left(y - \frac{1}{z}
ight),\; 4\!\left(z - \frac{1}{x}
ight)
ight\}.
\]","From the hypothesis, $xyz \leq \dfrac{4(xy-1)}{y}$, $xyz \leq \dfrac{4(yz-1)}{z}$, and $xyz \leq \dfrac{4(zx-1)}{x}$.

Since $(xy - 2)^2 \geq 0$, we get $\dfrac{4(xy-1)}{y} \leq x^2 y$, from which $\dfrac{4(xy-1)}{y} \leq x^2 y \leq \dfrac{4(zx-1)}{z}$.

Then $x - \dfrac{1}{y} \leq x - \dfrac{1}{z}$, so $\dfrac{1}{z} \leq \dfrac{1}{y}$, i.e., $y \leq z$. \hfill \textbf{[3p]}

Analogously, $\dfrac{4(yz-1)}{z} \leq \dfrac{4(xy-1)}{x}$, so $z \leq x$, and $\dfrac{4(zx-1)}{x} \leq \dfrac{4(yz-1)}{y}$, i.e., $x \leq y$.

It follows that $x = y = z$. \hfill \textbf{[2p]}

Therefore $x^3 \leq \dfrac{4(x^2-1)}{x}$, so $(x^2 - 2)^2 \leq 0$, i.e., $x^2 = 2$. Since $x > 0$, the unique solution is $x = y = z = \sqrt{2}$. \hfill \textbf{[2p]}

\bigskip

\subsubsection*{Alternative Solution}

From the hypothesis, $xyz \leq \dfrac{4(xy-1)}{y}$, so $(xy)(yz) \leq 4(xy - 1)$, and analogously $(yz)(zx) \leq 4(yz-1)$, $(zx)(xy) \leq 4(zx-1)$.

Setting $xy = c$, $yz = a$, $zx = b$, we get $ab \leq 4(a-1)$, $bc \leq 4(b-1)$, $ca \leq 4(c-1)$. Dividing by $a$, $b$, $c$ respectively, these can be rewritten as $b + \dfrac{4}{a} \leq 4$, $c + \dfrac{4}{b} \leq 4$, $a + \dfrac{4}{c} \leq 4$, from which
\[
a + b + c + \frac{4}{a} + \frac{4}{b} + \frac{4}{c} \leq 12. \tag{1}
\]
\hfill \textbf{[3p]}

But $a + \dfrac{4}{a} \geq 2\sqrt{a \cdot \dfrac{4}{a}} = 4$ and analogously, so
\[
a + b + c + \frac{4}{a} + \frac{4}{b} + \frac{4}{c} \geq 12. \tag{2}
\]
\hfill \textbf{[2p]}

From (1) and (2), all inequalities hold with equality, giving $a = b = c = 2$. Then $xy = yz = zx = 2$, so $(xy \cdot yz \cdot zx)^2 = 8$. Since $x, y, z > 0$, $xyz = 2\sqrt{2}$, from which $x = y = z = \sqrt{2}$. \hfill \textbf{[2p]}",192,1705,Algebra,6
400,tst_jbmo_ro_2023_2_p2,tst_jbmo,2023,g,"Consider an acute triangle $ABC$ with $BC > AB$, such that points $A$, $H$, $I$, and $C$ are concyclic (where $H$ is the orthocenter and $I$ is the incenter of triangle $ABC$).

Line $AC$ intersects the circumscribed circle of triangle $BHC$ at point $T$, and line $BC$ intersects the circumscribed circle of triangle $AHC$ at point $P$. If lines $PT$ and $HI$ are parallel, find the angles of triangle $ABC$.","Let $U$ be the intersection of lines $CH$ and $PT$, and $A'$, $B'$, $C'$ be the feet of the altitudes from $A$, $B$, $C$ in triangle $ABC$.

The quadrilateral $AHIC$ is cyclic, so $\angle CHI = \angle CAI = \dfrac{\angle A}{2}$. From $PT \parallel HI$, $\angle HUT = \angle CHI = \dfrac{\angle A}{2}$ (alternate interior angles).

We get $\angle CUP = \angle HUT = \dfrac{\angle A}{2}$. \hfill (1)

The quadrilateral $AHIC$ is cyclic, so $\angle A'HI = \angle ACI = \dfrac{\angle C}{2}$. From triangle $A'HC$, $\angle A'CH = 90^\circ - \angle A'HC = 90^\circ - \dfrac{\angle A + \angle C}{2} = \dfrac{\angle B}{2}$. From right triangle $BCC'$, since $\angle CBC' + \angle BCC' = 90^\circ$, we get $\angle B + \dfrac{\angle B}{2} = 90^\circ$, so $\angle B = 60^\circ$. \hfill \textbf{[3p]}

From right triangles $ABB'$ and $ACC'$, $\angle ABB' = \angle ACC' = 90^\circ - \angle A$.

The quadrilateral $BCTH$ is cyclic, so $\angle B'BT = \angle TCH = 90^\circ - \angle A = \angle ABB'$.

Therefore $BB'$ is both bisector and altitude in triangle $ABT$, so $AB = BT$. \hfill \textbf{[1p]}

The quadrilateral $AHIC$ is cyclic, so $\angle BPI = \angle CAI = \dfrac{\angle A}{2}$ and $\angle API = \angle ACI = \dfrac{\angle C}{2}$.

We get $\angle APB = \dfrac{\angle A + \angle C}{2} = 60^\circ = \angle B$, so triangle $ABP$ is equilateral. \hfill \textbf{[1p]}

Consequently $BP = AB = BT$, and from isosceles triangle $BPT$:
\[
\angle BPT = \frac{180^\circ - \angle PBT}{2}.
\]
Triangle $ABT$ is isosceles, so $\angle ABT = 180^\circ - 2\angle A$. Since $\angle PBT = \angle B - \angle ABT$, we get $\angle PBT = 2\angle A - 120^\circ$ and $\angle BPT = 150^\circ - \angle A$. \hfill (2) \hfill \textbf{[1p]}

But $\angle BPT$ is exterior to triangle $CPU$, so $\angle BPT = \angle CUP + \angle PCU$. From (1) and (2):
\[
150^\circ - \angle A = \frac{\angle A}{2} + \frac{\angle B}{2} = \frac{\angle A}{2} + 30^\circ,
\]
so $\angle A = 80^\circ$ and $\angle C = 40^\circ$. \hfill \textbf{[1p]}",409,1992,Geometry,7
401,tst_jbmo_ro_2023_2_p3,tst_jbmo,2023,c,"Let a regular hexagon of side length $2$. Through the vertices and midpoints of its sides, we construct parallels to the sides, which divide it into $24$ congruent equilateral triangles whose vertices we call \emph{nodes}. We call a \emph{leaf} the surface of any equilateral triangle with vertices at these nodes. For each node $X$, we call a \emph{trio} of $X$ the figure formed by three leaves with common vertex $X$, such that the intersection of any two leaves is only the point $X$, and the leaves are pairwise non-congruent.

\begin{enumerate}[label=\alph*)]
\item Determine the maximum value of the area of a trio.
\item Show that there exists a node whose trios can cover the surface of the hexagon, and there exists a node whose trios cannot cover the entire surface of the hexagon.
\item Determine the total number of trios associated with the hexagon.
\end{enumerate}","\textbf{a)} Let the regular hexagon $ABCDEF$ with center $O$, and $T, U, V, X, Y, Z$ the midpoints of sides $AB, BC, CD, DE, EF, FA$ respectively. We observe that no node that has a trio can be on the sides of the hexagon. So the nodes that have trios are $O$, or nodes at distance $1$ from $O$, which we denote $M, N, P, Q, R, S$ as in Figure 1. \hfill \textbf{[1p]}

The leaves can be equilateral triangles of side length $1, \sqrt{3}, 2, \sqrt{7}, 3,$ or $2\sqrt{3}$. Leaves of side $3$ and $2\sqrt{3}$ cannot be part of a trio, since their vertices lie on the sides of the hexagon. From $O$ we cannot construct trios of side $\sqrt{7}$, and the only two leaves of side $\sqrt{7}$ with vertex at $R$ are $RAU$ and $RCT$. From $R$ we cannot construct leaves of side $2$ disjoint from the leaf of side $\sqrt{7}$.

The trio with leaves $RAU$, $RDP$, and $REY$ (as in Figure 1) has the maximum area. Its area equals
\[
\frac{\sqrt{3}}{4} + \frac{3\sqrt{3}}{4} + \frac{7\sqrt{3}}{4} = \frac{11\sqrt{3}}{4}.
\]
\hfill \textbf{[1p]}

\textbf{b)} In the colored trio from Figure 2, leaf $OAB$ covers one-sixth of the hexagon. Analogously, we find a trio of $O$ with leaves $OBC$, $OCD$, $ODE$, $OEF$, and $OFA$, and these six trios cover the hexagon. \hfill \textbf{[1p]}

There are exactly two leaves of maximum side with a vertex at $R$, namely $RAU$ and $RCT$, with side $\sqrt{7}$. Since $RB = 3 > \sqrt{7}$, not all points of segment $RB$ are covered by the trios of $R$. \hfill \textbf{[1p]}

\textbf{c)} We say a trio is of type $(x, y, z)$ if the side lengths of its leaves are $x, y, z$.

The center $O$ of the hexagon has only trios of type $(1, \sqrt{3}, 2)$, as in Figure 2. For the leaf $OQR$, the only leaves of side $2$ we can choose are $OAF$, $OAB$, and $OBC$, and in each case we can choose the leaf of side $\sqrt{3}$ in two ways, so there are $6$ trios containing leaf $OQR$. There are $6$ leaves of side $1$ with vertex $O$, so there are $36$ trios of $O$. \hfill \textbf{[1p]}

Point $R$ can only have trios of types $(1, \sqrt{3}, 2)$ and $(1, \sqrt{3}, \sqrt{7})$.

For leaves $RDP$ and $RNZ$ there are $2$ choices of the leaf of side $1$: $REX$ and $REY$, similarly for leaves $RFM$ and $RNV$, so node $R$ has $4$ trios of type $(1, \sqrt{3}, 2)$. For leaves $RAU$ and $RDP$, there are $3$ choices of the leaf of side $1$: $REX$, $REY$, and $RSY$, similarly for leaves $RCT$ and $RFM$, so node $R$ has $6$ trios of type $(1, \sqrt{3}, \sqrt{7})$. It follows that in the hexagon there are another $60$ trios. In total, there are $\boxed{96}$ trios associated with the hexagon. \hfill \textbf{[2p]}",879,2617,Combinatorics,8
402,tst_jbmo_ro_2023_2_p4,tst_jbmo,2023,n,"Let $M \geq 1$ be a real number. Determine all natural numbers $n$ for which there exist natural numbers $a, b, c > M$, pairwise distinct, such that
\[
n = (a,b)\cdot(b,c) + (b,c)\cdot(c,a) + (c,a)\cdot(a,b)
\]
(where $(x, y)$ denotes the greatest common divisor of $x$ and $y$).","Call a natural number $n$ \emph{good} if there exist $a, b, c \in \mathbb{N}^*$ such that $n = (a,b)(b,c) + (b,c)(c,a) + (c,a)(a,b)$.

We prove that the numbers $n$ for which there exist $a, b, c > M$ making $n$ good are exactly the numbers of the form $n = 2^{2t}(2k+1)$, where $t, k \in \mathbb{N}$, $k \geq 1$. \hfill (1) \hfill \textbf{[1p]}

For $k \geq 1$, let $p, q, r > M$ be distinct primes with $r > \max\{kp, kq\}$. For $a = kp$, $b = kq$, $c = r$, we get $(a,b) = k$ and $(b,c) = (c,a) = 1$, so $(a,b)(b,c) + (b,c)(c,a) + (c,a)(a,b) = 2k+1$, proving $2k+1$ is good. \hfill \textbf{[2p]}

For $t \in \mathbb{N}$ and $n$ a natural number for which there exist $a, b, c > M$ pairwise distinct with $n = (a,b)(b,c) + (b,c)(c,a) + (c,a)(a,b)$, we have $2^t a, 2^t b, 2^t c > M$ and $2^{2t} \cdot n = (2^t a, 2^t b)(2^t b, 2^t c) + (2^t b, 2^t c)(2^t c, 2^t a) + (2^t c, 2^t a)(2^t a, 2^t b)$, so the numbers of the form (1) are good. \hfill \textbf{[1p]}

We show that numbers of the form $n = 2^t$ or $n = 2^{2t+1}(2k+1)$ with $t, k \in \mathbb{N}$, $k \geq 1$, are not good.

First, we prove any even good number is divisible by $4$. Indeed, if $n_0$ is even and $n_0 = (a,b)(b,c) + (b,c)(c,a) + (c,a)(a,b)$, then $(a,b)$, $(b,c)$, $(c,a)$ cannot all be odd. If $2 \mid (a,b)$, then $2 \mid (b,c)(c,a)$, so $a, b, c$ are all even. It follows $4 \mid n_0$, and $\dfrac{n_0}{4}$ is also good with $a' = \dfrac{a}{2}$, $b' = \dfrac{b}{2}$, $c' = \dfrac{c}{2}$. \hfill \textbf{[1p]}

Suppose $n = 2^t$ with $t \geq 2$ is good; then $2^{t-2}, 2^{t-4}, \ldots$ are good, so $1$ or $2$ would be good, contradiction since $a, b, c \geq 1$ implies $(a,b)(b,c) + (b,c)(c,a) + (c,a)(a,b) \geq 3$.

Similarly, if $n = 2^{2t+1}(2k+1)$ with $t, k \in \mathbb{N}$, $k \geq 1$ is good, then $2^{2t-1}(2k+1), 2^{2t-3}(2k+1), \ldots, 2(2k+1)$ are good, contradiction since $2(2k+1)$ is not divisible by $4$. \hfill \textbf{[2p]}",279,1919,Number Theory,9
403,tst_jbmo_ro_2023_3_p1,tst_jbmo,2023,n,"Determine the natural numbers $n \geq 2$ with at least four natural divisors, having the property that for any two distinct proper divisors $d_1$ and $d_2$ of $n$, the integer $d_1 - d_2$ divides $n$.","If $n$ is odd and $d_1$, $d_2$ are two proper divisors of $n$, they are odd, so $d_1 - d_2$ is even, hence $d_1 - d_2 
mid n$. Therefore $n$ is even. \hfill \textbf{[1p]}

Let $n = 2^a \cdot b$ with $a, b \in \mathbb{N}^*$ and $b$ odd.

\textbf{Subcase $b = 1$:} Then $n = 2^a$ with $a \in \mathbb{N}^*$. For $a \in \{1, 2\}$ no solutions are obtained. If $a \geq 4$, choosing $d_1 = 2^{a-1}$ and $d_2 = 2^{a-3}$, we get $d_1 - d_2 = 3 \cdot 2^{a-3} 
mid n$, contradiction. So $a = 3$, giving $n = 8$, which is a solution. \hfill \textbf{[2p]}

If $b \geq 3$, we have the following cases:

\textbf{Case I:} $a = 1$, so $n = 2b$. Choose $d_1 = b$ and $d_2 = 2$ and obtain $b - 2 \mid 2b$, so $b - 2 \mid 2b - 2(b-2) = 4$. Since $b$ is odd, $b = 3$, so $n = 6$, which is a solution. \hfill \textbf{[1p]}

\textbf{Case II:} $a \geq 2$. Choose $d_1 = 2^{a-1} \cdot b$ and $d_2 = 2^a$ and obtain $2^{a-1}(b-2) \mid 2^a \cdot b$, so $b - 2 \mid 2b$. Since $b$ is odd, as in Case I, $b = 3$. So $n = 2^a \cdot 3$.

Choose $d_1 = 2^a$ and $d_2 = 3$ and get $2^a - 3 \mid 2^a \cdot 3$, so $2^a - 3 \mid 2^a \cdot 3 - 3(2^a - 3) = 9$, from which $a = 2$. So $n = 12$, which is a solution. \hfill \textbf{[3p]}

\bigskip

\subsubsection*{Alternative Solution and Marking Scheme}

If $n$ is odd and $d_1$, $d_2$ are two proper divisors of $n$, they are odd, so $d_1 - d_2$ is even, hence $d_1 - d_2 
mid n$. Therefore $n$ is even. \hfill \textbf{[1p]}

Let $n = 2^a \cdot b$ with $a, b \in \mathbb{N}^*$ and $b$ odd.

\textbf{Subcase $b = 1$:} As above, $n = 8$ is the only solution. \hfill \textbf{[2p]}

If $b$ is an odd number with $b \geq 3$, choose $d_1 = b$, $d_2 = 2^a$ and get $2^a - b \mid 2^a \cdot b$.

Suppose $|2^a - b| \geq 2$. Let $p$ be a prime such that $p \mid 2^a - b$. Since $2^a - b$ is odd, $p$ is odd. From $p \mid 2^a - b$ we deduce $p \mid 2^a \cdot b$, consequently $p \mid b$, so $p \mid (2^a - b) + b = 2^a$, contradiction. Therefore $2^a - b \in \{-1, 1\}$, so $b \in \{2^a - 1, 2^a + 1\}$. \hfill \textbf{[2p]}

\textbf{Subcase $n = 2^a(2^a + 1)$:} Choose $d_1 = 2^a + 1$ and $d_2 = 2$ and get $2^a - 1 = (2^a + 1) - 2 \mid n$. Since $\gcd(2^a - 1, 2^a) = 1$, it follows $2^a - 1 \mid 2^a + 1 = 2^a - 1 + 2$, so $2^a - 1 \mid 2$. Hence $a = 1$ and $n = 6$, a solution. \hfill \textbf{[1p]}

\textbf{Subcase $n = 2^a(2^a - 1)$:} Choose $d_1 = 2(2^a - 1)$ and $d_2 = 2^a$ and get $2^a - 2 = 2^{a+1} - 2 - 2^a \mid n$. Since $\gcd(2^a - 1, 2^a - 2) = 1$, it follows $2^a - 2 \mid 2^a = 2^a - 2 + 2$, so $2^a - 2 \mid 2$. Hence $a = 2$ and $n = 12$, a solution. \hfill \textbf{[1p]}",200,2596,Number Theory,10
404,tst_jbmo_ro_2023_3_p2,tst_jbmo,2023,c,"Let $n \in \mathbb{N}$, $n \geq 2$, and let $A$, $B$, $C$ be three pairwise disjoint sets of real numbers, each with $n$ elements.

Denote by $a$ the number of triples $(x, y, z) \in A \times B \times C$ with $x < y < z$, and by $b$ the number of triples $(x, y, z) \in A \times B \times C$ with $x > y > z$. Show that $n$ divides $a - b$.","Let $y \in B$. Denote by $A_y$ the number of pairs $(x, z) \in A \times C$ for which $x < y < z$, and by $B_y$ the number of pairs $(x, z) \in A \times C$ for which $x > y > z$.

We deduce that $a = \displaystyle\sum_{y \in B} A_y$ and $b = \displaystyle\sum_{y \in B} B_y$. \hfill \textbf{[2p]}

Let $A = \{a_1, a_2, \ldots, a_n\}$, $C = \{c_1, c_2, \ldots, c_n\}$, with $a_1 < a_2 < \cdots < a_n$ and $c_1 < c_2 < \cdots < c_n$.

On the real line, $y$ separates the numbers $a_1, a_2, \ldots, a_n$ into $k$ numbers less than $y$ and $n-k$ numbers greater than $y$, and the numbers $c_1, c_2, \ldots, c_n$ into $p$ numbers less than $y$ and $n-p$ numbers greater than $y$, with $k, p \in \{0, 1, 2, \ldots, n\}$. It follows that $A_y = k(n-p)$ and $B_y = p(n-k)$, so
\[
A_y - B_y = n(k - p).
\]
\hfill \textbf{[3p]}

Therefore $n \mid a - b = \displaystyle\sum_{y \in B}(A_y - B_y)$. \hfill \textbf{[2p]}",339,905,Combinatorics,11
405,tst_jbmo_ro_2023_3_p3,tst_jbmo,2023,g,"Let $ABC$ and $DEF$ be congruent equilateral triangles with centers $O_1$ and $O_2$ respectively, such that segment $AB$ intersects segments $DE$ and $DF$ at $M$ and $N$ respectively, and segment $AC$ intersects segments $DF$ and $EF$ at $P$ and $Q$ respectively. Denote by $I$ the intersection of the angle bisectors of $\angle EMN$ and $\angle DPQ$, and by $J$ the intersection of the angle bisectors of $\angle FNM$ and $\angle EQP$. Prove that $IJ$ is the perpendicular bisector of segment $O_1O_2$.","Let $\mathcal{C}_1$ be the circumscribed circle of triangle $ABC$ with center $O_1$, and $\mathcal{C}_2$ be the circumscribed circle of triangle $DEF$ with center $O_2$.

Since $\angle PAM = \angle PDM = 60^\circ$, the quadrilateral $APMD$ is cyclic, so $\angle APD = \angle AMD \overset{\text{def}}{=} x^\circ$.

From the quadrilateral $APIM$ we get:
\[
\angle PIM = 360^\circ - \angle MAP - \angle AMI - \angle API = 360^\circ - 60^\circ - 2 \cdot \frac{180^\circ - x^\circ}{2} - x^\circ = 120^\circ.
\]
Since $\angle PIM = 180^\circ - \angle PAM$, the quadrilateral $APIM$ is cyclic, so quadrilaterals $ADIP$ and $ADMI$ are also cyclic. Since $ADIP$ is cyclic, $\angle DAI = \angle DPI = \dfrac{180^\circ - x^\circ}{2}$.

Since $ADMI$ is cyclic, $\angle ADI = \angle AMI = \dfrac{180^\circ - x^\circ}{2}$.

It follows that $\angle ADI = \angle DAI$, so $AI = DI$. \hfill \textbf{[2p]}

Let $A'$ be the second intersection of ray $(AI$ with circle $\mathcal{C}_1$, and $D'$ be the second intersection of ray $(DI$ with circle $\mathcal{C}_2$.

Since $ADIP$ is cyclic, $\angle IAP = \angle IDP$, so arc $A'C$ in circle $\mathcal{C}_1$ and arc $D'F$ in circle $\mathcal{C}_2$ have the same measure. Since the minor arcs $AC$ and $DF$ in the two circles are both $120^\circ$, arcs $ACA'$ and $DFD'$ also have the same measure in the congruent circles $\mathcal{C}_1$ and $\mathcal{C}_2$, so $AA' = DD'$. \hfill \textbf{[2p]}

We obtain $A'I = AA' - AI = DD' - DI = D'I$, so $IA \cdot IA' = ID \cdot ID'$, meaning point $I$ has equal power with respect to circles $\mathcal{C}_1$ and $\mathcal{C}_2$. Therefore $I$ belongs to the radical axis $UV$ of the two circles. \hfill \textbf{[2p]}

Analogously, $J \in UV$, so lines $UV$ and $IJ$ coincide, meaning $IJ$ is the radical axis of the two circles, and consequently $IJ \perp O_1 O_2$. \hfill \textbf{[1p]}",503,1856,Geometry,12
406,tst_jbmo_ro_2023_3_p4,tst_jbmo,2023,n,"Let $a$ be a positive real number. Prove that there do not exist real numbers $b$ and $c$, with $b < c$, such that for all distinct $x, y \in (b, c)$ we have
\[
\left|\frac{x+y}{x-y}
ight| \leq a.
\]","Suppose there exist $b$ and $c$ as in the statement. Let $x \in (b, c)$, $x 
eq 0$.

\textbf{Case I:} $x > 0$. There exist infinitely many natural numbers $n$ with the property that $n > \dfrac{1}{x - b}$, so that $x - \dfrac{1}{n} > b$. For each such $n$, choose $y_n = x - \dfrac{1}{n} \in (b, c)$.

The hypothesis relation, applied to each pair $(x, y_n)$, leads to $x \leq \dfrac{1+a}{2n}$.

We obtain $0 < n \leq \dfrac{1+a}{2x}$ for infinitely many natural numbers $n$, contradiction. \hfill \textbf{[4p]}

\textbf{Case II:} $x < 0$. There exist infinitely many natural numbers $n$ with the property that $n > \dfrac{1}{c - x}$, so that $x + \dfrac{1}{n} < c$. For each such $n$, choose $z_n = x + \dfrac{1}{n} \in (b, c)$.

The hypothesis relation, applied to each pair $(x, z_n)$, leads to $\dfrac{-1-a}{2n} \leq x < 0$.

We obtain $0 < n \leq \dfrac{-1-a}{2x}$ for infinitely many natural numbers $n$, contradiction.

From Cases I and II the conclusion follows. \hfill \textbf{[3p]}",199,991,Number Theory,13
407,tst_jbmo_ro_2023_4_p1,tst_jbmo,2023,n,"Let $a$ and $b$ be two distinct nonzero natural numbers of the same parity. Prove that the fraction
\[
\frac{a! + b!}{2^a}
\]
cannot take integer values (for any nonzero natural number $n$, the notation $n!$ denotes the product $1 \cdot 2 \cdots n$).","First, we prove that for any nonzero natural number $n$, $2^n$ does not divide $n!$. Suppose there exists $n \in \mathbb{N}^*$ such that $2^n \mid n!$. Then, in the prime factorization of $n!$, at least $n$ factors equal to $2$ should appear. The exponent of $2$ in $n!$ equals
\[
\left\lfloor \frac{n}{2} 
ight
floor + \left\lfloor \frac{n}{2^2} 
ight
floor + \cdots + \left\lfloor \frac{n}{2^k} 
ight
floor,
\]
where $k \in \mathbb{N}^*$, $k < n$, with $\left\lfloor \frac{n}{2^k} 
ight
floor 
eq 0$ and $\left\lfloor \frac{n}{2^{k+1}} 
ight
floor = 0$. \hfill \textbf{[1p]}

It follows that
\[
\frac{n}{2} + \frac{n}{2^2} + \cdots + \frac{n}{2^k} \geq \left\lfloor \frac{n}{2} 
ight
floor + \left\lfloor \frac{n}{2^2} 
ight
floor + \cdots + \left\lfloor \frac{n}{2^k} 
ight
floor \geq n.
\]
Therefore $\frac{n}{2} + \frac{n}{2^2} + \cdots + \frac{n}{2^k} \geq n$, so $1 - \frac{1}{2^k} \geq 1$, which is false. Consequently, $2^n 
mid n!$ for all $n \in \mathbb{N}^*$. \hfill \textbf{[3p]}

Now suppose there exists $n \in \mathbb{N}^*$ such that $a! + b! = n \cdot 2^a$.

\textbf{Case 1:} If $a \geq b+2$, then
\[
a! + b! = b!\cdot\bigl(1 + (b+1)(b+2)\cdots a\bigr) = n \cdot 2^a,
\]
and since $1 + (b+1)(b+2)\cdots b$ is odd, we obtain $2^a \mid b!$. Since $b! \mid a!$, it follows that $2^a \mid a!$, contradiction. \hfill \textbf{[2p]}

\textbf{Case 2:} If $b \geq a+2$, then
\[
a! + b! = a!\cdot\bigl(1 + (a+1)(a+2)\cdots b\bigr) = n \cdot 2^a,
\]
and since $1 + (a+1)(a+2)\cdots b$ is odd, it follows that $2^a \mid a!$, contradiction. \hfill \textbf{[1p]}",250,1565,Number Theory,14
408,tst_jbmo_ro_2023_4_p2,tst_jbmo,2023,a,"If $a$, $b$, and $c$ are positive real numbers such that $a + b + c \geq \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}$, determine the maximum value of the expression
\[
E = \frac{a+b-c}{a^3+b^3+abc} + \frac{b+c-a}{b^3+c^3+abc} + \frac{c+a-b}{c^3+a^3+abc}.
\]","\textbf{Case 1:} We first consider the case where $a+b-c$, $b+c-a$, and $c+a-b$ are all non-negative.

Using the observation that $x^3 + y^3 \geq x^2 y + xy^2$ for all positive real numbers $x$ and $y$ (which follows from the evident inequality $(x+y)(x-y)^2 \geq 0$), we have:
\begin{align*}
\sum \frac{a+b-c}{a^3+b^3+abc} &\leq \sum \frac{a+b-c}{a^2 b + ab^2 + abc} = \frac{1}{a+b+c} \cdot \sum \frac{a+b-c}{ab} \\
&= \frac{1}{a+b+c} \cdot \frac{2(ab+bc+ca) - (a^2+b^2+c^2)}{abc} \\
&\leq \frac{1}{a+b+c} \cdot \frac{2(ab+bc+ca) - (ab+bc+ca)}{abc} \\
&= \frac{1}{a+b+c}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}
ight) \leq 1.
\end{align*}
\hfill \textbf{[3p]}

\textbf{Case 2:} Suppose one of $a+b-c$, $b+c-a$, $c+a-b$ is negative. Without loss of generality, suppose $a+b-c < 0$; then $c > a+b > |a-b|$, so both $b+c-a$ and $c+a-b$ are positive. We have:
\begin{align*}
\sum \frac{a+b-c}{a^3+b^3+abc} &\leq \frac{b+c-a}{b^3+c^3+abc} + \frac{c+a-b}{c^3+a^3+abc} \\
&\leq \frac{b+c-a}{bc(a+b+c)} + \frac{c+a-b}{ca(a+b+c)} \\
&= \frac{1}{a+b+c} \cdot \frac{ac+bc-(a-b)^2}{abc} \\
&< \frac{1}{a+b+c} \cdot \frac{ab+bc+ca}{abc} = \frac{1}{a+b+c}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}
ight) \leq 1.
\end{align*}

Therefore the maximum value of the expression is $\boxed{1}$, attained when $a = b = c = 1$. \hfill \textbf{[4p]}",256,1325,Algebra,15
409,tst_jbmo_ro_2023_4_p3,tst_jbmo,2023,c,"Consider a grid with $n$ rows and $m$ columns ($n, m \in \mathbb{N}$, $n, m \geq 2$) consisting of $n \cdot m$ unit squares $1 \times 1$, which we call \emph{cells}. We define a \emph{snake} as a sequence of cells with the following properties: the first cell is located in the first (top) row of the grid, the last cell is located in the last row of the grid, and starting from the second cell, each cell in the snake shares a side with the previous one and is not located in a row above the previous cell. We define the \emph{length} of a snake as the number of cells in it. Determine the arithmetic mean of the lengths of all snakes in the grid.","For a given snake and for each row $i \in \{1, 2, \ldots, n\}$ of the grid, denote by $a_i$ and $b_i$ the column numbers corresponding to the first and last cell, respectively, that the snake occupies in row $i$. We observe that $a_{i+1} = b_i$ for all $i \in \{1, 2, \ldots, n-1\}$, so the snake is uniquely determined by the values $a_1, b_1, b_2, \ldots, b_n \in \{1, 2, \ldots, m\}$. We deduce that the total number of snakes is $m^{n+1}$. \hfill \textbf{[3p]}

Let $(k, p)$ be any cell in the grid. The snakes containing this cell are those for which $a_k \leq p \leq b_k$. For these, we can choose $a_k$ and $b_k$ in $2p(m-p+1)-1$ ways, and the numbers $a_1, b_1, b_2, \ldots, b_n$ other than $a_k$ and $b_k$ in $m^{n-1}$ ways. Thus, cell $(k, p)$ is counted $m^{n-1}(2pm - 2p^2 + 2p - 1)$ times across all snakes in the grid. \hfill \textbf{[2p]}

It follows that the cells in column $p$ contribute $n \cdot m^{n-1}(2pm - 2p^2 + 2p - 1)$ to the total $T$ of the lengths of all snakes in the grid. Therefore,
\begin{align*}
T &= \sum_{p=1}^{m} n \cdot m^{n-1}(2pm - 2p^2 + 2p - 1) \\
&= n \cdot m^{n+1}(m+1) - \frac{n \cdot m^n(m+1)(2m+1)}{3} + n \cdot m^{n+1} \\
&= \frac{n \cdot m^n(m^2 + 3m - 1)}{3}.
\end{align*}

The arithmetic mean of the lengths of all snakes in the grid is
\[
\boxed{\dfrac{n(m^2+3m-1)}{3m}.}
\]
\hfill \textbf{[2p]}",648,1347,Combinatorics,16
410,tst_jbmo_ro_2023_4_p4,tst_jbmo,2023,g,"Consider an acute triangle $ABC$ with $\angle B > \angle C$. On the circumscribed circle $\mathcal{C}(O, R)$ of this triangle, choose points $D$, $E$, $J$, $K$, $S$ such that $A$, $E$, $J$, and $K$ are on the same side of line $BC$; $DE$ is the diameter perpendicular to chord $BC$; $S \in \arc{EK}$; and
\[
\arc{AE} = \arc{BJ} = \arc{CK} = \tfrac{1}{4}\arc{CE}.
\]
Let $\{F\} = AC \cap DE$, $\{M\} = BK \cap AD$, $\{P\} = BK \cap AC$, and $\{Q\} = CJ \cap BF$. If $\angle SMK = 30^\circ$ and $\angle AQP = 90^\circ$, show that line $MS$ is tangent to the circumscribed circle of triangle $AOF$.","Let $\{L\} = BF \cap \mathcal{C}(O, R)$ and $\{T\} = BK \cap DE$.

Since $DE$ is the perpendicular bisector of segment $BC$, triangle $BFC$ is isosceles and $\angle BFD = \angle CFD$, so
\[
\arc{BD} + \arc{EL} = \arc{CD} + \arc{AE}.
\]
Since $\arc{BD} = \arc{CD} = \angle A$, it follows that $\arc{AE} = \arc{EL} \overset{\text{def}}{=} x$.

From the hypothesis, $\arc{BJ} = \arc{CK} = x$ and $\arc{AJ} = \arc{KL} = 2x$.

Since $TD$ is the perpendicular bisector of $BC$, we have $BT = TC$, so
\[
\angle TCB = \angle TBC = \frac{\arc{CK}}{2} = \frac{x}{2}.
\]
Since $\dfrac{x}{2} = \dfrac{\arc{BJ}}{2} = \angle BCJ$, it follows that $T \in CJ$. \hfill \textbf{[1p]}

We have:
\[
\angle TFQ = \angle TFP = \tfrac{1}{2}(\angle A + x) \quad \text{and} \quad \angle QTF = \angle PTF = \tfrac{1}{2}(\angle A + 3x),
\]
so triangles $QTF$ and $PTF$ are congruent, and $QT = PT$. \hfill \textbf{[1p]}

We have $\angle ELB = \angle LQC = 2x$, so $EL \parallel QT$. From $\angle EAC = \angle APB = 2x$, we get $AE \parallel PT$. Consequently $\angle PTQ = \angle AEL = 180^\circ - x$. Triangles $EAL$ and $TPQ$ are isosceles, so $\angle TPQ = \angle EAL = \dfrac{x}{2}$, hence $AL \parallel PQ$, and $ALPQ$ is a trapezoid or rectangle. \hfill \textbf{[1p]}

From $\angle LAC = \angle ACB$ it follows that $AL \parallel BC$. Line $DE$ is the perpendicular bisector of segments $AL$ and $PQ$, so triangles $AFL$ and $FPQ$ are isosceles. It follows that triangles $AFQ$ and $LFP$ are congruent, so $AQ = LP$. Since $\angle AQP = 90^\circ$, $ALPQ$ cannot be a trapezoid, so $ALPQ$ is a rectangle, and $F$ is the midpoint of segment $AP$. \hfill \textbf{[1p]}

Let $\{X\} = AE \cap BL$. We have $\angle AXB = \tfrac{1}{2}(\arc{AB} - \arc{EL}) = x = \angle ABL$, so triangle $ABX$ is isosceles with $AB = AX$. Since $AF = FP$, $\angle XAF = \angle BPF$, and $\angle AFX = \angle PFB$, triangles $AFX$ and $PFB$ are congruent, so $AX = BP$, hence $AB = BP$. But $\angle ABK = \angle APB = 2x$, so $AB = AP$, and triangle $ABP$ is equilateral. \hfill \textbf{[1p]}

We obtain $2x = \angle ABP = 60^\circ$, so $x = 30^\circ$, $\angle ACB = 45^\circ$, and $\angle AFB = \angle AOB = \angle AMB = 90^\circ$, meaning points $A$, $F$, $O$, $M$, and $B$ all lie on the circle $\omega$ with diameter $AB$. \hfill \textbf{[1p]}

$AOMB$ is cyclic, so $\angle OMK = \angle OAB = 45^\circ$, and $\angle OMS = \angle OMK - \angle SMK = 15^\circ$. Since $\angle OAM = \angle OAB - \angle MAB = 15^\circ = \angle OMS$, it follows that line $MS$ is tangent to the circumscribed circle of triangle $AOM$, which is also the circumscribed circle of triangle $AOF$. \hfill \textbf{[1p]}",595,2649,Geometry,17
271,jbmo_2024_p1,jbmo,2024,a,"Let $a, b, c$ be positive real numbers such that
\[
a^2 + b^2 + c^2 = \frac{1}{4}.
\]
Prove that
\[
\frac{1}{\sqrt{b^2+c^2}} + \frac{1}{\sqrt{c^2+a^2}} + \frac{1}{\sqrt{a^2+b^2}}
\leq
\frac{\sqrt{2}}{(a+b)(b+c)(c+a)}.
\]","
oindent\textbf{Solution.}
Using the AM-QM and AM-GM inequalities, we have
\begin{align*}
\frac{1}{\sqrt{b^2+c^2}} + \frac{1}{\sqrt{c^2+a^2}} + \frac{1}{\sqrt{a^2+b^2}}
&\leq \frac{\sqrt{2}}{b+c} + \frac{\sqrt{2}}{c+a} + \frac{\sqrt{2}}{a+b} \\
&= \sqrt{2} \cdot \frac{a^2+b^2+c^2 + 3(ab+bc+ca)}{(a+b)(b+c)(c+a)} \\
&\leq \sqrt{2} \cdot
  \frac{a^2+b^2+c^2 + 3\!\left(\dfrac{a^2+b^2}{2}+\dfrac{b^2+c^2}{2}+\dfrac{c^2+a^2}{2}
ight)}
       {(a+b)(b+c)(c+a)} \\
&= \frac{\sqrt{2}}{2} \cdot \frac{8(a^2+b^2+c^2)}{(a+b)(b+c)(c+a)} \\
&= \frac{\sqrt{2}}{(a+b)(b+c)(c+a)}.
\end{align*}",220,579,Algebra,1
272,jbmo_2024_p2,jbmo,2024,g,"Let $ABC$ be a triangle such that $AB < AC$. Let the excircle opposite to $A$ be
tangent to the lines $AB$, $AC$ and $BC$ at points $D$, $E$ and $F$, respectively,
and let $J$ be its centre. Let $P$ be a point on the side $BC$. The circumcircles of
the triangles $BDP$ and $CEP$ intersect for the second time at $Q$. Let $R$ be the
foot of the perpendicular from $A$ to the line $FJ$. Prove that the points $P$, $Q$
and $R$ are collinear.

\smallskip

oindent(The \emph{excircle} of a triangle $ABC$ opposite to $A$ is the circle that
is tangent to the line segment $BC$, to the ray $AB$ beyond $B$, and to the ray $AC$
beyond $C$.)","
oindent\textbf{Solution.}
Since the quadrilateral $BDQP$ is cyclic, we have
\[
\angle DQP = 180^\circ - \angle DBP = \angle ABC.
\]
Analogously, from the cyclic quadrilateral $CEQP$, we obtain
$\angle EQP = \angle ACB$. Hence
\[
\angle DQE = \angle DQP + \angle EQP
           = \angle ABC + \angle ACB
           = 180^\circ - \angle BAC
           = 180^\circ - \angle DAE,
\]
so $Q$ lies on the circumcircle of $\triangle ADE$. On the other hand, since
\[
\angle ADJ = \angle AEJ = \angle ARJ = 90^\circ,
\]
we can conclude that $R$ and $J$ also lie on the circumcircle of $\triangle ADE$.

Note that the quadrilaterals $BDJF$ and $DJQR$ are cyclic, so we have
\[
\angle DQR = \angle DJR = \angle DJF = 180^\circ - \angle DBF = \angle ABC.
\]
Then, using $\angle DQP = \angle ABC$, we obtain $\angle DQR = \angle DQP$, so the
result follows. \hfill$\square$",632,861,Geometry,2
273,jbmo_2024_p3,jbmo,2024,n,"Find all triples of positive integers $(x, y, z)$ that satisfy the equation
\[
2020^x + 2^y = 2024^z.
\]","
oindent\textbf{Solution.}
Regarding the equation modulo 3, we obtain
\[
1 + (-1)^y \equiv (-1)^z \pmod{3},
\]
so we can conclude that $y$ is even and $z$ is odd. Let $y = 2y_1$, where $y_1$ is a
positive integer. Since the largest powers of $2$ in the factorisation of all three
terms in the equation are respectively $2^{2x}$, $2^{2y_1}$, $2^{3z}$, we rewrite
the equation as
\begin{equation}\label{eq1}
2^{2x} \cdot 505^x + 2^{2y_1} = 2^{3z} \cdot 253^z.
\end{equation}

We will now determine which one of the integers $2x$, $2y_1$ is larger, in order to
obtain the largest power of $2$ dividing the left-hand side.

\medskip

oindent\textbf{Case 1:} $2x > 2y_1$, i.e.\ $x > y_1$. Then the largest power of
$2$ dividing the left-hand side is $2^{2y_1}$, while the largest power of $2$
dividing the right-hand side is $2^{3z}$. Therefore $2y_1 = 3z$. Since $z$ is odd,
this is a contradiction.

\medskip

oindent\textbf{Case 2:} $2x < 2y_1$, i.e.\ $x < y_1$. Then the largest power of
$2$ dividing the left-hand side is $2^{2x}$, while the largest power of $2$
dividing the right-hand side is $2^{3z}$. Therefore $2x = 3z$. Since $z$ is odd,
this is a contradiction.

\medskip
Thus we must have $2x = 2y_1$, which implies $x = y_1$. Equation~\eqref{eq1} then
becomes
\begin{equation}\label{eq2}
4^x \cdot (505^x + 1) = 2020^x + 4^x = 2024^z = 2^{3z} \cdot 253^z.
\end{equation}
Since $2 \mid 505^x + 1$ but $4 
mid 505^x + 1$, the largest exponent of $2$
dividing $4^x(505^x+1)$ is $2x+1$, so
\begin{equation}\label{eq3}
2x + 1 = 3z.
\end{equation}
Plugging back into~\eqref{eq2}, it reduces to
\begin{equation}\label{eq4}
505^x + 1 = 2 \cdot 253^z.
\end{equation}
Observe that $x=1$, $z=1$ is a solution of the linear Diophantine
equation~\eqref{eq3}, so all solutions in positive integers are given by
\[
x = 3k+1, \quad z = 2k+1, \quad k \in \mathbb{N}_0.
\]
Now~\eqref{eq4} becomes
\begin{equation}\label{eq5}
505^{3k+1} + 1 = 2 \cdot 253^{2k+1}.
\end{equation}
Obviously $k=0$ is a solution of~\eqref{eq5}. If $k$ is a positive integer, then
\[
505^{3k+1} + 1
> 505^{3k+1}
= 505^{k+1} \cdot 505^{2k}
> 506 \cdot 505^{2k}
> 506 \cdot 253^{2k}
= 2 \cdot 253^{2k+1},
\]
so there are no further solutions.

Hence the only solution of~\eqref{eq5} is $k=0$, giving $x = y_1 = 1$ and $z = 1$,
and the only solution of the given equation is
\[
\boxed{(x, y, z) = (1, 2, 1)}.
\]",104,2376,Number Theory,3
274,jbmo_2024_p4,jbmo,2024,c,"Three friends Archie, Billie and Charlie play a game. At the beginning of the game,
each of them has a pile of $2024$ pebbles. Archie makes the first move, Billie makes
the second, Charlie makes the third and they continue to make moves in the same
order. In each move, the player making the move must choose a positive integer $n$
greater than any previously chosen number by any player, take $2n$ pebbles from his
pile and distribute them equally to the other two players. If a player cannot make a
move, the game ends and that player loses the game.

Determine all the players who have a strategy such that, regardless of how the other
two players play, they will not lose the game.","
oindent\textbf{Solution.}
We will prove that only Charlie has a non-losing strategy. First we discuss what
happens right before a player loses the game. Let $t$ be the number chosen in the
last move and let the losing player have $s$ pebbles in his pile before the move. In
order for the player to lose, $2t+2$ must be larger than $s+t$ so that he cannot
make the next move. This implies $t \geq s-1$, i.e.\ before the move the previous
player must have at least $2s-2$ pebbles. This means that if a player before his
move has at least $2s-2$ pebbles and the next player has $s$ pebbles, then he can
choose $s-1$ to make the next player lose (this will leave the next player with
$2s-1$ pebbles, disabling him from making a move). Also, if player $X$ has $s$
pebbles while player $Y$ has at most $2s-3$ pebbles right before player $Y$ plays,
then player $X$ guarantees not to lose in his turn.

Assume that at some point in the game the consecutive players have $x$, $y$, and $z$
pebbles in their piles and they choose numbers $u$, $u+v$, and $u+v+w$ respectively.
In these three moves we have:
\begin{align*}
(x,y,z)
&\mapsto (x-2u,\; y+u,\; z+u) \\
&\mapsto (x-u+v,\; y-u-2v,\; z+2u+v) \\
&\mapsto (x+2v+w,\; y-v+w,\; z-v-2w).
\end{align*}
Considering this, the following two statements hold:
\begin{enumerate}[label=(
oman*)]
\item If the player that plays second (after the position) plays $v=1$, then after
      the three moves the number of pebbles in his pile does not decrease (since $w$
      is a positive integer, $y - 1 + w \geq y$).
\item If the player that plays third (after the position) plays $w=1$, then after
      the three moves the number of pebbles in the second player's pile does not
      increase (since $v$ is a positive integer, $y - v + 1 \leq y$).
\end{enumerate}

Let $a_i$, $b_i$, $c_i$ be the numbers chosen by Archie, Billie, Charlie in their
$i$-th turns, respectively.

\medskip

oindent\textbf{Claim 1: Charlie has a non-losing strategy.}


oindent\textit{Proof.}
After Archie's first move, Charlie has at least $2025$ pebbles in his pile. If
Charlie chooses $c_i = b_i + 1$, then by~(i) (with Charlie in the role of the
middle player) he has at least $2025$ pebbles after $3i+1$ moves. In order for
Charlie to lose, Billie must have at least $2 \cdot 2024$ pebbles at some point.
However the total number of pebbles $3 \cdot 2024$ is conserved throughout the game,
and $3 \cdot 2024 < 2 \cdot 2024 + 2025$, so this is impossible. Hence Charlie has a
non-losing strategy. \hfill$\square$

\medskip

oindent\textbf{Claim 2: Billie does not have a non-losing strategy.}


oindent\textit{Proof.}
Assume Archie and Charlie play $a_{i+1} = c_i + 1$ and $c_i = b_i + 1$ (with
$c_0 = 0$ by definition). By~(ii) the number of pebbles in Billie's pile after $3i$
moves does not increase, so he cannot have more than $2024$ pebbles at any point
after $3i$ moves. If at some point Billie chooses $b_i > a_i + 1$, this increases
Archie's pile to at least $2025$ before move $3i+2$, so Archie cannot lose, and
since Charlie is playing a non-losing strategy, Billie will lose. Otherwise the play
proceeds as:
\begin{align*}
&(2024, 2024, 2024)
  \xrightarrow{1} (2022, 2025, 2025)
  \xrightarrow{2} (2024, 2021, 2027)
  \xrightarrow{3} \\
&(2027, 2024, 2021)
  \xrightarrow{4} \cdots
  \xrightarrow{2019} (4043, 2024, 5)
  \xrightarrow{2020} (3, 4044, 2025).
\end{align*}
After this Billie can choose $2021$ or $2022$, but both lead to his loss:
\begin{align*}
(3, 4044, 2025)
  &\xrightarrow{2021} (2024, 2, 4046)
   \xrightarrow{2022} (4046, 2024, 2)
   \xrightarrow{2023} (0, 4047, 2025), \\
(3, 4044, 2025)
  &\xrightarrow{2022} (2025, 0, 4047)
   \xrightarrow{2023} (4048, 2023, 1)
   \xrightarrow{2024} (0, 4047, 2025).
\end{align*}
Hence Billie does not have a non-losing strategy. \hfill$\square$

\medskip

oindent\textbf{Claim 3: Archie does not have a non-losing strategy.}


oindent\textit{Proof.}
Let Billie and Charlie choose $b_i = a_i + 1$ and $c_i = b_i + 1$ until the last
turn. If Archie chooses $a_i = c_{i-1} + 1$ in all his turns (with $c_0 = 0$), this
again leads to the case $(4043, 2024, 5)$. After this Archie can choose $2020$ or
$2021$, and in both cases Billie and Charlie can make Archie lose:
\begin{align*}
(4043, 2024, 5)
  &\xrightarrow{2020} (3, 4044, 2025)
   \xrightarrow{2021} (2024, 2, 4046)
   \xrightarrow{2023} (4047, 2025, 0), \\
(4043, 2024, 5)
  &\xrightarrow{2021} (1, 4045, 2026)
   \xrightarrow{2022} (2023, 1, 4048)
   \xrightarrow{2023} (4046, 2024, 2).
\end{align*}

Now assume Archie chooses $a_i > c_{i-1} + 1$ before this case happens. Let Billie
choose $b_i = a_i + 1$ and Charlie choose $c_i = b_i + a_i - c_{i-1}$. In this case,
before Archie moves, the players have $(2024 + 3k,\, 2024,\, 2024 - 3k)$ pebbles. If
Archie chooses $L$ with
\[
\frac{2024+3k}{2} > L > 3k+1,
\]
then with the described moves the pebble counts after this turn become
\[
(2024 + L + 2,\;\; 2024 + L - 3k - 1,\;\; 2024 - 2L + 3k - 1),
\]
and all moves are valid. Then Billie has $2024 + L - 3k - 1 \geq 2025$ pebbles when
Archie next moves, and Charlie has at least $2025$ pebbles when Billie next moves.
By~(i), choosing $b_i = a_i + 1$ and $c_i = b_i + 1$ guarantees Billie and Charlie
will not lose, so Archie loses.

If Archie chooses $L = \frac{2024+3k}{2} < 2023$, then after his play we have the
case $(0,\, 2024+L,\, 4048-L)$. Billie can choose $L+1$, giving $(L+1,\, 2022-L,\,
4049)$. Then Charlie can choose $2024$, leaving Archie with no valid move, so he
loses.

Hence Archie does not have a non-losing strategy. \hfill$\square$

\bigskip

oindent We conclude that \textbf{only Charlie} has a non-losing strategy.",685,5684,Combinatorics,4
1,shl_jbmo_2024_a1,shl_jbmo,2024,a,"Let $a$, $b$, $c$ be positive real numbers such that
\[
a^2 + b^2 + c^2 = \frac{1}{4}.
\]
Prove that
\[
\frac{1}{\sqrt{b^2+c^2}} + \frac{1}{\sqrt{c^2+a^2}} + \frac{1}{\sqrt{a^2+b^2}} \leq \frac{\sqrt{2}}{(a+b)(b+c)(c+a)}.
\]","According to AM-QM and AM-GM, we have
\begin{align*}
\frac{1}{\sqrt{b^2+c^2}} + \frac{1}{\sqrt{c^2+a^2}} + \frac{1}{\sqrt{a^2+b^2}}
&\leq \frac{\sqrt{2}}{b+c} + \frac{\sqrt{2}}{c+a} + \frac{\sqrt{2}}{a+b} \\
&= \sqrt{2} \cdot \frac{a^2+b^2+c^2 + 3(ab+bc+ca)}{(a+b)(b+c)(c+a)} \\
&\leq \sqrt{2} \cdot \frac{a^2+b^2+c^2 + 3\!\left(\dfrac{a^2+b^2}{2}+\dfrac{b^2+c^2}{2}+\dfrac{c^2+a^2}{2}
ight)}{(a+b)(b+c)(c+a)} \\
&= \frac{\sqrt{2}}{2} \cdot \frac{8(a^2+b^2+c^2)}{(a+b)(b+c)(c+a)} \\
&= \frac{\sqrt{2}}{(a+b)(b+c)(c+a)}.
\end{align*}
Equality holds iff $a = b = c = \dfrac{\sqrt{3}}{6}$.\hfill$\square$",224,601,Algebra,1
2,shl_jbmo_2024_a2,shl_jbmo,2024,a,"Let $a$, $b$, $c$ be real numbers such that $a+b+c=0$ and $abc=-16$. Find the minimum value of the expression
\[
W = \frac{a^2+b^2}{c} + \frac{b^2+c^2}{a} + \frac{c^2+a^2}{b}.
\]","
oindent\textbf{Solution 1.} From $a+b+c=0$ we have $ab+bc+ca = -\dfrac{a^2+b^2+c^2}{2}$, so we get:
\begin{align*}
W &= \frac{a^2+b^2}{c} + \frac{b^2+c^2}{a} + \frac{c^2+a^2}{b} \\
  &= \frac{a^2+b^2}{c}+c + \frac{b^2+c^2}{a}+a + \frac{c^2+a^2}{b}+b \\
  &= (a^2+b^2+c^2)\cdot\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}
ight) \\
  &= (a^2+b^2+c^2)\cdot\frac{ab+bc+ca}{abc} \\
  &= -\frac{(a^2+b^2+c^2)^2}{2abc} \\
  &= \frac{(a^2+b^2+c^2)^2}{32}.
\end{align*}


oindent As $a^2+b^2+c^2>0$, the goal is to minimize the expression:
\[
I = a^2+b^2+c^2 = a^2+b^2+(a+b)^2 = 2\cdot(a^2+ab+b^2) = 2\cdot\bigl((a+b)^2 - ab\bigr).
\]


oindent WLOG assume that $a \geq b \geq c$, so $a+b \geq 0$. From $abc = -16$ we have $ab(a+b)=16$. As $(a+b)^2 \geq 4ab$, it follows that $16 = ab(a+b) \leq \dfrac{(a+b)^3}{4}$, from which we conclude $a+b \geq 4$. Now we get $I = 2\cdot\bigl((a+b)^2 - ab\bigr) = 2\bigl((a+b)^2 - \dfrac{16}{a+b}\bigr) \geq 2\bigl(4^2 - \dfrac{16}{4}\bigr) = 24$. Finally, we have
\[
W = \frac{(a^2+b^2+c^2)^2}{32} \geq \frac{24^2}{32} = 18.
\]
This value can be obtained when $(a,b,c) = (2,2,-4)$.

\medskip

oindent\textbf{Comment 1:} There are many different ways to get $W = \dfrac{(a^2+b^2+c^2)^2}{32} = \dfrac{(ab+bc+ca)^2}{8} = \dfrac{(a^2+b^2+ab)^2}{8}$. But in the case of $W = \dfrac{(ab+bc+ca)^2}{8}$, we need to maximize $ab+bc+ca$, because it is negative.

\medskip

oindent\textbf{Comment 2:} One can express $a^2+ab+b^2 = (a+b)^2-ab$ as a function of $ab$ only (as $\dfrac{256}{ab} - ab$), and then minimize this expression using the inequality $ab \leq 4$ (which is equivalent to $a+b \geq 4$ because of $ab(a+b)=16$), or alternatively use both of these estimations to conclude $(a+b)^2 - ab \geq 4^2 - 4 = 12$.

\medskip

oindent\textbf{Solution 2.} We have
\begin{align*}
W &= \frac{a^2+b^2}{c} + \frac{b^2+c^2}{a} + \frac{c^2+a^2}{b} \\
  &= \frac{(a+b)^2-2ab}{c} + \frac{(b+c)^2-2bc}{a} + \frac{(c+a)^2-2ca}{b} \\
  &= \frac{c^2+\frac{32}{c}}{c} + \frac{a^2+\frac{32}{a}}{a} + \frac{b^2+\frac{32}{b}}{b} \\
  &= a+b+c+32\cdot\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}
ight) \\
  &= 32\cdot\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}
ight).
\end{align*}


oindent But, from $AM \geq GM$ we get
\begin{align*}
L &= \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \\
  &= \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{(a+b)^2} \\
  &= \frac{1}{4a^2}+\frac{1}{4a^2}+\frac{1}{4a^2}+\frac{1}{4a^2}+\frac{1}{4b^2}+\frac{1}{4b^2}+\frac{1}{4b^2}+\frac{1}{4b^2}+\frac{1}{(a+b)^2} \\
  &\geq 9\sqrt[9]{\frac{1}{2^{16}a^8b^8(a+b)^2}} = 9\sqrt[9]{\frac{1}{2^{16}\cdot 2^8 \cdot (ab)^6}} = \frac{9}{16},
\end{align*}
where we used $a^2b^2(a+b)^2 = 16^2 = 256$ and $ab \leq 4$. Finally, we have $W = 32\cdot\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}
ight) \geq 32 \cdot \dfrac{9}{16} = 18$. This value can be obtained for $a=b=2$, $c=-4$.

\medskip

oindent\textbf{Comment:} There are a few solutions based on the same idea. For example, it is easy to get $W = -\dfrac{a^4+b^4+c^4}{abc} = \dfrac{a^4+b^4+c^4}{16} = \dfrac{a^3+b^4+(a+b)^4}{16}$, from where one can apply $AM \geq GM$ to 18 numbers: $a^4$, $b^4$ and 16 numbers equal to $\dfrac{(a+b)^4}{16}$, and finish the solution using the fact that $ab(a+b)=16$ and $a+b \geq 4$. Also, for minimizing the expression $a^2+b^2+c^2 = a^2+b^2+(a+b)^2$ one can use $AM \geq GM$ for 6 numbers: $a^2$, $b^2$ and 4 numbers equal to $\dfrac{(a+b)^2}{4}$, and then finish in a similar way.

\medskip

oindent\textbf{Solution 3.} WLOG assume $a$ and $b$ are positive and $c$ is negative. Then, we have $c=-(a+b)$ and $ab(a+b)=16$ and
\begin{align*}
W &= \frac{a^2+b^2}{c} + \frac{b^2+c^2}{a} + \frac{c^2+a^2}{b} \\
  &= -\frac{a^2+b^2}{a+b} + \frac{b^2+(a+b)^2}{a} + \frac{(c+a)^2-2ca}{b} \\
  &= \frac{2a(a+b)}{b} + \frac{2b(a+b)}{a} + \frac{2ab}{a+b} \\
  &= \frac{a(a+b)}{2b}+\frac{a(a+b)}{2b}+\frac{a(a+b)}{2b}+\frac{a(a+b)}{2b} \\
  &\quad +\frac{b(a+b)}{2a}+\frac{b(a+b)}{2a}+\frac{b(a+b)}{2a}+\frac{b(a+b)}{2a}+\frac{2ab}{a+b} \\
  &\geq 9\sqrt[9]{\frac{ab(a+b)^7}{2^7}} = 9\sqrt[9]{\frac{(a+b)^6}{8}} \geq 18.
\end{align*}


oindent The last inequality is due to $a+b = \dfrac{16}{ab} \geq \dfrac{64}{(a+b)^2} \Rightarrow a+b \geq 4$.\hfill$\square$",178,4217,Algebra,2
3,shl_jbmo_2024_a3,shl_jbmo,2024,a,"Anna and Bob are constructing quadratic polynomials $f_A$ and $f_B$ as follows: With Anna starting first, they take alternate turns in choosing one by one the coefficients of the polynomials, with Anna choosing the coefficients of $f_A$, and Bob the coefficients of $f_B$. In their turn, each player can choose whichever coefficient of their polynomial is not yet chosen, with the only restriction being that the coefficients have to be positive real numbers.

\medskip

oindent Bob wins if any of the following two cases occurs:
\begin{enumerate}[label=(\alph*)]
    \item The roots of $f_A(x)$ are not real.
    \item The roots of both polynomials are real numbers and furthermore each root of $f_B(x)$ is strictly larger than each root of $f_A(x)$.
\end{enumerate}
Otherwise Anna wins. Determine which player has a winning strategy.","We will show that Anna has a winning strategy. In her first two turns she chooses arbitrarily the coefficient of $x^2$ and the constant coefficient.

\medskip

oindent We will prove the following two lemmas from which it is immediate that Anna can choose in her last turn the coefficient of $x$ in order to win irrespective of Bob's last choice.

\medskip

oindent\textbf{Lemma 1:} There is a real number $r < 0$, depending only on the first two choices of Bob, such that if $f_B$ has real roots, then at least one of them is less than or equal to $r$.

\medskip

oindent\textbf{Lemma 2:} Given $a, c > 0$ and $r < 0$, there is a $b > 0$ such that $ax^2+bx+c$ has two real roots with one of them equal to $r$.

\medskip

oindent\textbf{Proof of Lemma 1:} Let $f_B(x) = a'x^2+b'x+c'$. We may assume that $f_B$ has real roots as otherwise Lemma 1 is vacuously true. Note that the roots have sum $-b'/a' < 0$ and product $c'/a' > 0$, so they are both negative.

\medskip

oindent If Bob chose $a'$ and $b'$ in his first two moves, then we can take $r = -\dfrac{b'}{2a'}$.

\medskip

oindent If Bob chose $a'$ and $c'$ in his first two moves, then we can take $r = -\sqrt{c'/a'}$.

\medskip

oindent If Bob chose $b'$ and $c'$ in his first two moves, then the two roots $p$, $q$ of $f_B(x)$ satisfy
\[
\frac{1}{p}+\frac{1}{q} = \frac{p+q}{pq} = \frac{-b'/a'}{c'/a'} = -\frac{b'}{c'}.
\]
So we can take $r = -2c'/b'$ which works since if $r < p, q$, then
\[
-\frac{b'}{c'} = \frac{1}{p}+\frac{1}{q} < \frac{2}{r} = -\frac{b'}{c'},
\]
a contradiction.

\medskip

oindent\textbf{Proof of Lemma 2:} Just choose
\[
b = -\frac{ar^2+c}{r} > 0. \qquad \square
\]",835,1650,Algebra,3
4,shl_jbmo_2024_a4,shl_jbmo,2024,a,"If $a$, $b$, $c$, $d$ are positive real numbers such that $(a+c)^2 = 4(ad+bc)$ then prove that
\[
\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} + \frac{4bd}{ac} \geq 6.
\]
When does the equality hold?","
oindent\textbf{Solution 1.} Taking the LHS and writing on the following form and then using AM-GM we have,
\[
\frac{a}{b}+\frac{c}{d}+\frac{ab+cd+4bd}{ca} \geq 3\sqrt[3]{\frac{a}{b}\cdot\frac{c}{d}\cdot\frac{ab+cd+4bd}{ca}} = 3\sqrt[3]{\frac{a}{d}+\frac{c}{b}+4}.
\]
Hence in order to prove the inequality it is enough to prove,
\[
\frac{a}{d}+\frac{c}{b} \geq 4.
\]
But using Cauchy-Schwarz inequality and the condition we have,
\[
(ad+bc)\left(\frac{a}{d}+\frac{c}{b}
ight) \geq (a+c)^2 = 4(ad+bc) \Rightarrow \frac{a}{d}+\frac{c}{b} \geq 4,
\]
as desired.

\medskip

oindent Now to prove the equality, on the part we use Cauchy-Schwarz we have that the equality holds only if,
\[
\frac{ad}{\frac{a}{d}} = \frac{bc}{\frac{c}{b}} \Rightarrow d^2 = b^2 \Rightarrow b = d.
\]
On the part when we used AM-GM equality holds only if,
\[
\frac{a}{b} = \frac{c}{d} \Rightarrow a = c.
\]
Now from condition we learn that we must have $a = 2b$, hence equality holds only when we have $a = 2b = c = 2d$. We can check easily in all cases that inequality transforms to equality. So, equality holds iff $a:b:c:d = 2:1:2:1$.

\medskip

oindent\textbf{Solution 2.} If we let $a = 2x$ and $c = 2y$ and rearrange the inequality, this is equivalent to showing under the condition $(x+y)^2 = 2(xd+yb)$ that
\[
4x^2yd + xb^2d + 4xby^2 + byd^2 + 2b^2d^2 \geq 12xbyd.
\]
Immediately using $AM-GM$ for the 12 terms on LHS does not work since we get $x^{13}$, $y^{13}$, $b^{11}$, $d^{11}$. Therefore, we will first show that $xb^2d + byd^2 \geq 2b^2d^2$ and then $AM-GM$ will complete the solution. For that we need to show that $\dfrac{x}{d}+\dfrac{y}{b} \geq 2$ and it comes from Cauchy-Schwarz inequality
\[
\left(\frac{x}{d}+\frac{y}{b}
ight)(xd+yb) \geq (x+y)^2.
\]
Equality holds when $x = b = y = d$ from the $AM-GM$ part. This also satisfies the Cauchy-Schwarz equality condition.\hfill$\square$",208,1881,Algebra,4
5,shl_jbmo_2024_a5,shl_jbmo,2024,a,"Find all triples $(a,b,c)$ of positive real numbers that satisfy the system of equations
\[
a + b + c = \frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3},
\]
\[
ab + bc + ca = \sqrt{a} + \sqrt{b} + \sqrt{c}.
\]","From the following well-known inequality, which holds for all reals $x$, $y$, $z$
\[
3(x^2+y^2+z^2) \geq (x+y+z)^2 \geq 3(xy+yz+zx),
\]
it can be easily derived
\[
3(a+b+c) \geq (\sqrt{a}+\sqrt{b}+\sqrt{c})^2, \qquad (ab+bc+ca)^2 \geq 3abc(a+b+c).
\]
Using the second equation of the given system, we get
\[
3(a+b+c) \geq 3abc(a+b+c),
\]
so
\begin{equation}
abc \leq 1. \tag{1}
\end{equation}


oindent Furthermore, by AM-GM inequality, for all positive reals $x$ it holds
\begin{equation}
\frac{3}{x^3}+x = \frac{1}{x^3}+\frac{1}{x^3}+\frac{1}{x^3}+x \geq 4\sqrt[4]{\frac{1}{x^3}\cdot\frac{1}{x^3}\cdot\frac{1}{x^3}\cdot x} = \frac{4}{x^2}. \tag{2}
\end{equation}


oindent Using (2) and the first equation of the given system, we obtain
\begin{equation}
\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3} \geq \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}. \tag{3}
\end{equation}


oindent Then,
\[
a+b+c = \frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3} \geq \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \geq \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca} = \frac{a+b+c}{abc},
\]
so
\begin{equation}
abc \geq 1. \tag{4}
\end{equation}


oindent Therefore, from (1) and (4) it follows that $abc = 1$ and the equality must hold in each of the inequalities we have proved. The only solution of the given system is
\[
(a,b,c) = (1,1,1). \qquad\square
\]",206,1322,Algebra,5
6,shl_jbmo_2024_a6,shl_jbmo,2024,a,"Consider the function $f : \mathbb{R} \setminus \left\{-\dfrac{1}{1012},\, 1
ight\} \to \mathbb{R}$ defined by
\[
f(x) = \frac{2x-1}{(1012x+1)(1-x)}.
\]
Compute
\[
\left\lfloor \sum_{k=0}^{2023} f\!\left(\frac{k}{2024}
ight) 
ight
floor.
\]","
oindent\textbf{Solution 1.} Let $g(x) = 2x-1$ and $h(x) = (1012x+1)(1-x) = -1012x^2+1011x+1$. Note that $h(x)$ is a quadratic function which is symmetric about $\dfrac{1011}{2024}$. Therefore
\[
f(x)+f\!\left(\frac{1011}{1012}-x
ight) = \frac{g(x)+g\!\left(\frac{1011}{1012}-x
ight)}{h(x)} = \frac{-1}{506\,h(x)}.
\]


oindent So
\begin{align*}
\sum_{k=0}^{2023}f\!\left(\frac{k}{2024}
ight)
&= \sum_{k=0}^{1010}\left[f\!\left(\frac{k}{2024}
ight)+f\!\left(\frac{2022-k}{2024}
ight)
ight] + f\!\left(\frac{1011}{2024}
ight)+f\!\left(\frac{2023}{2024}
ight) \\
&= -\frac{1}{506}\sum_{k=0}^{1010}\frac{1}{h\!\left(\frac{k}{2024}
ight)} - \frac{1/1012}{\frac{1013}{2}\cdot\frac{1013}{2024}} + \frac{1011/1012}{\frac{2025}{2}\cdot\frac{1}{2024}} \\
&= -\frac{1}{506}\sum_{k=0}^{1010}\frac{1}{h\!\left(\frac{k}{2024}
ight)} - \frac{4}{1013^2} + \frac{4044}{2025}.
\end{align*}


oindent We now observe that
\[
S = \frac{1}{506}\sum_{k=0}^{1010}\frac{1}{h\!\left(\frac{k}{2024}
ight)} = \frac{1}{506}\sum_{k=0}^{1010}\frac{4048}{(k+2)(2024-k)} = \sum_{k=0}^{1010}\frac{8}{(k+2)(2024-k)}.
\]


oindent We have $S > 0$ since all summands are positive. Furthermore, since $h(x)$ is increasing on $\left[0,\dfrac{1011}{2024}
ight]$, then $(k+2)(2024-k)$ is decreasing on $[0,1010]$. Thus
\[
0 < S = \sum_{k=0}^{5}\frac{8}{(k+2)(2024-k)} + \sum_{k=6}^{1010}\frac{8}{(k+2)(2024-k)} \leq \frac{6\cdot8}{2\cdot2024}+\frac{8\cdot1005}{8\cdot2018} < \frac{3}{253}+\frac{1}{2}
\]
and so
\[
2 > \sum_{k=0}^{2023}f\!\left(\frac{k}{2024}
ight) > 2 - \frac{6}{2025} - \frac{4}{1013^2} - \frac{3}{253} - \frac{1}{2} > 1.
\]
Thus
\[
\left\lfloor\sum_{k=0}^{2023}f\!\left(\frac{k}{2024}
ight)
ight
floor = 1.
\]

\medskip

oindent\textbf{Solution 2.} Let $g(x) = \dfrac{1}{(1012x+1)(1-x)}$, $p(x) = \dfrac{1}{1-x}$, $q(x) = \dfrac{1}{1012x+1}$. Also let
\[
S = \sum_{k=0}^{2023}f\!\left(\frac{k}{2024}
ight),\quad T = \sum_{k=0}^{2023}g\!\left(\frac{k}{2024}
ight),\quad A = \sum_{k=0}^{2023}p\!\left(\frac{k}{2024}
ight),\quad B = \sum_{k=0}^{2023}q\!\left(\frac{k}{2024}
ight).
\]


oindent Since $(1012x+1)(1-x) = -1012x^2+1011x+1$ is symmetric about $\dfrac{1011}{2024}$, we have
\begin{align*}
S = \sum_{k=0}^{2023}f\!\left(\frac{k}{2024}
ight)
&= \sum_{k=0}^{1010}\left[f\!\left(\frac{k}{2024}
ight)+f\!\left(\frac{2022-k}{2024}
ight)
ight]+f\!\left(\frac{1011}{2024}
ight)+f\!\left(\frac{2023}{2024}
ight)\\
&= \frac{-1}{506}\sum_{k=0}^{1010}g\!\left(\frac{k}{2024}
ight) + f\!\left(\frac{1011}{2024}
ight)+f\!\left(\frac{2023}{2024}
ight)\\
&= \frac{-1}{1012}\sum_{k=0}^{2023}g\!\left(\frac{k}{2024}
ight) + f\!\left(\frac{1011}{2024}
ight)+f\!\left(\frac{2023}{2024}
ight)-g\!\left(\frac{1011}{2024}
ight)-g\!\left(\frac{2023}{2024}
ight)\\
&= -\frac{T}{1012} - \frac{4}{1013^2} + \frac{4044}{2025} + \frac{4}{1013^2} - \frac{4}{2025}\\
&= -\frac{T}{1012} + \frac{4048}{2025}.
\end{align*}


oindent From the expressions
\[
\frac{2x-1}{(1012x+1)(1-x)} = \frac{1}{1013}\left(\frac{-1014}{1012x+1}+\frac{1}{1-x}
ight),
\]
\[
\frac{1}{(1012x+1)(1-x)} = \frac{1}{1013}\left(\frac{1012}{1012x+1}+\frac{1}{1-x}
ight),
\]
we get
\[
S = \frac{A-1014B}{1013},\quad T = \frac{A+1012B}{1013}.
\]
This means $1012S+1014T = 2A$. Combining with $S = -\dfrac{T}{1012}+\dfrac{4048}{2025}$, we get
\[
S = \frac{1014\cdot4048}{1013\cdot2025} - \frac{A}{506\cdot1013}.
\]


oindent We can estimate $A$ by
\[
A = \sum_{k=0}^{2023}\frac{1}{1-\frac{k}{2024}} = 2024\left(\sum_{k=1}^{2024}\frac{1}{k}
ight),
\]
and
\[
\left(\sum_{k=1}^{2024}\frac{1}{k}
ight) \in \left(1+10\tfrac{1}{2},\, 1+10
ight) = (6,11).
\]


oindent Therefore,
\[
S > \frac{1014\cdot4048}{1013\cdot2025} - \frac{11\cdot2024}{506\cdot1013} = \frac{2048}{2025}\cdot\frac{2024}{1013} - \frac{4}{1013} > 1\cdot\frac{3}{2} - \frac{4}{1013} > 1,
\]
and
\[
S < \frac{1014\cdot4048}{1013\cdot2025} - \frac{6\cdot2024}{506\cdot1013} < \frac{2024}{1013}\cdot\left(1+\frac{3}{2025}-\frac{6}{506}
ight) < 2\cdot1 = 2.
\]
Hence $\left\lfloor S
ight
floor = 1$.\hfill$\square$",240,3998,Algebra,6
7,shl_jbmo_2024_a7,shl_jbmo,2024,a,"Show that for any real number $k>2$, the inequality
\[
1 - \frac{1}{2\sqrt{n}} < \left\{\frac{1+\sqrt{4n+1}}{2}
ight\} < 1 - \frac{1}{k\sqrt{n}}
\]
holds for infinitely many positive integers $n$.","Firstly, we show that there is a constant $N_k$ depending on $k$ such that
\[
\left\{\frac{1+\sqrt{4n+1}}{2}
ight\} < 1 - \frac{1}{k\sqrt{n}}
\]
holds for all $n > N_k$.

\medskip

oindent Write $\sqrt{4n+1} = 2m+1+2a$ where $m \geq 3$ is an integer and $a \in [0,1)$ is a real number. Then $n = (m+a)(m+a+1) > m^2$, the left-hand side is equal to $a$ and the right-hand side is at least $1-\dfrac{1}{km}$. So if $a = 1-b$, $b \in (0,1]$, then we have $n = (m+1-b)(m+2-b)$ for integers $m$ and $n$ and wish to prove $b > \dfrac{1}{km}$. Suppose the contrary, i.e.\ $b \leq \dfrac{1}{km}$. From $n = (m+1-b)(m+2-b) = (m+1)(m+2)-(2m+3)b+b^2$ we obtain that $(2m+3)b-b^2$ is necessarily an integer. But $(2m+3)b-b^2 = b(2m+3-b) > 0$ and $(2m+3)b - b^2 < (2m+3)b \leq \dfrac{2m+3}{km} \leq 1$, contradiction for $m > \dfrac{3}{k-2}$ (this corresponds to a threshold for $n$).

\medskip

oindent Now we check that for $n$ of the form $n = c^2+c-1$, where $c$ is a positive integer, the inequality
\[
1 - \frac{1}{2\sqrt{n}} < \left\{\frac{1+\sqrt{4n+1}}{2}
ight\}
\]
holds. This choice of $n$ makes $4n+1$ very close to, but below by a very little, from an odd perfect square.

\medskip

oindent We have $c < \dfrac{1+\sqrt{4c^2+4c-3}}{2} < c+1$, so
\begin{align*}
\left\{\frac{1+\sqrt{4n+1}}{2}
ight\} &= \frac{1+\sqrt{4n+1}}{2} - c = \frac{1-2c+\sqrt{4c^2+4c-3}}{2} \\
&= 1 - \frac{2c+1-\sqrt{4c^2+4c-3}}{2} = 1 - \frac{2}{2c+1+\sqrt{4c^2+4c-3}} \\
&> 1 - \frac{2}{\sqrt{4c^2+4c-4}+\sqrt{4c^2+4c-4}} = 1 - \frac{1}{2\sqrt{n}}.
\end{align*}


oindent Hence, the inequality in the statement holds for sufficiently large positive integers $n$ of the form $c^2+c-1$.

\medskip

oindent\textbf{Remark.} Original statement of the problem is: Determine the smallest positive integer $k$ such that the inequality
\[
\left\{\frac{1+\sqrt{4n+1}}{2}
ight\} < 1 - \frac{1}{k\sqrt{n}}
\]
holds for all positive integers $n \geq 100$.

\medskip

oindent\textbf{Solution to the original version.} Firstly we show that $k=3$ works, by proving the inequality for all $n \geq 12$. (One can verify by hand that it is also true when $n \leq 11$, though this is not needed here.) Write $\sqrt{4n+1} = 2m+1+2a$ where $m \geq 3$ is an integer and $a \in [0,1)$ is a real number. Then $n = (m+a)(m+a+1) > m^2$, the left-hand side is equal to $a$ and the right-hand side is at least $1-\dfrac{1}{3m}$. So if $a = 1-b$, $b \in (0,1]$, then we have $n = (m+1-b)(m+2-b)$ for integers $m$ and $n$ and wish to prove $b > \dfrac{1}{3m}$. Suppose the contrary, i.e.\ $b \leq \dfrac{1}{3m}$. From $n = (m+1-b)(m+2-b) = (m+1)(m+2)-(2m+3)b+b^2$ we obtain that $(2m+3)b - b^2$ is necessarily an integer. But $(2m+3)b-b^2 = b(2m+3-b) > 0$ and $(2m+3)b-b^2 < (2m+3)b \leq \dfrac{2m+3}{3m} = \dfrac{2}{3}+\dfrac{1}{m} \leq 1$, contradiction!

\medskip

oindent Now we check that $k=2$ (and hence $k=1$, too) does not work. It is natural to aim to make the left-hand side close to 1, by picking $n$ such that $4n+1$ is close to, but below by a very little, from an odd perfect square. So let $n = c^2+c-1 > c^2$. Then from $c < \dfrac{1+\sqrt{4c^2+4c-3}}{2} < c+1$, we get that
\begin{align*}
\left\{\frac{1+\sqrt{4n+1}}{2}
ight\} &= \frac{1-2c+\sqrt{4c^2+4c-3}}{2} = 1 - \frac{2c+1-\sqrt{4c^2+4c-3}}{2} \\
&= 1 - \frac{2}{2c+1+\sqrt{4c^2+4c-3}} > 1 - \frac{2}{\sqrt{4c^2+4c-4}+\sqrt{4c^2+4c-4}} = 1 - \frac{1}{2\sqrt{n}}.
\end{align*}

\medskip

oindent\textbf{Remark for the original version.} The first part of the proof can be easily mimicked to show that the inequality holds for all real numbers $k>2$ and sufficiently large $n$ (with lower bound depending on $k$), while the second part shows that when $k=2$ the inequality fails for infinitely many $n$. The proposed phrasing is just for a little ease, but of course feel free to change it if you like.\hfill$\square$",196,3832,Algebra,7
8,shl_jbmo_2024_c1,shl_jbmo,2024,c,"Determine the smallest positive integer $k$ with the following property: For any subset $S$ of the set $\{1,2,\ldots,2024\}$ with $|S|=k$, there are two distinct elements $a, b \in S$ such that $ab+1$ is a perfect square.","The answer is $k=1013$. Consider the 1012 pairs of the form $(4a+1,4a+3)$ and $(4a+2,4a+4)$ for $a=0,1,\ldots,505$. If $S$ has at least 1013 elements, then by the Pigeonhole principle there is a pair such that both of its elements are contained in $S$. For this pair, say $(m,m+2)$, we have $m(m+2)+1=(m+1)^2$ which is a perfect square. So, $k=1013$ satisfies the problem condition.

\medskip

oindent For $k=1012$, consider the set $S$ consisting of all the numbers of the form $4k+1$ and $4k+2$. Here, $|S|=1012$ and we will show that no two numbers $a$, $b$ in this set satisfy the problem condition. If $a\equiv 1\pmod{4}$ and $b\equiv 2\pmod{4}$ then $ab+1\equiv 3\pmod{4}$, can't be a perfect square. If $a\equiv 1\pmod{4}$ and $b\equiv 1\pmod{4}$ then $ab+1\equiv 2\pmod{4}$, can't be a perfect square. If $a\equiv 2\pmod{4}$ and $b\equiv 2\pmod{4}$ then $ab+1\equiv 5\pmod{8}$, can't be a square hence we're done.\hfill$\square$",221,936,Combinatorics,8
9,shl_jbmo_2024_c2,shl_jbmo,2024,c,"Let $n$ be a positive integer such that $n^2-1$ is divisible by 6. An $n\times n$ board is given. Prove that it is possible to place $\dfrac{n^2-1}{6}$ non-overlapping right triangles on this board with side lengths of 3, 4, and 5.","It is convenient to tile the board with $3\times4$ rectangles, because each of them can be split into two 3-4-5 right triangles. Also, we can tile the $5\times5$ and $7\times7$ boards in the following way: [for $5\times5$: place four 3-4-5 triangles leaving one unit square; for $7\times7$: place triangles as shown in the diagram.]

\medskip

oindent Note that the condition $6\mid n^2-1$, implies that $n$ must be of the form $6k-1$ or $6k+1$. We will use induction on $k$ to prove the problem statement. First, we will cover the base cases.

\medskip

oindent The $11\times11$ board can be tiled with four $8\times3$ rectangles and a $5\times5$ square in the middle. We place four 3-4-5 triangles into the $5\times5$ square, leaving one unit square unoccupied, as we have previously shown. Each $8\times3$ rectangle can be dissected into two $4\times3$ rectangles.

\medskip

oindent The $13\times13$ board can be tiled with four $9\times4$ rectangles and a $5\times5$ square in the middle. We place four 3-4-5 triangles into the $5\times5$ square and each $9\times4$ rectangle can be dissected into three $4\times3$ rectangles.

\medskip

oindent The $17\times17$ board can be tiled with four $8\times9$ rectangles and one unit square in the middle. Each $8\times9$ rectangle can be dissected into six $4\times3$ rectangles.

\medskip

oindent Suppose that we can place $\dfrac{n^2-1}{6}$ 3-4-5 right triangles on an $n\times n$ board for some $n>5$ while leaving exactly one unit square uncovered. Then we will prove that we can cover the $(n+12)\times(n+12)$ board leaving one unit square away to complete the induction. We can cover the top-left $n\times n$ square as in the induction hypothesis. Then we have an area consisting of two $12\times n$ rectangles and one $12\times12$ square is left. We can cover the $12\times12$ square with rectangles of the size $4\times3$. For the other rectangles, we observe that for any $n>5$, we can write $n$ as the sum of several 3s and 4s, since $6=3+3$, $7=3+4$, $8=4+4$ and we can add 3 to these sums to achieve the required sum for any number $n \geq 9$. Then, the $12\times n$ rectangles can be dissected into several $12\times4$ and $12\times3$ rectangles and each of these parts can be covered by $4\times3$ rectangles. This completes the entire covering of the $(n+12)\times(n+12)$ as desired and we're done.\hfill$\square$",231,2378,Combinatorics,9
10,shl_jbmo_2024_c3,shl_jbmo,2024,c,"A set of positive integers is called \emph{arithmetic} if it contains three distinct elements which form an arithmetic sequence. Prove that at least $51\%$ of the subsets of the set $\{1,2,3,\ldots,2024\}$ are arithmetic.","Let $A = \{1,2,3,\ldots,2024\}$ and $\mathcal{P} = \{X \mid X \subseteq A\}$. For each $X \in \mathcal{P}$, let $\overline{X} = A\setminus X \in \mathcal{P}$. Our main claim is that in each pair $(X,\overline{X})$ we find at least one arithmetic set.

\medskip

oindent Assume the contrary and let $X\in\mathcal{P}$ for which $X$ and $\overline{X}$ are not arithmetic. Among $1,2,3,4,5,6$ we find two consecutive numbers in the same set among $X$ and $\overline{X}$, otherwise $1,3,5$ are in one of them and $2,4,6$ are in the other, so both would be arithmetic. Let these numbers be $a$, $a+1\in X$. Then $a+2\in\overline{X}$.

\medskip

oindent If $a+3\in X$, then $a\in X$ implies $a+6\in\overline{X}$, and $a+1\in X$ implies $a+5\in\overline{X}$. Then $a+4,a+7\in X$, and as $a+1\in X$, the set $X$ is arithmetic.

\medskip

oindent If $a+3\in\overline{X}$, then $a+2\in\overline{X}$ implies $a+4\in X$. As $a+1\in X$, we have $a+7\in\overline{X}$, and as $a+3\in\overline{X}$, we have $a+5\in X$. Then, as $a+4\in X$, we have $a+6\in\overline{X}$. But $a+7\in\overline{X}$, so $a+8\in X$. Then $a$, $a+4$, $a+8\in X$, so $X$ is arithmetic.

\medskip

oindent For every $Y\subseteq\{7,8,\ldots,2024\}$, the pair
\[
\bigl(\{1,2,3\}\cup Y,\quad \{4,5,6\}\cup(\{7,8,\ldots,2024\}\setminus Y)\bigr)
\]
is of type $(X,\overline{X})$ and both of its components are arithmetic sets. These pairs count up to $2^{2024-6}=2^{2018}$. The total number of pairs is $\dfrac{2^{2024}}{2}$, so we are left with $2^{2023}-2^{2018}$ pairs which contain each at least one arithmetic set.

\medskip

oindent Then the number of arithmetic sets is at least $2^{2024}-2^{2023}+2^{2018} = 2^{2024}\cdot\left(\dfrac{1}{2}+\dfrac{1}{2^6}
ight)$. As $2^6<100$, we have $\dfrac{1}{2}+\dfrac{1}{2^6}>0.51$.\hfill$\square$",221,1796,Combinatorics,10
11,shl_jbmo_2024_c4,shl_jbmo,2024,c,All the unit cubes of a $5\times 5\times 3$ box is colored white initially. We call two unit cubes adjacent to each other if they share a common face. What is the largest number of unit cubes we can color in black such that each black unit cube is adjacent to at most one other black unit cube?,"The answer is 39. Consider the following colourings of a $5\times5\times1$ cube consisting of 14 and 11 black unit cubes respectively. We will use the colouring with 14 black squares in the first and third layers of size $1\times5\times5$ and we will use the colouring with 11 black squares in the second layer. Therefore, we have $2\cdot14+11=39$ black unit cubes in total and it can be checked that each black cube is only adjacent to black cubes on the same layer and there are at most one such black cube. Hence this colouring of 39 black squares satisfies the requirements.

\medskip

oindent Now we will prove that it is not possible to have more than 39 black unit cubes while satisfying the requirements. Assume that we have a valid colouring of the $5\times5\times3$ box.

\medskip

oindent\textit{Claim 1:} A $2\times2\times1$ box---and its all possible rotations---can contain at most two black cubes.

\medskip

oindent\textit{Proof:} Assume otherwise, we would have at least three black cubes, which means there is a row containing only black cubes, say it is the first row. Also, there must be another black cube on the other row, hence in the first row there is a black cube which is adjacent to two black squares, one in the same row and one in the same column, contradiction. $\square$

\medskip

oindent\textit{Claim 2:} A $3\times3\times1$ box---and its all possible rotations---can contain at most five black cubes.

\medskip

oindent\textit{Proof:} Assume otherwise, we would have at least six black cubes. If there is a row containing three black cubes, then the black cube in the middle is adjacent to two black cubes in the same row. So, we must have two black cubes in each row.

\medskip

oindent Assume that there is a row $R_1$ containing two black cubes that are adjacent to each other. Then, consider a row $R_2$ that is next to $R_1$, there are two unit cubes that are adjacent to the black cubes of $R_1$. They cannot be white at the same time, because there are two black cubes in each row. Hence, one of them is black and this would mean that the black square it is adjacent to in $R_1$ is violating the condition. Hence, all the rows must contain the black cubes in its first and third unit cubes. But in this case, we have three black squares in the first (and third) column, contradiction. $\square$

\medskip

oindent\textit{Claim 3:} Consider the L-shaped 9 squares formed by the 3 cubes lying at one of the edges of the $5\times5\times3$ box, and all the squares adjacent to them. These 9 squares can contain at most five black cubes.

\medskip

oindent\textit{Proof:} Exactly the same proof as Claim 2 works as this shape is essentially the same as the $3\times3\times1$ cube, in terms of adjacency. $\square$

\medskip

oindent Now, we are ready to connect the above observations for the main argument. We will dissect the $5\times5\times3$ box into several pieces. First, consider the $4\times4\times3$ box on the top-left corner. We can dissect it into twelve $2\times2\times1$ boxes hence it can contain at most $12\cdot2=24$ black cubes by Claim 1. Then, we can dissect the remaining shape into two $3\times3\times1$ cubes and one L-shaped figure described in Claim 3, and from Claim 2 and Claim 3 these parts can contain at most $3\cdot5=15$ black squares. Finally, in the entire box we can have at most $24+15=39$ black unit cubes hence we're done.\hfill$\square$",294,3411,Combinatorics,11
12,shl_jbmo_2024_c5,shl_jbmo,2024,c,"Let $n$ be a positive integer. Find the largest positive integer $M$ that satisfies the following condition: Whenever $a_1, a_2, \ldots, a_n$ are distinct positive integers with $a_1 + a_2 + \cdots + a_n = M$, there exists $b_1, b_2, \ldots, b_n \in \{-2,-1,0,1,2\}$, not all equal to 0, such that
\[
a_1 b_1 + a_2 b_2 + \cdots + a_n b_n = 0.
\]","The answer is $M = \dfrac{3^n-3}{2}$.

\medskip

oindent First, we will show that no positive integer $M$ greater than or equal to $\dfrac{3^n-1}{2}$ satisfies the problem condition. Let $M = \dfrac{3^n-1}{2}+k$ where $k$ is a non-negative integer. Consider the values $a_1=1$, $a_2=3$, $a_3=9$, $\ldots$, $a_{n-1}=3^{n-2}$ and $a_n=3^{n-1}+k$. Clearly we have $a_1+a_2+\cdots+a_n=M$. Assume that $a_1b_1+a_2b_2+\cdots+a_nb_n=0$ for some choice of $b_i$s and $j$ is the largest index where $b_j$ is non-zero. WLOG assume that $b_j$ is positive (otherwise, multiply all $b_i$s by $-1$). Then, we have $b_j - 2b_{j-1} - \cdots - 2b_1 > 0$ so there are no possible choice that satisfies the condition.

\medskip

oindent For the second part, assume that $M = \dfrac{3^n-3}{2}$. Consider $x = a_1c_1+a_2c_2+\cdots+a_nc_n$ for every choice of $c_1,c_2,\ldots,c_n\in\{-1,0,1\}$. These choices give $3^n$ possible sums, denote by $S$ the set of these possible sums. We see that $-M\leq x\leq M$, hence $x$ can achieve at most $2M+1$ different values where $2M+1<3^n$. Thus, by the Pigeonhole Principle, there must be two distinct sums such that
\[
a_1c_1+a_2c_2+\cdots+a_nc_n = a_1c_1'+a_2c_2'+\cdots+a_nc_n'.
\]
Hence we have $a_1(c_1-c_1')+\cdots+a_n(c_n-c_n')=0$ and $c_i-c_i'$ values are in the set $\{-2,-1,0,1,2\}$ and not all zero, so we're done.\hfill$\square$",345,1361,Combinatorics,12
13,shl_jbmo_2024_c6,shl_jbmo,2024,c,"Three friends Archie, Billie and Charlie play a game. At the beginning of the game, each of them has a pile of 2024 pebbles. Archie makes the first move, Billie makes the second, Charlie makes the third and they continue to make moves in the same order. In each move, the player who is making the move chooses a positive integer $n$ which is greater than any previously chosen number, removes $2n$ pebbles from his pile and gives $n$ pebbles to each of the other two players. The game ends when a player does not have enough pebbles to make a move, and that player loses the game. Determine the players who has a strategy such that, regardless of how the other two players play, he won't lose the game.","We will prove that only Charlie has a non-losing strategy. First we discuss what happens right before a player loses the game. Let $t$ be the number chosen in the last move and let the losing player have $s$ pebbles in his pile before the move. In order for the player to lose $2t+2$ must be larger than $s+t$ so that he can't make the next move. This implies that $t \geq s-1$, i.e.\ before the move the previous player must have at least $2s-2$ pebbles. This means that if a player before his move has at least $2s-2$ pebbles and the next player has $s$ pebbles, then he can choose $s-1$ to make the next player lose. This will leave the next player with $2s-1$ pebble, disabling him to make a move. Also, if player $X$ has $s$ pebbles while the player $Y$ has at most $2s-3$ pebbles right before player $Y$ plays, then player $X$ guarantees not to lose the game in his turn.

\medskip

oindent Assume that at some point in the game the consecutive players have $x$, $y$, and $z$ pebbles in their piles and they choose numbers $u$, $u+v$, and $u+v+w$ respectively. In these three moves we have:
\begin{align*}
(x,y,z) &\mapsto (x-2u,\, y+u,\, z+u) \mapsto \\
(x-u+v,\, y-u-2v,\, z+2u+v) &\mapsto \\
(x+2v+w,\, y-v+w,\, z-v-2w).
\end{align*}


oindent Considering this the following two statements are true:
\begin{enumerate}[label=(
oman*)]
    \item If the player that plays second (after the position) plays $v=1$, then after the three moves the number of pebbles in his pile does not decrease (since $w$ is positive integer $y-1+w\geq y$).
    \item If the player that plays third (after the position) plays $w=1$, then after the three moves the number of pebbles in the second players pile does not increase (since $v$ is positive integer $y-v+1\leq y$).
\end{enumerate}


oindent Now we will look at the cases for each player. Let $a_i$, $b_i$, $c_i$ be the number Archie, Billie, Charlie chooses in their $i$-th turns, respectively.

\medskip

oindent\textit{Claim 1: Charlie has a non-losing strategy.}

\medskip

oindent\textit{Proof:} After the first move of Archie, Charlie has at least 2025 pebbles in his pile. If Charlie chooses $c_i=b_i+1$, using (i) we conclude that after $3i+1$ moves Charlie has at least 2025 pebbles in his pile since he's in the position of the middle player. In order for Charlie to lose, Billie must have at least $2\cdot2024$ pebbles at some point. However the total number of pebbles $3\cdot2024$ doesn't change throughout the game, hence this situation is impossible $(3\cdot2024 < 2\cdot2024+2025)$. This means that Charlie has a non-losing strategy (choosing $c_i=b_i+1$). $\square$

\medskip

oindent\textit{Claim 2: Billie does not have a non-losing strategy.}

\medskip

oindent\textit{Proof:} Assume that Archie and Charlie play the strategies to choose $a_{i+1}=c_i+1$ and $c_i=b_i+1$ (where $c_0=0$ by definition). By (ii) the number of pebbles in Billie's pile after $3i$ moves does not increase. Since before the first move he has 2024 pebbles, he can't have more than 2024 pebbles at any point after $3i$ moves. If at some point (before his last move) Billie chooses $b_i>a_i+1$ this increases the number of pebbles in Archie's pile before the $3i+2$-nd move, so $a_i > a_{i+1}+1$ this increases the number of pebbles in Archie's pile before the $3i+2$-nd move. Billie is also playing a non-losing strategy, hence in this case Billie will lose. Otherwise, we have the situation:
\begin{align*}
(2024,2024,2024) &\xrightarrow{1} (2022,2025,2025) \xrightarrow{2} (2024,2021,2027) \xrightarrow{3}\\
(2027,2024,2021) &\xrightarrow{4} \cdots \xrightarrow{2019} (4043,2024,5) \xrightarrow{2020} (3,4044,2025).
\end{align*}


oindent After this Billie can choose 2021 or 2022, but both cases lead to him losing:
\begin{align*}
(3,4044,2025) &\xrightarrow{2021} (2024,2,4046) \xrightarrow{2022} (4046,2024,2) \xrightarrow{2023} (0,4047,2025),\\
(3,4044,2025) &\xrightarrow{2022} (2025,0,4047) \xrightarrow{2023} (4048,2023,1) \xrightarrow{2024} (0,4047,2025).
\end{align*}


oindent This means that Billie doesn't have a non-losing strategy. $\square$

\medskip

oindent\textit{Claim 3: Archie does not have a non-losing strategy.}

\medskip

oindent\textit{Proof:} Let Billie and Charlie choose $b_i=a_i+1$ and $c_i=b_i+1$ until the last turn. If Archie chooses $a_i=c_{i-1}+1$ in all his turns (while $c_0=0$), then again it leads to the case $(4043,2024,5)$. After this Archie can choose 2020 or 2021 and in both cases Billie and Charlie can make Archie lose as follows:
\begin{align*}
(4043,2024,5) &\xrightarrow{2020} (3,4044,2025) \xrightarrow{2021} (2024,2,4046) \xrightarrow{2023} (4047,2025,0),\\
(4043,2024,5) &\xrightarrow{2021} (1,4045,2026) \xrightarrow{2022} (2023,1,4048) \xrightarrow{2023} (4046,2024,2).
\end{align*}


oindent Now, assume that Archie chooses $a_i > c_{i-1}+1$ before this case happens. Then let Billie choose $b_i=a_i+1$ and Charlie choose $c_i=b_i+a_i-c_{i-1}$. In this case, before Archie moves, the players have $(2024+3k, 2024, 2024-3k)$ pebbles. If Archie chooses $\dfrac{2024+3k}{2} > L > 3k+1$, with the described moves, the number of pebbles at the end of this turn will be
\[
(2024+L+2,\quad 2024+L-3k-1,\quad 2024-2L+3k-1)
\]
respectively for Archie, Billie and Charlie, and we see that all the moves are valid. Then Billie have $2024+L-3k-1\geq2025$ pebbles when Archie moves, and Charlie have at least 2025 pebbles when Billie moves. Then, by (i), choosing $b_i=a_i+1$ and $c_i=b_i+1$ will guarantee that Billie and Charlie will not lose the game, hence in this situation Archie loses.

\medskip

oindent If Archie chooses $\dfrac{2024+3k}{2}=L<2023$, then, after he plays we have the case $(0, 2024+L, 4048-L)$. Let Billie choose $L+1$, which he can, then we have the case $(L+1, 2022-L, 4049)$. Here, Charlie can choose 2024 and Archie has no move, so he loses.

\medskip

oindent Hence, Archie does not have a non-losing strategy. $\square$

\medskip

oindent Finally, we conclude that only Charlie has a non-losing strategy.\hfill$\square$",702,6035,Combinatorics,13
14,shl_jbmo_2024_g1,shl_jbmo,2024,g,"Let $\triangle ABC$ be an acute-angled triangle with $AB = BC$. The perpendicular bisector of $AB$ meets the side $BC$ at the point $D$, and the circumcircle $\omega$ of $\triangle ADC$ again at the point $E$. Let $F$ is diametrically opposite point of $E$ in $\omega$. Prove that $BD = DF$.","Let $\angle ABC=\beta$. Since $\triangle ADB$ is isosceles, we have $\angle AEC=\angle ADC=2\beta$. On the other hand, $E$ lies on the perpendicular bisector of $AB$, hence $E$ is the circumcenter of $\triangle ABC$. This implies that $BE$ is the perpendicular bisector of $AC$, hence it passes through the circumcenter of $\triangle ADC$. Therefore, $F$ lies on $BE$. Thus, using $AB=BC$, we have
\[
\angle FBD = \angle EBC = \frac{\beta}{2}.
\]


oindent On the other hand, $\angle EDF=90°$ and $\angle EDB=90°-\beta$ together imply $\angle FDB=180°-\beta$. This gives $\angle BFD=\dfrac{\beta}{2}$, and the result follows.\hfill$\square$

\medskip

oindent\textbf{Alternative Version.} Let $\triangle ABC$ be an acute-angled triangle with $AB=BC$. The perpendicular bisector of $AB$ meets the side $BC$ at the point $D$, and the circumcircle $\omega$ of $\triangle ADC$ again at the point $E$. Let $F$ is diametrically opposite point of $E$ in $\omega$. Prove that the circumcircle of $\triangle BFC$ is the excenter of $\triangle DFC$ opposite to $F$.

\medskip

oindent\textbf{Adapted Solution.} In addition to the solution above, we have $\angle FDC=180°-\angle FDB=\beta$. Let $J$ be the excenter of $\triangle DFC$ opposite to $F$. It is well-known that
\[
\angle FJC = \frac{\angle FDC}{2} = \frac{\beta}{2},
\]
so we have $\angle FJC=\angle FBD=\angle FBC$, which completes the proof.\hfill$\square$",291,1409,Geometry,14
15,shl_jbmo_2024_g2,shl_jbmo,2024,g,"Let $ABC$ be an acute-angled triangle, and $A'$ and $B'$ be the feet of the altitudes from $A$ and $B$, respectively. Let $K$ and $L$ be the reflections of $A'$ with respect to $AB$ and $AC$, respectively. Let $M$ and $N$ be the reflections of $B'$ with respect to $AB$ and $BC$, respectively. Prove that $KL = MN$.","Let $C'$ be the foot of the altitude from $C$ in $\triangle ABC$. Note that quadrilaterals $A'C'AC$ and $B'C'BC$ are cyclic, hence we have $\angle AC'B'=\angle C=\angle BC'A'$. Since $K$ is the reflection of $A'$ with respect to $AB$, we obtain
\[
\angle BC'K = \angle BC'A' = \angle AC'B' = \angle AC'B',
\]
which shows $K$ lies on $B'C'$. Due to the symmetry, we can conclude that $L$ also lies $B'C'$.

\medskip

oindent Hence, $KC'=C'A'$ and $LB'=A'B'$ gives
\[
KL = KC'+C'B'+B'L = A'C'+C'B'+B'A'.
\]


oindent Again, due to the symmetry, we find that $M$ and $N$ lie on $A'C'$. Similarly, $MC'=C'B'$ and $NA'=A'B'$ gives
\[
MN = MC'+C'A'+A'N = B'C'+C'A'+A'B',
\]
which completes the proof.\hfill$\square$",315,709,Geometry,15
16,shl_jbmo_2024_g3,shl_jbmo,2024,g,"Let $ABC$ be a scalene triangle. The excircle opposite to $A$ with center $J$ is tangent to the lines $AB$, $AC$ and $BC$ at points $D$, $E$ and $F$, respectively. Let $P$ be a point on the side $BC$. The circumcircles of the triangles $BDP$ and $CEP$ intersect for the second time at $Q$. Let $R$ be the foot of the perpendicular from $A$ to the line $FJ$. Prove that the points $P$, $Q$ and $R$ are collinear.","Since the quadrilateral $BDQP$ is cyclic, we have
\[
\angle DQP = 180°-\angle DBP = \angle ABC.
\]
Analogously from the cyclic quadrilateral $CEQP$, we obtain $\angle EQP=\angle ACB$. Hence
\[
\angle DQE = \angle DQP+\angle EQP = \angle ABC+\angle ACB = 180°-\angle BAC = 180°-\angle DAE,
\]
so $Q$ lies on the circumcircle of $\triangle ADE$. On the other hand, since
\[
\angle ADJ = \angle AEJ = \angle ARJ = 90°,
\]
we can conclude that $R$ and $J$ also lie on the circumcircle of $\triangle ADE$.

\medskip

oindent Note that the quadrilateral $BDJF$ and $DJQR$ are cyclic, so we have
\[
\angle DQR = \angle DJR = \angle DJF = 180°-\angle DBF = \angle ABC.
\]
Then, using $\angle DQP=\angle ABC$, we obtain $\angle DQR=\angle DQP$, so the result follows.\hfill$\square$

\medskip

oindent\textbf{Alternative Version.} Let $ABC$ be a scalene triangle. The excircle opposite to $A$ is tangent to the lines $AB$ and $AC$ at points $D$ and $E$, respectively. Let $P$ be an arbitrary point on the side $BC$. The circumcircles of the triangles $BDP$ and $CEP$ intersect for the second time at $Q$. Prove that the line $PQ$ passes through a fixed point, independent from the choice of $P$.

\medskip

oindent\textbf{Adapted Solution.} Since the quadrilateral $BDQP$ is cyclic, we have
\[
\angle DQP = 180°-\angle DBP = \angle ABC.
\]
Analogously from the cyclic quadrilateral $CEQP$, we obtain $\angle EQP=\angle ACB$. Hence
\[
\angle DQE = \angle DQP+\angle EQP = \angle ABC+\angle ACB = 180°-\angle BAC = 180°-\angle DAE,
\]
so $Q$ lies on the circumcircle of $\triangle ADE$. Let $R$ be the second intersection point of $PQ$ with the circumcircle of $ADQE$. Then, we obtain
\[
\angle BAR = \angle DAR = \angle DQR = \angle DQP = \angle ABC,
\]
which implies $AR\parallel BC$. This means that $R$ is the second intersection point of the line parallel to $BC$ from $A$ with the circumcircle of $\triangle ADE$, so it is a fixed point.\hfill$\square$",411,1947,Geometry,16
17,shl_jbmo_2024_g4,shl_jbmo,2024,g,"Let $ABCD$ be a circumscribed quadrilateral with circumcircle $\omega$ such that $AE = EC$ where $E$ is the intersection point of the diagonals $AC$ and $BD$. Point $F$ is taken on $\omega$ such that $BF \parallel AC$. If $G$ is the reflection of $F$ with respect to $A$, prove that the circumcircle of $\triangle ADG$ is tangent to the line $AC$.","
oindent\textbf{Solution 1.} Let $R$ be the reflection of $B$ with respect to $E$. Since $BF\parallel AC$ and $AF=AG$, we have $AC$ bisects the segment $BG$, which implies that $GR\parallel AC$ as $BE=ER$. This gives
\[
\angle GAD = 180°-\angle FAD = 180°-\angle FBD = 180°-\angle FBR = \angle GRD,
\]
so the quadrilateral $GARD$ is cyclic.

\medskip

oindent On the other hand, we have $AR\parallel BC$ as $AE=EC$ and $RE=EB$, which implies that
\[
\angle ARD = 180°-\angle ARE = 180°-\angle EBC = 180°-\angle DBC = 180°-\angle DAC.
\]
As a result, we obtain $\angle AGD = 180°-\angle ARD = \angle DAC$, so the result follows.\hfill$\square$

\medskip

oindent\textbf{Solution 2.} Let $R$ be the point on $BD$ such that $FR\perp AC$. Note that we have $BE=EF$ as $ABCP$ is an isosceles trapezoid and $E$ is the midpoint of $AC$. Since $\angle BFR=90°$, we obtain $ER=BE=EF$, which implies that $AC$ is the perpendicular bisector of $FR$. Then, we have $AC\parallel GR$ using $GA=AF$, which means $\angle FRG=90°$ and so $AR=AG=AF$. Since $GAR$ is an isosceles triangle with $GA=AR$ and $AC\parallel GR$, we obtain the circumcircle of $\triangle GAR$ is tangent to the line $AC$. On the other hand, we have
\[
\angle GAD = 180°-\angle FAD = 180°-\angle FBD = 180°-\angle FBR = \angle GRD,
\]
where the last equality holds as $GR\parallel BF$. This implies that $D$ lies on the circumcircle of $\triangle GAR$, the result follows.\hfill$\square$

\medskip

oindent\textbf{Alternative Version.} Let $ABC$ be an acute-angled triangle with circumcircle $\omega$, and $D$ be a point on $BC$ such that $AD\perp BC$. Let $X$ be an arbitrary point on the minor arc $BC$ of $\omega$. Point $Y$ is taken on $\omega$ such that $XY\perp BC$. If $Q$ is the reflection of $Y$ with respect to $A$, and $XD$ meets $\omega$ again at $P$, prove that the circumcircle of $\triangle APQ$ passes through a fixed point, independent from the choice of $X$.

\medskip

oindent\textbf{Adapted Solution.} Let $R$ be the reflection of $X$ with respect to $D$. Note that we have $QR\parallel AD$ since $QA=AY$, $RD=DX$, and $AD\parallel XY$. Let $H$ be the orthocenter of $\triangle ABC$, and $S$ be the second intersection point of $AH$ with $\omega$. We will prove that $H$ lies on the circumcircle of $\triangle APQ$.

\medskip

oindent It is well-known that $HD=DS$, which implies that $R$ lies on the circumcircle of $\triangle APQ$. It is well-known that $RHXS$ is a parallelogram using $XD=DR$, so we obtain $RH=SX$. Moreover, $AYXS$ is a cyclic and $AS\parallel XY$, so it is an isosceles trapezoid, which gives $AY=SX$. As a result, we have
\[
QA=AY=SX=RH.
\]


oindent Then, using $QR\parallel AD$, we find that $QAHR$ is an isosceles trapezoid, so $QAHR$ is cyclic. On the other hand, since $APXY$ is cyclic and $QR\parallel XY$, we have
\[
\angle RQA = 180°-\angle XYA = \angle APX,
\]
which implies $RQAP$ is cyclic. As a result, $Q$, $A$, $H$, $R$, $P$ lie on the same circle, the result follows.\hfill$\square$",347,2997,Geometry,17
18,shl_jbmo_2024_g5,shl_jbmo,2024,g,"Let $ABCD$ be a rectangle, and $H$ be the midpoint of the side $AB$. Point $K$ is taken on $DH$ such that $\angle BKD = 90°$. Let $F$ be a point on the diagonal $AC$. The perpendicular line to $AB$ through $F$ meets $AB$ at $G$, and the parallel line to $AB$ through $F$ meets $DH$ at $L$. If $M$ is the midpoint of $GB$, prove that the angles $\angle AKF$ and $\angle LKM$ are equal.","
oindent\textbf{Solution 1.} Note that $K$ lies on the circumcircle of the rectangle $ABCD$ as $\angle BKD=90°$. Then, using $LF\parallel CD$, we have
\[
\angle KAF = \angle KAC = \angle KDC = \angle KLF,
\]
which implies that the quadrilateral $KALF$ is cyclic.

\medskip

oindent On the other hand, let $T$ and $S$ be the intersection points of $LF$ with $CH$ and $CB$, respectively. Since $AH=HB$ and $FS\parallel AB$, we have $FT=TS$. Moreover, $FGBS$ is a rectangle, then $FMBT$ is a parallelogram and so $FM=BT$. Note that $\triangle AHL$ and $\triangle BHT$ are congruent due to the symmetry. Hence, we obtain $FM=AL$ and so $AMFL$ is an isosceles trapezoid. Therefore, $ALFM$ is cyclic and $AF=LM$. Thus, $A$, $K$, $M$, $F$, $L$ lie on the same circle and we get $\angle AKF=\angle LKM$.\hfill$\square$

\medskip

oindent\textbf{Solution 2.} Let $P$ be the point on $AB$ such that $LP\perp AB$. It is clear that $\triangle ABC\sim\triangle AGF$ and $\triangle LPH\sim\triangle ADH$.

\medskip

oindent Then, using $GF=PL$ and $AD=BC$, we have
\[
\frac{AG}{AB} = \frac{GF}{BC} = \frac{PL}{AD} = \frac{LP}{AD} = \frac{PH}{AH},
\]
which implies that $AG=2\cdot PH$ as $AB=2\cdot AH$. Hence, using $AB=2\cdot BH$ and $BG=2\cdot BM$, we obtain
\[
MH = BH-BM = \frac{AB-BG}{2} = \frac{AG}{2} = PH.
\]
This implies that
\[
AP = AH-PH = BH-MH = BM = MG,
\]
so we find that $ALFM$ is an isosceles trapezoid. Hence, $ALFM$ is cyclic, and then
\[
\angle LMA = \angle FAM = \angle CAB = \angle DBA,
\]
which implies $LM\parallel BD$.

\medskip

oindent On the other hand, $K$ lies on the circumcircle of the rectangle $ABCD$ since $\angle BKD=90°$. Therefore, using $LM\parallel BD$, we have
\[
\angle KLM = \angle KDB = \angle KAB = \angle KAM,
\]
which gives $K$ also lies on the circumcircle of the quadrilateral $ALFM$. Thus, the result follows as $ALFM$ is an isosceles trapezoid and so $AF=LM$.\hfill$\square$",384,1911,Geometry,18
19,shl_jbmo_2024_g6,shl_jbmo,2024,g,Let $ABCD$ be a trapezoid with $AB \parallel CD$. Let $E$ and $F$ be the points on $CD$ such that $AE \perp CD$ and $AF \perp AD$. Let $G$ be the point on $AE$ such that $BG \parallel AD$. Prove that the perpendicular line from $A$ to $BD$ bisects the segment $FG$.,"Let $H$ be the reflection of $G$ with respect to $A$. Let $K$ be the intersection of the perpendicular line to $BD$ through $A$. Since $GA=AH$, it suffices to prove that $KA\parallel HF$.

\medskip

oindent Let $\angle DAE=\theta$. Since $AD\parallel BG$, we have
\[
\angle AGB = \angle DAG = \angle DAE = \theta.
\]


oindent Also, using $AE\perp CD$ and $AF\perp AD$, we obtain $\angle AFD=\theta$. Moreover, $\angle BAG=90°$ as $AB\parallel CD$. Hence, we see that $\triangle BAG\sim\triangle DAF$. Therefore, using $AH=AG$, we have
\begin{equation}
\frac{AB}{AH} = \frac{AB}{AG} = \frac{AD}{AF}. \tag{*}
\end{equation}


oindent On the other hand, $AB\parallel CF$ implies $\angle BAF=\angle AFD=\theta$, which gives $\angle HAF=90°+\theta$. Then, we have
\begin{equation}
\angle BAD = \angle BAG+\angle DAE = 90°+\theta = \angle HAF. \tag{**}
\end{equation}


oindent Hence, by $(*)$ and $(**)$, we find that $\triangle HAF\sim\triangle BAD$. It follows that $\angle FHA=\angle DBA$. Since $AK\perp BD$ and $\angle BAG=90°$, we obtain
\[
\angle KAG = \angle KBA = \angle DBA = \angle FHA.
\]
As a result, we have $\angle KAG=\angle FHA$, which implies $KA\parallel HF$.\hfill$\square$",265,1189,Geometry,19
20,shl_jbmo_2024_g7,shl_jbmo,2024,g,"Let $ABC$ be an acute-angled and scalene triangle, and $D$ be a point on the side $BC$. Points $E$ and $F$ are taken on $AD$ such that $EB \perp AB$ and $FC \perp AC$. Points $S$ and $T$ are taken on $BC$ such that $SE \parallel AC$ and $TF \parallel AB$. The circumcircle of $\triangle BSE$ intersects $AB$ for the second time at $M$, and the circumcircle of $\triangle CTF$ intersects $AC$ for the second time at $N$. Prove that the lines $MS$, $NT$, and $AD$ are concurrent.","First note that $\angle MSE=\angle MBE=90°$, so we have $MS\perp AC$. Analogously, we can see that $\angle NTF=\angle NCF=90°$ and so $NT\perp AB$. Then, $MS$ and $NT$ intersect at the orthocenter of $\triangle AMN$. Therefore, it suffices to show that $AD\perp MN$.

\medskip

oindent Let $P$ be the second intersection point of $AD$ with the circumcircle of $\triangle ABC$. We will prove that $AP\perp MN$. Using $SE\parallel AC$, we have
\[
180°-\angle BSE = \angle DSE = \angle DCA = \angle BCA = \angle BPA = \angle BPE,
\]
which implies that $P$ lies on the circumcircle of $\triangle BSE$. This gives
\[
\angle APM = \angle EPM = 180°-\angle EBM = 90°.
\]


oindent Analogously, using $TF\parallel AB$, we have
\[
180°-\angle CTF = \angle DTF = \angle DBA = \angle CBA = \angle CPA = \angle BPE = 180°-\angle CPF,
\]
which implies that $P$ lies on the circumcircle of $\triangle CTF$. This gives
\[
\angle APN = \angle180°-\angle FPN = 180°-\angle FCN = 90°.
\]


oindent As a result, we obtain $\angle APM=90°=\angle APN$, so $P$ lies on $MN$ and we get $AP\perp MN$.\hfill$\square$

\medskip

oindent\textbf{Alternative Version.} Let $ABC$ be an acute-angled and scalene triangle, and $D$ be a point on the side $BC$. Points $E$ and $F$ are taken on $AD$ such that $EB\perp AB$ and $FC\perp AC$. Points $S$ and $T$ are taken on $BC$ such that $SE\parallel AC$ and $TF\parallel AB$. The circumcircle of $\triangle BSE$ intersects $AB$ for the second time at $M$, and the circumcircle of $\triangle CTF$ intersects $AC$ for the second time at $N$. Prove that the lines $SE$, $TF$, and $MN$ are concurrent.

\medskip

oindent\textbf{Adapted Solution.} In addition to the solution above, it suffices to show that $\angle HPA=90°$ where $H$ is the intersection point of $SE$ and $TF$. Let $A'$ be the intersection point of $BE$ and $CF$. Since $SH\perp A'F$ and $TH\perp A'E$, we see that $H$ is the orthocenter of the triangle $A'EF$. On the other hand, $AA'$ is a diameter of the circumcircle of $\triangle ABC$, so we have $\angle A'PA=90°$. This implies $\angle HPA=90°$, and the result follows.\hfill$\square$",477,2119,Geometry,20
21,shl_jbmo_2024_g8,shl_jbmo,2024,g,"Let $ABC$ be a scalene triangle with smallest side $BC$, and $D$ be a point on the side $BC$ such that $\angle BAD = \angle CAD$. Let $\omega_1$ and $\omega_2$ be the circumcircles of $\triangle ABD$ and $\triangle ACD$, respectively. The line $AC$ meets $\omega_1$ again at $F$, and the line $AB$ meets $\omega_2$ again at $E$. The line $DE$ meets $\omega_1$ again at $G$, and the line $DF$ meets $\omega_1$ again at $H$. Prove that circumcircles of $\triangle ABC$, $\triangle AEF$, and $\triangle AGH$ have a common point other than $A$.","Let $\angle BAD=\angle CAD=\alpha$. Since $G$, $A$, $F$, $D$ lie on $\omega_1$, we have
\[
\angle EGF = \angle DGF = \angle DAF = \angle DAC = \alpha.
\]
Similarly, since $H$, $A$, $E$, $D$ lie on $\omega_2$, we find $\angle EHF=\alpha$. Then, the quadrilateral $GEFH$ is cyclic. Let $M$ be the center of this cycle. Since
\[
\angle EMF = 2\cdot\angle EGH = 2\alpha = \angle EAF,
\]
it follows that $M$ lies on the circumcircle of $\triangle AEF$. Also, since $G$, $A$, $D$, $B$ lie on $\omega_1$ and $A$, $E$, $D$, $C$ lie on $\omega_2$, we have $\angle GAB=\angle GDB=180°-\angle EDC=\angle EAC=2\alpha$. Similarly, we can see that $\angle HAC=\angle HDC=2\alpha$, so we obtain $\angle GAH=\angle GAB+\angle BAC+\angle HAC=6\alpha$. Moreover, we find
\[
\angle GMH = 360°-2\cdot\angle GFH = 2\cdot\angle GFD = 2\cdot\angle GAD = 2\cdot(\angle GAB+\angle BAD) = 6\alpha,
\]
which implies $\angle GMH=\angle GAH$ and so $M$ lies on the circumcircle of $\triangle AGH$. Therefore, it suffices to show that $M$ lies on the circumcircle of $\triangle ABC$.

\medskip

oindent Since $M$ lies on the circumcircle of $\triangle AEF$, we have $\angle AEM=\angle AFM$, which implies $\angle MEB=\angle MFC$. Recall that we already found that $\angle GDB=2\alpha=\angle HDC$. On the other hand, since $E$, $A$, $D$, $C$ lie on $\omega_2$ and $\angle EAD=\alpha=\angle CAD$, we have $ED=DC$. Similarly, $FD=DB$. Also, we know that $ME=MF$ as $M$ is the center of the cyclic quadrilateral $GEFH$. Thus, $\triangle MEB$ and $\triangle MFC$ are congruent, which implies that
\[
\angle ABM = \angle EBM = \angle FCM = \angle ACM.
\]
Hence, $M$ lies on the circumcircle of $\triangle ABC$, the result follows.\hfill$\square$

\medskip

oindent\textbf{Remark.} It can be seen that the common point $M$ is the midpoint of the arc $BC$ in the circumcircle of $\triangle ABC$ that contains $A$.",540,1875,Geometry,21
22,shl_jbmo_2024_n1,shl_jbmo,2024,n,"Find all pairs of positive integers $(m,n)$ such that $|4^m - 7^n|$ is a prime number.","Considering modulo 3, we have $|4^m-7^n|\equiv 0\pmod{3}$ and since it is a prime number, it means that is equal to 3. So, either $4^m-7^n=3$ or $4^m-7^n=-3$.

\medskip

oindent\textit{Case 1:} $4^m-7^n=3$.

\medskip

oindent Considering modulo 4 we have $3\equiv 4^m-7^n\equiv -7^n\pmod{4}$, hence $n$ must be even. Let $n=2k$ for some positive integer $k$. So,
\[
4^m-7^n=3 \Rightarrow (2^m)^2-(7^k)^2=3 \Rightarrow (2^m-7^k)(2^m+7^k)=3.
\]
Hence we must have $2^m-7^k=1$ and $2^m+7^k=3$. Which clearly isn't possible because $3=2^m+7^k\geq 9$.

\medskip

oindent\textit{Case 2:} $4^m-7^n=-3$.

\medskip

oindent If $m=1$ then clearly $n=1$. If $m\geq2$ then considering modulo 8 we have
\[
-3\equiv 4^m-7^n\equiv -(-1)^n\equiv -1,1\pmod{8},
\]
which is a contradiction.

\medskip

oindent In conclusion the only pair that satisfies the condition is $(m,n)=(1,1)$.\hfill$\square$",86,881,Number Theory,22
23,shl_jbmo_2024_n2,shl_jbmo,2024,n,"Find all the pairs $(p,q)$ of distinct prime numbers such that
\[
q^p \mid p + p^q + p^{q^p}.
\]","Using the Fermat's Little Theorem, we obtain
\[
p^q \equiv p^{q^p} \equiv p \pmod{q},
\]
so
\[
p+p^q+p^{q^p} \equiv 3p \pmod{q}.
\]
Since the condition of the problem $q^p\mid p+p^q+p^{q^p}$ implies that also $q\mid p+p^q+p^{q^p}$, we conclude $q\mid 3p$. Since $p$ and $q$ are distinct, we have $q\mid 3$, so $q=3$. The problem now reduces to determining all primes $p
eq3$ such that
\[
3^p\mid p+p^3+p^{3^p}.
\]


oindent Since we have $\varphi(3^p)=2\cdot3^{p-1}$, the Euler Theorem gives us $p^{2\cdot3^{p-1}}\equiv1\pmod{3^p}$. Let $a=p^{3^{p-1}}$, then $a^2\equiv1\pmod{3^p}$, so $3^p\mid a^2-1=(a-1)(a+1)$. Since $\gcd(a-1,a+1)=1,2$ we have either $3^p\mid a-1$ or $3^p\mid a+1$, i.e.\ $p^{3^{p-1}}\equiv\pm1\pmod{3^p}$.

\medskip

oindent Therefore,
\[
p+p^3+p^{3^p} = p+p^3+p^{2\cdot3^{p-1}}\cdot p^{3^{p-1}} \equiv p+p^3+(\pm1)\cdot1 = p^3+p\pm1\pmod{3^p}.
\]


oindent But, it is straightforward to prove (by induction) that for $p\geq5$ it holds
\[
p^3+p-1 < p^3+p+1 < 3^p.
\]
The only case left is $p=2$ and a simple check shows that it is a solution since
\[
3^2=9\mid 522=2+2^3+2^{3^2}.
\]


oindent The only solution of the problem is the ordered pair
\[
(p,q)=(2,3). \qquad\square
\]",96,1200,Number Theory,23
24,shl_jbmo_2024_n3,shl_jbmo,2024,n,Let $c \in \mathbb{Z}^+$ be a Turkish number if there is a positive integer $m$ such that $m^3 - m = c!$ and $m^2-1$ has less than 12 positive divisors. Determine all Turkish numbers.,"We will divide the solution into three cases. Let $\tau(n)$ denote the number of positive divisors of $n$.

\medskip

oindent\textit{Case 1:} One of $m-1$ or $m+1$ is a prime.

\medskip

oindent Call this prime $p$. Then since $p\mid c!$ we must have $c\geq p$. On the other hand, we have $m^3-m=(m-1)m(m+1)\leq p(p+1)(p+2)$ so we have $(p+2)(p+1)p\geq p!$. It is easy to show by induction that this inequality does not hold for any prime (or any integer) $p\geq7$, hence we must have $p\in\{2,3,5\}$. This means that $m\in\{1,2,3,4,6\}$ and by checking these values, we see that only $m=2,3$ gives $c=3,4$ respectively as Turkish numbers.

\medskip

oindent\textit{Case 2:} $m-1$ and $m+1$ are both composite and $m$ is even.

\medskip

oindent In this case we have $\gcd(m-1,m+1)=1$. Since both numbers are composite, each number must have at least 3 distinct positive divisors. Moreover, since they are relatively prime we have $\tau(m^2-1)=\tau(m-1)\tau(m+1)<12$ so none of the numbers $m-1$, $m+1$ can have 4 positive divisors. Hence $m-1=p^2$, $m+1=q^2$ for some prime numbers $p$, $q$; which is clearly impossible since $q^2-p^2=2$ has no solutions in positive integers.

\medskip

oindent\textit{Case 3:} $m-1$ and $m+1$ are both composite and $m$ is odd.

\medskip

oindent In this case we have $\gcd(m-1,m+1)=2$ and one of these integers must be divisible by 4. Let $m^2-1=2^s t$ where $t$ is odd. Then, $s\geq3$ and $\tau(m^2-1)=(s+1)\tau(t)<12$. Since $s+1\geq4$, we must have $\tau(t)<3$ hence $t$ must be an odd prime. This also implies $s+1<6$ so the only cases are $s=3$, $s=4$ which means $(m-1)(m+1)=8t$ or $(m-1)(m+1)=16t$ where $t$ is an odd prime. Then we can easily find that $m-1=4$, $m+1=2t=6$ is the only solution for the first case. Similarly, $m-1=8$, $m+1=2t=10$ and $m+1=8$, $m-1=2t=6$ are the only solutions for the second case. Hence the only possibilities are $m=5,7,9$ and here only $m=5,9$ yields the solutions $c=5,6$ as Turkish numbers.

\medskip

oindent So, all the Turkish numbers are $c=3,4,5,6$.\hfill$\square$",183,2051,Number Theory,24
25,shl_jbmo_2024_n4,shl_jbmo,2024,n,"For any positive integer $n$, let $s(n) = 1 + 2 + \cdots + n$. Define a strictly increasing sequence of positive integers $\{a_n\}_{n=1}^{\infty}$ such that $a_1 = 1$ and
\[
a_{n+1} = \min\{m \mid s(m) - s(a_n) \text{ is a perfect square}\}
\]
for all positive integers $n$. Find the value of $a_{2024}$.","We will prove by induction that $a_n=\dfrac{3^n-1}{2}$ for all positive integers $n$. We can verify that it is true for $n=1,2$. Assume that it is true for $n=k$. Define $b_n=2a_n+1$ for all $n$. Then, $b_k=3^k$ and we have
\[
s(a_{k+1})-s(a_k)=x^2 \Rightarrow b_{k+1}^2-3^{2k}=8x^2
\]
by the induction hypothesis. Let $b_{k+1}=3^\alpha\beta$ for some non-negative integer $\alpha$ and $(\beta,3)=1$.

\medskip

oindent Assume that $\alpha<k$, then we have $\beta^2=3^{2k-2\alpha}+8x^2$. Since $3
mid\beta$, we must have $3
mid x$ hence $\beta^2\equiv8x^2\equiv2\pmod{3}$, which is a contradiction.

\medskip

oindent Thus, we must have $\alpha\geq k$, say $b_{k+1}=3^k\ell$ for some positive integer $\ell$. So, by the minimality condition, $\ell$ is the minimum positive integer greater than 1 such that $3^{2k}(\ell^2-1)=8x^2$. Since $\ell=3$ is the smallest number satisfying this condition, we must have $b_{k+1}=3^{k+1}$ and $a_{k+1}=\dfrac{3^{k+1}-1}{2}$ as desired.

\medskip

oindent Hence, the claim is proven by induction and $a_{2024}=\dfrac{3^{2024}-1}{2}$.\hfill$\square$",304,1085,Number Theory,25
26,shl_jbmo_2024_n5,shl_jbmo,2024,n,"Does there exist a positive integer $k$, such that the number of quadruples $(a,b,c,n)$ of positive integers satisfying
\[
a^5 + b^6 + c^{15} - n! = k
\]
is non-zero and finite?","Consider the equation mod 31. We have $a^5\equiv0,1,5,6,25,26,30\pmod{31}$ and $b^6\equiv0,1,2,4,8,16\pmod{31}$ by computing all possible values. Also, since $c^{30}=(c^{15}-1)(c^{15}+1)\equiv0,1\pmod{31}$ by Fermat's Little Theorem and 31 is a prime we have $c^{15}\equiv0,1,30\pmod{31}$.

\medskip

oindent Hence
\begin{align*}
a^5+b^6 &\equiv 0,1,5,6,30,26,25\\
&\quad 1,2,6,7,0,27,26\\
&\quad 2,3,7,8,1,28,27\\
&\quad 4,5,9,10,3,30,29\\
&\quad 8,9,13,14,7,3,2\\
&\quad 16,17,22,15,11,10 \pmod{31}
\end{align*}
and in particular, none of the values 18, 19, 20 appears. Having in mind the values for $c^{15}$, we deduce that the congruence $a^5+b^6+c^{15}\equiv19\pmod{31}$ has no solutions.

\medskip

oindent Therefore, it makes sense to restrict to $k\equiv19\pmod{31}$. For $n\geq31$ we have no solutions. For each $n\leq30$ and fixed $k$ we have $a^5,b^6,c^{15}<n!+k$, so there are finitely many possibilities for each of $a$, $b$, $c$ and hence finitely many possible $(a,b,c,n)$.

\medskip

oindent It remains to find a $k$ with $k\equiv19\pmod{31}$ for which the equation has at least one solution. For this we could aim for $b=c=1$ and $a^5-n!\equiv17\pmod{31}$. This is satisfied e.g.\ by $a=11$ and $n=8$, corresponding to $k=11^5-8!+2>0$.\hfill$\square$",177,1267,Number Theory,26
27,shl_jbmo_2024_n6,shl_jbmo,2024,n,"Find all positive integers $x$, $y$, $z$ which satisfy the equation
\[
2020^x + 2^y = 2024^z.
\]","Regarding the equation modulo 3, we obtain
\[
1+(-1)^y \equiv (-1)^z \pmod{3},
\]
so we can conclude that $y$ is even and $z$ is odd. Let $y=2y_1$, where $y_1$ is a positive integer. Since the largest powers of 2 in the factorisation of all three terms in the equation are respectively $2^{2x}$, $2^{2y_1}$, $2^{3z}$, hence we rewrite the equation as
\begin{equation}
2^{2x}\cdot505^x + 2^{2y_1} = 2^{3z}\cdot253^z. \tag{1}
\end{equation}


oindent We will now determine which one of the integers $2x$, $2y_1$ is larger, in order to obtain the largest power of 2 dividing LHS.

\medskip

oindent\textit{Case 1:} $2x>2y_1$, i.e.\ $x>y_1$. Then, the largest power of 2 dividing LHS is $2^{2y_1}$, while the largest power of 2 dividing RHS is $2^{3z}$. Therefore, we derive $2y_1=3z$. Since $z$ is odd, this is a contradiction.

\medskip

oindent\textit{Case 2:} $2x<2y_1$, i.e.\ $x<y_1$. Then, the largest power of 2 dividing LHS is $2^{2x}$, while the largest power of 2 dividing RHS is $2^{3z}$. Therefore, we derive $2x=3z$. Since $z$ is odd, this is a contradiction.

\medskip

oindent Thus, we must have $2x=2y_1$, which implies $x=y_1$. The equation (1) then becomes
\begin{equation}
4^x\cdot(505^x+1) = 2020^x+4^x = 2024^x = 2^{3z}\cdot253^z. \tag{2}
\end{equation}


oindent Since $2\mid505^x+1$, but $4
mid505^x+1$, the largest exponent of 2 dividing $4^x(505^x+1)$ is $2x+1$, so we derive
\begin{equation}
2x+1=3z. \tag{3}
\end{equation}


oindent Plugging it back to (2), it reduces to
\begin{equation}
505^x+1=2\cdot253^z. \tag{4}
\end{equation}


oindent Observe that $x=1$, $z=1$ is a solution of the linear Diophantine equation (3), so all the solutions in positive integers are given by
\[
x=3k+1, \quad z=2k+1, \quad k\in\mathbb{N}_0.
\]


oindent Now (4) becomes
\begin{equation}
505^{3k+1}+1=2\cdot253^{2k+1}. \tag{5}
\end{equation}


oindent Obviously $k=0$ is a solution of (5). If $k$ is positive integer, then from the sequence of inequalities
\[
505^{3k+1}+1 > 505^{3k+1} = 505^{k+1}\cdot505^{2k} > 506\cdot505^{2k} > 506\cdot253^{2k} = 2\cdot253^{2k+1}
\]
which implies that there are no more solutions.

\medskip

oindent Hence, the only solution of (5) is $k=0$, so $x=y_1=1$ and $z=1$ and the only solution of the given equation is
\[
(x,y,z)=(1,2,1). \qquad\square
\]",96,2294,Number Theory,27
360,tst_jbmo_ro_2024_1_p1,tst_jbmo,2024,n,Find all nonzero natural numbers $a$ and $b$ for which $\dfrac{7^a - 5^b}{8}$ is a prime natural number.,"\begin{solution}
Since $5^{2k} \equiv 1 \pmod{8}$, $5^{2k+1} \equiv 5 \pmod{8}$, $7^{2k} \equiv 1 \pmod{8}$, and $7^{2k+1} \equiv 7 \pmod{8}$ for all $k \in \mathbb{N}$, the condition $8 \mid 7^a - 5^b$ implies $a$ and $b$ are both even. \hfill \textbf{[2p]}

Write $a = 2m$, $b = 2n$ with $m, n \in \mathbb{N}^*$. Then $7^{2m} - 5^{2n} = 8p$ for some prime $p$, so $(7^m - 5^n)(7^m + 5^n) = 8p$.

If $p = 2$: the equation $(7^m - 5^n)(7^m + 5^n) = 16$ has no solutions (by direct case analysis). \hfill \textbf{[1p]}

If $p \geq 3$: since $7^m - 5^n < 7^m + 5^n$ and both are even, the cases are:
\[
(1^\circ)\; \begin{cases} 7^m - 5^n = 4 \\ 7^m + 5^n = 2p \end{cases}
\qquad
(2^\circ)\; \begin{cases} 7^m - 5^n = 2 \\ 7^m + 5^n = 4p \end{cases}
\]

In case $(1^\circ)$: since $7^m \equiv 1 \pmod{3}$ and $5^n \equiv \pm 1 \pmod{3}$, we get $7^m - 5^n \equiv 0$ or $2 \pmod{3}$. But $4 \equiv 1 \pmod{3}$, so no solutions arise. \hfill \textbf{[2p]}

In case $(2^\circ)$: we get $2p = 7^m + 5^n$ and subtracting, $2 \cdot 5^n = 4p - 2$, leading to $2p = 7^m - 1$. Since $3 \mid 7^m - 1$, we need $3 \mid 2p$, so $p = 3$. Then $m = n = 1$, giving $(a, b) = (2, 2)$. \hfill \textbf{[2p]}

\medskip

oindent\textbf{Answer:} $(a, b) = (2, 2)$.
\end{solution}",104,1256,Number Theory,1
361,tst_jbmo_ro_2024_1_p2,tst_jbmo,2024,g,"Consider a square $ABCD$ and the midpoint $M$ of side $AD$. Let $DFM$ and $BFE$ be equilateral triangles, with $F$ in the interior of square $ABCD$ and with $E$ and $F$ on opposite sides of line $BC$. Let $P$ be the midpoint of segment $ME$. Prove that:
\begin{enumerate}[label=\alph*)]
    \item point $P$ lies on line $AC$;
    \item ray $PM$ is the angle bisector of $\angle APF$.
\end{enumerate}","\begin{solution}
\textbf{a)} Construct equilateral triangle $BDQ$ so that $C$ lies in its interior. \hfill \textbf{[1p]}

Point $Q$ lies on the perpendicular bisector of diagonal $BD$, so $Q, C, A$ are collinear. \hfill \textbf{[1p]}

Since $\angle EBQ = \angle FBD = 60^\circ - \angle FBQ$, $BQ = BD$, and $BE = BF$, we get $\triangle BEQ \cong \triangle BFD$ (SAS), so $QE = DF = DM = AM$ and $\angle BQE = \angle BDF = \angle ADF - \angle ADB = 15^\circ$. \hfill \textbf{[1p]}

Since $\angle QBC = \angle QBD - \angle CBD = 15^\circ = \angle BQE$, it follows that $QE \parallel CB \parallel AD$ and $QE = AM$. Therefore $AMQE$ is a parallelogram, so the midpoint $P$ of $ME$ also lies on $AQ$, i.e.\ $P$ lies on $AC$. \hfill \textbf{[1p]}

\textbf{b)} From $QE = DM$ and $QE \parallel DM$, quadrilateral $DMEQ$ is a parallelogram, so $\angle DME = \angle DQE = 75^\circ$. Thus $\angle FMP = \angle DMP - \angle DMF = 15^\circ$. \hfill \textbf{[1p]}

Triangle $FAD$ is right-angled with $\angle FAD = 30^\circ$, so $\angle FAP = \angle DAP - \angle DAF = 15^\circ$, and $\angle MFA = \angle DFA - \angle DFM = 30^\circ$.

Since $\angle FMP = \angle FAP$, quadrilateral $AMFP$ is cyclic, with $\angle MPA = \angle MFA = 30^\circ$. \hfill \textbf{[1p]}

Since $\angle FPM = \angle FAM = 30^\circ$, we have $\angle FPM = \angle APM$, so $PM$ is the angle bisector of $\angle FPA$. \hfill \textbf{[1p]}
\end{solution}",399,1415,Geometry,2
362,tst_jbmo_ro_2024_1_p3,tst_jbmo,2024,a,"Find all nonzero natural numbers $a, b, c, d, e, f$ with the property that for any two chosen from among them, $x$ and $y$, among the remaining four numbers there exist two, $z$ and $t$, such that $\dfrac{x}{y} = \dfrac{z}{t}$.","\begin{solution}
If all six numbers are equal, the condition is trivially satisfied. \hfill \textbf{[0p]}

We seek solutions where the numbers are not all equal. Suppose $a \leq b \leq c \leq d \leq e \leq f$, so $a < f$.

Taking $x = a$ and $y = f$: for any $z, t \in \{b, c, d, e\}$ we have $\dfrac{x}{y} = \dfrac{a}{f} \leq \dfrac{z}{t}$, with equality only if $z = a$ and $t = f$. Since $z \geq b \geq a$, we need $a = b$. By symmetry, $e = f$. \hfill \textbf{[2p]}

Taking $x = c$, $y = d$: then $\dfrac{c}{d} = \dfrac{z}{t}$ with $z, t \in \{a, f\}$. Since $c \leq d$, either $c = d$ or $\dfrac{c}{d} = \dfrac{a}{f}$. \hfill \textbf{[1p]}

If $c = d$: the condition holds because each value appears twice. For $x 
eq y$, choose $z$ and $t$ as their equal counterparts; for $x = y$, choose any other equal pair. Thus all $(a, a, c, c, f, f)$ with $a \leq c \leq f$ are solutions. \hfill \textbf{[2p]}

If $c 
eq d$: then $\dfrac{c}{d} = \dfrac{a}{f}$, giving $d = \dfrac{cf}{a} \geq f$, so $a = c$ and $d = f$, and $(a, a, a, f, f, f)$ is also a solution (a special case of the above with $c = a$).

\medskip

oindent\textbf{Answer:} The solutions are all multisets of the form $(a, a, b, b, c, c)$ with $a \leq b \leq c$ and $(a, a, a, b, b, b)$ with $a \leq b$, and all their permutations. \hfill \textbf{[2p]}
\end{solution}",227,1332,Algebra,3
363,tst_jbmo_ro_2024_1_p4,tst_jbmo,2024,g,"Let $ABC$ be a triangle. An arbitrary circle through vertices $B$ and $C$ meets sides $AC$ and $AB$ again at points $D$ and $E$, respectively. The circumcircle of triangle $AEC$ meets line $BD$ at points $P$ and $Q$, and the circumcircle of triangle $ABD$ meets line $CE$ at points $R$ and $S$, such that $P$ lies between $B$ and $D$, and $R$ lies between $C$ and $E$. Prove that:
\begin{enumerate}[label=\alph*)]
    \item points $P, Q, R, S$ are concyclic;
    \item triangle $APQ$ is isosceles.
\end{enumerate}","\begin{solution}
\textbf{a)} Let $X$ be the intersection of lines $CE$ and $BD$. Using the power of point $X$ with respect to each of the three circles:
\[
XR \cdot XS = XB \cdot XD = XE \cdot XC = XP \cdot XQ.
\]
\hfill \textbf{[2p]}

Since $\{X\} = PQ \cap RS$, by the converse of the power of a point, the points $P, Q, R, S$ are concyclic. \hfill \textbf{[1p]}

\textbf{b)} We show the centre of the circle through $P, Q, R, S$ is the vertex $A$.

We have $\triangle ADP \sim \triangle APC$ (AA), so $AP^2 = AD \cdot AC$. Similarly $AQ^2 = AE \cdot AB$. \hfill \textbf{[2p]}

By the power of point $A$ with respect to the circumcircle of $BCDE$:
\[
AD \cdot AC = AE \cdot AB = 
ho(A),
\]
so $AP = AQ = \sqrt{
ho(A)}$. Analogously $AR = AS = \sqrt{
ho(A)}$, hence triangle $APQ$ is isosceles. \hfill \textbf{[2p]}
\end{solution}",513,831,Geometry,4
364,tst_jbmo_ro_2024_1_p5,tst_jbmo,2024,c,"A \textit{tile triangle of type} $n$ (where $n \geq 2$) is the configuration of cells of a $(2n+1) \times (2n+1)$ grid lying below both diagonals of the grid. (For example, a tile triangle of type 3 has rows of widths $1, 2, 3$ from top to bottom, with a total of $1 + 2 + 3 = 6$ cells.)

Find the maximum length of a sequence of pairwise distinct cells in a tile triangle of type $n$, such that starting from the second cell, each cell in the sequence shares a side with the preceding cell.","\begin{solution}
Colour the cells alternately (like a chessboard). The tile triangle of type $n$ has:
\[
b_n = 1 + 2 + \cdots + n = \frac{n(n+1)}{2} \text{ black cells}, \quad a_n = 1 + 2 + \cdots + (n-1) = \frac{n(n-1)}{2} \text{ white cells}.
\]

In any valid sequence of cells, colours alternate, so the length cannot reach $2(a_n + 1)$. Therefore the maximum length satisfies:
\[
L \leq 2a_n + 1 = n^2 - n + 1.
\]
\hfill \textbf{[4p]}

To show $L = n^2 - n + 1$ is achievable, consider the sequence starting at the top cell, then at each step descending to the next row and traversing to one end:
\[
1, \underbrace{2, 3}_{\text{row 2}}, \underbrace{7, 6, 5, 4}_{\text{row 3}}, \ldots
\]
This sequence has length $1 + 2 + 4 + \cdots + 2(n-1) = 1 + n(n-1) = n^2 - n + 1$. \hfill \textbf{[3p]}

\medskip

oindent\textbf{Answer:} The maximum length is $L = n^2 - n + 1$.
\end{solution}",491,885,Combinatorics,5
365,tst_jbmo_ro_2024_2_p1,tst_jbmo,2024,n,"In a $7 \times 7$ grid, the natural numbers from $1$ to $49$ are written such that for every $k \in \{1, 2, \ldots, 7\}$, the product of the numbers on row $k$ equals the product of the numbers on row $8 - k$.
\begin{enumerate}[label=\alph*)]
    \item Prove that there exists a row of the grid whose entries have a prime sum.
    \item Give an example of such a grid.
\end{enumerate}","\begin{solution}
\textbf{a)} Let $p_1, p_2, \ldots, p_7$ be the products of the numbers in rows $1, 2, \ldots, 7$. We have $p_1 = p_7$, $p_2 = p_6$, $p_3 = p_5$, so $49! = (p_1 p_2 p_3)^2 \cdot p_4$.

Note that $49! = 2^{46} \cdot 3^{22} \cdot 5^{10} \cdot 7^8 \cdot 11^4 \cdot 13^3 \cdot 17^2 \cdot 19^2 \cdot 23^2 \cdot 29 \cdot 31 \cdot 37 \cdot 41 \cdot 43 \cdot 47$.

The primes appearing to odd powers in $49!$ must appear at least once in the factorisation of $p_4$, so $13 \cdot 29 \cdot 31 \cdot 37 \cdot 41 \cdot 43 \cdot 47 \mid p_4$. Any product of two of these seven primes exceeds $49$, so each divides exactly one of the seven entries on row 4. \hfill \textbf{[2p]}

Since $\dfrac{49!}{p_4}$ is a perfect square, if $p_4 > 13 \cdot 29 \cdot 31 \cdot 37 \cdot 41 \cdot 43 \cdot 47$ then $p_4 \geq 2^2 \cdot 13 \cdot 29 \cdot 31 \cdot 37 \cdot 41 \cdot 43 \cdot 47$, forcing one cell to contain at least $\min\{4 \cdot 13, 2 \cdot 29\} > 49$, a contradiction. Therefore the central row contains exactly the numbers $13, 29, 31, 37, 41, 43, 47$, whose sum is $\mathbf{241}$, a prime. \hfill \textbf{[2p]}

\textbf{b)} The following grid satisfies the conditions:
\[
\begin{array}{|c|c|c|c|c|c|c|}
\hline
26 & 34 & 19 & 46 & 33 & 44 & 6 \\
\hline
7 & 35 & 12 & 49 & 24 & 25 & 8 \\
\hline
15 & 30 & 2 & 27 & 18 & 4 & 16 \\
\hline
13 & 29 & 31 & 37 & 41 & 43 & 47 \\
\hline
45 & 20 & 3 & 9 & 36 & 32 & 1 \\
\hline
14 & 21 & 10 & 28 & 42 & 5 & 40 \\
\hline
39 & 17 & 38 & 23 & 11 & 22 & 48 \\
\hline
\end{array}
\]
\hfill \textbf{[3p]}
\end{solution}",384,1558,Number Theory,6
366,tst_jbmo_ro_2024_2_p2,tst_jbmo,2024,a,"For every nonzero natural number $n$, let $a_n = \left\{\dfrac{n}{s(n)}
ight\}$, where $s(k)$ denotes the digit sum of the natural number $k$, and $\{x\}$ denotes the fractional part of the real number $x$.
\begin{enumerate}[label=\alph*)]
    \item Prove that there are infinitely many nonzero natural numbers $n$ with $a_n = \dfrac{1}{2}$.
    \item Find the smallest nonzero natural number $n$ for which $a_n = \dfrac{1}{6}$.
\end{enumerate}","\begin{solution}
\textbf{a)} Since $n$ is a natural number, for $a_n = \dfrac{1}{2}$ it suffices to have $s(n) = 2$ and $n$ odd. The numbers $n = 10^k + 1$, $k \in \mathbb{N}^*$, are an infinite family with this property. \hfill \textbf{[2p]}

\textbf{b)} Let $n \in \mathbb{N}^*$ with $a_n = \left\{\dfrac{n}{s(n)}
ight\} = \dfrac{1}{6}$.

From $\dfrac{n}{s(n)} - \left\lfloor\dfrac{n}{s(n)}
ight
floor = \dfrac{1}{6}$ we get $6n - 6s(n)\left\lfloor\dfrac{n}{s(n)}
ight
floor = s(n)$, so $s(n) \mid 6$, hence $3 \mid n$. Setting $n = 3k$ and $s(n) = 6m$, from (1) we get $3k - 6m\left\lfloor\dfrac{k}{2m}
ight
floor = m$, so $3 \mid m$. \hfill \textbf{[1p]}

Thus $m = 3u$ and $s(n) = 18u$, so $9 \mid n$. \hfill \textbf{[1p]}

Setting $n = 9v$: $3v - 6u\left\lfloor\dfrac{v}{2u}
ight
floor = u$, so $3 \mid u$. With $u = 3t$: $m = 9t$, $s(n) = 54t$. The minimum digit sum is $54$. \hfill \textbf{[1p]}

The smallest number with digit sum $54$ is $n = 999999$. But $a_{999999} = \left\{\dfrac{999999}{54}
ight\} = \left\{18518.5
ight\} = \dfrac{1}{2} 
eq \dfrac{1}{6}$, so it fails.

The next number with digit sum $54$ is $n = 1899999$, for which $a_{1899999} = \left\{\dfrac{1899999}{54}
ight\} = \left\{35185\tfrac{1}{6}
ight\} = \dfrac{1}{6}$.

Therefore $n_{\min} = \boxed{1899999}$. \hfill \textbf{[2p]}
\end{solution}",444,1325,Algebra,7
367,tst_jbmo_ro_2024_2_p3,tst_jbmo,2024,g,"On the exterior of acute triangle $ABC$, construct isosceles triangles $DAB$ and $EAC$ with bases $AB$ and $AC$ respectively, such that $\angle DBC = \angle ECB = 90^\circ$.

Let $M$ and $N$ be the reflections of point $A$ over $D$ and $E$, respectively. Prove that line $MN$ passes through the orthocenter of triangle $ABC$.","\begin{solution}
Let $H$ be the orthocenter of $\triangle ABC$ and $I$ the midpoint of $AH$. Let $B'$ be the reflection of $B$ over $D$, and $C'$ the reflection of $C$ over $E$.

Since $AD = DB = DB'$, triangle $ABB'$ is right-angled at $A$, so $AB' \perp AB$. \hfill \textbf{[1p]}

Since $CH \perp AB$, we get $AB' \parallel CH$, so $AP \parallel CH$ (where $P = AB' \cap CE$). From $AH \perp BC$ and $CP \perp BC$, we get $AH \parallel CP$. Hence $AHCP$ is a parallelogram and $CP = AH$. \hfill \textbf{[2p]}

Similarly $AHFC'$ is a parallelogram, so $C'F = AH = CP$. Since $C'E = CE$, we get $EP = EF$. \hfill \textbf{[1p]}

Therefore, in trapezoid $BB'PF$, $D$ and $E$ are the midpoints of the bases, and setting $\{Q\} = B'P \cap BF$, the points $D, E, Q$ are collinear. \hfill \textbf{[1p]}

Since $AHBB'$ is a trapezoid with $D$ and $I$ as midpoints of the bases and $\{Q\} = B'A \cap BH$, the points $D, I, Q$ are also collinear, so $I \in DE$. \hfill \textbf{[1p]}

$DI$ and $IE$ are midlines in triangles $AMH$ and $ANH$ respectively, so $DI \parallel MH$ and $IE \parallel NH$. Therefore $H \in MN$. \hfill \textbf{[1p]}
\end{solution}",325,1146,Geometry,8
368,tst_jbmo_ro_2024_2_p4,tst_jbmo,2024,c,"Let $n \geq 2$ be a natural number. A \textit{Welsh dartboard} is a disc divided into $2n$ equal sectors, half of them coloured red and the other half white. Two Welsh dartboards \textit{match} if they have the same radius and can be placed one on top of the other so that each sector of the first aligns exactly with one sector of the second. Consider two Welsh dartboards that can be matched such that more than half of the overlapping sector pairs have different colours. Prove that these dartboards can be matched so that at least $2\left\lfloor \dfrac{n}{2} 
ight
floor + 2$ of the overlapping sector pairs have the same colour.","\begin{solution}
For any matching, call a pair of overlapping sectors a \textit{concordance} if they share the same colour, and a \textit{discordance} otherwise.

Write $+1$ on each red sector and $-1$ on each white sector of both boards. For any matching, the product of overlapping values is $+1$ for a concordance and $-1$ for a discordance. If $t$ is the total number of concordances, the sum $S$ of all products equals $S = t \cdot 1 + (2n - t)(-1) = 2(t - n)$. \hfill \textbf{[1p]}

At any matching, let $k$ be the number of white-white concordances and $j$ the number of red-red concordances. Then $n - k$ white sectors of the first board overlap $n - k$ red sectors of the second, and $n - j$ red sectors of the first overlap $n - j$ white sectors of the second. The second board thus has $k + (n-j)$ white and $j + (n-k)$ red sectors, so $k + (n-j) = n$, giving $k = j$. Hence the total $t = k + j$ is always even. \hfill \textbf{[2p]}

Since the boards have more than $n$ discordances in the given matching, we have $S_1 < 0$. Fix the first board and rotate the second to obtain all $2n$ matchings, giving sums $S_1, S_2, \ldots, S_{2n}$ with:
\[
S_1 + S_2 + \cdots + S_{2n} = (u_1 + u_2 + \cdots + u_{2n})(v_1 + v_2 + \cdots + v_{2n}) = 0.
\]
\hfill \textbf{[2p]}

Since $S_1 < 0$ and the total is $0$, there exists $j$ with $S_j > 0$. Let $t$ be the concordance count for this matching. Then $2(t - n) = S_j \geq 2$ (since $S_j$ is a positive even integer), so $t \geq n + 1$. Since $t$ is even, $t \geq 2\left\lfloor \dfrac{n}{2} 
ight
floor + 2$. \hfill \textbf{[2p]}
\end{solution}",633,1596,Combinatorics,9
369,tst_jbmo_ro_2024_3_p1,tst_jbmo,2024,a,"Let $a, b, c > 0$ such that $ab + bc + ca = 1$. Show that:
\[
\frac{2}{abc} + 9abc \geq 7(a + b + c).
\]","\begin{solution}
The inequality to be proved can be rewritten as $2 + 9(abc)^2 \geq 7abc(a + b + c)$.

Setting $bc = x$, $ca = y$, $ab = z$, it suffices to show that for all $x, y, z > 0$ with $x + y + z = 1$:
\[
9xyz + 2 \geq 7(xy + yz + zx). \tag{1}
\]
\hfill \textbf{[2p]}

Inequality (1) can be rewritten as
\[
9xyz + 2(x+y+z)^3 \geq 7(x+y+z)(xy+yz+zx),
\]
which after expanding and collecting like terms reduces to:
\[
2(x^3 + y^3 + z^3) \geq x^2y + xy^2 + y^2z + yz^2 + z^2x + xz^2. \tag{2}
\]
\hfill \textbf{[3p]}

From $(x - y)^2 \geq 0$ we get $x^2 - xy + y^2 \geq xy$; multiplying by $x + y$ gives $x^3 + y^3 \geq x^2y + xy^2$. Adding the analogous inequalities yields (2), and the conclusion follows. \hfill \textbf{[2p]}
\end{solution}",104,747,Algebra,10
370,tst_jbmo_ro_2024_3_p2,tst_jbmo,2024,g,"Let $ABC$ be a triangle with $AB < AC$ and $\omega$ its circumscribed circle. The tangent to $\omega$ at $A$ meets line $BC$ at $D$, and the line through $B$ parallel to $AD$ meets $\omega$ again at $E$. Line $DE$ meets $AB$ at $F$ and $\omega$ again at $G$. On line $BE$ consider the point $N$ such that $B, G, F, N$ are concyclic.

Let $S$ and $T$ be the intersection points of line $FN$ with $AD$ and $AE$, respectively. Prove that quadrilateral $STDG$ is cyclic.","\begin{solution}
Since $ABCE$ and $BNFG$ are cyclic quadrilaterals, we have:
\[
\angle TAB = \angle BCE = 180^\circ - \angle BGE = 180^\circ - \angle BGF = \angle BNF = \angle BNT,
\]
so quadrilateral $ANBT$ is cyclic. \hfill \textbf{[2p]}

From this, using also $AD \parallel BE$, we deduce $\angle ATS = \angle ABN = \angle BAD = \angle FAS$, so line $AB$ is tangent at $A$ to the circumcircle of triangle $AST$. \hfill \textbf{[2p]} \hfill (1)

From $AD \parallel BE$ we get $\triangle FAD \sim \triangle FBE$, so $\dfrac{FA}{FB} = \dfrac{FD}{FE}$.

Using the power of point $F$ with respect to $\omega$: $FA \cdot FB = FE \cdot FG$.

Multiplying these two equalities gives $FA^2 = FD \cdot FG$. \hfill \textbf{[2p]} \hfill (2)

From (1): $FS \cdot FT = FA^2 = FD \cdot FG$, so $S, T, D, G$ are concyclic, i.e.\ $STDG$ is a cyclic quadrilateral. \hfill \textbf{[1p]}
\end{solution}",466,884,Geometry,11
371,tst_jbmo_ro_2024_3_p3,tst_jbmo,2024,n,"For each nonzero natural number $x$, let $\sigma(x)$ denote the sum of divisors and $\tau(x)$ the number of natural divisors of $x$.

Find all nonzero natural numbers $n$ with the property that $\sigma(\tau(n)) = n$.","\begin{solution}
Let $k = \tau(n)$ and let $d$ be a divisor of $n$ with $d \leq \sqrt{n}$. Since $\tfrac{n}{d}$ is also a divisor of $n$ with $\tfrac{n}{d} \geq \sqrt{n}$, the number of divisors of $n$ is at most $2\sqrt{n}$, i.e.\ $\tau(n) \leq 2\sqrt{n}$ for all $n \in \mathbb{N}^*$. \hfill \textbf{[1p]}

Also, if $1 = d_1 < d_2 < \cdots < d_k = n$ are the divisors of $n$, then
\[
2\sigma(n) = (d_1 + d_k) + (d_2 + d_{k-1}) + \cdots = \sum_i \left(d_i + \frac{n}{d_i}
ight).
\]
Since $1 + n - d - \tfrac{n}{d} = \tfrac{(n-d)(d-1)}{d} \geq 0$ for any divisor $d$, it follows that $\sigma(n) \leq \tfrac{1}{2}\tau(n)(n+1)$. \hfill \textbf{[1p]}

Now:
\[
n = \sigma(\tau(n)) \leq \tfrac{1}{2}\,\tau(\tau(n))\cdot(\tau(n)+1) \leq \tfrac{1}{2}\cdot 2\sqrt{\tau(n)}\cdot\left(2\sqrt{n}+1
ight) \leq \sqrt{2\sqrt{n}}\cdot\left(2\sqrt{n}+1
ight).
\]
Squaring gives $n^2 \leq 2\sqrt{n}(4n + 4\sqrt{n}+1)$, which simplifies to $\sqrt{n} \leq 8 + \dfrac{8}{\sqrt{n}} + \dfrac{2}{n}$.

Assuming for contradiction that $n \geq 81$, then $\sqrt{n} \geq 9$ and $8 + \dfrac{8}{\sqrt{n}} + \dfrac{2}{n} \leq 8 + \dfrac{8}{9} + \dfrac{2}{81} < 9 \leq \sqrt{n}$, a contradiction. \hfill \textbf{[3p]}

Therefore $n \leq 80$. Checking all cases with up to 12 divisors, the solutions are:

\begin{center}
\begin{tabular}{ccl}
\toprule
$\tau(n)$ & $n = \sigma(\tau(n))$ & Verdict \\
\midrule
1 & $\sigma(1) = 1$ & \checkmark \\
2 & $\sigma(2) = 3$ & \checkmark\ ($\tau(3) = 2$) \\
3 & $\sigma(3) = 4$ & \checkmark\ ($\tau(4) = 3$) \\
4 & $\sigma(4) = 7$ & $\tau(7) = 2 
eq 4$, fail \\
5 & $\sigma(5) = 6$ & $\tau(6) = 4 
eq 5$, fail \\
6 & $\sigma(6) = 12$ & \checkmark\ ($\tau(12) = 6$) \\
7 & $\sigma(7) = 8$ & $\tau(8) = 4 
eq 7$, fail \\
$\vdots$ & $\vdots$ & fail \\
\bottomrule
\end{tabular}
\end{center}

The solutions are $n \in \{1, 3, 4, 12\}$. \hfill \textbf{[2p]}
\end{solution}",216,1873,Number Theory,12
372,tst_jbmo_ro_2024_3_p4,tst_jbmo,2024,c,"Let $n \in \mathbb{N}$, $n \geq 3$, $M = \{1, 2, \ldots, n\}$, and $k > 0$ a real number. We assign to each nonempty subset of $M$ a distinct point in the plane (different subsets receive different points). If the absolute difference between the arithmetic means of two distinct nonempty subsets of $M$ is at most $k$, we connect their corresponding points with a segment.

Find the minimum value of $k$ such that the points corresponding to any two distinct nonempty subsets of $M$ are connected by a segment or a polygonal path.","\begin{solution}
Let $m_A$ denote the arithmetic mean of the elements of set $A$.

For any nonempty $A \subset M$ different from $\{1\}$, either $m_A \geq 2$ (if $A \subset \{2, 3, \ldots, n\}$), or $m_A = \dfrac{1 + |B| \cdot m_B}{1 + |B|} \geq \dfrac{1 + 2|B|}{1 + |B|} \geq \dfrac{3}{2}$ if $A = \{1\} \cup B$ with $B \subset \{2, \ldots, n\}$.

Since $\{1\}$ must be connected to at least one other subset $A 
eq \{1\}$, we need $k \geq \dfrac{1}{2}$. \hfill \textbf{[3p]}

To show $k = \dfrac{1}{2}$ is the minimum:

(1) For any nonempty $A \subset M$: $1 \leq m_A \leq n$.

(2) Any two single-element subsets $\{k\}$ and $\{p\}$ (with $k < p$) are connected by the chain
\[
\{k\},\ \{k, k+1\},\ \{k+1\},\ \{k+1, k+2\},\ \ldots,\ \{p-1, p\},\ \{p\},
\]
where consecutive means differ by $\dfrac{1}{2}$.

For any two nonempty subsets $A$ and $B$ with means $m_A$ and $m_B$, let $n_A$ and $n_B$ be the integers nearest to $m_A$ and $m_B$ respectively. By (1), $n_A, n_B \in [1, n]$, and we connect $A$ to $B$ via the chain $A, \{n_A\}, \ldots, \{n_B\}, B$ using the path described in (2). \hfill \textbf{[4p]}

\medskip

oindent\textbf{Answer:} The minimum value is $k = \dfrac{1}{2}$.
\end{solution}",530,1203,Combinatorics,13
373,tst_jbmo_ro_2024_4_p1,tst_jbmo,2024,a,"Let $n \in \mathbb{N}$, $n \geq 3$, and let $a_1, a_2, \ldots, a_n$ be pairwise distinct, strictly positive real numbers with the property that there exists a permutation $b_1, b_2, \ldots, b_n$ of these numbers such that
\[
\frac{a_1}{b_1} = \frac{a_2}{b_2} = \cdots = \frac{a_{n-1}}{b_{n-1}} 
eq 1.
\]
Show that there exist $a, b > 0$ such that $\{a_1, a_2, \ldots, a_n\} = \{ab, ab^2, \ldots, ab^n\}$.","\begin{solution}
Denote $r = \dfrac{a_1}{b_1}$. Without loss of generality, we may assume that $a_1 < a_2 < \cdots < a_{n-1}$. Since $a_k = r \cdot b_k$ for all $k \in \{1, 2, \ldots, n-1\}$, it follows that $b_1 < b_2 < \cdots < b_{n-1}$. \hfill \textbf{[1p]}

We assume $r > 1$ (the case $r < 1$ is entirely analogous, swapping the roles of $a_k$ and $b_k$, $1 \leq k \leq n$).

From $r > 1$ it follows that $a_k > b_k$ for all $k \in \{1, 2, \ldots, n-1\}$. Therefore $b_1 < a_1 < a_2 < \cdots < a_{n-1}$, and since $b_1 \in \{a_1, a_2, \ldots, a_n\}$, we conclude that $b_1 = a_n$. \hfill \textbf{[2p]}

Since we also have $b_1 < b_2 < \cdots < b_{n-1} < a_{n-1}$, it follows that $a_1 = b_2$, $a_2 = b_3$, $\ldots$, $a_{n-2} = b_{n-1}$. \hfill \textbf{[2p]}

Using $a_k = r \cdot b_k$ for all $k \in \{1, 2, \ldots, n-1\}$, we obtain successively:
\[
a_2 = r \cdot b_2 = r \cdot a_1,\quad a_3 = r \cdot b_3 = r \cdot a_2 = r^2 \cdot a_1,\quad \ldots,\quad a_{n-1} = r \cdot b_{n-1} = r^{n-2} \cdot a_1.
\]
Moreover, since $a_1 = r \cdot b_1 = r \cdot a_n$, we get
\[
\{a_1, a_2, \ldots, a_n\} = \{a_n,\; a_n \cdot r,\; a_n \cdot r^2,\; \ldots,\; a_n \cdot r^{n-1}\},
\]
and the conclusion follows for $a = \dfrac{a_n}{r}$ and $b = r$. \hfill \textbf{[2p]}
\end{solution}",404,1275,Algebra,14
374,tst_jbmo_ro_2024_4_p2,tst_jbmo,2024,g,"Let $ABC$ be a scalene triangle inscribed in circle $\omega$, and let $I$ be the incenter of the triangle. The tangent to $\omega$ at $C$ meets line $AB$ at $D$. The angle bisector of $\angle BDC$ meets lines $BI$ and $AI$ at $P$ and $Q$, respectively.

Let $M$ be the midpoint of segment $PQ$. Prove that line $IM$ passes through the midpoint of arc $ACB$ of circle $\omega$.","\begin{solution}
Let $E$ and $F$ be the second intersections of rays $(AI$ and $(BI$ with circle $\omega$, and let $N$ be the midpoint of arc $ACB$.

Since $CD$ is tangent to $\omega$ at $C$, we have $\angle ACD = \angle B$.

Then $\angle ADC = \angle BDC = 180^\circ - \angle DBC - \angle BCD = \angle A - \angle B$, so
\[
\angle IQP = \angle AQD = \angle ABE - \angle ADQ = \frac{\angle B}{2}.
\]
But $\angle AEF = \dfrac{\widehat{AF}}{2} = \dfrac{\angle B}{2}$, so the corresponding angles $\angle AQP$ and $\angle AEF$ are equal, hence $PQ \parallel EF$. \hfill \textbf{[2p]}

\[
\widehat{FN} = \widehat{AN} - \widehat{AF} = \frac{360^\circ - \widehat{AB}}{2} - \widehat{AF} = 180^\circ - \angle C - \angle B = \angle A = \widehat{BE},
\]
so $BF \parallel EN$, hence $IF \parallel EN$. \hfill \textbf{[2p]}

\[
\widehat{EN} = \widehat{BN} - \widehat{BE} = \frac{360^\circ - \widehat{AB}}{2} - \widehat{BE} = 180^\circ - \angle C - \angle A = \angle B = \widehat{AF},
\]
so $AE \parallel FN$, hence $IE \parallel FN$ and $IENF$ is a parallelogram. \hfill \textbf{[2p]}

Let $\{T\} = EF \cap IN$. Since $T$ is the midpoint of $EF$ and $PQ \parallel EF$, it follows that $IT$ passes through the midpoint of $PQ$, i.e.\ the points $I, M, T, N$ are collinear. \hfill \textbf{[1p]}
\end{solution}",376,1294,Geometry,15
375,tst_jbmo_ro_2024_4_p3,tst_jbmo,2024,n,"Find all prime numbers $p$ with the following properties:
\begin{enumerate}[label=\alph*)]
    \item $\dfrac{p+1}{2}$ is a prime number;
    \item there exist at least three distinct nonzero natural numbers $n$ such that $\dfrac{p^2 + n}{p + n^2}$ is a natural number.
\end{enumerate}","\begin{solution}
Note that $n = p$ satisfies condition (b). \hfill \textbf{[0p]}

Let $r = \dfrac{p+1}{2}$. Let $n 
eq p$ be a nonzero natural number such that $\dfrac{p^2+n}{p+n^2} \in \mathbb{N}$.

We deduce $p^2 + n \geq n^2 + p$, from which $(p-n)(p+n-1) \geq 0$, so $n < p$.

In particular, condition (b) implies $p > 3$.

Since $p + n^2 \mid p^2 + n$, we deduce that $p + n^2$ divides
\[
(p^2 + n) - p(p + n^2) + n^2(p + n^2) = n^4 + n.
\]
Since $n^4 + n = n(n+1)(n^2 - n + 1)$ and $\gcd(p, n) = 1$, it follows that
\[
p + n^2 \mid (n+1)(n^2 - n + 1). \tag{1}
\]
\hfill \textbf{[1p]}

Let $d = \gcd(p + n^2,\; n+1)$. Then $d$ divides $p + n^2 - (n^2 - 1) = p + 1 = 2r$. Since $d \leq n + 1 < p + 1 = 2r$ and $r$ is prime, we deduce $d \in \{1, 2\}$ (for any $p$) or $d = r$ (for $p > 3$). \hfill \textbf{[2p]}

If $d = 1$ or $d = 2$, from (1) we get $p + n^2 \mid 2(n^2 - n + 1)$. Since $n < p$, we have $2(p + n^2) > 2(n + n^2) > 2(n^2 - n + 1)$, so $p + n^2 = 2(n^2 - n + 1)$, giving $p = (n-1)^2 + 1$. \hfill \textbf{[2p]}

If $d = r$, from $r \mid n+1$ and $n + 1 < p + 1 = 2r$, we get $r = n + 1$. Substituting $n = r - 1$ and $p = 2r - 1$ into (1) gives $r^2 \mid r(r^2 - 3r + 3)$, i.e.\ $r \mid r^2 - 3r + 3$. Therefore $r \mid 3$, so $r = 3$, giving $p = 5$ and $n = 2$.

In conclusion, if $\dfrac{p^2 + n}{p + n^2} \in \mathbb{N}$, then one of the following holds: (i) $p = 5$ and $n = 2$; (ii) $p = n$; (iii) $p = (n-1)^2 + 1$.

Since $5 = (3-1)^2 + 1$, the answer is $p = 5$ (for which $n_1 = 2$, $n_2 = 3$, $n_3 = 5$ all satisfy (b)). \hfill \textbf{[2p]}
\end{solution}",284,1588,Number Theory,16
376,tst_jbmo_ro_2024_4_p4,tst_jbmo,2024,c,"Let $n \geq 2$ be a natural number and $A$ a set of $n$ points in the plane. Determine all natural numbers $k \in \{1, 2, \ldots, n-1\}$ for which the following property holds:

\textit{For any two circles $\mathcal{C}_1$ and $\mathcal{C}_2$ in the plane with $A \cap \mathrm{Int}(\mathcal{C}_1) 
eq A \cap \mathrm{Int}(\mathcal{C}_2)$ and $|A \cap \mathrm{Int}(\mathcal{C}_1)| = |A \cap \mathrm{Int}(\mathcal{C}_2)| = k$, the two circles have at least one point in common.}","\begin{solution}
For $k \leq \left\lfloor \dfrac{n}{2} 
ight
floor$, consider a line $d$ not parallel to any line determined by two points of $A$. Without loss of generality, call $d$ ``vertical.''

Choose a line $g \parallel d$ such that exactly $k$ points of $A$ lie to the right of $g$ (so at least $k$ lie to the left). Choose another line $h 
eq g$, $h \parallel d$, such that exactly $k$ points of $A$ lie to the left of $h$. \hfill \textbf{[2p]}

Choose $M \in g$ and construct a circle $\mathcal{C}_1$ tangent to $g$ at $M$ with sufficiently large radius so that all $k$ points of $A$ to the right of $g$ lie inside $\mathcal{C}_1$. Choose $N \in h$ and construct $\mathcal{C}_2$ tangent to $h$ at $N$ with sufficiently large radius so that all $k$ points of $A$ to the left of $h$ lie inside $\mathcal{C}_2$. Clearly $A \cap \mathrm{Int}(\mathcal{C}_1) 
eq A \cap \mathrm{Int}(\mathcal{C}_2)$, $|A \cap \mathrm{Int}(\mathcal{C}_1)| = |A \cap \mathrm{Int}(\mathcal{C}_2)| = k$, and $\mathcal{C}_1 \cap \mathcal{C}_2 = \emptyset$. Therefore no $k \leq \left\lfloor \dfrac{n}{2} 
ight
floor$ is a solution. \hfill \textbf{[2p]}

If $\left\lfloor \dfrac{n}{2} 
ight
floor < k < n$, let $\mathcal{C}_1$ and $\mathcal{C}_2$ be circles with $A \cap \mathrm{Int}(\mathcal{C}_1) 
eq A \cap \mathrm{Int}(\mathcal{C}_2)$ and $|A \cap \mathrm{Int}(\mathcal{C}_1)| = |A \cap \mathrm{Int}(\mathcal{C}_2)| = k$.

Since $\mathrm{Int}(\mathcal{C}_1)$ contains exactly $k$ points of $A$, fewer than $k$ points of $A$ remain outside it, so $\mathrm{Int}(\mathcal{C}_2)$ must contain at least one point that also lies in $\mathrm{Int}(\mathcal{C}_1)$. Hence $\mathrm{Int}(\mathcal{C}_1) \cap \mathrm{Int}(\mathcal{C}_2) 
eq \emptyset$. \hfill \textbf{[2p]}

Furthermore, since each of $\mathrm{Int}(\mathcal{C}_1)$ and $\mathrm{Int}(\mathcal{C}_2)$ contains exactly $k$ points of $A$ and $A \cap \mathrm{Int}(\mathcal{C}_1) 
eq A \cap \mathrm{Int}(\mathcal{C}_2)$, neither circle is contained in the other. Therefore $\mathcal{C}_1 \cap \mathcal{C}_2 
eq \emptyset$.

Thus every natural number $k$ with $\left\lfloor \dfrac{n}{2} 
ight
floor < k < n$ is a solution. \hfill \textbf{[1p]}

\medskip

oindent\textbf{Answer:} $k \in \left\{\left\lfloor \dfrac{n}{2} 
ight
floor + 1,\; \left\lfloor \dfrac{n}{2} 
ight
floor + 2,\; \ldots,\; n-1
ight\}$.
\end{solution}",474,2352,Combinatorics,17
275,jbmo_2025_p1,jbmo,2025,a,"For all positive real numbers $a, b, c$, prove that
\[
\frac{(a^2 + bc)^2}{b + c} + \frac{(b^2 + ca)^2}{c + a} + \frac{(c^2 + ab)^2}{a + b}
\geq \frac{2abc(a + b + c)^2}{ab + bc + ca}.
\]","\subsection*{Solutions}

\textbf{Solution 1.} Apply Cauchy-Schwarz inequality.
\[
\left((b+c)+(c+a)+(a+b)
ight)\left(\frac{(a^2+bc)^2}{b+c}+\frac{(b^2+ca)^2}{c+a}+\frac{(c^2+ab)^2}{a+b}
ight)
\geq \left(a^2+b^2+c^2+ab+bc+ca
ight)^2.
\]

\begin{lemma}
$3(a^2 + b^2 + c^2 + ab + bc + ca) \geq 2(a+b+c)^2$.
\end{lemma}

\begin{proof}
\begin{align*}
3(a^2 + b^2 + c^2 + ab + bc + ca)
&= 2(a^2 + b^2 + c^2 + 2ab + 2bc + 2ca) + (a^2 + b^2 + c^2 - ab - bc - ca) \\
&= 2(a+b+c)^2 + \frac{1}{2}(a-b)^2 + \frac{1}{2}(b-c)^2 + \frac{1}{2}(c-a)^2 \\
&\geq 2(a+b+c)^2,
\end{align*}
as desired.
\end{proof}

Then we have
\[
\frac{(a^2+bc)^2}{b+c}+\frac{(b^2+ca)^2}{c+a}+\frac{(c^2+ab)^2}{a+b}
\geq \frac{4}{9} \cdot \frac{(a+b+c)^4}{2(a+b+c)} = \frac{2}{9}(a+b+c)^3.
\]

Again by Cauchy-Schwarz inequality we have
\[
(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}
ight) \geq 9.
\]

Finally, using these we get
\[
\frac{(a^2+bc)^2}{b+c}+\frac{(b^2+ca)^2}{c+a}+\frac{(c^2+ab)^2}{a+b}
\geq \frac{2}{9}(a+b+c)^3 \geq \frac{2(a+b+c)^2}{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}}
= \frac{2abc(a+b+c)^2}{ab+bc+ca},
\]
as desired. \qed

\medskip

\textbf{Solution 2.} By using AM-GM we have:
\[
\sum_{\mathrm{cyc}} \frac{(a^2+bc)^2}{b+c}
\geq \sum_{\mathrm{cyc}} \frac{\left(2\sqrt{a^2 bc}
ight)^2}{b+c}
= 4abc \sum_{\mathrm{cyc}} \frac{a}{b+c}.
\]

From here, by using Titu's Lemma, we obtain
\[
4abc \sum_{\mathrm{cyc}} \frac{a}{b+c}
= 4abc \sum_{\mathrm{cyc}} \frac{a^2}{ab+ac}
\geq 4abc \cdot \frac{(a+b+c)^2}{2(ab+bc+ca)}
= \frac{2abc(a+b+c)^2}{ab+bc+ca},
\]
as desired. \qed

\medskip

\textbf{Solution 3.} Using Titu's Lemma and Schur's Inequality, we have
\[
\sum_{\mathrm{cyc}} \frac{(a^2+bc)^2}{b+c}
= \sum_{\mathrm{cyc}} \frac{(a^3+abc)^2}{a^2(b+c)}
\overset{\mathrm{Titu}}{\geq}
\frac{\left(\sum a^3 + 3abc
ight)^2}{\sum a^2(b+c)}
\overset{\mathrm{Schur}}{\geq}
\sum_{\mathrm{cyc}} a^2(b+c).
\]

Using the inequality $(ab+bc+ca)^2 \geq 3abc(a+b+c)$, we get:
\[
\sum_{\mathrm{cyc}} a^2(b+c) = -3abc + (a+b+c)(ab+bc+ca)
\geq -3abc + (a+b+c)\cdot\frac{3abc(a+b+c)}{ab+bc+ca}.
\]

Finally, by $(a+b+c)^2 \geq 3(ab+bc+ca)$:
\begin{align*}
-3abc + (a+b+c)\cdot\frac{3abc(a+b+c)}{ab+bc+ca}
&= abc\left(-3 + \frac{(a+b+c)^2}{ab+bc+ca}
ight) + \frac{2abc(a+b+c)^2}{ab+bc+ca} \\
&\geq \frac{2abc(a+b+c)^2}{ab+bc+ca}.
\end{align*}
\qed",187,2313,Algebra,1
276,jbmo_2025_p2,jbmo,2025,n,"Determine all numbers of the form
\[
\underbrace{20252025\ldots2025}_{n \text{ blocks}}
\]
(where the block $2025$ is repeated one or more times) that are perfect squares of positive integers.","\subsection*{Solution}

We will prove that the only solution is $2025 = 45^2$.

First, observe that the given number is equal to
\[
2025\left(1 + 10^4 + 10^8 + \cdots + 10^{4(n-1)}
ight)
\]
for some $n \in \mathbb{N}$. Since $2025$ is a perfect square, it must hold that
\[
1 + 10^4 + 10^8 + \cdots + 10^{4(n-1)} = x^2 \tag{$\star$}
\]
for some $x \in \mathbb{N}$. Multiplying both sides by $10^4 - 1$ yields
\[
10^{4n} - 1 = 9999x^2 = 11 \cdot 101 \cdot (3x)^2.
\]

Applying the difference of squares twice gives
\[
(10^n - 1)(10^n + 1)(10^{2n} + 1) = 11 \cdot 101 \cdot (3x)^2. \tag{$\star\star$}
\]

Using the Euclidean algorithm, we see that $10^n - 1$ and $10^n + 1$ are coprime (their difference is $2$ and they are both odd). Additionally, $10^{2n}+1$ is coprime with both $10^n - 1$ and $10^n + 1$ since it is coprime with their product $10^{2n} - 1$ (same reasoning as before). Hence, all three factors on the left-hand side of $(\star\star)$ are pairwise coprime. Since the numbers $11$ and $101$ are prime, at least one of the factors on the left-hand side of $(\star\star)$ must be a perfect square.

\begin{lemma}
The equation $10^n + 1 = m^2$ has no solutions in the set of positive integers.
\end{lemma}
\begin{proof}
Reducing modulo $3$ implies $m^2 \equiv 2 \pmod{3}$, which cannot hold.
\end{proof}

Hence, neither $10^{2n}+1$ nor $10^n+1$ can be perfect squares.

\begin{lemma}
The only solution to the equation $10^n - 1 = m^2$ in the set of positive integers is $(n, m) = (1, 3)$.
\end{lemma}
\begin{proof}
Consider the equation modulo $4$. For $n \geq 2$, $10^n \equiv 0 \pmod{4}$, hence the right-hand side is congruent to $3$ modulo $4$. On the other hand, $n = 1$ gives $m = 3$ as a solution.
\end{proof}

Lemmas 1 and 2 imply that the only solution to the given equation is $(n, x) = (1, 1)$. Therefore, the only solution is the number $\mathbf{2025}$. \qed",192,1883,Number Theory,2
277,jbmo_2025_p3,jbmo,2025,g,Let $\triangle ABC$ be right-angled at $A$ and let $D$ be the foot of the altitude from $A$ to $BC$ and let $E$ be the midpoint of $DC$. The circumcircle of $\triangle ABD$ intersects $AE$ again at point $F$. Let $X$ be the intersection of $AB$ and $DF$. Prove that $XD = XC$.,"\subsection*{Solutions}

\textbf{Solution 1.} Since $E$ is the midpoint of $DC$, it is sufficient to show that $XE \perp DC$, which is equivalent to proving that $XE \parallel AD$. Let $H$ be the intersection of $BF$ and $AD$. Since $\angle ADB = 90^\circ \Rightarrow \angle AFB = 90^\circ \Rightarrow H$ is the orthocenter of $\triangle ABE \Rightarrow EH \perp AB \Rightarrow EH \parallel AC$. Since $E$ is the midpoint of $DC$ we have that $H$ is also the midpoint of $AD$. Apply (unoriented) Menelaus' theorem in $\triangle ABH$ for points $D, F, X$ and in $\triangle BHD$ for points $A, F, E$. We get that
\[
\frac{AX}{XB} \cdot \frac{BF}{FH} \cdot \frac{HD}{DA} = 1 = \frac{DE}{EB} \cdot \frac{BF}{FH} \cdot \frac{HA}{AD}.
\]
Since we have that $AH = HD$ the ratios cancel and we get that $\dfrac{AX}{XB} = \dfrac{DE}{EB}$, which implies that $AD \parallel XE$ by Thales. \qed

\medskip

\textbf{Solution 2.} Let $M$ be the midpoint of $AC$. As a midsegment in $\triangle ADC$, $ME \parallel AD$. Also, by Thales' Theorem in the right triangle $ADC$, we get $MA = MD = MC$.

\begin{claim}
$DFME$ is cyclic. \hfill\ldots(1)
\end{claim}

\textbf{Proof 1.} $\angle DFE \overset{(ABDF)}{=} \angle DBA = 90^\circ - \angle DCA = \angle DAC \overset{AD \parallel ME}{=} \angle EMC = \angle DME$, so $DFME$ is cyclic.

\textbf{Proof 2.} First we prove that $MD$ is tangent to $(ABDF)$. Let $O$ be the midpoint of $AB$ and therefore center of $(ABDF)$. Then, $OA = OD$, so by criterion SSS, we get $\triangle OAM \cong \triangle ODM$, thus $\angle ODM = \angle OAM = 90^\circ$.

Now, $\angle MEF \overset{ME \parallel AD}{=} \angle FAD = \angle MDF$, so $DFME$ is cyclic.

\begin{claim}
$X, M, E$ are collinear.
\end{claim}

\textbf{Proof 1.} $\angle MFX \overset{(1)}{=} \angle MED = 90^\circ = \angle MAX$, so $MFAX$ is cyclic. \hfill\ldots(2)

$\angle XMA \overset{(2)}{=} \angle XFA = \angle DFE \overset{(1)}{=} \angle DME = \angle CME$ and since $A, M, C$ are collinear, so are $X, M, E$.

\textbf{Proof 2.} Same as in Proof 1, $MFAX$ is cyclic.
\[
\angle XMF \overset{(2)}{=} \angle FAB \overset{(ABDF)}{=} \angle FDE \overset{(1)}{=} 180^\circ - \angle FME,
\]
so $\angle XME = 180^\circ$.

\textbf{Proof 3.} $\angle MAB + \angle MEB = 90^\circ + 90^\circ = 180^\circ$, so $MABE$ is cyclic. The three pairwise radical axes of the circles $(MABE)$, $(ABDF)$ and $(DFME)$ are $AB$, $DF$ and $ME$. Then $X = AB \cap DF$ is their radical center, so $X \in ME$.

Now, $X, M, E$ are collinear, i.e.\ $X$ lies on the side bisector of $DC$, so $XD = XC$. \qed

\medskip

\textbf{Solution 3.} Let $Y$ be the intersection of the side bisector of $DC$ with line $AB$. Then, $YD = YC$ and therefore $\angle YDC = \angle YCD$. Also, $\angle YEC = 90^\circ = \angle YAC$, so $YAEC$ is cyclic. From there,
\[
\angle BAE = \angle YCE \equiv \angle YCD = \angle YDC.
\]
On the other hand, from $(ABDF)$, $\angle FDC = \angle BAF \equiv \angle BAE$.

Therefore, $\angle YDC = \angle BAE = \angle FDC$, thus points $D, F, Y$ are collinear, i.e.\ $Y \in DF$. Since $X = AB \cap DF$, we get $Y \equiv X$, and therefore $XD = XC$. \qed",276,3118,Geometry,3
278,jbmo_2025_p4,jbmo,2025,c,"Let $n$ be a positive integer. The integers from $1$ to $n$ are written in the cells of an $n \times n$ table (one integer per cell) so that each of them appears exactly once in each row and exactly once in each column. Denote by $r_i$ the number of pairs $(a, b)$ of numbers in the $i$th row ($1 \leq i \leq n$), such that $a > b$, but $a$ is written to the left of $b$ (not necessarily next to it). Denote by $c_j$ the number of pairs $(a, b)$ of numbers in the $j$th column ($1 \leq j \leq n$), such that $a > b$, but $a$ is written above $b$ (not necessarily next to it). Determine the largest possible value of the sum
\[
r_1 + r_2 + \cdots + r_n + c_1 + c_2 + \cdots + c_n.
\]


oindent\textit{Note}: In the $n \times n$ table we label the rows $1$ to $n$ from top to bottom, and we label the columns $1$ to $n$ from left to right.","\subsection*{Solution}

\textbf{Answer:} $\dfrac{n(n-1)(2n-1)}{3}$.

Suppose $x$ is in position $i$ in some row/column. Then after it there could be at most $\min(n - i,\, x - 1)$ smaller numbers. Having in mind that this bound is separately for the row of $x$ and for the column of $x$, as well as that $i$ is different for the different appearances of $x$ (as no row/column has a number appearing more than once, by the problem condition), we deduce that the contribution of $x$ to the overall sum is at most
\begin{align*}
2\sum_{i=1}^{n} \min(n - i,\, x - 1)
&= 2\sum_{i=n-x+1}^{n}(n - i) + 2\sum_{i=1}^{n-x}(x - 1) \\
&= 2\sum_{i=0}^{x-1} i + 2(n - x)(x - 1) \\
&= x(x-1) + 2(n-x)(x-1) = (2n - x)(x - 1).
\end{align*}

Summing through $x = 1, \ldots, n$ now gives the following upper bound for the sum:
\begin{align*}
\sum_{x=1}^{n}(2n - x)(x - 1)
&= (2n + 1)\sum_{x=1}^{n} x - \sum_{x=1}^{n} x^2 - \sum_{x=1}^{n} 2n \\
&= \frac{n(n+1)(2n+1)}{2} - \frac{n(n+1)(2n+1)}{6} - 2n^2 \\
&= \frac{n(n-1)(2n-1)}{3}.
\end{align*}

Equality holds e.g.\ for the table in which the $s$-th row is
\[
n+1-s,\ n-s,\ \ldots,\ 1,\ n,\ n-1,\ \ldots,\ n+2-s,
\]
since for each $x$ in position $i$ in a row or column there are indeed exactly $\min(n - i,\, x - 1)$ smaller numbers after it. \qed",837,1280,Combinatorics,4
343,tst_jbmo_ro_2025_1_p1,tst_jbmo,2025,n,"We say that a natural number $n \geq 3$ is \emph{almost squarefree} if there exists a prime $p$, $p \equiv 1 \pmod{3}$, such that $n$ is divisible by $p^2$ and $\dfrac{n}{p}$ is squarefree (i.e.\ not divisible by the square of any prime).

Show that, for any almost squarefree natural number $n$, the ratio of twice the sum of divisors of $n$ to the number of divisors of $n$ is a natural number.","\textbf{Solution.}
Let $\tau(n)$ and $\sigma(n)$ denote the number and sum of divisors of $n$.

By hypothesis $n = p_1 \cdots p_{k-1} \cdot p_k^2 \cdot p_{k+1} \cdots p_s$ with $p_1 < p_2 < \cdots < p_s$ prime, $p = p_k \equiv 1 \pmod{3}$. \hfill\textbf{[1p]}

Then $p$ is odd and $p^2 + p + 1 \equiv 0 \pmod{3}$, so:
\[
\tau(n) = 2^{s-1} \cdot 3, \qquad
\sigma(n) = (p_1+1)\cdots(p_{k-1}+1)\cdot(p_k^2+p_k+1)\cdot(p_{k+1}+1)\cdots(p_s+1). \hfill\textbf{[2p]}
\]

\textbf{Case I: $n$ even ($p_1 = 2$).} Then:
\[
2\sigma(n) = 2\cdot 3 \cdot (p_2+1)\cdots(p_{k-1}+1)(p_k^2+p_k+1)(p_{k+1}+1)\cdots(p_s+1),
\]
which is divisible by $2 \cdot 3^2 \cdot 2^{s-2} = 3 \cdot \tau(n)$. \hfill\textbf{[2p]}

\textbf{Case II: $n$ odd.} All $p_i$ are odd, so each $(p_i+1)$ is even and $(p_k^2+p_k+1)$ is divisible by $3$. Thus $\sigma(n)$ is divisible by $3 \cdot 2^{s-1} = \tau(n)$, so $2\sigma(n)/\tau(n)$ is a natural number. \hfill\textbf{[2p]}",396,935,Number Theory,1
344,tst_jbmo_ro_2025_1_p2,tst_jbmo,2025,g,"Consider an isosceles triangle $ABC$ with $\angle BAC > 90^\circ$, and the circle $\mathcal{C}$ with center $A$ and radius $AB$. Let $M$ be the midpoint of side $AC$. Line $BM$ meets circle $\mathcal{C}$ again at $D$. Let $E$ be a point on $\mathcal{C}$ such that $BE \perp AC$ and $DE \cap AC = \{N\}$. Show that $AN = 2 \cdot AB$.","\textbf{Solution.}
From $AC \perp BE$, $AC$ is the perpendicular bisector of $BE$, so $MB = ME$. Since $AB = AE$, we get $\triangle MAB \equiv \triangle MAE$. \hfill\textbf{[2p]}

Thus $\angle MAB = \angle MAE$ \quad (1) and $\angle MBA = \angle MEA$ \quad (2).

From $AB = AD$, triangle $ABD$ is isosceles, so $\angle ABD = \angle ADB$ \quad (3).

From (2) and (3): $\angle MAE = \angle MDA$, so quadrilateral $MAED$ is cyclic. \hfill\textbf{[2p]}

Then $\angle AED = \angle DMN$, and since $\angle AMB = \angle DMN$, we get $\angle AED = \angle AMB$ \quad (4).

From (1) and (4): $\triangle MAB \sim \triangle AEN$, giving:
\[
\frac{AN}{AB} = \frac{AE}{AM} = \frac{AB}{AB/2} = 2,
\]
so $AN = 2 \cdot AB$. \hfill\textbf{[3p]}",332,726,Geometry,2
345,tst_jbmo_ro_2025_1_p3,tst_jbmo,2025,a,"Let $n \geq 3$. Ana chooses positive integers $a_1, a_2, \ldots, a_n$ and, for each nonempty $A \subseteq \{1,2,\ldots,n\}$, computes the sum $s_A = \sum_{k \in A} a_k$, then sorts these sums in nondecreasing order: $s_1 \leq s_2 \leq \cdots \leq s_{2^n - 1}$.

Show that there exists a subset $B \subseteq \{1, 2, \ldots, 2^n - 1\}$ with $2^{n-2}+1$ elements such that, regardless of which numbers $a_1, \ldots, a_n$ Ana chose, they can be determined knowing all values $s_i$, $i \in B$.","\textbf{Solution.}
Assume $a_1 \leq a_2 \leq \cdots \leq a_n$. Take $B' = \{1, 2, \ldots, 2^{n-2}\}$. Among $s_1, \ldots, s_{2^{n-2}}$, the values $a_1, \ldots, a_t$ appear for some $t \geq 1$ (since $s_1 = a_1$ and each $a_{i-1} < a_i$ implies $a_{i-1}$ appears before $a_i$). If $t \leq n-2$, at most $2^{n-2}-1$ sums come from nonempty subsets of $\{a_1,\ldots,a_{n-2}\}$, so $t \geq n-1$. \hfill\textbf{[3p]}

Thus the first two sums give $a_1, a_2$. Inductively: having determined $a_1, \ldots, a_k$, compute all sums from nonempty subsets of $\{a_1,\ldots,a_k\}$, remove them from the sorted list, and the smallest remaining sum is $a_{k+1}$. This recovers $a_1, \ldots, a_{n-1}$. \hfill\textbf{[3p]}

Adding index $2^n - 1$ to $B'$ gives $B$ with $2^{n-2}+1$ elements. Knowing $s_{2^n-1} = a_1+\cdots+a_n$ and $a_1+\cdots+a_{n-1}$, we recover $a_n$. \hfill\textbf{[1p]}",488,876,Algebra,3
346,tst_jbmo_ro_2025_1_p4,tst_jbmo,2025,g,"Let $ABCDEF$ be a convex hexagon in which triangles $ABC$ and $DEF$ are equilateral, and diagonals $AD$, $BE$, $CF$ are concurrent. Prove that $AC \parallel DF$ or $BE = AD + CF$.","\textbf{Solution.}
Let $X \in AD$ and $Y \in CF$ such that $EX \parallel AB$ and $EY \parallel BC$. Let $\{T\} = AD \cap BE \cap CF$. Then $\angle XEY = 60^\circ$. \hfill\textbf{[1p]}

\textbf{Case 1: $X = D$.} From $\angle XEY = 60^\circ$ we get $Y = F$, and then $\dfrac{CT}{TF} = \dfrac{BT}{TE} = \dfrac{AT}{TD}$, so $AC \parallel DF$. \hfill\textbf{[2p]}

\textbf{Case 2: $X 
eq D$ (and hence $Y 
eq F$).} Then $\dfrac{XE}{AB} = \dfrac{ET}{TB} = \dfrac{YE}{BC}$, giving $XE = YE$.

Since $\angle XEF = \angle YEF$, we get $\triangle XED \equiv \triangle YEF$. Thus $\angle TFE = \angle XDE = 180^\circ - \angle TDE$, so $T$ lies on the circumcircle of $\triangle DEF$. \hfill\textbf{[2p]}

By symmetry $T$ also lies on the circumcircle of $\triangle ABC$. Applying Ptolemy's theorem: $TE = TD + TF$ and $TB = TA + TC$. Adding these gives $BE = AD + CF$. \hfill\textbf{[2p]}",179,877,Geometry,4
347,tst_jbmo_ro_2025_1_p5,tst_jbmo,2025,c,"Let $n \geq 3$ and let $\mathcal{F}$ be a family of at most $n$ distinct subsets of $\{1, 2, \ldots, n\}$ with the property that we can place $n$ distinct points in the plane labeled $1, 2, \ldots, n$ and draw segments between some of them such that for any distinct $i, j \in \{1, 2, \ldots, n\}$, points $i$ and $j$ are connected if and only if the number $i$ belongs to exactly $j$ sets of $\mathcal{F}$.

Find the maximum value of the sum of the sizes of all sets in $\mathcal{F}$.","\textbf{Solution.}
For $k \in \{1,\ldots,n\}$, let $a_k = |\{F \in \mathcal{F} \mid k \in F\}|$. Then $a_k \in \{0,1,\ldots,n\}$ and the sum to maximize is:
\[
s = \sum_{F \in \mathcal{F}} |F| = \sum_{k=1}^n a_k. \hfill\textbf{[1p]}
\]
The connectivity condition means: if $a_k 
eq k$, then $a_{a_k} = k$, for all $k$. Conversely, any such function gives a valid family. \hfill\textbf{[1p]}

Any vertex can be connected to at most one other (otherwise some $i$ would belong to both $j_1$ and $j_2$ subsets). Thus the values $a_k$ are pairwise distinct (if $a_i = a_j = a$, then $a$ is connected to both $i$ and $j$, forcing $i = j$). \hfill\textbf{[2p]}

Hence $s \leq n + (n-1) + \cdots + 1 = \dfrac{n(n+1)}{2}$. \hfill\textbf{[1p]}

This maximum is achieved by taking any family $\mathcal{F}$ where $a_{a_k} = k$ for all $k$ (a fixed-point-free involution on $\{1,\ldots,n\}$ extended appropriately). \hfill\textbf{[2p]}",485,922,Combinatorics,5
348,tst_jbmo_ro_2025_2_p1,tst_jbmo,2025,a,"Let $n \geq 2$ be a natural number and $a_1, a_2, \ldots, a_n$ nonzero integers such that
\[
a_1 + a_2 + \cdots + a_n = a_1 a_2 \cdots a_n.
\]
Show that $(a_1^2 - 1)(a_2^2 - 1)\cdots(a_n^2 - 1)$ is a perfect square.","\textbf{Solution.}
For $n = 2$: $a_1 + a_2 = a_1 a_2$ rewrites as $(a_1-1)(a_2-1) = 1$, giving $a_1 = a_2 = 2$, so $(a_1^2-1)(a_2^2-1) = 9 = 3^2$. \hfill\textbf{[1p]}

For $n \geq 3$: we show at least one $|a_i| = 1$, so the product is $0$. Using the hypothesis:
\[
|a_1 a_2 \cdots a_n| = |a_1 + a_2 + \cdots + a_n| \leq |a_1| + |a_2| + \cdots + |a_n|. \hfill\textbf{[1p]}
\]
Dividing by $|a_1 a_2 \cdots a_n| > 0$:
\[
1 \leq \frac{|a_1|}{|a_1 \cdots a_n|} + \cdots + \frac{|a_n|}{|a_1 \cdots a_n|}.
\]
If all $|a_i| \geq 2$, the right side is $\leq \dfrac{n}{2^{n-1}}$, i.e., $2^{n-1} \leq n$. \hfill\textbf{[2p]}

For $n \geq 3$, by induction (or Bernoulli's inequality) $2^{n-1} > n$, a contradiction. \hfill\textbf{[2p]}

Thus some $|a_i| = 1$, so $(a_1^2-1)\cdots(a_n^2-1) = 0$, which is a perfect square. \hfill\textbf{[1p]}",215,830,Algebra,6
349,tst_jbmo_ro_2025_2_p2,tst_jbmo,2025,g,"Consider a triangle $ABC$ with incenter $I$, and let $I_a$, $I_b$, $I_c$ be the excenters opposite to $A$, $B$, $C$ respectively. Let $E$, $F$, $G$ be the points of tangency of the incircle with sides $BC$, $CA$, $AB$ respectively.

Prove that the circumcircles of triangles $IEI_a$, $IFI_b$, $IGI_c$ have a common point other than $I$.","\textbf{Solution.}
Let $J_a$, $J_b$, $J_c$ be the circumcenters of triangles $IEI_a$, $IFI_b$, $IGI_c$. We prove $J_a$, $J_b$, $J_c$ are collinear. \hfill\textbf{[1p]}

Let $L_a$, $L_b$, $L_c$ be the points diametrically opposite to $I$ in the three circumcircles. Since $J_aJ_b$, $J_bJ_c$, $J_cJ_a$ are midlines of triangles $IL_aL_b$, $IL_bL_c$, $IL_cL_a$, it suffices to show $L_a$, $L_b$, $L_c$ are collinear. \hfill\textbf{[1p]}

Since $IL_a$ is a diameter of the circumcircle of $\triangle IEI_a$, $\angle II_aL_a = 90^\circ$, so $I_aL_a \perp II_a$. But $II_a \perp I_bI_c$, giving $I_aL_a \parallel I_bI_c$, hence $\angle L_aI_aB = \angle I_aI_cI_b$ and $\angle L_aI_aC = 180^\circ - \angle I_aI_bI_c$. \hfill\textbf{[1p]}

Let $J$ be the midpoint of $IE$. Then $JJ_a \parallel EL_a$. But $JJ_a \perp IE$ and $BE \perp IE$, so $JJ_a \parallel BE$, hence $L_a \in BE \subset BC$. Similarly $L_b \in CA$ and $L_c \in AB$. \hfill\textbf{[1p]}

Computing:
\[
\frac{BL_a}{CL_a} = \frac{BI_a \cdot \sin(\angle I_aI_cI_b)}{CI_a \cdot \sin(\angle I_aI_bI_c)}, \quad
\frac{CL_b}{AL_b} = \frac{CI_b \cdot \sin(\angle I_bI_aI_c)}{AI_b \cdot \sin(\angle I_bI_cI_a)}, \quad
\frac{AL_c}{BL_c} = \frac{AI_c \cdot \sin(\angle I_cI_bI_a)}{BI_c \cdot \sin(\angle I_cI_aI_b)}. \hfill\textbf{[1p]}
\]
Their product equals $1$, so by the converse of Menelaus' theorem, $L_a$, $L_b$, $L_c$ are collinear, hence $J_a$, $J_b$, $J_c$ are collinear. \hfill\textbf{[1p]}

The radical axes of each pair of the three circles are parallel or coincide. Since $I$ lies on all three, they share the same radical axis. As $I 
otin J_aJ_b$, the circles are not tangent, so their common radical axis contains a second common point. \hfill\textbf{[1p]}",336,1723,Geometry,7
350,tst_jbmo_ro_2025_2_p3,tst_jbmo,2025,c,"Let $n \geq 3$ be a natural number. Ion draws a regular $n$-gon and connects all pairs of vertices. On each segment, Ion writes a positive integer such that in every triangle (formed by three vertices), one of the three numbers equals the sum of the other two.

Find the minimum number of distinct values Ion can write.","\textbf{Solution.}
The answer is $n-1$ for $n 
eq 4$, and $2$ for $n = 4$.

For $n = 3$ and $n = 4$, one verifies directly that the answer is $2$. \hfill\textbf{[1p]}

For $n \geq 5$: label vertices $A_1, \ldots, A_n$ and set $a_{A_iA_j} = |i-j|$. The triangle condition is satisfied, so the minimum is at most $n-1$. \hfill\textbf{[2p]}

We show the minimum cannot be $n-2$. Suppose for contradiction it is. Let $d$ be the maximum value written, achieved on edge $AB$.

\textbf{Obs.\ 1:} No three vertices $A_1,A_2,A_3$ can all have equal labels from $A$, as the triangle $A_1A_2A_3$ would violate the sum rule.

\textbf{Obs.\ 2:} For any $X 
eq A,B$: $d = a_{AX} + a_{BX}$ (maximality), so $a_{AX} < d$.

By assumption and the pigeonhole principle, there exist $C,D$ with $a_{AC} = a_{AD} = x < d$. In triangle $ACD$: $a_{CD} = 2x$. In $ABD$: $a_{BD} = d-x$. In $ABC$: $a_{BC} = d-x$. Then in $\triangle BCD$: $2x = (d-x)+(d-x)$, giving $x = d/2$.

Among the $n-2$ values on edges from $A$, the only repeated value is $d/2$ (which can appear at most twice by Obs.\ 1). So there are $n-2$ distinct values used overall.

Since $n \geq 5$, pick a fifth vertex $E$, $y = a_{AE} 
eq d/2$. WLOG $y > d/2$. In $\triangle ACE$: $a_{CE} = y - d/2$, and similarly $a_{DE} = y-d/2$. In $\triangle CDE$: $a_{CD} = 2(y - d/2) < d$, but $a_{CD} = d$ --- contradiction.

Hence the minimum is $n-1$. \hfill\textbf{[4p]}",319,1405,Combinatorics,8
351,tst_jbmo_ro_2025_2_p4,tst_jbmo,2025,n,Find all natural numbers $n$ such that $2^n - n^2 + 1$ is a perfect square.,"\textbf{Solution.}
By checking, $n = 2, 3, 4$ are solutions. We show these are the only ones, assuming $n \geq 5$.

\textbf{Case I: $n$ even.} Write $n = 2t$ ($t \geq 3$). One shows $(2^{t-1})^2 < 2^n - n^2 + 1 < (2^t)^2$, so $2^n - n^2 + 1$ is strictly between two consecutive perfect squares and cannot itself be a perfect square. The upper bound is clear; the lower bound is proved by induction on $t \geq 7$, with cases $t = 3,4,5,6$ checked directly. \hfill\textbf{[2p]}

\textbf{Case II: $n$ odd.} Let $p$ be the largest odd prime dividing $n$. If $p = 3$, then $n = 3^k$ with $k \geq 2$, so $9 \mid n$. Then $2^9 + 1 \mid 2^n + 1$, and since $513 = 27 \cdot 19$, we get $19 \mid 2^n + 1$. If $2^n - n^2 + 1 = a^2$, then $n^2 + a^2 = 2^n + 1$, and since $19 \equiv 3 \pmod{4}$, we get $19 \mid n^2 + a^2$ implies $19 \mid n$, contradicting $n = 3^k$.

So $p > 3$. Since $2^p + 1 \equiv 1 \pmod{4}$, $3 \mid 2^p+1$, and $9 
mid 2^p+1$, there exists a prime $q > 3$, $q \equiv 3 \pmod{4}$, with $q \mid 2^p+1$. By the claim below $q > p$. Since $q \mid 2^p+1 \mid 2^n+1$, if $2^n-n^2+1 = a^2$, the same argument gives $q \mid n$, contradicting the maximality of $p$.

\textit{Claim:} If $p$ is prime, $p \geq 5$, every prime factor $q > 3$ of $2^p+1$ satisfies $q > p$.

\textit{Proof:} $2^p \equiv -1 \pmod{q}$ means $\text{ord}_q(2) = 2p$. By Fermat's little theorem $2p \mid q-1$, so $q > p$. \hfill\textbf{[4p]}",75,1419,Number Theory,9
352,tst_jbmo_ro_2025_3_p1,tst_jbmo,2025,g,"Let $ABC$ be a triangle with $\angle ABC = 2 \cdot \angle ACB$. Denote by $X$ and $Y$ the midpoints of arcs $AB$ and $BC$ of the circumcircle of triangle $ABC$ not containing $C$ and $A$, respectively. Let $BL$ be the angle bisector of $\angle ABC$, with $L \in AC$. Given that $\angle XLY = 90^\circ$, find the measures of the angles of triangle $ABC$.","\textbf{Solution.}
Since $BL$ bisects $\angle ABC$ and $\angle ABC = 2\angle ACB$, we have $\angle ABL = \angle LBC = \angle ACB$. From $\angle ACB = \angle LBC$ we get that $\triangle LBC$ is isosceles, so $LB = LC$ \quad(1). Since $Y$ is the midpoint of arc $\widehat{BC}$, $BY = CY$ \quad(2). \hfill\textbf{[1p]}

From (1) and (2), $\triangle LBY \equiv \triangle LCY$ (SSS), so $\angle BLY = \angle CLY$, meaning $LY$ bisects $\angle BLC$ \quad(3). Also $\widehat{AX} = \widehat{BX}$ implies $\angle ACX = \angle XCB$, so $XC$ bisects $\angle LCB$ \quad(4).

Let $I$ be the intersection of $LY$ and $CX$. From (3) and (4), $I$ is the incenter of $\triangle BLC$, so $BI$ bisects $\angle LBC$ \quad(5). \hfill\textbf{[2p]}

By hypothesis $\angle XLY = 90^\circ$, and $LY$ bisects $\angle BLC$ by (3), so $LX$ is the external bisector of $\angle BLC$ \quad(6). From (4) and (6), $X$ is the excenter of $\triangle BLC$ opposite $C$, so $XB$ is the external bisector of $\angle LBC$. Combined with (5), $XB \perp BI$, giving $\angle XBI = 90^\circ$ \quad(7). \hfill\textbf{[1p]}

Let $D$ be the second intersection of $BI$ with the circumcircle $\Gamma$, and let $\alpha = \angle IBC = \angle IBL$. Then $\angle ABL = \angle LBC = 2\alpha$, and $\angle LBC = \angle LCB = 2\alpha$, so $\angle LCX = \angle XCB = \alpha$.

We compute:
\[
\widehat{AD} = 2\angle ABD = 2(\angle ABL + \angle LBD) = 2(2\alpha + \alpha) = 6\alpha,
\]
and $\widehat{AX} = 2\angle ACX = 2\alpha$, so $\widehat{DX} = 8\alpha$.

On the other hand, $\widehat{DX} = 2\angle XBD = 180^\circ$ by (7). Combining: $8\alpha = 180^\circ$, so $\alpha = \dfrac{45^\circ}{2}$, giving $\angle ACB = 2\alpha = 45^\circ$, $\angle ABC = 4\alpha = 90^\circ$, and $\angle BAC = 45^\circ$. \hfill\textbf{[3p]}",353,1765,Geometry,10
353,tst_jbmo_ro_2025_3_p2,tst_jbmo,2025,a,"Let $n \geq 2$ be a natural number and $a_1, a_2, \ldots, a_n$ be real numbers. Define
\[
S = \sum_{1 \leq i < j \leq n} |a_j - a_i|
\quad\text{and}\quad
d = \max\{|a_j - a_i| \mid 1 \leq i \leq j \leq n\}.
\]
Prove that
\[
(n-1)d \leq S \leq \frac{n^2 d}{4},
\]
and state when equality holds in each inequality.","\textbf{Solution.}
WLOG assume $a_1 \leq a_2 \leq \cdots \leq a_n$ and set $d_k = a_{k+1} - a_k$ for $k = 1, \ldots, n-1$. Then:
\[
d_1 + d_2 + \cdots + d_{n-1} = d, \qquad |a_j - a_i| = d_i + d_{i+1} + \cdots + d_{j-1} \text{ for } i < j.
\]
Let $n_k$ be the number of pairs $(i,j)$ with $i < j$ such that $[a_k, a_{k+1}] \subset [a_i, a_j]$. Then $n_k = k(n-k)$, and:
\[
S = \sum_{k=1}^{n-1} n_k d_k = \sum_{k=1}^{n-1} k(n-k)\, d_k. \hfill\textbf{[2p]}
\]

\textbf{Left inequality:} For all $k \in \{1,\ldots,n-1\}$:
\[
k(n-k) \geq n-1 \iff (k-1)(n-k-1) \geq 0,
\]
which is always true. Hence $S \geq \sum_{k=1}^{n-1}(n-1)d_k = (n-1)d$. \hfill\textbf{[1p]}

Equality holds when $a_1 \leq a_2 = a_3 = \cdots = a_{n-1} \leq a_n$. \hfill\textbf{[1p]}

\textbf{Right inequality:} By AM-GM, $k(n-k) \leq \dfrac{n^2}{4}$, so:
\[
S = \sum_{k=1}^{n-1} k(n-k)\, d_k \leq \frac{n^2}{4} \sum_{k=1}^{n-1} d_k = \frac{n^2 d}{4}. \hfill\textbf{[2p]}
\]

Equality holds if and only if all $a_i$ are equal (for any $n \geq 2$), or if $n$ is even ($n = 2t$) and $a_1 = \cdots = a_t \leq a_{t+1} = \cdots = a_{2t}$ with $a_{t+1} - a_t = d$. If $n$ is odd and not all $a_i$ are equal, the right inequality is strict. \hfill\textbf{[1p]}",312,1219,Algebra,11
354,tst_jbmo_ro_2025_3_p3,tst_jbmo,2025,c,"Let $n \geq 2$ and consider a table with $n+1$ rows and $n$ columns, in which, on each of the first $n$ rows, Nicu\c{s}or writes the numbers $1, 2, \ldots, n$ in some order. Then he chooses a permutation $a_1, a_2, \ldots, a_n$ of $1, 2, \ldots, n$ and fills the last row as follows: for each $j \in \{1, 2, \ldots, n\}$, he writes in column $j$ the number of occurrences of $a_j$ in the cells at the intersection of column $j$ with rows $1, 2, \ldots, n$.

Find all $n$ for which Nicu\c{s}or can fill the table and choose $a_1, a_2, \ldots, a_n$ such that the last row also contains (in some order) the numbers $1, 2, \ldots, n$.","\textbf{Solution.}
We prove the answer is all $n \geq 4$.

For $n = 2$: regardless of how $a_1, a_2$ are chosen, the third row will have two equal values. For $n = 3$: assume WLOG $a_3 = 3$. Then column $3$ has only the value $3$. Since $(1,2)$ has only two permutations, two of the first three rows must be equal, so no $a_i$ can equal $1$ --- contradiction. \hfill\textbf{[2p]}

For $n = 4$, taking $a_1=1,a_2=2,a_3=3,a_4=4$, the following construction works:
\[
\begin{array}{|c|c|c|c|}
\hline
1 & 2 & 3 & 4 \\\hline
2 & 1 & 3 & 4 \\\hline
2 & 1 & 3 & 4 \\\hline
3 & 2 & 1 & 4 \\\hline
1 & 2 & 3 & 4 \\\hline
\end{array}
\] \hfill\textbf{[1p]}

For $n \geq 4$, we prove by induction that a valid construction exists with $a_i = i$ for all $i$ and last row $1, 2, \ldots, n$.

\textit{Inductive step:} Given a valid $(n+1) \times n$ table $T_n$, construct a valid $(n+2)\times(n+1)$ table $T_{n+1}$ by appending a column at the right and a new row between the last and second-to-last rows of $T_n$. Fill all cells of the new rightmost column with $n+1$, and fill the first $n$ cells of the new row with $2, 3, \ldots, n, 1$ (any permutation consistent with $a_1,\ldots,a_n$). \hfill\textbf{[1p + 2p + 1p]}",630,1207,Combinatorics,12
355,tst_jbmo_ro_2025_3_p4,tst_jbmo,2025,n,"Find all positive integers $a, b, c$ such that $ab+c$, $bc+a$, and $ca+b$ are all powers of $2$.","\textbf{Solution.}
We prove that the only solution families are:
\[
(1,\; 1,\; 2^x - 1) \quad \text{and} \quad (1,\; 2^x-1,\; 2^x+1),
\]
for any positive integer $x$ (and permutations thereof).

\textbf{Step I: $a, b, c$ are odd and pairwise coprime.}

It is easy to see $a \equiv b \equiv c \pmod{2}$. If all are even, let $d \leq e \leq f$ be the 2-adic valuations of $a, b, c$. Then $v_2(bc + a) = d$, so $bc + a = 2^d \leq a$, a contradiction. Hence $a, b, c$ are odd.

If $a, b$ share an odd prime $p$, then $p \mid ac + b$, so $ac+b$ is not a power of $2$. Similarly $(a,c) = (b,c) = 1$. \hfill\textbf{[2p]}

\textbf{Step II: $a = 1$.}

Assume $1 < a \leq b \leq c$. Write $ab+c = 2^k$, $ac+b = 2^m$, $bc+a = 2^n$ with $3 \leq k < m < n$. Then:
\[
ab \equiv -c \pmod{2^k},\quad ac \equiv -b \pmod{2^k},\quad bc \equiv -a \pmod{2^k}.
\]
Multiplying all three: $a^2b^2c^2 \equiv -1 \pmod{2^k}$, so $abc \equiv -1 \pmod{2^k}$. Multiplying the first congruence by $c$ gives $a^2 \equiv 1\pmod{2^k}$, and similarly $b^2 \equiv c^2 \equiv 1 \pmod{2^k}$. Since $\gcd(x-1, x+1) = 2$ for odd $x$, we get $a,b,c \geq 2^{k-1}-1$. Thus:
\[
2^k = ab + c \geq (2^{k-1}-1)^2 + 2^{k-1}-1 = 2^{2k-2} - 2^{k-1},
\]
giving $2 \geq 2^{k-1} - 1 \geq 3$ for $k > 3$, a contradiction. So $a = 1$. \hfill\textbf{[4p]}

\textbf{Solutions.} If $b = 1$, then $c = 2^x - 1$. If $a = 1 < b < c$: from Step II, $b, c \geq 2^{k-1}-1$ and $b + c = 2^k$ with $b, c$ odd, so $b = 2^{k-1}-1$ and $c = 2^{k-1}+1$. \hfill\textbf{[1p]}",96,1503,Number Theory,13
356,tst_jbmo_ro_2025_4_p1,tst_jbmo,2025,g,"Circles $\Gamma_1$, $\Gamma_2$, $\Gamma_3$ in the plane are pairwise externally tangent. Let $P_2$ be the point of tangency between circles $\Gamma_1$ and $\Gamma_3$, and $P_1$ be the point of tangency between circles $\Gamma_2$ and $\Gamma_3$. Consider points $A$ and $B$ on circle $\Gamma_3$, diametrically opposite, such that $ABP_1P_2$ is a convex quadrilateral.

Line $AP_2$ meets circle $\Gamma_1$ again at point $X$, line $BP_1$ meets circle $\Gamma_2$ again at point $Y$, and lines $AP_1$ and $BP_2$ meet at point $Z$.

Prove that points $X$, $Y$, and $Z$ are collinear.","\textbf{Solution.}
Let $\{P_3\} = \Gamma_1 \cap \Gamma_2$ and let $O_1, O_2, O_3$ be the centers of circles $\Gamma_1, \Gamma_2, \Gamma_3$ respectively. Denote by $O_4$ the intersection of the common external tangents to $\Gamma_1$ and $\Gamma_2$. Then $O_4P_2 = O_4P_3$ (tangents to $\Gamma_1$) and $O_4P_3 = O_4P_1$ (tangents to $\Gamma_2$), so $O_4P_1 = O_4P_2 = O_4P_3$. Since $O_4P_1 \perp O_2O_3$, $O_4P_2 \perp O_3O_1$, and $O_4P_3 \perp O_1O_2$, we deduce that $O_4$ is the incenter of triangle $O_1O_2O_3$. \hfill\textbf{[1p]}

We have:
\[
\widehat{P_1ZP_2} = \frac{1}{2}\left(\widehat{AB} + \widehat{P_1P_2}
ight)
= \frac{1}{2}\left(180^\circ + \widehat{P_1O_3P_2}
ight)
= \frac{1}{2}\left(180^\circ + 180^\circ - \widehat{P_1O_2P_3} - \widehat{P_2O_1P_3}
ight)
\]
\[
= \frac{1}{2}\left(180^\circ - \widehat{P_1O_2P_3}
ight) + \frac{1}{2}\left(180^\circ - \widehat{P_2O_1P_3}
ight)
= \widehat{O_1P_3P_2} + \widehat{O_2P_3P_1} = 180^\circ - \widehat{P_1P_3P_2},
\]
so quadrilateral $ZP_1P_3P_2$ is cyclic. \hfill\textbf{[2p]}

We show that $X$, $P_3$, and $Z$ are collinear. It suffices to show $\widehat{XP_3O_1} = \widehat{ZP_3O_2}$.

Since triangle $O_1XP_3$ is isosceles and $BP_2 \perp AX$ ($AB$ is a diameter), we have:
\[
\widehat{XP_3O_1} = \frac{1}{2}\left(180^\circ - \widehat{XO_1P_3}
ight)
= 90^\circ - \frac{1}{2}\widehat{XP_3}
= \widehat{XP_2B} - \widehat{XP_2P_3}
= \widehat{BP_2P_3} = \widehat{ZP_2P_3}
= \frac{1}{2}\widehat{ZP_1P_3}.
\]
Since $O_1O_2$ is tangent to the circumcircle of triangle $P_1P_2P_3$, we have $\widehat{P_1P_3O_2} = \frac{1}{2}\widehat{P_1P_3}$, so
\[
\widehat{ZP_3O_2} = \widehat{ZP_3P_1} + \widehat{P_1P_3O_2}
= \frac{1}{2}\widehat{ZP_1} + \frac{1}{2}\widehat{P_1P_3}
= \frac{1}{2}\widehat{ZP_1P_3}
= \widehat{XP_3O_1}. \hfill\textbf{[3p]}
\]
Therefore, points $X$, $Z$, $P_3$ are collinear. By symmetry, $Y$, $Z$, $P_3$ are also collinear, hence $X$, $Y$, $Z$ are collinear. \hfill\textbf{[1p]}",578,1946,Geometry,14
357,tst_jbmo_ro_2025_4_p2,tst_jbmo,2025,n,"Let $k, n \in \mathbb{N}^*$ and let $A$ be an arbitrary set of $2n$ positive integers. For each $a \in A$, denote by $p_a$ the number of elements of the set $P_a = \{b \in A \mid b - a \equiv k \pmod{2k}\}$.

Find the maximum value of the sum
\[
E_A = \sum_{a \in A} p_a^2,
\]
as $A$ ranges over all subsets of $\mathbb{N}^*$ with $2n$ elements.","\textbf{Solution.}
Denote $A_i = \{a \in A \mid a \equiv i \pmod{2k}\}$ and let $a_i = |A_i|$ for $i = 0, \ldots, 2k-1$. Note that $A = \bigcup_{i=0}^{2k-1} A_i$, these sets are pairwise disjoint, so $\sum_{i=0}^{2k-1} a_i = 2n$. For $a \in A_i$:
\[
P_a = \{b \in A \mid b - a \equiv k \pmod{2k}\} = \{b \in A \mid b \equiv k + i \pmod{2k}\} = A_{k+i},
\]
(indices mod $2k$). So $p_a = a_{k+i}$ for all $a \in A_i$. Therefore:
\[
E_A = \sum_{a \in A} p_a^2 = \sum_{i=0}^{2k-1} \sum_{a \in A_i} p_a^2
= \sum_{i=0}^{2k-1} a_i a_{k+i}^2
= \sum_{i=0}^{k-1} a_i a_{k+i}^2 + \sum_{i=k}^{2k-1} a_i a_{k+i}^2.
\]
Re-indexing the second sum gives $\sum_{i=k}^{2k-1} a_i a_{k+i}^2 = \sum_{i=0}^{k-1} a_{k+i} a_i^2$, hence:
\[
E_A = \sum_{i=0}^{k-1} a_i a_{k+i}(a_i + a_{k+i}). \hfill\textbf{[4p]}
\]
We claim that replacing two distinct pairs $(a_i, a_{i+k})$ and $(a_j, a_{j+k})$ by $(a_i + a_j,\, a_{i+k} + a_{j+k})$ and $(0,0)$ does not decrease $E_A$, since:
\[
(a_i + a_j)(a_{i+k} + a_{j+k})(a_i + a_j + a_{i+k} + a_{j+k})
\geq a_i a_{i+k}(a_i + a_{i+k}) + a_j a_{j+k}(a_j + a_{j+k}). \hfill\textbf{[2p]}
\]
Thus the maximum is achieved when only one pair $(a_i, a_{i+k})$ is nonzero, with $a_i + a_{i+k} = 2n$. By the AM-GM inequality:
\[
E_A = a_i \cdot a_{i+k} \cdot (a_i + a_{i+k}) \leq \frac{(a_i + a_{i+k})^3}{4} = \frac{(2n)^3}{4} = 2n^3.
\]
The maximum $2n^3$ is achieved, for example, by any set $A$ containing $n$ multiples of $2k$ and $n$ numbers congruent to $k \pmod{2k}$. \hfill\textbf{[1p]}",345,1500,Number Theory,15
358,tst_jbmo_ro_2025_4_p3,tst_jbmo,2025,a,"Let $n \in \mathbb{N}$, $n \geq 2$, and let $x_1, x_2, \ldots, x_n$ be positive real numbers with sum equal to $1$. Denote by $m$ and $M$ the minimum and maximum, respectively, among the numbers:
\[
\frac{x_1}{1+x_1},\quad \frac{x_2}{1+x_1+x_2},\quad \ldots,\quad \frac{x_n}{1+x_1+\cdots+x_n}.
\]
Find:
\begin{enumerate}[label=\alph*)]
  \item the largest possible value of $m$;
  \item the smallest possible value of $M$,
\end{enumerate}
over all possible choices of $x_1, x_2, \ldots, x_n$.","\textbf{Solution.}
Set $a_k = \dfrac{x_k}{1+x_1+\cdots+x_k}$ and $b_k = 1 - a_k = \dfrac{1+x_1+\cdots+x_{k-1}}{1+x_1+\cdots+x_k}$ for $1 \leq k \leq n$.

\textbf{Part a)} By AM-GM:
\[
b_1 + b_2 + \cdots + b_n \geq n\sqrt[n]{b_1 b_2 \cdots b_n} = n\sqrt[n]{\frac{1}{1+x_1+\cdots+x_n}} = \frac{n}{\sqrt[n]{2}}.
\]
Then:
\[
n \cdot m \leq a_1 + \cdots + a_n = n - (b_1 + \cdots + b_n) \leq n\!\left(1 - \frac{1}{\sqrt[n]{2}}
ight),
\]
so $m \leq 1 - \dfrac{1}{\sqrt[n]{2}}$. \hfill\textbf{[2p]}

We show there exists a sequence $(x_1, \ldots, x_n)$ of positive reals summing to $1$ with $a_1 = a_2 = \cdots = a_n = 1 - \dfrac{1}{\sqrt[n]{2}} =: m_0$. The condition $a_1 = a_i$ for each $i$ gives $x_i = x_1(1+x_1)^{i-1}$ by induction. Summing to $1$ yields $(1+x_1)^n = 2$, so $x_1 = \sqrt[n]{2} - 1$ and:
\[
x_k = \sqrt[n]{2}^{\,k-1}\!\left(\sqrt[n]{2}-1
ight), \quad k = 1, \ldots, n. \hfill\textbf{[1p (A)]}
\]

\textbf{Part b)} We show the minimum of $M$ is also $M_0 = 1 - \dfrac{1}{\sqrt[n]{2}}$, achieved when $a_1 = a_2 = \cdots = a_n$, i.e.\ exactly when $x_k = \sqrt[n]{2}^{\,k-1}(\sqrt[n]{2}-1)$ for all $k$. \hfill\textbf{[1p (B)]}

Let $\alpha_k = \sqrt[n]{2}^{\,k-1}(\sqrt[n]{2}-1)$. Consider any sequence $(x_1, \ldots, x_n) 
eq (\alpha_1, \ldots, \alpha_n)$ summing to $1$. There exists a smallest $k_0 \in \{1,\ldots,n\}$ with $\alpha_{k_0} < x_{k_0}$; thus $x_i \leq \alpha_i$ for $i < k_0$. Then:
\[
a_{k_0} = \frac{x_{k_0}}{1+x_1+\cdots+x_{k_0}}
= \frac{1}{\dfrac{1+x_1+\cdots+x_{k_0-1}}{x_{k_0}}+1}
> \frac{1}{\dfrac{1+\alpha_1+\cdots+\alpha_{k_0-1}}{\alpha_{k_0}}+1}
= \frac{\alpha_{k_0}}{1+\alpha_1+\cdots+\alpha_{k_0}} = M_0,
\]
so $M = \max_k a_k \geq a_{k_0} > M_0$.

In conclusion, $M_0 = 1 - \dfrac{1}{\sqrt[n]{2}}$ is the minimum sought. \hfill\textbf{[4p]}",492,1783,Algebra,16
359,tst_jbmo_ro_2025_4_p4,tst_jbmo,2025,c,"Find the smallest positive integer $k$ for which there exists a set $A$ of $10$ points in the plane, no three collinear, with the following property: we can color the $45$ segments with endpoints in $A$ using $k$ colors, such that any two segments of the same color intersect either in their interior or at an endpoint.","\textbf{Solution.}
We prove $k = 6$ by constructing a configuration achieving $k = 6$ (Step 1) and showing $k \geq 6$ (Step 2).

\textbf{Step 1.} Consider a regular $10$-gon and color its vertices alternately with 2 colors: red and blue. For each red vertex $v$, take the $9$ segments containing it, together with the diagonal formed by the two adjacent blue vertices. Any two of these segments intersect (in interior or at endpoints), so they can receive the same color. This way we use $5$ colors to cover all segments except the diagonals of the blue pentagon. Since those $5$ diagonals pairwise intersect, we color them with the $6$th color. \hfill\textbf{[2p]}

\textbf{Step 2.} We show $k \geq 6$ using the lemma below. Given $A$ as in the problem, construct a line $\ell_0$ not parallel to any of the $45$ segments. Order the $10$ points by distance to $\ell_0$; find $\ell \parallel \ell_0$ with exactly $5$ points on each side. The $\binom{5}{2} = 10$ segments on one side require at least $3$ colors by the lemma. No segment on one side can share a color with a segment on the other side (they do not intersect), so we need at least $6$ colors total. \hfill\textbf{[2p]}

\textbf{Lemma.} \textit{Given $5$ points in the plane, no three collinear, we cannot color the $10$ segments using $2$ colors so that same-color segments always intersect.}

\textit{Proof.} Assume for contradiction this is possible with colors red and blue. We first show any $4$ of the points determine a monochromatic triangle. The $4$ points form either a convex or concave quadrilateral; label them $A,B,C,D$ so $AB,CD$ and $AD,BC$ are non-intersecting pairs. Assume WLOG $AB,BC$ are red and $CD,AD$ are blue. Then the color of $AC$ (red or blue) forces a monochromatic triangle.

Next, no two distinct monochromatic triangles can share the same color. If there were two red triangles, one being $ABC$, every red segment would have an endpoint in $\{A,B,C\}$, so WLOG $ABD$ is the other. Then pairs $AD,BC$ and $AC,BD$ must intersect --- a contradiction.

Thus the $5$ monochromatic triangles determined by the $5$ subsets of size $4$ must all be the same triangle, meaning the same $3$ points appear in every $4$-element subset --- a contradiction. \hfill\textbf{[3p]}",319,2255,Combinatorics,17