question stringlengths 12 7.47k | solution stringlengths 0 13.5k | openr1_source stringclasses 8
values | id stringlengths 3 36 ⌀ | dataset stringclasses 2
values | choices listlengths 1 1 |
|---|---|---|---|---|---|
Use the method of contradiction to prove that the correct assumption for the proposition "At least one of the internal angles of a triangle is not greater than 60°" is ( ):
A: Assume that all three internal angles are not greater than 60°
B: Assume that all three internal angles are greater than 60°
C: Assume that at... | Proof: To prove the proposition "At least one of the internal angles of a triangle is not greater than 60°" using the method of contradiction, we should assume the negation of the proposition is true. The negation of the proposition "At least one of the internal angles of a triangle is not greater than 60°" is: All thr... | cn_k12 | 3519db43-ce27-5d63-ab64-e971807bee90 | open-r1/OpenR1-Math-220k | [
""
] |
I4.3 In Figure 1 , there are eight positive numbers in series. Starting from the $3^{\text {rd }}$ number, each number is the product of the previous two numbers. Given that the $5^{\text {th }}$ number is $\frac{1}{r}$ and the $8^{\text {th }}$ number is $\frac{1}{r^{4}}$. If the first number is $s$, find the value of... | Let the $2^{\text {nd }}$ number be $a$, then other numbers are as follows:
$s$
$s^{2} a^{3}=\frac{1}{r}$ $\square$
$\square$ $s^{8} a^{13}=\frac{1}{r^{4}}$
$s^{8} a^{13}=\frac{1}{r^{4}}$
$s^{2} a^{3}=\frac{1}{r}$ $\qquad$
$s^{8} a^{13}=\frac{1}{r^{4}}$ $\qquad$
$(2) \div(1)^{4}: a=1$
Sub. $a=1$ into (1): $s^{2}=\frac{... | olympiads | 4b050bc9-e2a8-58b8-9aeb-26d072283ca6 | open-r1/OpenR1-Math-220k | [
""
] |
Continuously toss a fair coin 3 times. Given that there is at least one head facing up, the probability of exactly one tail facing up is ______. | To solve this problem, we first consider the total number of outcomes when tossing a fair coin 3 times. There are $2^3 = 8$ possible outcomes, which are: HHH, HHT, HTH, HTT, THH, THT, TTH, and TTT, where H represents heads and T represents tails.
Given that there is at least one head facing up, we exclude the outcome ... | cn_k12 | 379190e0-6dde-5982-9032-2068456c218b | open-r1/OpenR1-Math-220k | [
""
] |
Solve the equations:$(1)\frac{1}{x-2}+3=\frac{1-x}{2-x}$.$(2)\frac{x}{x-1}-1=\frac{3}{{x}^{2}-1}$. | ### Problem 1 Solution:
Given: $\frac{1}{x-2}+3=\frac{1-x}{2-x}$
Step 1: Multiply both sides by $(x-2)(2-x)$ to eliminate the denominators:
\[ (x-2)(2-x) \left( \frac{1}{x-2}+3 \right) = (x-2)(2-x) \left( \frac{1-x}{2-x} \right) \]
Step 2: Simplify both sides:
\[ (2-x)(1+3(x-2)) = (x-2)(x-1) \]
\[ -1-3(x-2) = x-1 \]... | cn_k12 | 14558a12-7864-5a7d-af16-866b5b371b92 | open-r1/OpenR1-Math-220k | [
""
] |
A company has newly recruited 8 employees, who are to be evenly distributed between two departments, A and B. Among them, two English translators cannot be in the same department, and likewise, the three computer programmers cannot all be in the same department. The total number of different distribution schemes is ( ... | Since the 8 employees are to be evenly distributed between the two departments, A and B, and considering the constraints that the two English translators cannot be in the same department, and the three computer programmers cannot all be in the same department, we prioritize the special elements and complete the distrib... | cn_k12 | 85863b9c-e84a-575f-9ff3-87863aaccf6d | open-r1/OpenR1-Math-220k | [
""
] |
One afternoon at the park there were twice as many dogs as there were people, and there were twice as many people as there were snakes. The sum of the number of eyes plus the number of legs on all of these dogs, people, and snakes was $510$. Find the number of dogs that were at the park. | 1. Let \( D \) be the number of dogs, \( P \) be the number of people, and \( S \) be the number of snakes.
2. According to the problem, we have the following relationships:
\[
D = 2P \quad \text{(twice as many dogs as people)}
\]
\[
P = 2S \quad \text{(twice as many people as snakes)}
\]
3. Each dog ... | aops_forum | 0f20d545-64b5-5f92-9619-1d26eae06f2b | open-r1/OpenR1-Math-220k | [
""
] |
Example 8 Let $R_{n}=\frac{1}{2}\left(a^{n}+b^{n}\right), a=3+2 \sqrt{2}, b=3-2 \sqrt{2}, n=1,2, \cdots$ What is the units digit of $R_{12345}$?
(41st AHSME Problem) | Given that $a$ and $b$ are the roots of the equation $x^{2}=6 x-1$, according to Newton's formula, we have
$$
R_{n+2}=6 R_{n+1}-R_{n} \text {. }
$$
It is easy to see that $R_{1}=\frac{1}{2}(a+b)=3, R_{2}=\frac{1}{2}\left(a^{2}+b^{2}\right)=17$. For convenience in writing, let $D_{n}$ represent the unit digit of $R_{n}... | olympiads | 1a7f3560-a003-5ff8-ad1b-d4f8e67862ef | open-r1/OpenR1-Math-220k | [
""
] |
Calculate the inverse of 36 modulo 101.
Theorem 2 (Chinese Remainder Theorem (CRT)).
Let $m_{1}, m_{2} \in$ be two coprime integers. Let $r_{1}, r_{2} \in \mathbb{N}$. There exists a unique $r \in$ $\left\{0, \ldots, m_{1} m_{2}-1\right\}$ such that:
$$
x \equiv r_{1}\left[m_{1}\right] \text { and } x \equiv r_{2}\l... | $$
\begin{aligned}
0 \cdot 36+1 \cdot 101 & =101 \\
1 \cdot 36+0 \cdot 101 & =36 \\
-2 \cdot 36+1 \cdot 101 & =29 \\
3 \cdot 36-1 \cdot 101 & =7 \\
-14 \cdot 36+5 \cdot 101 & =1
\end{aligned}
$$
The inverse of 36 modulo 101 is therefore $-14 \equiv 87[101]$. | olympiads | 83e4a294-2ab7-5f8f-96ee-71477e51482c | open-r1/OpenR1-Math-220k | [
""
] |
Two cyclists: $A$ and $B$ start from two locations, $M$ and $N$, heading towards $P$ at the same time. The distance $NP$ is $30 \mathrm{~km}$ longer than the distance $MP$. Each cyclist takes 10 hours to complete their journey.
What is the distance $MP$, if for every $42 \mathrm{~km}$ that $B$ travels, it takes $B$ 20... | Let the sought $M P=x \mathrm{~km}$. For a $1 \mathrm{~km}$ distance, it takes $A \frac{10}{x}$ hours and $B \frac{10}{x+30}$ hours. For 42 $\mathrm{km}$, $A$ takes $\frac{420}{x}$ hours, and $B$ takes $\frac{420}{x+30}$ hours, which, according to the problem, is $\frac{1}{3}$ hour less than the former. Therefore,
$$
... | olympiads | f8df4488-ec83-56df-84a7-17086b2bbc74 | open-r1/OpenR1-Math-220k | [
""
] |
The "One Helmet, One Belt" safety protection campaign is a safety protection campaign launched by the Ministry of Public Security nationwide. It is also an important standard for creating a civilized city and being a civilized citizen. "One helmet" refers to a safety helmet. Drivers and passengers of electric bicycles ... | ### Step-by-Step Solution
#### Part 1: Cost of Each Helmet Type
Let's denote the cost of purchasing 1 type A helmet as $x$ dollars and 1 type B helmet as $y$ dollars.
Given the information, we can set up the following system of equations:
1. For 8 type A helmets and 6 type B helmets, the cost is $630: 8x + 6y = 6... | cn_k12 | 50c86bec-bf9d-55e4-99c7-1f3ed0a0025d | open-r1/OpenR1-Math-220k | [
""
] |
To obtain the graph of the function $y=\sin \left(3x- \frac{\pi}{3}\right)$, one only needs to shift the graph of the function $y=\sin 3x$ by ( )
A: shifting it to the right by $\frac{\pi}{9}$ units
B: shifting it to the left by $\frac{\pi}{9}$ units
C: shifting it to the right by $\frac{\pi}{3}$ units
D: shifting i... | **Analysis**
This question mainly examines the transformation rules of the graph of the function $y=A\sin (\omega x+\varphi)$, which is a basic question.
**Solution**
Solution: By shifting the graph of the function $y=\sin 3x$ to the right by $\frac{\pi}{9}$ units, we can obtain the graph of $y=\sin 3(x- \frac{\pi}{... | cn_k12 | 47c194ab-1ea0-5920-a8af-14782c8f8e7a | open-r1/OpenR1-Math-220k | [
""
] |
4. Let the set $A=\left\{\frac{1}{2}, \frac{1}{7}, \frac{1}{11}, \frac{1}{13}, \frac{1}{15}, \frac{1}{32}\right\}$ have non-empty subsets $A_{1}, A_{2}, \cdots, A_{63}$. Denote the product of all elements in the set $A_{i}$ as $p_{i}$ (the product of elements in a singleton set is the element itself), $i=1,2, \cdots, 6... | Rongge $\frac{79}{65}$.
Analysis $\left(\frac{1}{2}+1\right)\left(\frac{1}{7}+1\right)\left(\frac{1}{11}+1\right)\left(\frac{1}{13}+1\right)\left(\frac{1}{15}+1\right)\left(\frac{1}{32}+1\right)-1=\frac{79}{65}$. | olympiads | 376043a5-aa7e-554a-b379-236469cb547f | open-r1/OpenR1-Math-220k | [
""
] |
3. Find the period length of the sequence of pseudo-random numbers generated by the linear congruential method with $x_{0}=2$ and $x_{n+1} \equiv 4 x_{n}+7(\bmod 25)$. | 3. 10 | number_theory | c4993657-b442-5c43-8c84-e2e0de09a097 | open-r1/OpenR1-Math-220k | [
""
] |
16. In a mentoring program at a school, several first-grade students can become friends with one sixth-grade student, but a sixth-grade student cannot be friends with multiple first-grade students. It is known that $\frac{1}{3}$ of the sixth-grade students and $\frac{2}{5}$ of the first-grade students become friends. T... | 16. B.
Let there be $x$ people in the first grade and $y$ people in the sixth grade. Then $\frac{1}{3} x=\frac{2}{5} y \Rightarrow x=\frac{6}{5} y$.
Thus, the required value is $\frac{\frac{1}{3} x+\frac{2}{5} y}{x+y}=\frac{4}{11}$. | olympiads | 08159e44-1d39-56fd-be8d-92c64ca50a0a | open-r1/OpenR1-Math-220k | [
""
] |
Example 9. Solve the inequality
$$
\frac{\lg ^{2} x-3 \lg x+3}{\lg x-1}<1
$$ | Solution. Let $y=\lg x$, then the given inequality takes the form
$$
\frac{y^{2}-3 y+3}{y-1}<1
$$
This inequality is equivalent to the inequality
$$
\frac{y^{2}-3 y+3}{y-1}-1<0 \Leftrightarrow \frac{y^{2}-4 y+4}{y-1}<0
$$
i.e., the inequality
$$
\frac{(y-2)^{2}}{y-1}<0
$$
The solution to the last inequality is th... | olympiads | 8b251bf6-1f8c-500b-b93d-46e195b0583e | open-r1/OpenR1-Math-220k | [
""
] |
1.64 Given that for any integer $k \neq 27$, the number $a-k^{1964}$ is divisible by $27-k$. Find $a$.
(27th Moscow Mathematical Olympiad, 1964) | ［Solution］ Let $27-k=m$, then $k=27-m$.
$$
\begin{aligned}
& \frac{a-k^{1964}}{27-k} \\
= & \frac{a-(27-m)^{1964}}{m} \\
= & \frac{a-27^{1964}}{m}+C_{1964}^{1} 27^{1963}-C_{1964}^{2} 27^{1962} m+\cdots-m^{1963} .
\end{aligned}
$$
Since all terms from the second term onward in the above expression are integers, it must... | olympiads | 83862258-85c6-5d13-b382-f334478bb240 | open-r1/OpenR1-Math-220k | [
""
] |
Three, (50 points) $n$ people exchange greetings by phone during a holiday. It is known that each person makes calls to at most three friends, any two people make at most one call to each other, and among any three people, at least two of them, one person has called the other. Find the maximum value of $n$.
| Three, first prove the lemma.
Lemma In a simple graph $G$ of order $n$ without $K_{3}$, then
$$
f(G) \leqslant\left[\frac{n^{2}}{4}\right],
$$
where $f(G)$ represents the number of edges in $G$.
Proof of the lemma: Let $A$ be the vertex with the maximum degree, and let the set of vertices adjacent to $A$ be $M=\left\{... | cn_contest | cb4a72c8-e33c-5a79-89d2-59ec220b1f29 | open-r1/OpenR1-Math-220k | [
""
] |
6. As shown in Figure 1, the cube $A B C D-E F G H$ has an edge length of 1, and point $M$ is on the surface $E F G H$ of the cube. Define a path as a route where every point lies on the surface of the cube. It is known that the shortest path length from point $M$ to $A$, $l(M, A)$, is equal to the shortest path length... | 6. $\frac{5 \sqrt{2}}{6}$.
By symmetry, without loss of generality, assume point $M$ is inside $\triangle E F G$ (including the boundary).
Unfold the square $A B F E$ and the square $E F G H$ into a plane, as shown in Figure 3. Then point $M$ lies on the perpendicular bisector of $A G$.
When $M$ is the intersection ... | olympiads | f226e8f7-5368-5d78-81af-bd9fa42372ca | open-r1/OpenR1-Math-220k | [
""
] |
Example 4. Solve the equation
$$
\left|\frac{x^{2}-6 \sqrt{x}+7}{x^{2}+6 \sqrt{x}+7}\right|=1
$$ | Solution. The given equation is equivalent to the system of equations
$\left[\begin{array}{l}\frac{x^{2}-6 \sqrt{x}+7}{x^{2}+6 \sqrt{x}+7}=1, \\ \frac{x^{2}-6 \sqrt{x}+7}{x^{2}+6 \sqrt{x}+7}=-1\end{array} \Leftrightarrow\right.$
$$
\Leftrightarrow\left[\begin{array}{l}
\frac{-12 \sqrt{x}}{x^{2}+6 \sqrt{x}+7}=0 \\
\fr... | olympiads | 89fd9391-a5a7-577e-819d-aebbbb116ad3 | open-r1/OpenR1-Math-220k | [
""
] |
In a bag, there are 100 balls of the same size, including red, white, and black balls. There are 45 red balls. If the probability of drawing a white ball from the bag is 0.23, then the probability of drawing a black ball is ( )
A: 0.32
B: 0.07
C: 0.64
D: 0.45 | Given that there are 100 balls of the same size in the bag, including red, white, and black balls, and the probability of drawing a white ball is 0.23,
Therefore, the number of white balls in the bag is $0.23 \times 100 = 23$.
Since there are 45 red balls, the number of black balls is $100 - 45 - 23 = 32$.
The pr... | cn_k12 | acd38dbd-587b-500e-a609-531058873d92 | open-r1/OpenR1-Math-220k | [
""
] |
6. There are 100 equally divided points on a circle. The number of obtuse triangles formed by these points as vertices is $\qquad$ . | $6.50 \times 49 \times 48$ (or 117600).
Let $A_{t}$ be the vertex of the obtuse angle. $\angle A_{\text{s}}$
intercepts an arc of $x$ equal parts, and the other two angles intercept arcs of $y, z$ equal parts, respectively, such that
$$
\left\{\begin{array}{l}
x+y+z=100, \\
x \geqslant 51, y \geqslant 1, z \geqslant 1 ... | cn_contest | 8e9279e0-89f0-5598-aa95-dfca7d059d58 | open-r1/OpenR1-Math-220k | [
""
] |
Andover has a special weather forecast this week. On Monday, there is a $\frac{1}{2}$ chance of rain. On Tuesday, there is a $\frac{1}{3}$ chance of rain. This pattern continues all the way to Sunday, when there is a $\frac{1}{8}$ chance of rain. The probability that it doesn't rain in Andover all week can be expressed... | 1. The probability that it doesn't rain on a given day is the complement of the probability that it rains. For Monday, the probability that it doesn't rain is:
\[
1 - \frac{1}{2} = \frac{1}{2}
\]
2. For Tuesday, the probability that it doesn't rain is:
\[
1 - \frac{1}{3} = \frac{2}{3}
\]
3. This pat... | aops_forum | a7ce9095-1de9-58f7-9873-ad3b4e27aa4e | open-r1/OpenR1-Math-220k | [
""
] |
Two pieces of bread are needed for one regular sandwich. 3 pieces of bread are needed for a double meat sandwich. How many pieces of bread are needed for 14 regular sandwiches and 12 double meat sandwiches? | Regular = 2 * 14 = <<2*14=28>>28 pieces
Double = 3 * 12 = <<3*12=36>>36 pieces
28 + 36 = <<28+36=64>>64 pieces
64 pieces of bread are needed for those sandwiches.
#### 64 | null | null | openai/gsm8k | [
""
] |
Given an ellipse C: $$\frac {x^{2}}{a^{2}}+ \frac {y^{2}}{b^{2}}=1$$ ($a>b>0$) with its left and right foci as $F_1$ and $F_2$ respectively. A line passing through point $F_1$ intersects the ellipse at points A and B. The minimum value of $|AB|$ is 3, and the perimeter of $\triangle ABF_2$ is 8.
1. Find the equation of... | Solution:
1. Since AB is a chord passing through focus $F_1$, $|AB|$ is minimized when $AB \perp x$-axis, and the minimum value is $$\frac {2b^{2}}{a}$$.
According to the problem, $$\frac {2b^{2}}{a}=3$$. From the definition of an ellipse, the perimeter of $\triangle ABF_2$ is $4a$, so $$a=2, b= \sqrt {3}$$.
Therefore,... | cn_k12 | ecef9603-9bea-5ba0-af49-db23495d1ac0 | open-r1/OpenR1-Math-220k | [
""
] |
$35 \cdot 3$ A domino is composed of two squares, one black and one white, arranged like this: 民. Among the following chessboards, which one cannot be completely covered without repetition using only this type of domino?
(A) $3 \times 4$.
(B) $3 \times 5$.
(C) $4 \times 4$.
(D) $4 \times 5$.
(E) $6 \times 3$. | ［Solution］Dominoes consist of one white square and one black square, so the figures they can completely cover should have the same number of black and white squares.
However, in (B), the number of black and white squares is different.
Of course, we can also consider the parity of the number of small squares, $(B)$ is t... | olympiads | 4ed445bc-293d-5d08-9082-464c5487475f | open-r1/OpenR1-Math-220k | [
""
] |
Three. (20 points) As shown in Figure 2, $\triangle A O B$ is an equilateral triangle, and the coordinates of point $B$ are $(2,0)$. A line $l$ is drawn through point $C$ $(-2,0)$, intersecting $A O$ at $D$ and $A B$ at $E$, such that the areas of $\triangle A D E$ and $\triangle D C O$ are equal. Find the function exp... | Three, $\because \triangle A D E$ and $\triangle D C O$ have equal areas,
$\therefore \angle A O B$ and $\triangle C B E$ have equal areas.
The area of $\triangle A O B$ is $\sqrt{3}$. Let the coordinates of point $E$ be $\left(x_{0}, y_{0}\right)$, then the area of $\triangle C B E$ is $2 y_{0}$. From $2 y_{0}=\sqrt{3... | cn_contest | 58bbf99b-f7ab-5c5c-b866-3bb7e7b7b227 | open-r1/OpenR1-Math-220k | [
""
] |
Use the comparison method to prove the following inequality for $x, y \in \mathbb{R}, x \neq y$:
$$x^4 + y^4 > x^3y + xy^3.$$ | **Answer**: To prove $x^4 + y^4 > x^3y + xy^3$, according to the comparison method, it suffices to show that $(a^4 + b^4) - (a^3b + ab^3) > 0$, which can be proven by factorization.
Proof: $(a^4 + b^4) - (a^3b + ab^3) = (a^5 - a^3b^2) + (b^5 - a^2b^3)$
$= a^3(a^2 - b^2) + b^3(a^2 - b^2) = (a^2 - b^2)(a^3 - b^3)$
$... | cn_k12 | 1d9c97e0-ec3f-5faa-ab04-b5579b8bab7c | open-r1/OpenR1-Math-220k | [
""
] |
Given $\sin \alpha = 2\cos \alpha$, the value of $\tan \left( \frac{\pi}{4} + \alpha \right)$ is equal to \_\_\_\_\_\_. | **Analysis:**
This question mainly examines the basic relationships of trigonometric functions for the same angle and the application of the tangent formula for the sum of two angles. It is a basic question.
From the given condition, we find $\tan \alpha = 2$, and then we calculate the result using $\tan \left( \fr... | cn_k12 | 7e84ffd2-ad08-57b3-a269-b54ac52043bc | open-r1/OpenR1-Math-220k | [
""
] |
A number is a perfect square if it is equal to an integer raised to the square. For example, $25=5^{2}, 49=7^{2}$ and $125=25^{2}$ are perfect squares. What is the smallest number by which we must multiply 120 to obtain a perfect square?
(a) 10
(b) 15
(c) 20
(d) 30
(e) 35 | The correct option is (d).
Factoring 120 we get $120=2^{3} \times 3 \times 5$. To obtain a perfect square, all the exponents in this factorization must be even, so it is enough to multiply 120 by
$$
2 \times 3 \times 5=30
$$
Indeed, we have, $120 \times 30=2^{3} \times 3 \times 5 \times 2 \times 3 \times 5=2^{4} \ti... | olympiads | 3cbf64ef-c75a-5798-8df7-ca2f41ee5ff8 | open-r1/OpenR1-Math-220k | [
""
] |
Given the function $f(x) = \log_{2}(1+\frac{2}{x-1})$.
(1) Use the definition of monotonicity of functions to prove: $f(x)$ is a decreasing function on $(1,+\infty)$;
(2) If for any $x \in [3,4]$, the inequality $f(x)-m+1 \leqslant 0$ always holds, find the range of the real number $m$. | (1) Proof: For any $x_{1}, x_{2} \in (1,+\infty)$ and $x_{1} 0$, $(x_{1}+1)(x_{2}-1) > 0$, $(x_{1}-1)(x_{2}+1) > 0$,
thus $x_{1}x_{2}+(x_{2}-x_{1})-1 > x_{1}x_{2}-(x_{2}-x_{1})-1 > 0$,
then $\frac{x_{1}x_{2}+(x_{2}-x_{1})-1}{x_{1}x_{2}-(x_{2}-x_{1})-1} > 1$,
thus $f(x_{1})-f(x_{2}) > 0$, i.e., $f(x_{1}) > f(x_{2}... | cn_k12 | e2ca87c4-e3b6-55a8-82c1-0c94417866ec | open-r1/OpenR1-Math-220k | [
""
] |
Compute the number of nonempty subsets $S \subseteq\{-10,-9,-8, . . . , 8, 9, 10\}$ that satisfy $$|S| +\ min(S) \cdot \max (S) = 0.$$ | 1. We start by noting that the set \( S \) must contain both negative and positive elements because \(\min(S) \cdot \max(S)\) must be negative. Let \(\min(S) = -a\) and \(\max(S) = b\) where \(a > 0\) and \(b > 0\).
2. The given condition is:
\[
|S| + \min(S) \cdot \max(S) = 0
\]
Substituting \(\min(S) = -... | aops_forum | f54b4d99-7ad0-5b1a-8eb0-f45e32b47da1 | open-r1/OpenR1-Math-220k | [
""
] |
Andrew eats 14 donuts on Monday, and half as many on Tuesday. On Wednesday Andrew eats 4 times as many as he did on Monday. How many donuts did Andrew eat total in the three days? | Monday:14
Tuesday:14/2=7
Wednesday:4(7)=28
Total:14+7+28=<<14+7+28=49>>49 donuts
#### 49 | null | null | openai/gsm8k | [
""
] |
A stationery store sells a certain type of pen bag for $18$ yuan each. Xiao Hua went to buy this pen bag. When checking out, the clerk said, "If you buy one more, you can get a 10% discount, which is $36 cheaper than now." Xiao Hua said, "Then I'll buy one more, thank you." According to the conversation between the two... | To solve this problem, let's denote the number of pen bags Xiao Hua initially intended to buy as $x$. The statement "If you buy one more, you can get a 10% discount, which is $36 cheaper than now" allows us to set up an equation based on the cost.
First, without the discount, the cost for $x$ pen bags is $18x$ yuan. I... | cn_k12 | d5bbf66a-1cde-5013-9fe8-f28460a61b6c | open-r1/OpenR1-Math-220k | [
""
] |
The radii of two spheres are 5 and 3 units, and the distance between their centers is 4 units. Let's calculate the volume of the common part of the two spheres. | I. Solution. Consider an arbitrary plane section passing through the centers of the two spheres. Let the centers of the two circles visible in the plane section be $O_{1}$ and $O_{2}$, and the intersection points be $P_{1}$ and $P_{2}$.
![](https://cdn.mathpix.com/cropped/2024_05_02_0e9839144bda379391a0g-1.jpg?height=... | olympiads | 63153714-4852-536f-9e33-4444d5b07225 | open-r1/OpenR1-Math-220k | [
""
] |
2. The integer solutions of the inequality $0 \leqslant a x+5 \leqslant 4$ are 1, 2, 3, 4. Then the range of values for $a$ is ( ).
(A) $a \leqslant-\frac{5}{4}$
(B) $a<-1$
(C) $-\frac{5}{4} \leqslant a<-1$
(D) $a \geqslant-\frac{5}{4}$ | 2.C.
It is known that $-\frac{1}{a} \leqslant x \leqslant-\frac{5}{a}$.
According to the problem, we should have
$$
\begin{array}{l}
0-\frac{5}{a} \geqslant 4 . \\
\text { Therefore, }-\frac{5}{4} \leqslant a<-1 .
\end{array}
$$ | cn_contest | 29401160-9ed1-5698-a9a1-9f57e816d3f8 | open-r1/OpenR1-Math-220k | [
""
] |
4. Let the natural number $n \geqslant 5, n$ different natural numbers $a_{1}, a_{2}, \cdots$, $a_{n}$ have the following property: for any two different non-empty subsets $A, B$ of the set $S=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$, the sum of the numbers in $A$ and the sum of the numbers in $B$ will not be equal.... | 4. Let's assume $a_{1}c_{2}>\cdots >c_{n}$, and $D_{k}=\sum_{i=1}^{t} a_{i}-\left(1+2+\cdots+2^{k-1}\right)=\sum_{i=1}^{k} a_{i}-\left(2^{k}-1\right) \geqslant 0$. Then,
$$
\begin{array}{l}
1+\frac{1}{2}+\cdots+\frac{1}{2^{n-1}}-\left(\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}\right)=\sum_{i=1}^{n} c_{i} d_{i} \\
= c_{1} ... | cn_contest | 29064f9e-d463-58b4-a753-532e23cd983c | open-r1/OpenR1-Math-220k | [
""
] |
How many squares with sides along the grid lines can be drawn on an $8 \times 8$ board?
# | Squares of $1 \times 1$ can be drawn 64. Squares of $2 \times 2$ can be drawn 49 (take the square in the top right corner and first slide it along the top strip - this gives seven squares, and then all these squares can be moved down seven times). Squares of $3 \times 3$ can be drawn 36, and so on. The total number of ... | olympiads | 42aed6ae-087c-5f0e-b182-024f50994dd9 | open-r1/OpenR1-Math-220k | [
""
] |
For each real number $x, f(x)$ is defined to be the minimum of the values of $2 x+3,3 x-2$ and $25-x$. What is the maximum value of $f(x)$ ? | We begin by comparing pairs of the given expressions:
- $2 x+3 \leq 3 x-2$ exactly when $5 \leq x$
- $2 x+3 \leq 25-x$ exactly when $3 x \leq 22$ or $x \leq \frac{22}{3}$
- $25-x \leq 3 x-2$ exactly when $27 \leq 4 x$ or $\frac{27}{4} \leq x$
Therefore,
- when $x \leq 5$, we have $3 x-2 \leq 2 x+3$ and when $x \geq ... | olympiads | 266135dd-843b-5797-9703-5b54c89e034a | open-r1/OpenR1-Math-220k | [
""
] |
Given that the first three terms of a geometric sequence $\{{a}_{n}\}$ are respectively $a-1$, $a+1$, and $a+4$, find the expression for ${a}_{n}$ which is equal to $(\ \ \ )$.
A: $4\cdot ({\frac{3}{2})}^{n}$
B: $4\cdot ({\frac{2}{3})}^{n}$
C: $4\cdot ({\frac{3}{2})}^{n-1}$
D: $4\cdot ({\frac{2}{3})}^{n-1}$ | First, let's find the common ratio $(r)$ of the geometric sequence by using the formula:
$$r=\frac{{t}_{2}}{{t}_{1}}=\frac{{t}_{3}}{{t}_{2}},$$
where ${t}_{1}$, ${t}_{2}$, and ${t}_{3}$ represent the first, second, and third terms of the sequence, respectively.
1. Calculate the common ratio $(r)$:
$$r=\frac{a+1}{a-... | cn_k12 | 75396a5b-754d-5a27-8cfc-5f12f5f5639b | open-r1/OpenR1-Math-220k | [
""
] |
Casey is trying to decide which employee she wants to hire. One employee works for $20 an hour. The other employee works for $22 an hour, but Casey would also get a $6/hour subsidy from the government for hiring a disabled worker. How much money per week would Casey save by hiring the cheaper employee, if they both wor... | First find the weekly cost of the first employee: $20/hour * 40 hours/week = $<<20*40=800>>800/week
Then find the hourly cost of the second employee after the government subsidy: $22/hour - $6/hour = $<<22-6=16>>16/hour
Then find the weekly cost of the second employee: $16/hour * 40 hours/week = $<<16*40=640>>640/week
... | null | null | openai/gsm8k | [
""
] |
Let $$N=2^5+2^{5^2}+2^{5^3}+\cdots{}+2^{5^{2015}}.$$Wrotten in the usual decimal form, find the last two digits of the number $N$.
| To find the last two digits of \( N = 2^5 + 2^{5^2} + 2^{5^3} + \cdots + 2^{5^{2015}} \), we need to compute \( N \mod 100 \).
1. **Compute \( N \mod 4 \):**
\[
2^5 \equiv 0 \pmod{4}, \quad 2^{5^2} \equiv 0 \pmod{4}, \quad \ldots, \quad 2^{5^{2015}} \equiv 0 \pmod{4}
\]
Since each term in the sum is divisi... | aops_forum | 73b161aa-194f-5d42-b6d9-e104adc9c914 | open-r1/OpenR1-Math-220k | [
""
] |
2. Let $a$ be a real number, and the sequence $a_{1}, a_{2}, \cdots$ satisfies:
$$
\begin{array}{l}
a_{1}=a, \\
a_{n+1}=\left\{\begin{array}{ll}
a_{n}-\frac{1}{a_{n}}, & a_{n} \neq 0 ; \\
0, & a_{n}=0
\end{array}(n=1,2, \cdots) .\right.
\end{array}
$$
Find all real numbers $a$ such that for any positive integer $n$, w... | 2. Discuss in two cases.
(1) The sequence $\left\{a_{n}\right\}$ contains a term that is zero. Let $m$ be the smallest positive integer such that $a_{m}=0$.
If $m \geqslant 2$, then $a_{m-1} \neq 0$. Therefore,
$$
a_{m-1}-\frac{1}{a_{m-1}}=a_{m}=0 \Rightarrow a_{m-1}= \pm 1 \text {, }
$$
which contradicts the problem ... | olympiads | 026ca491-2ee7-5883-9402-81605e23d547 | open-r1/OpenR1-Math-220k | [
""
] |
Define the sequence $(w_n)_{n\ge0}$ by the recurrence relation
$$w_{n+2}=2w_{n+1}+3w_n,\enspace\enspace w_0=1,w_1=i,\enspace n=0,1,\ldots$$
(1) Find the general formula for $w_n$ and compute the first $9$ terms.
(2) Show that $|\Re w_n-\Im w_n|=1$ for all $n\ge1$.
[i]Proposed by Ovidiu Bagdasar[/i] | ### Part 1: Finding the General Formula and First 9 Terms
1. **Initial Terms**:
Given the initial terms:
\[
w_0 = 1, \quad w_1 = i
\]
2. **Recurrence Relation**:
Using the recurrence relation:
\[
w_{n+2} = 2w_{n+1} + 3w_n
\]
3. **Compute the First 9 Terms**:
\[
\begin{aligned}
w_2 &=... | aops_forum | eb551ad3-d2c5-582e-a9ba-45f755eac48a | open-r1/OpenR1-Math-220k | [
""
] |
5. (20 points) Students' written work is evaluated on a two-point grading system, i.e., a work will either be credited if it is done well, or not credited if it is done poorly. The works are first checked by a neural network, which gives an incorrect answer in $10 \%$ of cases, and then all works deemed uncredited are ... | Answer: 66.
Solution. We will represent the students' works in the diagram below.
Neural network errors can:
a) classify all 10% of good works as bad;
b) classify part of the good works as bad and part of the bad works as good;
c) classify all 10% of bad works as good.
![](https://cdn.mathpix.com/cropped/2024_05_0... | olympiads | d80512b3-6243-5ce0-8aab-04c006d2ee15 | open-r1/OpenR1-Math-220k | [
""
] |
14.4. How many necklaces can be made from five white beads and two black ones? | 14.4. Answer: 3. Black beads divide the white beads into two groups (one of these groups may contain 0 beads). The appearance of the necklace is completely determined by the number of beads in the smaller group. This number can be 0, 1, or 2. | olympiads | fc5fcb97-1aa4-54fe-840d-7d4dfc0f0e86 | open-r1/OpenR1-Math-220k | [
""
] |
14. (12 points) The price of 2 cherries is the same as the price of 3 apples, but the size of one apple is 12 times that of one cherry. If Mom uses the money to buy 1 box of cherries to buy the same size box of apples, she can buy ( ) boxes.
A. 4
B. 6
C. 18
D. 27 | 【Answer】Solution: According to the problem:
The price of 2 cherries $\times 6 = $ the price of 3 apples $\times 6$, which means the money for 12 cherries can buy 18 apples;
Also, the size of one apple is 12 times that of one cherry, so the cherries of the same size as one apple can buy 18 apples, and 1 box of cherries... | olympiads | 65bcc5e4-bb90-51c9-9b1e-7e710802d730 | open-r1/OpenR1-Math-220k | [
""
] |
Problem 7.4. In triangle $ABC$, the median $CM$ and the bisector $BL$ were drawn. Then, all segments and points were erased from the drawing, except for points $A(2 ; 8)$, $M(4 ; 11)$, and $L(6 ; 6)$. What were the coordinates of point $C$?
$.
Solution. Since $M$ is the midpoint of $A B$, point $B$ has coordinates (6;14). Since $\angle A B L=\angle C B L$, point $C$ lies on the line symmetric to line $A M$ with respect to the vertical line $B L$. Also, point $C$ lies on line $A L$. Carefully finding the intersection point of these lines ... | olympiads | 8898a4c5-c4fa-5a8f-9a52-c405484c4d8f | open-r1/OpenR1-Math-220k | [
""
] |
Mr. Williams bought 10 gallons of juice for a party. Each gallon has 10 cups. At the party, 5 cups of juice were left. How many cups of juice were drunk? | Mr. Williams bought 10 x 10 = <<10*10=100>>100 cups of juice.
Since 5 cups were left, then 100 - 5 = <<100-5=95>>95 cups were drunk.
#### 95 | null | null | openai/gsm8k | [
""
] |
Marta is arranging floral centerpieces for a reception party. Each arrangement needs to have 8 roses,12 daisies, 3 snapdragons and twice as many lilies. The reception will have 10 tables. How many flowers will she need in total to fill this order? | She needs twice as many lilies as the 3 snapdragons in the arrangement so she needs 2*3 = <<2*3=6>>6 lilies
She needs 8 roses, 12 daisies, 3 snapdragons and 6 lilies for a total of 8+12+3+6 = <<8+12+3+6=29>>29 flowers per arrangement
She needs 29 flowers per each of the 10 tables for a total of 29*10 = <<29*10=290>>290... | null | null | openai/gsm8k | [
""
] |
Given the parabola $f(x)=ax^{2}+2x+c$, its lowest point is $(-1,-2)$.
(1) Find the solution set for the inequality $f(x) > 7$.
(2) If for any $x\in[2,4]$, the inequality $f(x-t)\leqslant x-2$ always holds, find the range of values for the real number $t$. | (1) According to the given conditions, we have $-\dfrac{2}{2a}=-1$, $(1)$
and $f(-1)=a-2+c=-2$, $(2)$
Solving equations $(1)$ and $(2)$, we get $a=1$ and $c=-1$.
Hence, $f(x)=x^{2}+2x-1$,
Then the original inequality can be simplified to $x^{2}+2x-8 > 0$, which yields $x 2$.
Thus, the solution set for the inequality $... | cn_k12 | 17edbc4b-d836-5c13-a93e-d5c0ca9e0e9b | open-r1/OpenR1-Math-220k | [
""
] |
1.019. $\frac{0.125: 0.25+1 \frac{9}{16}: 2.5}{(10-22: 2.3) \cdot 0.46+1.6}+\left(\frac{17}{20}+1.9\right) \cdot 0.5$. | ## Solution.
$$
\begin{aligned}
& \frac{0.125: 0.25+1 \frac{9}{16}: 2.5}{(10-22: 2.3): 0.46+1.6}+\left(\frac{17}{20}+1.9\right) \cdot 0.5=\frac{\frac{1}{2}+\frac{5}{8}}{\left(10-\frac{220}{23}\right) \cdot \frac{23}{50}+\frac{8}{5}}+\frac{17}{40}+\frac{19}{20}= \\
& =\frac{\frac{9}{8}}{\frac{1}{5}+\frac{8}{5}}+\frac{1... | olympiads | 42d79486-cac1-517b-a90f-fcc724831979 | open-r1/OpenR1-Math-220k | [
""
] |
An insurance company wishes to determine the probability of car windshields shattering within a year. The company collected data from 20,000 vehicles, starting from May 1st of a certain year to April 30th of the following year, and found that 600 vehicles had experienced shattered windshields. Therefore, the approximat... | To calculate the probability of a car's windshield shattering within a year, we divide the number of cars that experienced shattered windshields by the total number of cars observed:
$$ p = \frac{\text{Number of shattered windshields}}{\text{Total number of cars}} = \frac{600}{20,000} $$
Upon simplification, we get:
... | cn_k12 | af00fbda-173b-5f16-a297-6477802792fa | open-r1/OpenR1-Math-220k | [
""
] |
$15 \cdot 21$ The proposition "If $n$ is a positive odd number, and the sum of its digits is 4, and none of them are 0, then $n$ is a prime number." How many numbers can serve as counterexamples to the above proposition?
(A) 0.
(B) 1.
(C) 2.
(D) 3.
(E) 4.
(45th American High School Mathematics Examination, 1994) | [Solution] The odd numbers that meet the proposition conditions are as follows:
$$
13,31,121,211,1111 .
$$
Among them, 121 and 1111 are not prime numbers.
Therefore, the answer is $(C)$. | olympiads | 03532afe-e55e-572f-9755-befba28f402d | open-r1/OpenR1-Math-220k | [
""
] |
Given that the function $f(x)$ is an even function defined on $\mathbb{R}$, and when $x < 0$, $f(x) = x^3$, then the value of $f(2)$ is ( )
A: 8
B: -8
C: $\frac {1}{8}$
D: $- \frac {1}{8}$ | Since when $x < 0$, $f(x) = x^3$,
then $f(-2) = -8$,
and since $f(x)$ is an even function defined on $\mathbb{R}$,
then $f(2) = f(-2) = -8$,
therefore, the correct answer is $\boxed{\text{B}}$.
From the given information, we can deduce that $f(2) = f(-2)$, combined with when $x < 0$, $f(x) = x^3$, we can obta... | cn_k12 | 9d541625-fd61-5fdf-a1cc-25188981634f | open-r1/OpenR1-Math-220k | [
""
] |
Problem 3. Determine the numbers $a, b$, and $c$, if their sum is $\frac{5}{2}$ greater than the number $a$, $\frac{59}{6}$ greater than the number $b$, and $\frac{5}{3}$ greater than the number $c$. | Solution. From the condition of the problem, the following equalities are written:
$$
a+b+c=a+\frac{5}{2}, a+b+c=b+\frac{59}{6} \text { and } a+b+c=c+\frac{5}{3} \text {. }
$$
By rearranging the equations, we get $b+c=\frac{5}{2}, a+c=\frac{59}{6}$ and $a+b=\frac{5}{3}$. By adding the three equations, we get $2 a+2 b... | olympiads | 58195e3a-aa1a-5165-a04e-c7f8c5d86ae1 | open-r1/OpenR1-Math-220k | [
""
] |
9.3 Two parabolas with different vertices are the graphs of quadratic trinomials with leading coefficients $p$ and $q$. It is known that the vertex of each parabola lies on the other parabola. What can $p+q$ be? | 9.3 Two parabolas with different vertices are the graphs of quadratic trinomials with leading coefficients $p$ and $q$. It is known that the vertex of each parabola lies on the other parabola. What can $p+q$ be equal to?
Answer: 0.
Solution: Let $\left(x_{1}, y_{1}\right)$ be the coordinates of the vertex of one para... | olympiads | 61a16d1b-6baf-516d-8589-80c8564b9fe6 | open-r1/OpenR1-Math-220k | [
""
] |
Suppose $a$ and $b$ are real numbers, then "$a+b \gt 0$" is ( ) "$ab \gt 0$".
A: Sufficient but not necessary condition
B: Necessary but not sufficient condition
C: Necessary and sufficient condition
D: Neither sufficient nor necessary condition | Given that $a$ and $b$ are real numbers, let's examine the relationship between the conditions "$a+b \gt 0$" and "$ab \gt 0$".
**Case 1:**
- Let $a = -1$ and $b = 2$.
- Then, $a+b = -1 + 2 = 1 \gt 0$, which satisfies the condition "$a+b \gt 0$".
- However, $ab = -1 \times 2 = -2 \not\gt 0$, which does not satisfy the ... | cn_k12 | 5898a094-a2d3-5d5d-bc2b-db44b480bf7e | open-r1/OpenR1-Math-220k | [
""
] |
In the figure below ,$4$ of the $6$ disks are to be colored black and $2$ are to be colored white. Two colorings that can be obtained from one another by rotation or reflection of the entire figure are considered the same. [img]https://i.imgur.com/57nQwBI.jpg[/img]
There are only four such colorings for the given two... | To solve this problem, we will use Burnside's Lemma, which is a tool in group theory for counting the number of distinct objects under group actions, such as rotations and reflections.
1. **Identify the Symmetry Group:**
The symmetry group of the hexagon consists of 6 elements: 3 rotations (0°, 120°, 240°) and 3 re... | aops_forum | 47bb9816-8599-5719-b739-ab3727856a03 | open-r1/OpenR1-Math-220k | [
""
] |
6. Find the sum of the integers that belong to the set of values of the function $f(x)=\log _{2}(5 \cos 2 x+11)$ for $x \in[1,25(\operatorname{arctg}(1 / 3)) \cos (\pi+\arcsin (-0.6)) ; \operatorname{arctg} 2] \quad$ (10 points) | # Solution:
Since $\quad$ $\cos (\pi+\arcsin (-0.6))=\cos (\pi-\arcsin 0.6)=-\cos (\arcsin 0.6)=-0.8, \quad$ then $x \in[1.25(\operatorname{arctg}(1 / 3)) \cos (\pi+\arcsin (-0.6)) ; \operatorname{arctg} 2]=[-\operatorname{arctg}(1 / 3) ; \operatorname{arctg} 2]$ Therefore, $2 x \in[-2 \operatorname{arctg}(1 / 3) ; 2 ... | olympiads | 756c31fa-6283-5f42-8cd9-2b3a69c9694b | open-r1/OpenR1-Math-220k | [
""
] |
4. In the city of liars and knights, there are 366 residents, all born on different days of a leap year. All residents of the city answered two questions. To the question “Were you born in February?” 100 people answered affirmatively, and to the question “Were you born on the 30th?” 60 people answered affirmatively. Ho... | Answer: 29. Solution: To the first question, knights born in February and liars born in other months answered affirmatively. Let $x$ be the number of knights born in February, where $x$ does not exceed 29. Then, $29-x$ liars were born in February, and $100-x$ liars were born in other months. The total number of liars i... | olympiads | cb710771-33ac-5679-bcab-b9a027624ad7 | open-r1/OpenR1-Math-220k | [
""
] |
Simplify first, then evaluate: $\left(\frac{{a}^{2}}{a-2}-\frac{1}{a-2}\right) \div \frac{{a}^{2}-2a+1}{a-2}$, where $a=\sqrt{3}+1$. | To simplify and then evaluate the given expression $\left(\frac{{a}^{2}}{a-2}-\frac{1}{a-2}\right) \div \frac{{a}^{2}-2a+1}{a-2}$ for $a=\sqrt{3}+1$, we follow these steps:
1. Combine the fractions in the numerator of the division:
\[
\left(\frac{a^2}{a-2}-\frac{1}{a-2}\right) = \frac{a^2 - 1}{a-2}
\]
2. Notice that ... | cn_k12 | f44003b4-3f50-5095-8a29-cef6bb22a591 | open-r1/OpenR1-Math-220k | [
""
] |
Let $n>0$ be an integer. We are given a balance and $n$ weights of weight $2^{0}, 2^{1}, \ldots, 2^{n-1}$. In a sequence of $n$ moves we place all weights on the balance. In the first move we choose a weight and put it on the left pan. In each of the following moves we choose one of the remaining weights and we add it ... | Assume $n \geq 2$. We claim $$ f(n)=(2 n-1) f(n-1) . $$ Firstly, note that after the first move the left pan is always at least 1 heavier than the right one. Hence, any valid way of placing the $n$ weights on the scale gives rise, by not considering weight 1, to a valid way of placing the weights $2, 2^{2}, \ldots, 2... | olympiads_ref | NaN | open-r1/OpenR1-Math-220k | [
""
] |
## Task B-1.5.
Ivo and Ana both drank lemonade at the cinema and watched a movie. Ivo took a medium size, and Ana a large one, which is $50\%$ larger than the medium. After both had drunk $\frac{3}{4}$ of their lemonade, Ana gave Ivo one third of what was left to her and an additional 0.5 dl. After the movie ended and... | ## Solution.
Let $x$ be the average amount of lemonade Ivo took at the beginning. Ana took $1.5x$.
Ivo received from Ana $\frac{1}{3} \cdot \frac{1}{4} \cdot 1.5 x + 0.5 = \frac{1}{8} x + 0.5$,
which means he drank a total of $x + \frac{1}{8} x + 0.5 = \frac{9}{8} x + 0.5 \text{ dl}$ of lemonade.
Ana drank a total ... | olympiads | 00dae79f-0c20-51f7-9849-7ff377bbd30a | open-r1/OpenR1-Math-220k | [
""
] |
The approximate interval where the zeros of the function $f\left(x\right)=2^{x}+x^{2}-6x-1\left(x \gt 0\right)$ lie is ( ).
A: $(0,1)$
B: $(1,2)$
C: $(2,3)$
D: $(3,4)$ | To solve for the approximate interval where the zeros of the function $f(x) = 2^x + x^2 - 6x - 1$ lie, given that $x > 0$, we use the given information that the function is continuous on its domain $(0, +\infty)$.
First, we evaluate the function at $x=3$:
\begin{align*}
f(3) &= 2^3 + 3^2 - 6\cdot3 - 1 \\
&= 8 + 9 - 1... | cn_k12 | ba8a661b-2cf5-5d5e-9718-5ce4fc40a4e1 | open-r1/OpenR1-Math-220k | [
""
] |
The monotonic decreasing interval of the function $f(x)=x^{2}-2\ln x$ is $(\quad)$
A: $(0,1]$
B: $[1,+\infty)$
C: $(-\infty,-1]\cup(0,1]$
D: $[-1,0)\cup(0,1]$ | Solution: $f'(x)=2x- \dfrac {2}{x}= \dfrac {2x^{2}-2}{x}$, $(x > 0)$,
Let $f'(x)\leqslant 0$, solving this gives: $0 < x\leqslant 1$,
Therefore, the choice is: $\boxed{\text{A}}$.
First, find the derivative of the function $f(x)$, solve the inequality regarding the derivative, thereby determining the function's d... | cn_k12 | 373fd1ca-8487-598a-ad95-1d7476da7bf1 | open-r1/OpenR1-Math-220k | [
""
] |
Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to the real number $x$, such as $\lfloor 3.2 \rfloor = 3$, $\lfloor -4.5 \rfloor = -5$. The area of the shape formed by points $(x, y)$ on the plane that satisfy $\lfloor x \rfloor^2 + \lfloor y \rfloor^2 = 50$ is. | From the problem, we have the equation: $\lfloor x \rfloor^2 + \lfloor y \rfloor^2 = 50$
When $x, y \geq 0$, the integer solutions for $\lfloor x \rfloor, \lfloor y \rfloor$ are three sets: $(7, 1)$, $(5, 5)$, $(1, 7)$. Therefore, the possible values for $|\lfloor x \rfloor|$ are: $7, 5, 1$.
When $|\lfloor x \rfloor|... | cn_k12 | 007c4300-4576-5aa4-9c8b-2740834f2dc5 | open-r1/OpenR1-Math-220k | [
""
] |
In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let $N$ be the number of shadings with this property. Find the remainder when $N$ is divided by 1000.
[AIME I 2007-10.png](https://artofproblemsolving.co... | Solution 1
Consider the first column. There are ${6\choose3} = 20$ ways that the rows could be chosen, but without loss of generality let them be the first three rows. (Change the order of the rows to make this true.) We will multiply whatever answer we get by 20 to get our final answer.
Now consider the 3x3 that is... | amc_aime | cb59916b-a097-5b82-a8d1-473a23654d41 | open-r1/OpenR1-Math-220k | [
""
] |
3. (20 points) At the Journalism Faculty of the University of Enchanted Commonwealth, 4 chickens are applying. The faculty has 2 places in the daytime program and 3 places in the evening program. Assuming all 4 chickens will be admitted to the faculty, determine the number of outcomes in which exactly two chickens will... | Answer: 6.
Solution. The faculty has a total of 5 places, to which 4 applicants are applying. Since exactly two chickens will be admitted to the evening department, the other two will be admitted to the daytime department. The number of ways to choose 2 out of 4 applicants to be admitted to the daytime department is $... | olympiads | 82cd9c19-7479-5ffd-9dca-1a013e53e252 | open-r1/OpenR1-Math-220k | [
""
] |
The zero of the function $f(x)=x-3+\log_{3}x$ is located in the interval $(\quad)$
A: $(0,1)$
B: $(1,2)$
C: $(2,3)$
D: $(3,+\infty)$ | The function $f(x)$ is increasing on the interval $(0,+\infty)$,
and since $f(1)=-2 0$,
$\therefore f(2)f(3) < 0$,
$\therefore$ the zero of $f(x)$ is located in the interval $(2,3)$.
Therefore, the correct choice is: $\boxed{C}$.
The function values at the endpoints of each interval are calculated, and the e... | cn_k12 | cfe794db-cb56-5ec7-87c9-1cd4671877a8 | open-r1/OpenR1-Math-220k | [
""
] |
\section*{Problem 4 - 051044}
Calculate the difference \(D\) between the sum of the squares of all even natural numbers \(\leq 100\) and the sum of the squares of all odd natural numbers \(<100\)! | }
It holds: Sum of the squares of odd numbers \(<100=\sum_{i=1}^{50}(2 i-1)^{2}\) and sum of the squares of even numbers \(\leq 100=\sum_{i=1}^{50}(2 i)^{2}\). Therefore, it follows that
\[
\begin{gathered}
D=\sum_{i=1}^{50}(2 i)^{2}-\sum_{i=1}^{50}(2 i-1)^{2}=\sum_{i=1}^{50} 4 i^{2}-\sum_{i=1}^{50}\left(4 i^{2}-4 i+... | olympiads | 87d1ca59-9976-5253-bb86-18770309c546 | open-r1/OpenR1-Math-220k | [
""
] |
1. Given that $a$, $b$, and $c$ are non-zero real numbers, satisfying
$$
\frac{b+c-a}{a}=\frac{c+a-b}{b}=\frac{a+b-c}{c} \text {. }
$$
Then the value of $\frac{(a+b)(b+c)(c+a)}{a b c}$ is $\qquad$ | 2. -1 or 8.
Let $\frac{b+c-a}{a}=\frac{c+a-b}{b}=\frac{a+b-c}{c}=k$, then
$$
\begin{array}{l}
b+c=(k+1) a, c+a=(k+1) b, \\
a+b=(k+1) c .
\end{array}
$$
Thus, $2(a+b+c)=(k+1)(a+b+c)$.
Hence $a+b+c=0$ or $b+c=2 a, c+a=2 b, a+b$ $=2 c$.
Therefore, $\frac{(a+b)(b+c)(c+a)}{a b c}=-1$ or 8. | cn_contest | 6fa44e10-8575-5ca7-b822-359caae4b002 | open-r1/OpenR1-Math-220k | [
""
] |
| For the anniversary of the St. Petersburg Mathematical Olympiads, the mint has struck three commemorative coins. One coin turned out to be correct, on the second coin, both sides ended up with two eagles, and on the third coin, both sides were tails. The director of the mint randomly selected one of these three coins... | It is known that heads came up. Therefore, either the first or the second coin was chosen. In total, heads are depicted three times on these coins. In one of these three equally possible cases, tails is on the opposite side of the coin, and in the other two cases - heads.
## Answer
$2 / 3$. | olympiads | 2f96b272-4ec6-53c3-8f6a-3a6ec8e96ada | open-r1/OpenR1-Math-220k | [
""
] |
7. Find all real values of $a$ for which the equation $x^{4}-2 a x^{2}+x+a^{2}-a=0$ has all its roots real.
|
Solution: Let us consider $x^{4}-2 a x^{2}+x+a^{2}-a=0$ as a quadratic equation in $a$. We see that thee roots are
$$
a=x^{2}+x, \quad a=x^{2}-x+1
$$
Thus we get a factorisation
$$
\left(a-x^{2}-x\right)\left(a-x^{2}+x-1\right)=0
$$
It follows that $x^{2}+x=a$ or $x^{2}-x+1=a$. Solving these we get
$$
x=\frac{-1 ... | olympiads | 4f8b6cc8-57bd-56d7-9a62-6ca882eab8d3 | open-r1/OpenR1-Math-220k | [
""
] |
4.3. Through the vertex $A$ of the parallelogram $A B C D$, a line is drawn intersecting the diagonal $B D$, the side $C D$, and the line $B C$ at points $E, F$, and $G$ respectively. Find $E D$, if $F G$ : $F E=7, B E=8$. Round your answer to the nearest hundredth if necessary. | Answer: 2.83 (exact value: $2 \sqrt{2}$ ). | olympiads | 92e90922-0a88-5ef6-8071-c3c93707f7e3 | open-r1/OpenR1-Math-220k | [
""
] |
4. The sum $b_{7}+b_{6}+\ldots+b_{2019}$ of the terms of the geometric progression $\left\{b_{n}\right\}, b_{n}>0$ is 27, and the sum of their reciprocals $\frac{1}{b_{7}}+\frac{1}{b_{6}}+\ldots+\frac{1}{b_{2019}}$ is 3. Find the product $b_{7} \cdot b_{6} \cdot \ldots \cdot b_{2019}$. | Answer: $b_{7} \cdot b_{6} \cdot \ldots \cdot b_{2019}=3^{2013}$. | olympiads | 9546428a-f703-5a68-a9bd-5d3bbab55492 | open-r1/OpenR1-Math-220k | [
""
] |
Translate the parabola $y=2x^{2}$ first to the left by $3$ units, then up by $4$ units, the resulting function expression of the parabola is ( )
A: $y=2\left(x+3\right)^{2}+4$
B: $y=2\left(x+3\right)^{2}-4$
C: $y=2\left(x-3\right)^{2}-4$
D: $y=2\left(x-3\right)^{2}+4$ | To translate the parabola $y=2x^{2}$ first to the left by $3$ units, we replace $x$ with $x+3$. This gives us the intermediate equation $y=2(x+3)^{2}$.
Next, to translate this intermediate parabola up by $4$ units, we add $4$ to the entire equation, resulting in $y=2(x+3)^{2}+4$.
Therefore, the final expression for... | cn_k12 | 3586c97b-c40d-50cf-b7d9-88d1f8082505 | open-r1/OpenR1-Math-220k | [
""
] |
For some constant $k$ the polynomial $p(x) = 3x^2 + kx + 117$ has the property that $p(1) = p(10)$. Evaluate $p(20)$. | 1. Given the polynomial \( p(x) = 3x^2 + kx + 117 \) and the condition \( p(1) = p(10) \), we need to find the value of \( k \) and then evaluate \( p(20) \).
2. First, we calculate \( p(1) \):
\[
p(1) = 3(1)^2 + k(1) + 117 = 3 + k + 117 = k + 120
\]
3. Next, we calculate \( p(10) \):
\[
p(10) = 3(10)^... | aops_forum | dead72e5-815c-5138-95bb-c7606be32b0d | open-r1/OpenR1-Math-220k | [
""
] |
Point $P(m+2,3)$ is moved $3$ units to the right to point $P'$, and $P'$ lies on the $y$-axis. The value of $m$ is ( )
A: $-5$
B: $1$
C: $-1$
D: $-3$ | To solve the problem, let's follow the steps based on the given solution:
1. The initial coordinates of point $P$ are $(m+2,3)$. When point $P$ is moved $3$ units to the right, the $x$-coordinate increases by $3$. Therefore, the new $x$-coordinate of point $P'$ is $(m+2)+3 = m+5$.
2. The new coordinates of point $P'$... | cn_k12 | f15d9c27-0f28-5406-b158-2a5321e73537 | open-r1/OpenR1-Math-220k | [
""
] |
Problem 2. The sum of three natural numbers is 2015. Find the three numbers, knowing that if we add 11 to the first, subtract 11 from the second, and divide the third number by 11, we get the same number each time. | Solution. Let a, b, c be the three natural numbers.
## 1p for free
$$
\begin{aligned}
& a+b+c=2015 \\
& a+11=x \rightarrow a=x-11 \\
& b-11=x \rightarrow b=x+11 \\
& c: 11=x \rightarrow c=11 x
\end{aligned}
$$
Substituting into the first relation, we get:
$$
x-11+x+11+11 x=2015
$$
$13 x=2015$
$x=155$
Finalizatio... | olympiads | 10700067-d953-5218-b910-13ac83192e7c | open-r1/OpenR1-Math-220k | [
""
] |
3. Find the smallest $n$ such that $n$ ! ends in 290 zeroes. | Solution:
1170
Each 0 represents a factor of $10=2 \cdot 5$. Thus, we wish to find the smallest factorial that contains at least 290 2's and 290 5's in its prime factorization. Let this number be $n$ !, so the factorization of $n$ ! contains 2 to the power $p$ and 5 to the power $q$, where
$$
p=\left\lfloor\frac{n}{2}... | olympiads | 601f3669-ac93-53ff-89fa-eab4244ad44f | open-r1/OpenR1-Math-220k | [
""
] |
3. Solve the system of equations $\left\{\begin{array}{l}x^{2} y+x y^{2}+3 x+3 y+24=0, \\ x^{3} y-x y^{3}+3 x^{2}-3 y^{2}-48=0 .\end{array}\right.$ | Answer: $(-3, -1)$.
Solution. The given system is equivalent to the following:
$$
\left\{\begin{array} { l }
{ x y ( x + y ) + 3 ( x + y ) + 24 = 0 , } \\
{ x y ( x ^ { 2 } - y ^ { 2 } ) + 3 ( x ^ { 2 } - y ^ { 2 } ) - 48 = 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(x y+3)(x+y)=-24 \\
(x y+3)(x-y)(x+y)... | olympiads | 8168d89b-76d3-5bd2-a793-6ef26e3ff458 | open-r1/OpenR1-Math-220k | [
""
] |
4.9 On a $10 \times 10$ grid paper, there are 11 horizontal grid lines and 11 vertical grid lines. A line segment connecting two adjacent nodes on the same straight line is called a "link segment". How many link segments need to be erased at least, so that at most 3 link segments remain at each node? | [Solution] Removing the dotted links in the right figure satisfies the requirements of the problem, and there are 41 dotted links in the figure, so the minimum value required does not exceed 41.
On the other hand, there are 81 internal nodes on the grid paper, each with 4 links. Therefore, at least one link must be er... | olympiads | 8253de0b-ede9-5630-a1d1-7d0ae1ae5cd2 | open-r1/OpenR1-Math-220k | [
""
] |
Find the maximum number of natural numbers $x_{1}, x_{2}, \ldots, x_{m}$ satisfying the conditions:
a) No $x_{i}-x_{j}, 1 \leq i<j \leq m$ is divisible by 11 ; and
b) 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. | The required maximum is 10.
According to a), the numbers $x_{i}, 1 \leq i \leq m$, are all different $(\bmod 11)$ (1)
Hence, the number of natural numbers satisfying the conditions is at most 11.
If $x_{j} \equiv 0(\bmod 11)$ for some $j$, then
$$
x_{2} x_{3} \ldots x_{m}+x_{1} x_{3} \ldots x_{m}+\cdots+x_{1} x_{2}... | olympiads_ref | NaN | open-r1/OpenR1-Math-220k | [
""
] |
Three. (25 points) Given that $n$ is a positive integer. If in any permutation of $1,2, \cdots, n$, there always exists a sum of six consecutive numbers greater than 2013, then $n$ is called a "sufficient number". Find the smallest sufficient number.
---
Please note that the term "温饱数" is translated as "sufficient nu... | Three, the smallest well-fed number is 671.
First, prove: 671 is a well-fed number.
For any permutation $b_{1}, b_{2}, \cdots, b_{671}$ of $1,2, \cdots, 671$, divide it into 112 groups, where each group contains six numbers, and the last two groups share one number, i.e.,
$$
\begin{array}{l}
A_{1}=\left(b_{1}, b_{2}, \... | cn_contest | 33a28194-d7ba-5d82-a587-c8d96cce5db5 | open-r1/OpenR1-Math-220k | [
""
] |
2. Let $x$ and $y$ both be positive numbers. Then the minimum value of $M=\frac{4 x}{x+3 y}+\frac{3 y}{x}$ is $\qquad$ . | 2.3.
Notice that, $M=\frac{4 x}{x+3 y}+\frac{x+3 y}{x}-1 \geqslant 3$. When $x=3 y$, $M=3$. | olympiads | 6a438b8a-b4e8-5629-9202-07ce97c6359b | open-r1/OpenR1-Math-220k | [
""
] |
How many sequences of zeros and ones of length 20 have all the zeros consecutive, or all the ones consecutive, or both?
$\textbf{(A)}\ 190\qquad\textbf{(B)}\ 192\qquad\textbf{(C)}\ 211\qquad\textbf{(D)}\ 380\qquad\textbf{(E)}\ 382$ | Solution 1
There are $\binom{20}{2}$ selections; however, we count these twice, therefore
$2\cdot\binom{20}{2} = \boxed{\textbf{(D)}\ 380}$. The wording of the question implies D, not E.
However, MAA decided to accept both D and E.
Solution 2
Consider the 20 term sequence of $0$'s and $1$'s. Keeping all other terms 1... | amc_aime | 0b2bf140-821c-5727-9ecc-ecbfb577a868 | open-r1/OpenR1-Math-220k | [
""
] |
Let set $A=\{m+1, -3\}$, and set $B=\{2m+1, m-3\}$. If $A \cap B = \{-3\}$, then the value of the real number $m$ is \_\_\_\_\_\_. | Since $A=\{m+1, -3\}$, $B=\{2m+1, m-3\}$, and $A \cap B = \{-3\}$,
it follows that $2m+1=-3$ or $m-3=-3$,
solving these gives $m=-2$ or $m=0$.
When $m=0$, $A=\{1, -3\}$, $B=\{1, -3\}$, which does not satisfy the condition, so it is discarded;
When $m=-2$, $A=\{-1, -3\}$, $B=\{-5, -3\}$, which satisfies the cond... | cn_k12 | be697d1f-d670-50b9-a30b-4e1fe974c6f5 | open-r1/OpenR1-Math-220k | [
""
] |
In an opaque bag, there are $4$ black balls and $2$ white balls that are completely identical except for their colors. If $3$ balls are randomly drawn from the bag, which of the following events is certain to happen?
A: All $3$ balls are black.
B: $2$ black balls and $1$ white ball are drawn.
C: $2$ white balls and ... | Given that in the bag, there are $4$ black balls and $2$ white balls, let's analyze each event to see which one is certain to happen when $3$ balls are drawn randomly.
- **Event A: All $3$ balls are black.**
- It's possible to draw a combination that includes white balls, such as $2$ white balls and $1$ black ball. ... | cn_k12 | 5292fd94-683e-5626-a125-8cfedc60bd3f | open-r1/OpenR1-Math-220k | [
""
] |
Evaluate the definite integral $\int_{0}^{\frac{\pi}{2}} (3x + \sin x) dx =$ \_\_\_\_\_\_\_\_\_\_. | **Step 1: Understanding the Problem**
This problem requires us to compute a definite integral using the fundamental theorem of calculus and the geometric meaning of definite integrals.
**Step 2: Computing the Antiderivative**
The antiderivative of $3x + \sin x$ is given by $\frac{3}{2}x^2 - \cos x$.
**Step 3: Applyin... | cn_k12 | cc6b92a6-1df0-5d18-bf48-9d00441756c2 | open-r1/OpenR1-Math-220k | [
""
] |
Given a circle $O$ and two points $A$ and $B$ in its plane. Connecting a point $M$ on the circle to points $A$ and $B$, the lines $AM$ and $BM$ intersect the circle again at points $C$ and $D$, respectively. Determine the point $M$ such that $CD \parallel AB$! | I. Solution. Let us assume that we have solved our problem: point $M$ meets the requirements of the problem, such that $C D \| A B$. Now let us consider the circumcircle $\gamma$ of $\triangle A M B$. Since $C D \| A B$, it is clear that $\frac{M A}{M C}=\frac{M B}{M D}$, meaning the two circles are in a similar positi... | olympiads | 2e074a63-1690-5231-9747-e3d9fa9f42d3 | open-r1/OpenR1-Math-220k | [
""
] |
Given that the area of $\triangle ABC$ is $360$, and point $P$ is a point on the plane of the triangle such that $\overrightarrow{AP} = \frac{1}{4} \overrightarrow{AB} + \frac{1}{4} \overrightarrow{AC}$, find the area of $\triangle PAB$. | Let $D$ be the midpoint of $AB$, and $E$ be the midpoint of $AC$. Then, $P$ is the midpoint of $DE$.
Since the area of $\triangle ABC$ is $360$, and $\triangle PAB$ is similar to $\triangle ADE$ with a scale factor of $\frac{1}{2}$, the area of $\triangle PAB$ is:
$$
\text{Area of } \triangle PAB = \text{Area of } \tr... | cn_k12 | 8b0d6455-197d-504b-b37c-8cc72d76d424 | open-r1/OpenR1-Math-220k | [
""
] |
1. Add parentheses to the numerical expression $36+144: 9-3 \cdot 2$ so that its value is:
a) 84 ;
b) 14 . | 1. a) $36+144:(9-3 \cdot 2)=36+144: 3=36+48=84$. 2 points
b) $(36+144): 9-3 \cdot 2=180: 9-3 \cdot 2=20-6=14$. 2 points
.TOTAL 4 POINTS | olympiads | 815dc526-618e-580b-9de3-5b059065c16d | open-r1/OpenR1-Math-220k | [
""
] |
Find the value of $ t$ such that ${ \frac{\int_0^{\frac{\pi}{2}}(\sin x+t\cos x)dx}{\sqrt{\int_0^{\frac{\pi}{2}} (\sin x+t\cos x)^2dx}}}$ is maximized. | 1. **Define the integral expressions:**
\[
A := \int_{0}^{\frac{\pi}{2}} (\sin x + t \cos x) \, dx
\]
\[
B := \int_{0}^{\frac{\pi}{2}} (\sin x + t \cos x)^2 \, dx
\]
2. **Evaluate \( A \):**
\[
A = \int_{0}^{\frac{\pi}{2}} (\sin x + t \cos x) \, dx
\]
Using the antiderivatives:
\[
\... | aops_forum | 8b909e03-8780-53c1-92bc-e0e9cdcb9b36 | open-r1/OpenR1-Math-220k | [
""
] |
3. The curve represented by the equation $\frac{x^{2}}{\sin \left(19^{2007}\right)^{\circ}}+\frac{y^{2}}{\cos \left(19^{2007}\right)^{\circ}}=1$ is ( ).
(A) Hyperbola
(B) Ellipse with foci on the $x$-axis
(C) Ellipse with foci on the $y$-axis
(D) None of the above answers is correct | 3.C.
$$
\begin{array}{l}
19^{2007}=19 \times\left(19^{2}\right)^{1003}=19(360+1)^{1003} \\
=19(360 n+1)\left(n \in \mathbf{N}_{+}\right) .
\end{array}
$$
Then $\sin \left(19^{2007}\right)^{\circ}=\sin (360 \times 19 n+19)^{\circ}=\sin 19^{\circ}$,
$$
\cos \left(19^{2007}\right)^{\circ}=\cos (360 \times 19 n+19)^{\circ... | cn_contest | bb03c6da-da93-519e-8b9a-73ce93ea292c | open-r1/OpenR1-Math-220k | [
""
] |
1. The solution to the equation $\left|\sqrt{(x-2)^{2}}-1\right|=x$ is ).
(A) $\frac{3}{2}$
(B) $\frac{1}{2}$
(C) $\frac{3}{2}$ or $\frac{1}{2}$
(D) No solution | $\begin{array}{l}\text { I. 1. B } \\ \text { 1. Hint: } 0 \leqslant x<2,|1-x|=x, x^{2}-2 x \\ +1=x^{2}, x=\frac{1}{2} .\end{array}$ | cn_contest | 2ec57aff-88fb-5077-ad91-1f82028409b1 | open-r1/OpenR1-Math-220k | [
""
] |
(9) Let $\left\{a_{n}\right\}$ be a geometric sequence with the sum of the first $n$ terms denoted as $S_{n}$, satisfying $S_{n}=\frac{\left(a_{n}+1\right)^{2}}{4}$. Then the value of $S_{20}$ is $\qquad$. | (9) 0 Hint: From the condition, we know $a_{1}=\frac{\left(a_{1}+1\right)^{2}}{4}$, solving this gives $a_{1}=1$.
When $n \geqslant 2$, from $S_{n}=\frac{\left(a_{n}+1\right)^{2}}{4}$ we know $S_{n-1}=\frac{\left(a_{n-1}+1\right)^{2}}{4}$, so
$$
a_{n}=\frac{1}{4}\left(a_{n}+1\right)^{2}-\frac{1}{4}\left(a_{n-1}+1\right... | olympiads | c602ea28-26f0-5deb-bd51-7ffd1908eb14 | open-r1/OpenR1-Math-220k | [
""
] |
14. Xiaoding, Xiaoming, and Xiaoqian are playing a game together and need to determine the order of play. They agree to use "scissors, paper, rock" to decide. What is the probability that all three of them will choose "paper" in one round? $\qquad$ . | 14. $\frac{1}{27}$.
Use the tree diagram shown in Figure 8 to list all possible outcomes of a round: the gestures made by the person.
Figure 8
Figure 8 only shows part of the tree diagram (lists 9 possible outcomes, please list 9 outcomes; finally, change “hammer” to list 9 outcomes as well. Therefore, there are a tot... | cn_contest | 34e4c8ea-5d49-56d6-b6bf-6e36e1a3b962 | open-r1/OpenR1-Math-220k | [
""
] |
9.2. Electronic clocks display time from 00.00.00 to 23.59.59. How much time during the day does the number on the display that reads the same from left to right and from right to left light up? | Answer: 96 seconds.
Solution. If the digits on the display are $a b . c d . m n$, then $a=0,1,2,0 \leq b \leq 9,0 \leq c \leq 5$, $0 \leq d \leq 9,0 \leq m \leq 5,0 \leq n \leq 9$. Therefore, if $a=n, b=m, c=d$, the symmetrical number on the display is uniquely determined by the digits $a, b$ and $c$, where $a=0,1,2,0... | olympiads | c838d5af-d08a-51d9-8cb7-c68d3133bf16 | open-r1/OpenR1-Math-220k | [
""
] |
A public holiday is always celebrated on the third Wednesday of a certain month. In that month, the holiday cannot occur on which of the following days?
(A) $16^{\text {th }}$
(B) $22^{\text {nd }}$
(C) $18^{\text {th }}$
(D) $19^{\text {th }}$
(E) $21^{\text {st }}$ | The first Wednesday of a month must occur on one of the first 7 days of a month, that is, the $1^{\text {st }}, 2^{\text {nd }}, 3^{\text {rd }}, 4^{\text {th }}, 5^{\text {th }}, 6^{\text {th }}$, or $7^{\text {th }}$ of the month.
The second Wednesday of a month occurs 7 days following the first Wednesday of that mo... | olympiads | dc471560-0362-56e5-aceb-e6cbd4f3c34e | open-r1/OpenR1-Math-220k | [
""
] |
5. In a right-angled triangle, the sines of the three interior angles form a geometric progression. Then the smallest angle of the triangle is equal to ( ).
(A) $\arcsin \frac{\sqrt{5}-1}{2}$
(B) $\arccos \frac{\sqrt{5}-1}{2}$
(C) $\arcsin \frac{\sqrt{5}+1}{4}$
(D) $\arccos \frac{\sqrt{5}+1}{4}$ | 5.A.
Let the three sides be $1$, $x$, and $x^{2}$ $(0<x<1)$. Then the sine of the smallest angle of the triangle is $x^{2}$.
From $1^{2}=x^{2}+\left(x^{2}\right)^{2}$, we solve to get $x^{2}=\frac{\sqrt{5}-1}{2}$. | cn_contest | cb78856c-32c8-5f83-9c7b-4ecddd37c0fe | open-r1/OpenR1-Math-220k | [
""
] |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.