TITLE: How find this maximum $S_{\Delta ABC}$
QUESTION [6 upvotes]: in $\Delta ABC$,and $\angle ABC=60$,such that $PA=10,PB=6,PC=7$,
find the maximum $S_{\Delta ABC}$.

My try:let $AB=c,BC=a,AC=b$, then
$$b^2=a^2+c^2-2ac\cos{\angle ABC}=a^2+c^2-2ac$$
then
$$S_{ABC}=\dfrac{1}{2}ac\sin{60}=\dfrac{\sqrt{3}}{4}ac$$
Then I can't

REPLY [6 votes]: The maximal area is $36 + 22\sqrt 3$,
attained by a triangle with
$$
AB = \sqrt{\frac{8128+3840\sqrt3}{73}},
\quad
BC = \sqrt{\frac{5323+3024\sqrt3}{73}}.
$$
Let $a$ and $c$ be the sides $BC$, $AB$ of the triangle.
By the Law of Cosines, side $AC$ has length $\sqrt{a^2-ac+c^2}$.
Now any four points $A,B,C,P$ in the plane determine six distances
that satisfy an algebraic relation, the vanishing of the

Cayley-Menger determinant
$$
\det\left|
\begin{array}{c&c&c&c&c}
  0 &  1   &  1   &  1   &  1   \\
  1 &  0   & AB^2 & AC^2 & AP^2 \\
  1 & AB^2 &  0   & BC^2 & BP^2 \\
  1 & AC^2 & BC^2 &  0   & CP^2 \\
  1 & AP^2 & BP^2 & CP^2 & 0
\end{array}
\right|
\, .
$$
Here we're given $PA = 10$, $PB = 6$, $PC = 7$.
So, with $AB^2=c^2$, $AC^2=a^2-ac+c^2$, and $BC^2=a^2$ we calculate
(after removing a common factor of $2$):
$$
c^2 a^4 + (-c^3 + 64 c) a^3 + (c^4 - 262 c^2 + 4096) a^2
 + (13 c^3 - 832 c) a + 169 c^2 = 0.
$$
We want to maximize the area of the triangle, which is
$\frac12 ac \sin 60^\circ = (\sqrt3/4) ac$; so we seek
the maximal value of $M := ac$.  Setting $c=M/a$ and multiplying by $a^2$,
we find
$$
(M^2+64M+4096) a^4 - (M^3+262M^2+832M) a^2 + (M^4+13M^3+169M^2) = 0,
$$
a quadratic equation in $a^2$ with negative linear coefficient.
Thus there is a positive solution iff the discriminant is nonnegative.
We find that this discriminant is $-3 M^2 (M^2-176M+832) (M^2+104M+832)$,
and thus that for $M>0$ there is a positive solution iff
$M^2-176M+832 \leq 0$.
The largest such $M$ is the larger root of this quadratic polynomial,
which is $88 + 48 \sqrt{3}$ (and agrees numerically with the value
$f_\max = 171.138\ldots$ computed by chenbai).
This gives the value $(\sqrt3/4)M = 36 + 22\sqrt 3$
announced in the first sentence.
It remains to check that this is consistent with the given lengths
$PA$, $PB$, $PC$: besides the vanishing of the Cayley-Menger determinant,
we must verify that each of $PBC$, $PCA$, $PAB$ satisfies the
triangle inequality.  We compute
$a^2 = (5323+3024\sqrt 3)/73$ and $c^2 = (8128+3840\sqrt 3)/73$,
which makes $AC^2 = a^2-ac+c^2 = (7027+3360\sqrt 3)/73$;
thus numerically our maximal-area triangle $ABC$ is to have sides
$12.03$, $13.76$, and $14.23$ opposite vertices $A,B,C$,
which is less than $6+7$, $7+10$, and $10+6$ respectively.
Thus these distances are relaized by an actual configuration $ABCP$,
so we're done.
[Added later: the same technique works for any given values 
$r,s,t$ of $PA$, $PB$, $PC$, finding that the maximal $M$ is
$$
rt + \frac{s^2}{2} + s \sqrt{r^2+rt+t^2-\frac34 s^2}
$$
and the maximal area is again $\sqrt3/4$ times that.]