Maximize the area of triangle $ABC$

Question

In triangle $ABC$, $\angle A=\frac{\pi}{3}$, $D$ is a point in the plane, which $DA=DB=8,DC=6$, find the maximum area of triangle $ABC$.

It seems no idea to use an Euclidean method. So I tried analytic method, but too complicated.

Any hint for finding the maximum value?

Answer 1

Let $\widehat{DAB}=\theta$. Then $AB=16\cos\theta$ and

$$AC_1=AK+KC_1=8\cos\left(\frac{\pi}{3}-\theta\right)+\sqrt{36-64\sin^2\left(\frac{\pi}{3}-\theta\right)}$$ and we have $[ABC_1]=\frac{\sqrt{3}}{4}AB\cdot AC_1$.

The maximum of such expression occurs at $\theta=\arccos\frac{5}{\sqrt{37}}$, for which $\color{red}{[ABC_1]=40\sqrt{3}}.$

Answer 2

Since $D$ lies on the bisector of $AB$ and the angle at $A$ is $\frac\pi3$, $ABD$ is an isosceles triangle. The area of $ABC$ is maximal if $C$ also lies on the bisector, on the other side of $D$ than $AB$. In this case the altitude is $6+\frac{\sqrt3}2\cdot8=6+4\sqrt3$, so the maximal area is

$$ \frac12\cdot8(6+4\sqrt3)=8(3+2\sqrt3)\;. $$