<s>[INST] A question about an equilateral triangle Suppose that $\triangle ABC$ is an equilateral triangle. Let $D$ be a point inside the triangle so that $\overline{DA}=13$, $\overline{DB}=12$, and $\overline{DC}=5$. Find the length of $\overline{AB}$.
 [/INST] 
Consider an equilateral  triangle ABC and a Point P such that $\overline{AP} = 12$, $\overline{BP} = 13$ and $\overline{CP} = 5$.
Let $P'$, $P'_1$ and $P'_2$ be the reflection of the Point $P$ along  $\overline{BC}$, $\overline{AC}$ and $\overline{AB}$ 
Then we have
$${\triangle APB} \cong {\triangle AP'_2B}$$
$${\triangle APC} \cong {\triangle AP'_1B}$$
$${\triangle BPC} \cong {\triangle BP'C}$$
Thus we can safely say,
The area of the hexagon $AP'_1CP'BP'_2 = 2. {\triangle ABC}$ 
${\triangle AP'_2B}$, ${\triangle AP'_1B}$ and ${\triangle BP'C}$ are isosceles with an apex angle of $120^o$ and composed of $30^o$ and $60^o$ right angle triangle
Thus we have
$$\overline {P'_1P'_2} = 12\sqrt3$$
$$\overline {P'P'_2} = 13\sqrt3$$
$$\overline {P'_1P'} = 5\sqrt3$$
which implies $\angle P'P'_1P'_2 = 90^o$
So the area of the hexagon $AP'_1CP'BP'_2$
$$= \triangle AP'_1P'_2 + \triangle P'P'_1C + \triangle P'P'_2B + \triangle P'P'_1P'_2$$
$$= 12^2.\frac{\sqrt3}{4} + 5^2.\frac{\sqrt3}{4} + 13^2.\frac{\sqrt3}{4} + \frac{1}{2}(5\sqrt3)(12\sqrt3)$$
$$=169\frac{\sqrt3}{2} + 90$$
$${\triangle ABC} =169\frac{\sqrt3}{4} + 45$$
Now, given the sides of the triangle $\overline {AB}$ = $\overline {BC}$ = $\overline {CA}$ = $a$, then
$$\frac{\sqrt3}{4}a^2 = {\triangle ABC} =169\frac{\sqrt3}{4} + 45$$
$$\frac{\sqrt3}{4}a^2 =  169\frac{\sqrt3}{4} + 45$$
$$a^2 =  169 + 45\frac{4}{\sqrt3}$$
$$a = \sqrt{169 + 45\frac{4}{\sqrt3}}$$
Simplifying
$$a = \sqrt{169 + 60\sqrt3}$$
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