[ Symmetry helps solve the problem.] Rearrangement of areas $\quad]$

A point located inside an equilateral triangle is at distances of 5, 6, and 7 from its vertices. Find the area of the triangle.
Display the point $M$ symmetrically relative to the sides of the given triangle.

## Solution

Let $M$ be a point inside an equilateral triangle $A B C$, such that $A M=5, B M=6, C M=7$. Consider the points $A_{1}$, $B_{1}$, and $C_{1}$, which are symmetric to point $M$ with respect to the lines $B C, A C$, and $A B$ respectively. In the isosceles triangles $A B_{1} C_{1}, B A_{1} C_{1}$, and $C A_{1} B_{1}$, the lateral sides and the angles at vertices $A, B$, and $C$ are known to be $120^{\circ}$. The areas of these triangles are respectively $\frac{25 \sqrt{3}}{4}, \frac{36 \sqrt{3}}{4}$, and $\frac{49 \sqrt{3}}{4}$. Using the cosine theorem, we find the bases of these triangles:

$$
B_{1} C_{1}=5 \sqrt{3}, A_{1} C_{1}=6 \sqrt{3}, A_{1} B_{1}=7 \sqrt{3} .
$$

Using Heron's formula, we find that the area of triangle $A_{1} B_{1} C_{1}$ is $18 \sqrt{6}$. Therefore, the area of the hexagon $A B_{1} C A_{1} B C_{1}$ is

$$
\frac{25 \sqrt{3}}{4}+\frac{36 \sqrt{3}}{4}+\frac{49 \sqrt{3}}{4}+18 \sqrt{6}=\frac{110 \sqrt{3}}{4}+18 \sqrt{6}
$$

Since triangles $A B_{1} C, C A_{1} B$, and $A C_{1} B$ are respectively equal to triangles $A M C, C M B$, and $A M B$, the area of the hexagon $A B_{1} C A_{1} B C_{1}$ is twice the area $S$ of triangle $A B C$. Therefore,

$$
S_{\Delta \mathrm{ABC}}=\frac{1}{2}\left(\frac{110 \sqrt{3}}{4}+18 \sqrt{6}\right)=\frac{55 \sqrt{3}+36 \sqrt{6}}{4}
$$