HMMT-Feb 2019 guts, Problem 17

Let $A B C$ be a triangle with $A B=3, B C=4$, and $C A=5$. Let $A_{1}, A_{2}$ be points on side $B C$, $B_{1}, B_{2}$ be points on side $C A$, and $C_{1}, C_{2}$ be points on side $A B$. Suppose that there exists a point $P$ such that $P A_{1} A_{2}, P B_{1} B_{2}$, and $P C_{1} C_{2}$ are congruent equilateral triangles. Find the area of convex hexagon $A_{1} A_{2} B_{1} B_{2} C_{1} C_{2}$.

Solution: Since $P$ is the shared vertex between the three equilateral triangles, we note that $P$ is the incenter of $A B C$ since it is equidistant to all three sides. Since the area is 6 and the semiperimeter is also 6 , we can calculate the inradius, i.e. the altitude, as 1, which in turn implies that the side length of the equilateral triangle is $\frac{2}{\sqrt{3}}$. Furthermore, since the incenter is the intersection of angle bisectors, it is easy to see that $A B_{2}=A C_{1}, B C_{2}=B A_{1}$, and $C A_{2}=C B_{1}$. Using the fact that the altitudes from $P$ to $A B$ and $C B$ form a square with the sides, we use the side lengths of the equilateral triangle to compute that $A B_{2}=A C_{1}=2-\frac{1}{\sqrt{3}}, B A_{1}=B C_{2}=1-\frac{1}{\sqrt{3}}$, and $C B_{1}=C A_{2}=3-\frac{1}{\sqrt{3}}$. We have that the area of the hexagon is therefore \[ 6-\left(\frac{1}{2}\left(2-\frac{1}{\sqrt{3}}\right)^{2} \cdot \frac{4}{5}+\frac{1}{2}\left(1-\frac{1}{\sqrt{3}}\right)^{2}+\frac{1}{2}\left(3-\frac{1}{\sqrt{3}}\right)^{2} \cdot \frac{3}{5}\right)=\frac{12+22 \sqrt{3}}{15} \] $\fbox{\frac{12+22 \sqrt{3}}{15}}$.

Answer: \frac{12+22 \sqrt{3}}{15}