, 2019

Solution:

Since $P$ is the shared vertex between the three equilateral triangles, we note that $P$ is the incenter of $ABC$ 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 $AB_2 = AC_1$, $BC_2 = BA_1$, and $CA_2 = CB_1$. Using the fact that the altitudes from $P$ to $AB$ and $CB$ form a square with the sides, we use the side lengths of the equilateral triangle to compute that $AB_2 = AC_1 = 2 - \frac{1}{\sqrt{3}}$, $BA_1 = BC_2 = 1 - \frac{1}{\sqrt{3}}$, and $CB_1 = CA_2 = 3 - \frac{1}{\sqrt{3}}$.

We have that the area of the hexagon is therefore
$$6 - \left(\frac{1}{2}\left(2 - \frac{1}{\sqrt{3}}\right)^2 \cdot \frac{4}{5} + \frac{1}{2}\left(1 - \frac{1}{\sqrt{3}}\right)^2 + \frac{1}{2}\left(3 - \frac{1}{\sqrt{3}}\right)^2 \cdot \frac{3}{5}\right) = \frac{12 + 22 \sqrt{3}}{15}.$$