25. $[\mathbf{1 7}]$ Let $A B C$ be an equilateral triangle of side length 6 inscribed in a circle $\omega$. Let $A_{1}, A_{2}$ be the points (distinct from $A$ ) where the lines through $A$ passing through the two trisection points of $B C$ meet $\omega$. Define $B_{1}, B_{2}, C_{1}, C_{2}$ similarly. Given that $A_{1}, A_{2}, B_{1}, B_{2}, C_{1}, C_{2}$ appear on $\omega$ in that order, find the area of hexagon $A_{1} A_{2} B_{1} B_{2} C_{1} C_{2}$.
$A A^{\prime} B$, we find that $A A^{\prime}=2 \sqrt{7}$. By power of a point, $A^{\prime} A_{1}=\frac{2 * 4}{2 \sqrt{7}}=\frac{4}{\sqrt{7}}$. Using side length ratios, $A_{1} A_{2}=2 \frac{A A_{1}}{A A^{\prime}}=2 \frac{2 \sqrt{7}+\frac{4}{\sqrt{7}}}{2 \sqrt{7}}=\frac{18}{7}$

Now our hexagon can be broken down into equilateral triangle $A_{1} B_{1} C_{1}$ and three copies of triangle $A_{1} C_{1} C_{2}$. Since our hexagon has rotational symmetry, $\angle C_{2}=120$, and using law of cosines on this triangle with side lengths $\frac{18}{7}$ and 6 , a little algebra yields $A_{1} C_{2}=\frac{30}{7}$ (this is a 3-5-7 triangle with an angle 120$)$.

The area of the hexagon is therefore $\frac{6^{2} \sqrt{3}}{4}+3 * \frac{1}{2} \frac{18}{7} \frac{30}{7} \frac{\sqrt{3}}{2}=\frac{846 \sqrt{3}}{49}$