8. Let $A B C$ be an equilateral triangle with side length 8 . Let $X$ be on side $A B$ so that $A X=5$ and $Y$ be on side $A C$ so that $A Y=3$. Let $Z$ be on side $B C$ so that $A Z, B Y, C X$ are concurrent. Let $Z X, Z Y$ intersect the circumcircle of $A X Y$ again at $P, Q$ respectively. Let $X Q$ and $Y P$ intersect at $K$. Compute $K X \cdot K Q$.
Answer: 304
Let $B Y$ and $C X$ meet at $O . O$ is on the circumcircle of $A X Y$, since $\triangle A X C \cong \triangle C Y B$.
We claim that $K A$ and $K O$ are tangent to the circumcircle of $A X Y$. Let $X Y$ and $B C$ meet at $L$. Then, $L B Z C$ is harmonic. A perspectivity at $X$ gives $A Y O P$ is harmonic. Similarly, a perspectivity at $Y$ gives $A X O Q$ is harmonic. Thus, $K$ is the pole of chord $A O$.
Now we compute. Denote $r$ as the radius and $\theta$ as $\angle A X O$. Then,
$$
\begin{array}{c}
r=\frac{X Y}{\sqrt{3}}=\frac{\sqrt{5^{2}+3^{2}-3 \cdot 5}}{\sqrt{3}}=\sqrt{\frac{19}{3}} \\
\sin \theta=\sin 60^{\circ} \cdot \frac{A C}{X C}=\frac{\sqrt{3}}{2} \cdot \frac{8}{\sqrt{5^{2}+8^{2}-5 \cdot 8}}=\frac{4}{7} \sqrt{3} \\
K X \cdot K Q=K A^{2}=(r \cdot \tan \theta)^{2}=\left(\sqrt{\frac{19}{3}} \cdot 4 \sqrt{3}\right)^{2}=304 .
\end{array}
$$