8. Let $A B C$ be a triangle with sides $A B=6, B C=10$, and $C A=8$. Let $M$ and $N$ be the midpoints of $B A$ and $B C$, respectively. Choose the point $Y$ on ray $C M$ so that the circumcircle of triangle $A M Y$ is tangent to $A N$. Find the area of triangle $N A Y$.
Answer: $600 / 73$ Let $G=A N \cap C M$ be the centroid of $A B C$. Then $G A=\frac{2}{3} G N=\frac{10}{3}$ and $G M=$ $\frac{1}{3} C M=\frac{1}{3} \sqrt{8^{2}+3^{2}}=\frac{\sqrt{73}}{3}$. By power of a point, $(G M)(G Y)=G A^{2}$, so $G Y=\frac{G A^{2}}{G Y}=\frac{(10 / 3)^{2}}{\frac{\sqrt{73}}{3}}=\frac{100}{3 \sqrt{73}}$. Thus
$$
\begin{aligned}
{[N A Y] } & =[G A M] \cdot \frac{[G A Y]}{[G A M]} \cdot \frac{[N A Y]}{[G A Y]} \\
& =\frac{1}{6}[A B C] \cdot \frac{G Y}{G M} \cdot \frac{N A}{G A} \\
& =4 \cdot \frac{100}{73} \cdot \frac{3}{2}=\frac{600}{73}
\end{aligned}
$$