10. Convex quadrilateral $M A T H$ is given with $H M / M T=3 / 4$, and $\angle A T M=\angle M A T=$ $\angle A H M=60^{\circ}$. $N$ is the midpoint of $M A$, and $O$ is a point on $T H$ such that lines $M T, A H, N O$ are concurrent. Find the ratio $H O / O T$.
Solution: $9 / 16$

Triangle $M A T$ is equilateral, so $H M / A T=H M / M T=3 / 4$. Also, $\angle A H M=\angle A T M$, so the quadrilateral is cyclic. Now, let $P$ be the intersection of $M T, A H, N O$. Extend $M H$ and $N O$ to intersect at point $Q$. Then by Menelaus's theorem, applied to triangle
$A H M$ and line $Q N P$, we have
$$
\frac{H Q}{Q M} \cdot \frac{M N}{N A} \cdot \frac{A P}{P H}=1,
$$
while applying the same theorem to triangle $T H M$ and line $Q P O$ gives
$$
\frac{H Q}{Q M} \cdot \frac{M P}{P T} \cdot \frac{T O}{O H}=1 .
$$

Combining gives $H O / O T=(M P / P T) \cdot(A N / N M) \cdot(H P / P A)=(M P / P A) \cdot(H P / P T)$ (because $A N / N M=1$ ). But since MATH is cyclic, $\triangle A P T \sim \triangle M P H$, so $M P / P A=$ $H P / P T=H M / A T=3 / 4$, and the answer is $(3 / 4)^{2}=9 / 16$. (See figure.)