Solution:

$\triangle M A T$ 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 $M A T H$ 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.)