, 2021

Solution:

Let $\omega$ be the circumcircle of $\triangle A B C$. Note that because $O N = O A$, $N$ is on $\omega$. Let $P$ be the reflection of $H$ over $M$. Then, $P$ is also on $\omega$. If $Q$ is the midpoint of $N P$, note that because
$$N H = H M = M P,$$
$Q$ is also the midpoint of $H M$. Since $O Q \perp N P$, we know that $O Q \perp H M$. As $Q$ is also the midpoint of $H M$,
$$O M = O H = 7$$
With this,
$$B M = \sqrt{O B^{2} - B M^{2}} = 6 \sqrt{2}$$
and $B C = 2 B M = 12 \sqrt{2}$. Therefore, $B C^{2} = 288$.