6. In triangle $A B C$, let $M$ be the midpoint of $B C, H$ be the orthocenter, and $O$ be the circumcenter. Let $N$ be the reflection of $M$ over $H$. Suppose that $O A=O N=11$ and $O H=7$. Compute $B C^{2}$.
Answer:
288
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, $B C^{2}=288$.