Let $x, y, z$ be integers such that for any triangle $ABC$ we have $16 H N^{2} = x BC^{2} + y CA^{2} + z AB^{2}$, where $H$ is the foot of the altitude from $A$, $M$ is the midpoint of $[BC]$, and $N$ is the midpoint of $[AM]$. Determine $x^{2} + y^{2} + z^{2}$.
Answer: 9. Since $A H M$ is a right triangle at $H$, $N$ is the center of the circumcircle of $A H M$ so $H N=A M / 2$. The median theorem then gives $16 H N^{2}=$ $4 A M^{2}=-A C^{2}+2\left(A B^{2}+B C^{2}\right)$, thus $x=2, y=2, z=-1$, and consequently $x^{2}+y^{2}+z^{2}=9$.