8. Given a triangle $A B C$ with sides $A B=13, B C=14$ and $A C=15$, let $H$ be the foot of the altitude relative to side $B C$, $M$ the midpoint of $B C$, and $N$ the midpoint of $A M$. What is the length of $H N$?
(A) $2+2 \sqrt{3}$
(B) 6
(C) $\sqrt{37}$
(D) $4+\sqrt{7}$
(E) $\sqrt{42}$
8. The answer is (C). Triangle $A H M$ is a right triangle, so we have $\overline{H N}=\overline{A M} / 2$ because $H N$ is the median relative to the hypotenuse. Setting $\overline{B H}=x$ we have $\overline{H C}=\overline{B C}-\overline{B H}=14-x$ and, by the Pythagorean theorem, $\overline{A H}^{2}=\overline{A B}^{2}-\overline{B H}^{2}=\overline{A C}^{2}-\overline{H C}^{2}$, that is $13^{2}-x^{2}=15^{2}-(14-x)^{2}$, from which we obtain $x=5$. We then have

$$
\overline{A H}=\sqrt{13^{2}-5^{2}}=12, \quad \overline{A M}=\sqrt{\overline{A H^{2}+\overline{H M}^{2}}}=\sqrt{12^{2}+2^{2}}=\sqrt{148}=2 \sqrt{37},
$$

thus $\overline{H N}=\sqrt{37}$.