10. $[7]$ Let $A B C$ be a triangle in the plane with $A B=13, B C=14, A C=15$. Let $M_{n}$ denote the smallest possible value of $\left(A P^{n}+B P^{n}+C P^{n}\right)^{\frac{1}{n}}$ over all points $P$ in the plane. Find $\lim _{n \rightarrow \infty} M_{n}$.
Answer:
$65 / 8$
Let $R$ denote the circumradius of triangle $A B C$. As $A B C$ is an acute triangle, it isn't hard to check that for any point $P$, we have either $A P \geq R, B P \geq R$, or $C P \geq R$. Also, note that if we choose $P=O$ (the circumcenter) then $\left(A P^{n}+B P^{n}+C P^{n}\right)=3 \cdot R^{n}$. Therefore, we have the inequality
$$
R \leq \min _{P \in \mathbb{R}^{2}}\left(A P^{n}+B P^{n}+C P^{n}\right)^{\frac{1}{n}} \leq\left(3 R^{n}\right)^{\frac{1}{n}}=R \cdot 3^{\frac{1}{n}}
$$

Taking $n \rightarrow \infty$ yields
$$
R \leq \lim _{n \rightarrow \infty} M_{n} \leq R
$$
(as $\lim _{n \rightarrow \infty} 3^{\frac{1}{n}}=1$ ), so the answer is $R=\frac{65}{8}$.