February, 2017

Geometry Difficulty 5.1 AIME, harder Prove it United States

Problem:

Let $ABC$ be a triangle in the plane with $AB = 13$ , $BC = 14$ , $AC = 15$ . Let $M_n$ denote the smallest possible value of $\left(AP^n + BP^n + CP^n\right)^{\frac{1}{n}}$ over all points $P$ in the plane. Find $\lim_{n \rightarrow \infty} M_n$ .

Solution

Solution:

Let $R$ denote the circumradius of triangle $ABC$ . As $ABC$ is an acute triangle, it isn't hard to check that for any point $P$ , we have either $AP \geq R$ , $BP \geq R$ , or $CP \geq R$ . Also, note that if we choose $P = O$ (the circumcenter) then $\left(AP^n + BP^n + CP^n\right) = 3 \cdot R^n$ . Therefore, we have the inequality
$R \leq \min_{P \in \mathbb{R}^2} \left(AP^n + BP^n + CP^n\right)^{\frac{1}{n}} \leq \left(3 R^n\right)^{\frac{1}{n}} = R \cdot 3^{\frac{1}{n}}$
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 $R = \frac{65}{8}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.