HMMT 二月 2017 · 冲刺赛 · 第 10 题

HMMT February 2017 — Guts Round — Problem 10

[ 7 ] Let ABC be a triangle in the plane with AB = 13 , BC = 14 , AC = 15 . Let M denote the smallest n 1 n n n n possible value of ( AP + BP + CP ) over all points P in the plane. Find lim M . n →∞ n

解析

[ 7 ] Let ABC be a triangle in the plane with AB = 13 , BC = 14 , AC = 15 . Let M denote the smallest n 1 n n n n possible value of ( AP + BP + CP ) over all points P in the plane. Find lim M . n →∞ n Proposed by: Yang Liu Answer: 65/8 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 ≥ R, BP ≥ R , or CP ≥ R. Also, note that if we choose n n n n P = O (the circumcenter) then ( AP + BP + CP ) = 3 · R . Therefore, we have the inequality 1 1 1 n n n n n n n R ≤ min ( AP + BP + CP ) ≤ (3 R ) = R · 3 . 2 P ∈ R Taking n → ∞ yields R ≤ lim M ≤ R n n →∞ 1 65 n (as lim 3 = 1), so the answer is R = . n →∞ 8