HMMT 十一月 2015 · 冲刺赛 · 第 12 题

HMMT November 2015 — Guts Round — Problem 12

[ 8 ] Let a and b be positive real numbers. Determine the minimum possible value of √ √ √ √ 2 2 2 2 2 2 2 2 a + b + ( a − 1) + b + a + ( b − 1) + ( a − 1) + ( b − 1) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2015, November 14, 2015 — GUTS ROUND Organization Team Team ID#

解析

[ 8 ] Let a and b be positive real numbers. Determine the minimum possible value of √ √ √ √ 2 2 2 2 2 2 2 2 a + b + ( a − 1) + b + a + ( b − 1) + ( a − 1) + ( b − 1) Proposed by: Alexander Katz √ Answer: 2 2 Let ABCD be a square with A = (0 , 0) , B = (1 , 0) , C = (1 , 1) , D = (0 , 1), and P be a point in the same plane as ABCD . Then the desired expression is equivalent to AP + BP + CP + DP . By the triangle √ inequality, AP + CP ≥ AC and BP + DP ≥ BD , so the minimum possible value is AC + BD = 2 2. 1 This is achievable when a = b = , so we are done. 2