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HMMT 十一月 2015 · 冲刺赛 · 第 12 题

HMMT November 2015 — Guts Round — Problem 12

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

  1. [ 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#

英文原题

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

解析

英文解析

  1. [ 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 sameplane 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.
    This is achievable when a = b = , so we are done.1 2