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HMMT 二月 2005 · 冲刺赛 · 第 18 题

HMMT February 2005 — Guts Round — Problem 18

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

  1. [8] If a , b , and c are random real numbers from 0 to 1, independently and uniformlychosen, what is the average (expected) value of the smallest of a , b , and c ?
    HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 19, 2005 — GUTS ROUND

英文原题

[8] If a , b , and c are random real numbers from 0 to 1, independently and uniformly
chosen, what is the average (expected) value of the smallest of a , b , and c ?
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
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HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 19, 2005 — GUTS ROUND

解析

英文解析

  1. These segments define a region containing the center of the hexagon. Find theratio of the area of this region to the area of the large hexagon.
    Solution: 9 / 13
    Let us assume all sides are of side length 3. Consider the triangle A A A . Let
    1 4 5
    P be the point of intersection of A A with A A . This is a vertex of the inner
    1 5 4 8
    hexagon. Then ∠ A A A = ∠ A A P , by symmetry. It follows that A A A ∼ A P A .
    4 1 5 5 4 1 4 5 4 5
    √
    °
    Also, ∠ A A A = 120 , so by the Law of Cosines A A = 13. It follows that
    1 4 5 1 5
    √
    P A = ( A A ) · ( A A ) / ( A A ) = 1 / 13. Let Q be the intersection of A A and
    5 4 5 4 5 1 5 1 5
    √ √
    A A . By similar reasoning, A Q = 3 / 13, so P Q = A A − A Q − P A = 9 / 13.
    16 2 1 1 5 1 5
    √
    By symmetry, the inner region is a regular hexagon with side length 9 / 13. Hence
    √
    the ratio of the area of the smaller to larger hexagon is (3 / 13) = 9 / 13.2 8