HMMT 二月 2005 · 冲刺赛 · 第 18 题

HMMT February 2005 — Guts Round — Problem 18

[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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 19, 2005 — GUTS ROUND

解析

These segments define a region containing the center of the hexagon. Find the ratio 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 √ 2 the ratio of the area of the smaller to larger hexagon is (3 / 13) = 9 / 13. 8