HMMT 十一月 2015 · 冲刺赛 · 第 31 题

HMMT November 2015 — Guts Round — Problem 31

[ 17 ] Two random points are chosen on a segment and the segment is divided at each of these two points. Of the three segments obtained, find the probability that the largest segment is more than three times longer than the smallest segment. x

解析

[ 17 ] Two random points are chosen on a segment and the segment is divided at each of these two points. Of the three segments obtained, find the probability that the largest segment is more than three times longer than the smallest segment. Proposed by: Sam Korsky 27 Answer: 35 We interpret the problem with geometric probability. Let the three segments have lengths x, y, 1 − x − y and assume WLOG that x ≥ y ≥ 1 − x − y . The every possible ( x, y ) can be found in the triangle ( ) 1 1 1 1 2 1 determined by the points , , ( , ), (1 , 0) in R , which has area . The line x = 3(1 − x − y ) 3 3 2 2 12 ( ) ( ) 3 3 3 1 intersects the lines x = y and y = 1 − x − y at the points , and , Hence x ≤ 3(1 − x − y ) 7 7 5 5 ( ) ( ) ( ) 1 1 3 3 3 1 2 if ( x, y ) is in the triangle determined by points , , , , , which by shoelace has area . 3 3 7 7 5 5 105 Hence the desired probability is given by 1 2 − 27 12 105 = 1 35 12 x