HMMT 二月 2007 · 冲刺赛 · 第 15 题

HMMT February 2007 — Guts Round — Problem 15

[ 9 ] Points A, B, and C lie in that order on line , such that AB = 3 and BC = 2 . Point H is such that CH is perpendicular to . Determine the length CH such that ∠ AHB is as large as possible. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 10 HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND

解析

[ 9 ] Points A, B, and C lie in that order on line , such that AB = 3 and BC = 2 . Point H is such that CH is perpendicular to . Determine the length CH such that ∠ AHB is as large as possible. √ Answer: 10 . Let ω denote the circumcircle of triangle ABH. Since AB is fixed, the smaller the radius of ω , the bigger the angle AHB . If ω crosses the line CH in more than one point, then ′ there exists a smaller circle that goes through A and B that crosses CH at a point H . But angle ′ AH B is greater than AHB , contradicting our assumption that H is the optimal spot. Thus the circle ω crosses the line CH at exactly one spot: ie, ω is tangent to CH at H . By Power of a Point, √ 2 CH = CA CB = 5 · 2 = 10, so CH = 10. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 10 HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND