HMMT 二月 2011 · 冲刺赛 · 第 16 题

HMMT February 2011 — Guts Round — Problem 16

[ 10 ] Let R be a semicircle with diameter XY . A trapezoid ABCD in which AB is parallel to CD is circumscribed about R such that AB contains XY . If AD = 4, CD = 5, and BC = 6, determine AB . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 14 HARVARD-MIT MATHEMATICS TOURNAMENT, 12 FEBRUARY 2011 — GUTS ROUND

解析

[ 10 ] Let R be a semicircle with diameter XY . A trapezoid ABCD in which AB is parallel to CD is circumscribed about R such that AB contains XY . If AD = 4, CD = 5, and BC = 6, determine AB . Answer: 10 We claim that AB = AD + BC . Let O denote the center of R . Since DA and DC are both tangent to R , we have ∠ ADO = ∠ ODC . Since CD is parallel to AB , we also have ∠ ODC = ∠ DOA . Thus ∠ ADO = ∠ DOA , and it follows that AD = AO . Similarly, we have BC = BO . Therefore, we AB = AO + BO = AD + BC = 10 .