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HMMT 二月 2011 · 冲刺赛 · 第 16 题

HMMT February 2011 — Guts Round — Problem 16

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

  1. [ 10 ] Let R be a semicircle with diameter XY . A trapezoid ABCD in which AB is parallel to CD iscircumscribed about R such that AB contains XY . If AD = 4, CD = 5, and BC = 6, determine AB .
    14 HARVARD-MIT MATHEMATICS TOURNAMENT, 12 FEBRUARY 2011 — GUTS ROUNDth

英文原题

[ 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 .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14 th HARVARD-MIT MATHEMATICS TOURNAMENT, 12 FEBRUARY 2011 — GUTS ROUND

解析

英文解析

  1. [ 10 ] Let R be a semicircle with diameter XY . A trapezoid ABCD in which AB is parallel to CD iscircumscribed 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 tangentto 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 .