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HMMT 二月 2008 · 冲刺赛 · 第 27 题

HMMT February 2008 — Guts Round — Problem 27

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

  1. [ 12 ] Cyclic pentagon ABCDE has a right angle ∠ ABC = 90 and side lengths AB = 15 and BC = 20 .
    Supposing that AB = DE = EA, find CD.
    11 HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUNDth

英文原题

[ 12 ] Cyclic pentagon ABCDE has a right angle ∠ ABC = 90 ◦ and side lengths AB = 15 and BC = 20 .
Supposing that AB = DE = EA, find CD.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11 th HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUND

解析

英文解析

  1. [ 12 ] Cyclic pentagon ABCDE has a right angle ∠ ABC = 90 and side lengths AB = 15 and BC = 20 .
    Supposing that AB = DE = EA, find CD.
    Answer: 7 By Pythagoras, AC = 25 . Since AC is a diameter, angles ∠ ADC and ∠ AEC are also
    2 2 2
    right, so that CE = 20 and AD + CD = AC as well. Beginning with Ptolemy’s theorem,
    ( )
    2 2 2 2 2 2
    ( AE · CD + AC · DE ) = AD · EC = AC − CD EC
    ( ) ( )
    2 2 2 2 2 2 2 = ⇒ CD AE + EC + 2 · CD · AE · AC + AC DE − EC = 0
    ( )
    AE2
    2 2 2 = ⇒ CD + 2 CD + DE − EC = 0 .
    2AC
    It follows that CD + 18 CD − 175 = 0 , from which CD = 7 .
    Remark: A simple trigonometric solution is possible. One writes α = ∠ ACE = ∠ ECD = ⇒ ∠ DAC =
    °
    90 − 2 α and applies double angle formula.
    th 7
    11 HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUND