HMMT 二月 2008 · 冲刺赛 · 第 27 题

HMMT February 2008 — Guts Round — Problem 27

[ 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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 11 HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUND

解析

[ 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 ( ) 2 AE 2 2 2 = ⇒ CD + 2 CD + DE − EC = 0 . AC 2 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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 th 11 HARVARD-MIT MATHEMATICS TOURNAMENT, 23 FEBRUARY 2008 — GUTS ROUND