HMMT 二月 2007 · 冲刺赛 · 第 30 题

HMMT February 2007 — Guts Round — Problem 30

[ 15 ] ABCD is a cyclic quadrilateral in which AB = 3 , BC = 5 , CD = 6, and AD = 10. M , I , and T are the feet of the perpendiculars from D to lines AB, AC , and BC respectively. Determine the value of M I/IT . 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 10 HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND 3 2

解析

[ 15 ] ABCD is a cyclic quadrilateral in which AB = 3 , BC = 5 , CD = 6, and AD = 10. M , I , and T are the feet of the perpendiculars from D to lines AB, AC , and BC respectively. Determine the value of M I/IT . 9 25 Answer: . Quadrilaterals AM ID and DICT are cyclic, having right angles ∠ AM D, ∠ AID , and 9 ∠ CID, ∠ CT D respectively. We see that M , I , and T are collinear. For, m ∠ M ID = π − m ∠ DAM = π − m ∠ DAB = m ∠ BCD = π − m ∠ DCT = π − m ∠ DIT . Therefore, Menelaus’ theorem applied to triangle M T B and line ICA gives M I T C BA · · = 1 IT CB AM ∼ ∼ On the other hand, triangle ADM is similar to triangle CDT since ∠ AM D ∠ CT D and ∠ DAM = = ∠ DCT and thus AM/CT = AD/CD . It follows that M I BC · AM BC · AD 5 · 10 25 = = = = IT AB · CT AB · CD 3 · 6 9 Remarks. The line M IT , constructed in this problem by taking perpendiculars from a point on the circumcircle of ABC , is known as the Simson line . It is often helpful for us to use directed angles while angle chasing to avoid supplementary configuration issues, such as those arising while establishing the collinearity of M, I , and T . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 10 HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND 3 2