HMMT 二月 2006 · 冲刺赛 · 第 18 题

HMMT February 2006 — Guts Round — Problem 18

[8] Cyclic quadrilateral ABCD has side lengths AB = 1 , BC = 2 , CD = 3 and DA = 4. 2 Points P and Q are the midpoints of BC and DA . Compute P Q . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th IX HARVARD-MIT MATHEMATICS TOURNAMENT, 25 FEBRUARY 2006 — GUTS ROUND

解析

Cyclic quadrilateral ABCD has side lengths AB = 1 , BC = 2 , CD = 3 and DA = 4. 2 Points P and Q are the midpoints of BC and DA . Compute P Q . 116 Answer: 35 Solution: Construct AC, AQ, BQ, BD , and let R denote the intersection of AC and BD . Because ABCD is cyclic, we have that 4 ABR ∼ 4 DCR and 4 ADR ∼ 4 BCR . Thus, we may write AR = 4 x, BR = 2 x, CR = 6 x, DR = 12 x . Now, Ptolemy 2 applied to ABCD yields 140 x = 1 · 3 + 2 · 4 = 11. Now BQ is a median in triangle 2 2 2 2 2 2 2 BA +2 BD − AD 2 CA +2 CD − DA 2 2 ABD . Hence, BQ = . Likewise, CQ = . But P Q is a 4 4 2 2 2 2 2 2 2 2 2 2 BQ +2 CQ − BC AB + BD + CD + CA − BC − AD 2 median in triangle BQC , so P Q = = = 4 4 11 2 2 2 2 2 2 148 · − 5 (196+100) x +1 +3 − 2 − 4 148 x − 5 116 140 = = = . 4 2 2 35 Another solution is possible. Extend AD and BC past A and B to their intersection S . Use similar triangles SAB and SCD , and similar triangles SAC and SBD to compute SA and SB , then apply the Law of Cosines twice, first to compute the cosine of ∠ A 2 and then to compute P Q . 5