HMMT 二月 2006 · 冲刺赛 · 第 18 题

英文解析

18. Cyclic quadrilateral ABCD has side lengths AB = 1 , BC = 2 , CD = 3 and DA = 4.
    Points P and Q are the midpoints of BC and DA . Compute P Q .2
    Answer:116
    Solution: Construct AC, AQ, BQ, BD , and let R denote the intersection of AC35
    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, Ptolemyapplied to ABCD yields 140 x = 1 · 3 + 2 · 4 = 11. Now BQ is a median in triangle 2
    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 − ADmedian in triangle BQC , so P Q = = =2
    4 4
    2 2 2 2 211
    148 · − 52
    (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 ∠ Aand then to compute P Q .2 5