HMMT 二月 2007 · TEAM1 赛 · 第 12 题

英文解析

12. [ 30 ] Let ABCD be a cyclic quadrilateral, and let P be the intersection of its two diagonals. Points
    R, S, T, and U are feet of the perpendiculars from P to sides AB, BC, CD, and AD, respectively. Showthat quadrilateral RST U is bicentric if and only if AC ⊥ BD. (Note that a quadrilateral is calledinscriptible if it has an incircle; a quadrilateral is called bicentric if it is both cyclic and inscriptible.)
    Solution. First we show that RST U is always inscriptible. Note that in addition to ABCD, we havecyclic quadrilaterals ARP U and BSP R . Thus,
    ∠ P RU = ∠ P AU = ∠ CAD = ∠ CBD = ∠ SBP = ∠ SRP ,
    and it follows that P lies on the bisector of ∠ SRU . Analogously, P lies on the bisectors of ∠ T SR and
    ∠ U T S, so is equidistant from lines U R, RS, ST, and T U, and RST U is inscriptible having incenter P.
    Now we show that RST U is cyclic if and only if the diagonals of ABCD are orthogonal. We have
    ∠ AP B = π − ∠ BAP − ∠ P BA = π − ∠ RAP − ∠ P BR = π − ∠ RU P − ∠ P SR = π − ( ∠ RU T + ∠ T SR ) .1
    π2
    It follows that ∠ AP B = if and only if ∠ RU T + ∠ T SR = π, as desired. 2