HMMT 二月 2007 · TEAM1 赛 · 第 12 题

HMMT February 2007 — TEAM1 Round — Problem 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. Show that quadrilateral RST U is bicentric if and only if AC ⊥ BD. (Note that a quadrilateral is called inscriptible if it has an incircle; a quadrilateral is called bicentric if it is both cyclic and inscriptible.)

解析

[ 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. Show that quadrilateral RST U is bicentric if and only if AC ⊥ BD. (Note that a quadrilateral is called inscriptible 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 have cyclic 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 1 = π − ( ∠ RU T + ∠ T SR ) . 2 π It follows that ∠ AP B = if and only if ∠ RU T + ∠ T SR = π, as desired. 2