HMMT 二月 2007 · TEAM2 赛 · 第 2 题

HMMT February 2007 — TEAM2 Round — Problem 2

[ 20 ] The four sides of quadrilateral ABCD are equal in length. Determine the perimeter of ABCD given that it has area 120 and AC = 10 .

解析

[ 40 ] Show that triangle P AI is similar to triangle BIC. Then conclude that P I P A = · BI. BC Solution. We have ∠ IBC = ∠ ABI because I lies on the angle bisector, and ∠ ABI = ∠ AP I because P AIB is cyclic. Additionally, ∠ BCI = ∠ ICD = ∠ P BA = ∠ P IA, by the angle bisector CI , that triangles P AB and IDC are similar, and the fact that P AIB is cyclic, respectively. It follows that triangles P AI and BIC are similar. In particular, it follows that IP/P A = BC/BI , as required. 3