HMMT 二月 2008 · TEAM1 赛 · 第 11 题
HMMT February 2008 — TEAM1 Round — Problem 11
题目详情
- [ 30 ] Let lines BI and EF meet at K . Show that I, K, E, C, D are concyclic.
解析
- [ 30 ] Let lines BI and EF meet at K . Show that I, K, E, C, D are concyclic. Solution: First, note that there are two possible configurations, as K could lie inside segment EF , or on its extension. The following proof works for both cases. We have A F K I E B C D 4 1 1 1 ◦ ∠ KIC = ∠ IBC + ∠ ICB = ∠ ABC + ∠ ACB = 90 − ∠ BAC = ∠ AEF. 2 2 2 It follows that I, K, E, C are concyclic. The point D also lies on this circle because ∠ IDC = ◦ ∠ IEC = 90 . Thus, all five points are concyclic.