HMMT 二月 2011 · 冲刺赛 · 第 20 题
HMMT February 2011 — Guts Round — Problem 20
题目详情
- [ 12 ] Let ABCD be a quadrilateral circumscribed about a circle with center O . Let O , O , O , and 1 2 3 ◦ ◦ O denote the circumcenters of 4 AOB, 4 BOC, 4 COD , and 4 DOA . If ∠ A = 120 , ∠ B = 80 , and 4 ◦ ∠ C = 45 , what is the acute angle formed by the two lines passing through O O and O O ? 1 3 2 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 14 HARVARD-MIT MATHEMATICS TOURNAMENT, 12 FEBRUARY 2011 — GUTS ROUND
解析
- [ 12 ] Let ABCD be a quadrilateral circumscribed about a circle with center O . Let O , O , O , and 1 2 3 ◦ ◦ O denote the circumcenters of 4 AOB, 4 BOC, 4 COD , and 4 DOA . If ∠ A = 120 , ∠ B = 80 , and 4 ◦ ∠ C = 45 , what is the acute angle formed by the two lines passing through O O and O O ? 1 3 2 4 Answer: 82 . 5 ˇ Lemma: Given a triangle 4 ABC , let I be the incenter, I be the excenter opposite A , and S be the A ˇ second intersection of AI with the circumcircle. Then S is the center of the circle through B , I , C , and I . A Proof. First, note ◦ ∠ ABC 180 − ∠ ABC ◦ ∠ IBI = ∠ IBC + ∠ CBI = + = 90 . A A 2 2 ◦ ˇ Similarly ∠ ICI = 90 . Therefore BICI is cyclic. Now note that A , I , S , and I are collinear A A A because they are all on the angle bisector of ∠ BAC . Hence ◦ ˇ ˇ ˇ ∠ CI S = 180 − ∠ CIA = ∠ CAI + ∠ ACI = ∠ BC S + ∠ ICB = ∠ IC S. ˇ ˇ ˇ ˇ ˇ (Note ∠ CAI = ∠ BA S = ∠ BC S since A , B , S , and C are concyclic.) Hence SC = SI . Similarly ˇ ˇ ˇ SB = SI . Thus S is the center of the circle passing through B , I , and C , and therefore I as well. A Let BA and CD intersect at E and DA and CB intersect at F . We first show that F , O , O , and O 1 3 are collinear. ′ ′ Let O and O denote the intersections of F O with the circumcircles of triangles F AB and F DC . 1 3 ′ Since O is the excenter of triangle F AB , by the lemma O is the circumcenter of 4 ABO ; since O 1 ′ ′ is incenter of triangle F DC , by the lemma O is the circumcenter of 4 DOC . Hence O = O and 1 3 1 ′ O = O . Thus, points F , O , O , and O are collinear, and similarly, we have E , O , O , and O are 3 1 3 2 3 3 ◦ ◦ collinear. Now ∠ BEC = 55 and ∠ DF C = 20 so considering quadrilateral EOF C , the angle between O O and O O is 1 3 2 4 ∠ EOF = ∠ OEC + ∠ OF C + ∠ F CE ∠ BEC ∠ DF C = + + ∠ F CE 2 2 ◦ ◦ ◦ ◦ = 27 . 5 + 10 + 45 = 82 . 5 . Guts Round