HMMT 二月 2011 · 冲刺赛 · 第 20 题
HMMT February 2011 — Guts Round — Problem 20
题目详情
英文原题
- [ 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
14 HARVARD-MIT MATHEMATICS TOURNAMENT, 12 FEBRUARY 2011 — GUTS ROUNDth
解析
英文解析
- [ 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
Lemma: Given a triangle 4 ABC , let I be the incenter, I be the excenter opposite A , and S be theˇsecond intersection of AI with the circumcircle. Then S is the center of the circle through B , I , C ,ˇAand I .
Proof. First, note a
°
∠ ABC 180 − ∠ ABC
°
∠ IBI = ∠ IBC + ∠ CBI = + = 90 .
A A
2 2
°
Similarly ∠ ICI = 90 . Therefore BICI is cyclic. Now note that A , I , S , and I are collinearˇ
A A Abecause 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.
Let BA and CD intersect at E and DA and CB intersect at F . We first show that F , O , O , and OA
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
3 11
′
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