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HMMT 二月 2011 · 冲刺赛 · 第 20 题

HMMT February 2011 — Guts Round — Problem 20

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. [ 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

  1. [ 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