HMMT 二月 2008 · TEAM1 赛 · 第 10 题

英文解析

10. On the circumcircle of ABC , let A be the midpoint of arc BC (not containing A ).
    ′
    (a) [ 10 ] Show that A, I, A are collinear.
    ′
    (b) [ 20 ] Show that A is the circumcenter of BIC .
    Solution:
    B CIA
    ′
    ′ ′ ′ ′A
    (a) Since A bisectors the arc BC , the two arcs A B and A C are equal, and so ∠ BAA =
    ′ ′
    ∠ CAA . Thus, A lies on the angle bisector of BAC . Since I also lies on the angle
    ′
    bisector of BAC , we see that A , I , A are collinear.
    (b) We have
    ′ ′ ′ ′ ′
    ∠ CIA = ∠ A AC + ∠ ICA = ∠ A AB + ∠ ICB = ∠ A CB + ∠ ICB = ∠ ICA .
    ′ ′ ′ ′ ′
    Therefore, A I = A C . By similar arguments, A I = A B . So, A is equidistant from
    B, I, C , and thus is its circumcenter.