HMMT 二月 2008 · TEAM1 赛 · 第 10 题

HMMT February 2008 — TEAM1 Round — Problem 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 .

解析

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: A I B C ′ 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.