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Harvard-MIT Mathematics Tournament

Geometry Difficulty 4.6 AIME Prove it United States

Problem:

On the circumcircle of $A B C$ , let $A'$ be the midpoint of arc $B C$ (not containing $A$ ).

a. Show that $A, I, A'$ are collinear.

b. Show that $A'$ is the circumcenter of $BIC$ .

Solution

Solution:

a. Since $A'$ bisects the arc $B C$ , the two arcs $A' B$ and $A' C$ are equal, and so $\angle B A A' = \angle C A A'$ . Thus, $A'$ lies on the angle bisector of $B A C$ . Since $I$ also lies on the angle bisector of $B A C$ , we see that $A, I, A'$ are collinear.

b. We have

$\angle C I A' = \angle A' A C + \angle I C A = \angle A' A B + \angle I C B = \angle A' C B + \angle I C B = \angle I C A'.$

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.

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