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National Math Olympiad - Final Round, 2013

Geometry Difficulty 5.7 AIME, harder Prove it Slovenia

Let $K_1$ be a circle with the center $S_1$ and the radius $r$ . Let $K_2$ be a circle with the center $S_2$ , lying on the circle $K_1$ , and the radius $\frac{2}{3}r$ . Let $A$ be the point of intersection of the line $S_1S_2$ and the circle $K_2$ that lies in the exterior of the circle $K_1$ . Let $C$ denote one of the intersection points of the circles $K_1$ and $K_2$ . The line $AC$ also intersects the circle $K_1$ at the point $D$ . Let $H$ be the orthogonal projection of the point $D$ onto the line $S_1S_2$ . Prove that the point $H$ lies on the circle $K_2$ .

Solution

The quadrilateral $ES_2CD$ is cyclic, so $\angle S_2ED = \angle S_2CA = \angle CAS_2$ and $EAD$ is an isosceles triangle with the apex at $D$ . This implies $|AH| = |EH|$ , or $|AH| = \frac{1}{2}|EA| = \frac{1}{2}(2r + \frac{2}{3}r) = \frac{4}{3}r$ . Since $\frac{4}{3}r$ is precisely the radius of the circle $K_2$ , we conclude that the point $H$ lies on $K_2$ .

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