Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 8.3 Shortlist Prove it Baltic Way

Let $ABC$ be an acute triangle, $H$ its orthocentre, and $M$ the midpoint of $BC$ . Furthermore, let $k_1$ and $k_2$ be the circle with diameter $AH$ and the circle with center $M$ that touches the circumcircle of triangle $ABC$ interiorly, respectively. Prove that $k_1$ and $k_2$ are touching circles.

Solution

Let $N$ be the midpoint of $AH$ (and of $k_1$ ), and let $X$ be the image of $H$ with respect to reflection about $M$ . Then $X$ lies on the circumcircle of $ABC$ , opposite to $A$ . As $OM$ and $AH$ are parallel, by the Intercept Theorem, we have $AH = 2OM$ . Hence, $AN = OM$ , i.e., $ANMO$ is a parallelogram. Let $r_1$ and $r_2$ be the radii of $k_1$ and $k_2$ , respectively, and let $R$ be the radius of $ABC$ 's circumcircle. Then $R - r_2 = OM = AN = r_1$ and, hence, $r_1 + r_2 = R = AO = NM$ . This means that the distance between the midpoints of $k_1$ and $k_2$ is the sum of their radii. Consequently, $k_1$ and $k_2$ touch each other.

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