Track / Stage 8 / 114 of 180 #1814 of 1964

Problem 1814

IMO Shortlist mid-range; USAMO P2/P5

Geometry Difficulty 8.3 Prove it Baltic Way Shortlist · 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.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official 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.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.