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Team Selection Test for IMO, 2023

Geometry Difficulty 8.4 Shortlist Prove it Turkey

In a scalene triangle $ABC$ let $O$ be the circumcenter, $I$ be the incenter and $H$ be the orthocenter. The second intersection point of the circle which passes through $O$ and is tangent to $IH$ at $I$ and the circle which passes through $H$ and is tangent to $IO$ at $I$ is $M$ . Show that $M$ lies on the circumcircle of the triangle $ABC$ .

Solution

First observe that $\angle MHI = \angle MIO$ and $\angle MIH = \angle MOI$ , hence the similarity $MIH \sim MOI$ ; thus $MI/MO = IH/IO$ . Now let $N$ be the midpoint of the segment $[OH]$ (so $N$ is the center of the 9-point circle), and let $S$ be the reflection of $I$ over $N$ . Thus $SOIH$ is a parallelogram; therefore $\angle IMO = \angle HIO = \angle IHS$ and $IM/MO = IH/IO = IH/HS$ , which implies the similarity $IMO \sim IHS$ . Consequently
$MO = \frac{HS \cdot IO}{IS} = \frac{IO^2}{2 \cdot IN} = \frac{R(R-2r)}{2(R/2-r)} = R.$
Note: The equality $IN = R/2 - r$ is Feuerbach's theorem.

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