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IMO Team Selection Test 2

Geometry Difficulty 7.8 National Olympiad, round 2 Prove it Netherlands

Let $\triangle ABC$ be a triangle with orthocentre $H$ and circumcircle $\Gamma$ . Let $D$ be the reflection of $A$ across the point $B$ , and let $E$ be the reflection of $A$ across the point $C$ . Let $M$ be the midpoint of segment $DE$ .
Prove that the tangent to $\Gamma$ at $A$ is perpendicular to $HM$ .

Solution

Let $A'$ be the reflection of $H$ across the midpoint of $BC$ and $A''$ the reflection of $H$ across $BC$ . Then by angle chasing, we find that $\angle BHC = \angle ABC + \angle BCA = 180^\circ - \angle CAB$ . This angle is also equal to $\angle BA'C$ and $\angle BA''C$ . Therefore, both $A'$ and $A''$ lie on the circle. Moreover, $A'A''$ is parallel to $BC$ , which in turn is perpendicular to $AA''$ . Hence, $\angle A'A''A = 90^\circ$ and $A'A$ is a diameter of the circle. Therefore $A'A$ is perpendicular to the tangent to $\Gamma$ at $A$ . Since $A$ is the reflection of $M$ across the midpoint of $BC$ , we note that $A'A$ is parallel to $HM$ (because these line segments are transformed to each other under the reflection). Therefore $HM$ is also perpendicular to the tangent to $\Gamma$ at $A$ . $\square$

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