Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.0 National olympiad Prove it Austria

Let $ABC$ be an acute triangle, with $AC \neq BC$ . Let $M$ be the midpoint of segment $AB$ . Let $H$ be the orthocenter of triangle $ABC$ , $D$ the footpoint of the altitude through $A$ on $BC$ and $E$ the footpoint of the altitude through $B$ on $AC$ .
Prove that lines $AB$ , $DE$ and the orthogonal to $MH$ through $C$ intersect in a point $S$ .
(Karl Czakler)

Solution

Let $\angle ACB = \gamma$ and $F$ be the foot of $C$ on $MH$ . We will first demonstrate that $F$ lies on the circumcircle $k$ of triangle $ABC$ .
Let $H_1$ denote the symmetric point to $H$ with respect to $M$ . The quadrilateral $AH_1BH$ is a parallelogram, and since we have $\angle AHB = \angle AH_1B = 180^\circ - \gamma$ , the point $H_1$ must lie on the circumcircle $k$ of $ABC$ . Reflecting point $H$ on triangle side $AB$ yields point $H_2$ , and it is well known that this point also lies on $k$ . The line $H_1H_2$ is parallel to $AB$ , and thus perpendicular to $CH_2$ . It follows that $CH_1$ is a diameter of the circumcircle $k$ , and it follows that $F$ lies on $k$ . In summary, we have:
* Points $A, B, D, E$ lie on a common circle $k_1$ .
* Points $C, E, H, D, F$ lie on a common circle $k_2$ .
* Points $A, B, F, C$ lie on the circumcircle $k$ .
The point $S$ is thus the radical center of these three circles, completing the proof.
(Karl Czakler, Josef Greilhuber) ☐

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