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Geometry Difficulty 8.2 Shortlist Prove it Saudi Arabia

Given an acute triangle $ABC$ with its altitudes $AD$ , $BE$ , $CF$ concurrent at $H$ . The point $K$ changes on the segment $AH$ . Let $M$ , $N$ be the projection of $H$ on the lines $KE$ , $KF$ . Prove that the line joining the circumcenter of triangles $HEN$ and $HFM$ always passes through a fixed point.

Solution

Let $S$ , $T$ be the intersection of the lines $KF$ , $KE$ with $CA$ , $AB$ and $P$ , $Q$ the midpoint of $HT$ , $HS$ . Since $\angle HFT = \angle HMT = 90^\circ$ so $HT$ is the diameter of $(HMF)$ and $P$ is the center of $(HMF)$ . Similarly, $Q$ is the center of $(HNE)$ .

Let $R$ be the intersection of $ST$ , $EF$ . Based on the basic properties of harmonic points, we have
$A(HR, EF) = -1.$
On the other hand, if $EF$ , $BC$ intersect at $R'$ , then $A(DR', EF) = -1$ , from which it follows that $R \equiv R'$ and resulting in $R$ being the fixed point. Since $H$ is fixed, the midpoint $L$ of $HR$ is also a fixed point. On the other hand, $PL$ , $QL$ are the medians of triangles $HTR$ , $HSR$ such that $R$ , $S$ , $T$ are collinear, so $L$ belongs to $PQ$ . So $PQ$ passes through the fixed $L$ . $\square$

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.