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Mongolian National Mathematical Olympiad

Geometry Difficulty 6.8 National Olympiad Prove it Mongolia

Let $H$ denote the intersection of the altitudes $AD$ and $CE$ of an acute triangle $ABC$ . Let $M$ and $N$ denote the midpoints of the sides $AB$ and $BC$ respectively. The rays $MH$ and $NH$ intersect the circumcircle $\omega$ of $ABC$ at points $K$ and $L$ respectively. If the circumcircles of the triangles $EHK$ and $DHL$ intersect $\omega$ again at $P$ and $Q$ , show that the points $D, E, P, Q$ lie on a common circle.
(Batzaya G.)

Solution

Let $O$ denote the center of the circumcircle $\omega$ and denote $\alpha := \angle BAC$ , $\beta := \angle ABC$ and $\gamma := \angle BCA$ . Let $A' := (AO) \cap \omega$ . First we show that $N \in A'H$ .

Indeed, let $N' := AH \cap BC$ . Since $O$ is the midpoint of $AA'$ , the centroid $G$ of the triangle $AHA'$ is the point on $OH$ satisfying $HG = 2GO$ . By Euler's theorem, $G$ is also the centroid of the triangle $ABC$ . It follows that $HBA'C$ is a parallelogram and thus $N' = N$ . Now using the fact that $PEHK$ , $BDHE$ and $PKCA$ are circumscribed, we compute

$\begin{align*} \angle PED &= \angle PEH - \angle DEH \\ &= (180^\circ - \angle PKH) - \angle DBH \\ &= (180^\circ - (\angle PKC - 90^\circ)) - (90^\circ - \gamma) \\ &= 180^\circ - \angle PKC + \gamma \\ &= \angle PAC + \gamma \\ &= \angle PAB + \alpha + \gamma. \end{align*}$

Similarly, using the fact that $PQAL$ , $QDHL$ and $LEHA$ are circumscribed, we compute

$\begin{align*} \angle PQD &= \angle PQL + \angle LQD \\ &= \angle PAL + (180^\circ - \angle LHD) \\ &= \angle PAL + (180^\circ - (\angle LHE + \angle EHD)) \\ &= \angle PAL - \angle LHE + \beta \\ &= \angle PAL - \angle LAE + \beta \\ &= \beta - \angle PAB. \end{align*}$

It follows that $\angle PED + \angle PQD = \alpha + \beta + \gamma = 180^\circ$ , hence $DEPQ$ is circumscribed.

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