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Problem 1791

IMO Shortlist mid-range; USAMO P2/P5

Geometry Difficulty 8.3 Prove it THE 68th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS · Romania

Let $ABC$ be a triangle such that $AB \neq AC$ , let $G$ be its centroid, and let $H$ be its orthocenter. Let $D$ be the orthogonal projection of $A$ on the line $BC$ , and let $M$ be the midpoint of the side $BC$ . The circle $ABC$ crosses the ray $MH$ emanating from $M$ at $P$ , and the ray $DG$ emanating from $D$ at $Q$ , outside the segment $DG$ . Show that the lines $DP$ and $MQ$ meet on the circle $ABC$ .
BMO 2017 Shortlist

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solutions — 2

Solution 1

*First solution.* We show that the angles $APD$ and $AQM$ are either equal or one is the supplement of the other.
To begin, recall that the reflection of $H$ across $M$ is the antipode of $A$ in the circle $ABC$ , to infer that the angle $APM$ is right, so $A$ , $D$ , $M$ , $P$ are concyclic. Hence the angles $APD$ and $AMD$ are either equal or one is the supplement of the other.
The circle $ABC$ is the image of the nine-point circle under a homothety of scale factor $-2$ centred at $G$ . This homothety sends $M$ to $A$ and $D$ to $Q$ , so $AQ$ and $BC$ are parallel.
Hence $\angle AMD = \angle MAQ$ , on the one hand; and $MA = MQ$ , on the other, so $\angle MAQ = \angle AQM$ . Consequently, $\angle AMD = \angle AQM$ , and the conclusion follows by the preceding paragraph.

Solution 2

*Second solution.* As in the previous solution, $A$ and $Q$ are reflections of one another in the perpendicular bisectrix of the segment $BC$ .
Let the line $MP$ cross the circle $ABC$ again at $P'$ , and let the lines $QP'$ and $BC$ cross at $S$ . Since $P'$ is the antipode of $A$ in the circle $ABC$ , the angle $AQS = AQP'$ is right, so the lines $QS$ and $BC$ are perpendicular.
Let the line $MQ$ cross the circle $ABC$ again at $Q'$ , let the lines $PQ'$ and $BC$ cross at $R$ , and refer to the butterfly theorem to infer that $R$ and $S$ are reflections of one another across $M$ .
The lines $AR$ and $QS$ are therefore reflections of one another in the perpendicular bisectrix of the segment $BC$ .
Finally, since $QS$ is perpendicular to $BC$ , so is $AR$ , and, consequently, the points $R$ and $D$ coincide. The conclusion follows.

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