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Problem 1858

IMO Shortlist mid-range; USAMO P2/P5

Geometry Difficulty 8.7 Prove it Balkan 2012 shortlist · Balkan Mathematical Olympiad · 2012

Let $P$ and $Q$ be points inside a triangle $ABC$ such that $\angle PAC = \angle QAB$ and $\angle PBC = \angle QBA$ . Let $D$ and $E$ be the feet of the perpendiculars from $P$ to the lines $BC$ and $AC$ , and $F$ be the foot of perpendicular from $Q$ to the line $AB$ . Let $M$ be the intersection of the lines $DE$ and $AB$ . Prove that $MP \perp CF$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Let $G$ be the foot of the perpendicular from $P$ to the line $AB$ , and $H$ and $I$ be the feet of the perpendiculars from $Q$ to the lines $CB$ and $CA$ , respectively. Observe that we also have $\angle PCA = \angle QCB$ by the trigonometric form of Ceva's Theorem.

The quadrilaterals $AEPG$ and $AFQI$ are similar, so $\angle AEG = \angle AFI$ , and therefore points $E, F, G, I$ lie on a circle $k_1$ . Similarly, points $D, E, I, H$ lie on a circle $k_2$ , and points $D, H, F, G$ on a circle $k_3$ .

If the circles $k_1, k_2$ and $k_3$ are all different, the radical axes of pairs of these circles are the lines $AB, BC$ and $CA$ , a contradiction. Therefore the points $D, E, F, G, H, I$ are cyclic.

Let $K$ and $L$ be the centers of the circles $CDPE$ and $PFG$ . Since $MD \cdot ME = MF \cdot MG$ , the line $MP$ is the radical axis of these two circles, and therefore perpendicular to $KL$ . Since $K$ and $L$ are the midpoints of segments $PC$ and $PF$ , the lines $KL$ and $CF$ are parallel, and therefore $MP \perp CF$ .

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