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Problem 2362

IMO Shortlist mid-range; USAMO P2/P5

Geometry Difficulty 8.1 Prove it Turkey — Team Selection Test · Turkey

Let $P$ be a point inside the triangle $ABC$ satisfying $\angle PAC = \angle PCB$ and $D$ be the midpoint of the line segment $PC$ . Let $AP$ intersect $BC$ at $E$ and $ED$ intersect $BP$ at $Q$ such that $P$ is on the line segment $BQ$ . Prove that $\angle BAP + \angle BCQ = 180^\circ$ .

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 the line passing through $B$ and parallel to $PC$ intersect the line $AE$ at $F$ and $CQ$ at $F'$ . Let $QD$ and $BF$ intersect at $K$ . Since the triangles $PEC$ and $FEB$ are similar, $PD = DC$ and $D, E, K$ are collinear, we obtain that $K$ is the midpoint of the line segment $BF$ . On the other hand, since the triangles $QPC$ and $QBF'$ are similar, $PD = DC$ and $Q, D, K$ are collinear, we also obtain that $K$ is the midpoint of the line segment $BF'$ . Therefore the points $F$ and $F'$ must coincide. Since $\angle EAC = \angle PCE = \angle CBF$ , we conclude that the points $A, B, F, C$ are concyclic and hence $\angle BAF = \angle BCF = 180^\circ - \angle BCQ$ which finishes the proof.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.