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Problem 1486

AIME late

Geometry Difficulty 5.5 Prove it Iranian Mathematical Olympiad · Iran

Square $ABCD$ is given. Points $N$ and $P$ are selected on sides $AB$ and $AD$ , respectively, such that $PN = NC$ , and point $Q$ is selected on segment $AN$ such that $\angle NCB = \angle QPN$ . Prove that $\angle BCQ = \frac{1}{2} \angle PQA$ .

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 $E$ be the intersection point of $PQ$ and $BC$ . According to the problem assumption, $PN = NC$ and so $\angle NPC = \angle PCN$ . On the other hand, we know $\angle QPN = \angle NCB$ . From these we conclude that $EPC$ is an isosceles triangle. Therefore, its altitudes $PS$ and $CK$ have equal length. So $CK = PS = AB = BC$ and therefore right-angled triangles $QBC$ and $QKC$ are congruent. So $QC$ is the bisector of angles $\angle KCB$ and $\angle KQB$ . Hence, $\angle BCQ = \frac{1}{2} \angle KCB$ (*).

On the other hand, since $\angle QBC + \angle QKC = 90^\circ + 90^\circ = 180^\circ$ , we get that $QBCK$ is a cyclic quadrilateral which implies $\angle BCK = \angle AQP$ . This together with (*) completes the proof.

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