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Problem 1751

National Olympiad, first round

Geometry Difficulty 6.5 Prove it THE 73rd NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD - FIRST SELECTION TEST · Romania

Let $\triangle ABC$ be a triangle and let $P$ and $Q$ be points on sides $AB$ and $AC$ , respectively, such that $AP = AQ$ and line $PQ$ passes through the incenter $I$ of triangle $ABC$ . Let $M$ be the second intersection point of the circumcircles of triangles $BPI$ and $CQI$ . The lines $PM$ and $BI$ intersect at $D$ and the lines $QM$ and $CI$ meet at $E$ . Prove that the line $MI$ passes through the midpoint of segment $DE$ .

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

Since triangle $APQ$ is isosceles, we have $\angle APQ = \angle AQP$ . $BMIP$ and $CMIQ$ are cyclic quadrilaterals, so $\angle APQ = \angle BMI$ and $\angle AQP = \angle CMI$ . It follows that $\angle BMI = \angle CMI$ . (1).

Let $K$ be the point where lines $IM$ and $BC$ meet. Since $BI$ is the bisector of $\angle ABC$ and $BMIP$ is a cyclic quadrilateral, we get $\angle IMP = \angle IBP = \angle IBK$ . From $\angle KMD = \angle KBD$ we infer that $BDKM$ is cyclic, so $\angle IDK = \angle BMK$ . Similarly, we prove that $CEKM$ is also cyclic and $\angle IEK = \angle CMK$ .

Using (1), it follows that $\angle IDK = \angle IEK$ .
The quadrilaterals $BDKM$ and $BMIP$ are cyclic, so $\angle BKD = \angle BMD = \angle BIP$ , (2).
Also, the quadrilaterals $CEKM$ and $CMIQ$ are cyclic, so $\angle CKE = \angle CME = \angle CIQ$ , (3).
From (2) and (3) we deduce that $\angle BKD + \angle CKE = \angle BIP + \angle CIQ$ . Consequently, we obtain $\angle DKE = 180^\circ - (\angle BKD + \angle CKE) = 180^\circ - (\angle BIP + \angle CIQ) = \angle DIE$ . Since $\angle IDK = \angle IEK$ and $\angle DKE = \angle DIE$ , it follows that $DIEK$ is a parallelogram, so line $KI$ bisects the segment $DE$ . As points $I$ , $K$ , and $M$ are collinear, the conclusion follows.

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