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Problem 1791

National Olympiad, first round

Geometry Difficulty 6.6 Prove it Croatia competition problems · Croatia

In an isosceles triangle $ABC$ with $|AB| = |AC|$ , points $M$ and $N$ are the midpoints of the sides $\overline{AB}$ and $\overline{BC}$ , respectively. The circle circumscribed to the triangle $AMC$ meets the line $AN$ at point $P$ different from $A$ . The line passing through $P$ parallel to the side $BC$ meets the circle circumscribed to the triangle $ABC$ at points $B_1$ and $C_1$ . Prove that the triangle $AB_1C_1$ is equilateral.

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

The points $A$ , $M$ , $P$ and $C$ lie on the same circle and $\angle MAP = \angle PAC$ . Therefore, $|MP| = |PC|$ because the corresponding subtended angles are equal. Since $P$ lies on the bisector of $\overline{BC}$ , we conclude that $|BP| = |CP|$ . Hence $|MP| = |BP|$ , which means that the point $P$ lies on the bisector of the segment $\overline{BM}$ .

Let $Q$ be the midpoint of the segment $\overline{BM}$ , and $O$ be the centre of the circumcircle of the triangle $ABC$ .

Notice that $PQ \perp AB$ and $OM \perp AB$ , which implies that $PQ \parallel OM$ . Since $M$ is the midpoint of $\overline{AB}$ , and $Q$ is the midpoint of $\overline{MB}$ , it follows that $|MQ| = \frac{1}{3}|AQ|$ . For the same reason we have $|OP| = \frac{1}{3}|AP|$ .

Consider a triangle $AB_1C_1$ . Its circumcentre is point $O$ , while $P$ is the foot of its altitude from vertex $A$ . Since $A$ , $O$ and $P$ are collinear, the triangle is isosceles.

Therefore, the segment $\overline{AP}$ is also a median of the triangle, so $|OP| = \frac{1}{3}|AP|$ implies that $O$ is the centroid. Finally, since the centroid coincides with the circumcentre, the triangle $AB_1C_1$ is isosceles.

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