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Junior Balkan Mathematical Olympiad

Geometry Difficulty 5.1 AIME, harder Prove it JBMO

Problem:
Let $AB$ be a diameter of a circle $\omega$ with center $O$ and $OC$ be a radius of $\omega$ which is perpendicular to $AB$ . Let $M$ be a point on the line segment $OC$ . Let $N$ be the second point of intersection of the line $AM$ with $\omega$ , and let $P$ be the point of intersection of the lines tangent to $\omega$ at $N$ and at $B$ . Show that the points $M, O, P, N$ are concyclic.

Solution

Solution:
Since the lines $PN$ and $BP$ are tangent to $\omega$ , $NP = PB$ and $OP$ is the bisector of $\angle NOB$ . Therefore the lines $OP$ and $NB$ are perpendicular. Since $\angle ANB = 90^\circ$ , it follows that the lines $AN$ and $OP$ are parallel. As $MO$ and $PB$ are also parallel and $AO = OB$ , the triangles $AMO$ and $OPB$ are congruent and $MO = PB$ . Hence $MO = NP$ . Therefore $MOPN$ is an isosceles trapezoid and therefore cyclic. Hence the points $M, O, P, N$ are concyclic.

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