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Brazilian Math Olympiad

Geometry Difficulty 5.7 AIME, harder Prove it Brazil

Let $O$ be the intersection point of the diagonals of the cyclic quadrilateral $ABCD$ . The circumcircles of triangles $AOB$ and $COD$ meet lines $BC$ and $AD$ again at $M$ , $N$ , $P$ and $Q$ . Prove that the quadrilateral $MNPQ$ is inscribed in a circle with center $O$ .

Solution

Consider angles oriented modulo $180^\circ$ . Since $\angle PCO = \angle BCA = \angle BDA = \angle ODN$ are inscribed in the circumcircle of the triangle $OCD$ , $OP = ON$ . Analogously, $OM = OQ$ .

Now, $\angle QOP = \angle CPO = \angle CDO = \angle CDB = \angle CAB = \angle OAB = \angle OQB = \angle OQP$ , so $OP = OQ$ , and analogously, $OM = ON$ .

Since $OM = ON = OP = OQ$ , the result follows.

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