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Bulgarian National Mathematical Olympiad

Geometry Difficulty 5.7 AIME, harder Prove it Bulgaria

The diagonals $AC$ and $BD$ of the convex quadrilateral $ABCD$ intersect at point $O$ . The points $A_1$ , $B_1$ , $C_1$ and $D_1$ from the segments $AO$ , $BO$ , $CO$ and $DO$ , respectively, are such that $AA_1 = CC_1$ and $BB_1 = DD_1$ . Let $M$ be the second intersection point of the circumcircles of $\angle AOB$ and $\angle COD$ ; $N$ be the second intersection point of the circumcircles of $\angle AOD$ and $\angle BOC$ ; $P$ be the second intersection point of the circumcircles of $\angle A_1OB_1$ and $\angle C_1OD_1$ , and $Q$ be the second intersection point of the circumcircles of $\angle A_1OD_1$ and $\angle B_1OC_1$ . Prove that the points $M$ , $N$ , $P$ and $Q$ are concyclic.

Solution

It follows from the condition of the problem that $\angle MAC = \angle MBD$ and $\angle MCA = \angle MDB$ . Therefore $\angle MAC \sim \angle MBD$ . Let $X$ and $Y$ be the midpoints of $AC$ and $BD$ , respectively. It follows that $\angle MXC = \angle MYD$ , which implies that the point $M$ lies on the circumcircle of $\angle OXY$ . Analogously, $N$ belongs to the same circle.

Since $X$ and $Y$ are also the midpoints of $A_1C_1$ and $B_1D_1$ , respectively, the above argument for the quadrilateral $A_1B_1C_1D_1$ yields that $P$ and $Q$ belong to the circumcircle of $\angle OXY$ .

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