Mathematica competitions in Croatia

Firstly note that the circumcircle of the triangle $MNC$ passes through the ortho-centre $H$ of the triangle $ABC$. The segment $CH$ is the diameter of that circle. Since the segment $CD$ is the common chord of circumcircles of triangles $ABC$ and $MNC$, the line $CD$ is perpendicular to the line $SO$ through their centres.



Since $\angle HDC=90^{\circ }$, we have $DH\parallel SO$. It is well-known that $|CH|=2|OP|$, so $|SH|=|OP|$. Since $CH\parallel OP$ (both lines are perpendicular to $AB$), it follows that $SHPO$ is a parallelogram, so $SO\parallel HP$. We can conclude that $DH\parallel HP$, and that means that the points $D$, $H$ and $P$ are collinear. Since $|OP|=|SH|=|SD|$, the quadrilateral $DPOS$ is an isosceles trapezium, hence it is inscribed in a circle.