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Croatian Mathematical Competitions

Geometry Difficulty 6.1 National Olympiad Prove it Croatia

In an acute triangle $ABC$ such that $|AC| < |BC|$ , points $M$ and $N$ are respectively the feet of altitudes from vertices $A$ and $B$ . The circumcircle of the triangle $ABC$ with the centre $O$ and the circumcircle of the triangle $MNC$ with the centre $S$ intersect in points $C$ and $D$ . If the point $P$ is the midpoint of the segment $\overline{AB}$ , prove that points $P$ , $O$ , $S$ and $D$ lie on the same circle. (Stipe Vidak)

Solution

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.

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