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Thailand Mathematical Olympiad

Geometry Difficulty 6.1 National Olympiad Prove it Thailand

Let $ABCD$ be a cyclic quadrilateral. Let $O$ be the circumcenter of the quadrilateral $ABCD$ . The diagonals $AC$ and $BD$ intersect at $G$ . Let $P, Q, R$ and $S$ be the circumcenters of triangles $AGB, BGC, CGD$ and $DGA$ respectively. The lines $PR$ and $QS$ intersect at $M$ . Show that $M$ is the midpoint of $G$ and $O$ .

Solution

First we show that $PORG$ is a parallelogram.

Since $R$ and $O$ lie on the perpendicular bisector of chord $CD$ , we have $OR$ is perpendicular to $CD$ .
Let $L$ be the intersection of $PG$ and $CD$ , and $X$ be the midpoint of $BG$ .
Since $\angle(LP, PX) = \angle(GP, PX) = \angle(GA, AB) = \angle(CA, AB) = \angle(CD, DB) = \angle(LD, DX)$ , the points $L, P, D, X$ are concyclic. From $PX \perp XG$ , we deduce $LP \perp CD$ and thus $GP \parallel OR$ .

A similar argument shows that $GR \parallel OP$ . Thus, $PORG$ is a parallelogram.
Since $PQ$ and $RS$ are perpendicular bisectors of segments $BG$ and $DG$ respectively, we have $PQ \parallel RS$ (since both are perpendicular to $BD$ ). Similarly we have $QR \parallel PS$ , and hence $PQRS$ is a parallelogram.
Since the diagonals of a parallelogram bisect each other, from the parallelogram $PQRS$ we deduce that $M$ is the midpoint of $PR$ . On the other hand, the parallelogram $PORG$ yields that $M$ is the midpoint of $GO$ as required.

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