Balti Tee 2021 · Ülesanne 14

Solution. Take $S$ to be the point such that $CPSQ$ is a parallelogram, as seen in figure 20. For points $X, Y, Z$ let $\text{rot } XYZ$ denote the morphism on translations induced by the rotation that takes line $XY$ to line $XZ$, modulo half turn. As $ABCD$ is a cyclic quadrilateral it follows that $\text{rot } BAD = \text{rot } BCD$. It follows that $\text{rot } PAD = \text{rot } BCP$ and hence $\text{rot } PAQ = \text{rot } QCP = \text{rot } PSQ$. It follows that $S$ lies on $\Gamma$. Using the cyclic quadrilateral AQRS it follows that $\text{rot } RSQ = \text{rot } RAQ$. Considering the cyclic quadrilateral ABCD it follows that $\text{rot } RAQ = \text{rot } RAD = \text{rot } RCD = \text{rot } RCP$. Hence, $\text{rot } RSQ = \text{rot } RCP$. As lines CP and SQ are parallel it follows that line SR is parallel to line CR. Now R is a common point so it follows that line SR = CR = CS. As lines CS and PQ are diagonals in a parallelogram CPSQ it follows that line CR = CS passes through M, the midpoint of linesegment PQ.