IMO Team Selection Test 3

2. Let $M,N,R,S$ be the midpoints of line segments $BC,CA,BD,AD$. Let $Z$ be the centroid of $△ABC$. Quadrilateral $QFMC$ is cyclic as $\angle QFC=90^{\circ }=\angle QMC$. Note that therefore $CQ$ is a diameter of the circumcircle of $QFMC$. Analogously, we see that $PNEC$ is a cyclic quadrilateral, with diameter $CP$. We show that $Z$ also lies on the circumcircles of these cyclic quadrilaterals. The similarity transforming triangle $BCA$ into triangle $BDC$ transforms triangle $CZM$ into $DFR$, as $C$ is mapped to $D$, the centroid $Z$ is mapped to the centroid $F$, and the midpoint $M$ of $BC$ is mapped to the midpoint of $BD$, which is $R$. So $△CZM\sim △DFR$, in particular $\angle CZM=\angle DFR=\angle CFM$ (opposite angles). Therefore $Z$ lies on the circumcircle of the cyclic quadrilateral $QFMC$. Analogously, we have $\angle CZN=\angle DES=\angle CEN$, from which follows that $Z$ lies on the circumcircle of the cyclic quadrilateral $PNEC$. We can now show that $Z$ lies on $PQ$. As $CQ$ is a diameter of the circle through $Q,F,M,C,Z$, we have $\angle QZC=90^{\circ }$. As $CP$ is a diameter of the circle through $P,N,E,Z,C$, we also have $\angle CZP=90^{\circ }$. Hence $P,Z$, and $Q$ are collinear. $\square$