Twentieth IMAR Mathematical Competition

Let $BY$ and $CX$ cross at $S$ and let circles $APX$ and $AQY$ cross again at $T$. We first prove that $A,S,T$ are collinear. Invert from $A$ with power $AP\cdot AB=AQ\cdot AC$. As $\angle ABY=\angle AXP$, the circle $APX$ is mapped to line $BY$. Similarly, the circle $AQY$ is mapped to line $CX$, so the inversion switches $S$ and $T$. Hence $A,S,T$ are collinear, as stated.

Next, we show that the lines $AT$, $PX$ and $QY$ are projectively concurrent. Let $PX$ and $QY$ cross projectively at $R$. Apply Pappus' theorem to the hexagram $BPXCQY$ to deduce that $A=BP\cap CQ$, $R=PX\cap QY$ and $S=XC\cap YB$ are collinear. The desired concurrence now follows by the preceding paragraph.

We now prove that $PQYX$ is cyclic. If $AT$, $PX$ and $QY$ are parallel, then $PQYX$ is an isosceles trapezoid, so it is cyclic. Otherwise, read the power of $R$ from circles $APX$ and $AQY$ to write $RX\cdot RP=RT\cdot RA=RY\cdot RQ$, so $PQYX$ is cyclic.

Finally, read angles from circle $PQYX$ and the given circle through $B$ and $C$ to write $\angle PYX=\angle PQX\equiv \angle PQB=\angle PCB$. Consequently, $XY$ and $BC$ are parallel, as required.