Kanada 2015

Let $\omega$ be the circumcircle of $BOC$. Let $M$ be the point diametrically opposite to $O$ on $\omega$ and let the line $AM$ intersect $\omega$ at $M$ and $K$. Since $O$ is the circumcenter of $ABC$, it follows that $OB=OC$ and therefore that $O$ is the midpoint of the arc $\widehat{BOC}$ of $\omega$. Since $M$ is diametrically opposite to $O$, it follows that $M$ is the midpoint of the arc $\widehat{BMC}$ of $\omega$. This implies since $K$ is on $\omega$ that $KM$ is the bisector of $\angle BKC$. Since $K$ is on $\omega$, this implies that $\angle BKM=\angle CKM$, i.e. $KM$ is the bisector of $\angle BKC$.

Since $O$ is the circumcenter of $ABC$, it follows that $\angle BOC=2\angle BAC$. Since $B$, $K$, $O$ and $C$ all lie on $\omega$, it also follows that $\angle BKC=\angle BOC=2\angle BAC$. Since $KM$ bisects $\angle BKC$, it follows that $\angle BKM=\angle CKM=\angle BAC$. The fact that $A$, $K$ and $M$ lie on a line therefore implies that $\angle AKB=\angle AKC=180^{\circ }-\angle BAC$. Now it follows that $$\angle KBA=180^{\circ }-\angle AKB-\angle KAB=\angle BAC-\angle KAB=\angle KAC.$$ This implies that triangles $KBA$ and $KAC$ are similar. Rearranging the condition in the problem statement yields that $BP/AP=AQ/CQ$ which, when combined with the fact that $KBA$ and $KAC$ are similar, implies that triangles $KPA$ and $KQC$ are similar. Therefore $\angle KPA=\angle KQC=180^{\circ }-\angle KQA$ which implies that $K$ lies on $\Gamma$.

Now let $S$ denote the centre of $\Gamma$ and let $T$ denote the centre of $\omega$. Note that $T$ is the midpoint of segment $OM$ and that $TM$ and $AS$, which are both perpendicular to $BC$, are parallel. This implies that $\angle KMT=\angle KAS$ since $A$, $K$ and $M$ are collinear. Further, since $KTM$ and $KSA$ are isosceles triangles, it follows that $\angle TKM=\angle KMT$ and $\angle SKA=\angle KSA$. Therefore $\angle TKM=\angle SKA$ which implies that $S$, $T$ and $K$ are collinear. Therefore $\Gamma$ and $\omega$ intersect at a point $K$ which lies on the line $ST$ connecting the centres of the two circles. This implies that the circles $\Gamma$ and $\omega$ are tangent at $K$. $\square$