XXIX Rioplatense Mathematical Olympiad

Let $P_1$ be the point where the circumcircle of $ACY$ intersects $BC$ for the second time. Let $D$ be the foot of the bisector of $\angle BAC$. Clearly, $D$ is interior to the circumcircle of $ACY$ and by power of a point, we have that $$D{P}_1\cdot DC=DA\cdot DY.$$ Since $BX$ and $CY$ are parallel, $\frac{DB}{DC}=\frac{DX}{DY}$, from where it is easily obtained that $D{P}_1\cdot DB=DA\cdot DX$ and as $D$ is external to segments $B{P}_1$ and $AX$, $AX{P}_{1}B$ is cyclic. For the two possible positions (see the diagrams below), using the cyclic quadrilaterals $ACY{P}_1$ and $AX{P}_{1}B$ we have that $\angle C{P}_{1}Z=\angle D{P}_{1}X=\angle BAX=\angle CAY=\angle C{P}_{1}Y$. Then $P_1$, $Z$ and $Y$ are collinear, and $P_1$ is the point where $YZ$ intersects $BC$, from where $P_1=P$ and $AXBP$ and $ACYP$ are cyclic. Now let $S$ and $T$ be the circumcenters of $ABX$ and $ACY$ respectively. Let us note that $SA=SP$ and $TA=TP$, from where the line $ST$ is the perpendicular bisector of $AP$. The problem then becomes equivalent to proving that $K$, $S$, and $T$ are collinear. But let us note that since $\angle BAX=\angle YAC$, by central angle we have that $\angle BSX=\angle YTC$, and since $BSX$ and $YTC$ are isosceles, these triangles are similar. Now, since $BX$ and $YC$ are parallel, we have by Thales that $\frac{KB}{KY}=\frac{KX}{KC}$. Then the dilation centered at $K$ that sends $B$ to $Y$ also sends $X$ to $C$. Let $T_1$ be the image of $S$ after applying this same dilation, we have that $BSX$ and $Y{T}_{1}C$ are similar, and since both $T$ and $T_1$ are in the same half plane as $A$ with respect to $CY$ we have that $T=T_1$. Then $K$, $S$ and $T$ are collinear and $KA=KP$.