, 2013

We need the following well-known result.
Lemma. In a triangle $KLM$ with $KL \neq KM$, $R$ is a point on the $LM$ such that $KR$ is perpendicular to $LM$. Let $U$ be a point on the line $KR$. Let the lines $LU$ and $MU$ intersect $KM$ and $KL$, respectively, at $S$ and $T$ respectively. Then $L, T, S, M$ are concyclic if and only if $U$ is the orthocenter of triangle $KLM$.

Suppose that $C, X, Y, A$ are concyclic. Let the lines $AP$ and $CQ$ intersect at $Z$. Since $BP = BQ$ we have $BQ^2 = BX \cdot BA$, so $BQ$ is tangent to the circumcircle of triangle $AXQ$. Hence $\angle BQX = \angle XAQ = \angle XCY$. Therefore the points $Q, X, P, C$ are concyclic. Further, $\angle XAP = \angle XPQ = \angle XCQ$. Adding the two we get $\angle YCZ = \angle YAZ$. This proves that the points $A, Y, Z, C$ are concyclic. Applying the lemma to triangle $QAC$ we get that $P$ is the orthocenter of triangle $QAC$. Hence $\angle CXA = \angle CYA = 90^\circ$.

For the converse, suppose that $CX$ is perpendicular to $AB$. Let the line $CP$ intersect the circumcircle of triangle $AXC$ at $Y'$, and let the lines $AY'$ and $BP$ intersect at $Q'$. Note that $P$ is the orthocenter of triangle $Q'AC$. Hence if $AP$ intersects $Q'C$ at $Z$, then $Z$ lies on the circle $\gamma$. Note that $\angle Q'CX = \angle Q'CY' - \angle XCY' = \angle ZAY' - \angle XAY' = \angle XAP = \angle Q'PX$. This shows that the points $Q', X, P, C$ are concyclic. Hence $\angle XQ'B = \angle XCP = \angle XAY'$, so $BQ'$ is tangent to the circumcircle of triangle $AXQ'$. Therefore $BQ'^2 = BX \cdot BA = BP^2$ and hence $Q = Q'$. This completes the solution. $\square$