, 2012

Let all the angles in the solution be directed. Given an acute triangle $ABC$, denote $G$ the foot (lying on the side $BC$) of the altitude from $A$. The powers of the point $B$ with respect to the circles through points $A$ and $F$ give us $|BP|^2 = |BA| \cdot |BF| = |BQ|^2$, hence $|BP| = |BQ|$.

According to the Tales' theorem, the points $A$, $E$, $G$ and $B$ are concyclic, the same holds for the points $A$, $F$, $G$ and $C$. From the theorem about the power of a point we derive
$$\begin{aligned} |CA| \cdot |CE| &= |CG| \cdot |CB| = |BC|^2 - |BC| \cdot |BG| = \\ &= |BC|^2 - |BA| \cdot |BF| = |BC|^2 - |BP|^2 = \\ &= (|BC| + |BP|)(|BC| - |BP|) = (|BC| + |BQ|)(|BC| - |BP|). \end{aligned}$$
The expression in the first brackets is positive, hence the expression in the second brackets is positive as well. We get $|BC| > |BP|$, which means that $P$ lies between $B$ and $C$, hence $|CA| \cdot |CE| = |CQ| \cdot |CP|$. The points $A$, $E$, $P$ and $Q$ are thus concyclic and $\angle CPE = \angle QAE$.

According to the theorem about the angle formed by a chord and a tangent, we have $\angle BQF = \angle QAF$, hence $\angle FAE = \angle QAE - \angle QAF = \angle CPE - \angle PQF$. Since the angle $\angle FAE$ is positively directed, we have $\angle CPE > \angle CQF$, which means that the lines $QF$ and $PE$ intersect. Denote $X$ their point of intersection. Then
$$\begin{aligned} \angle EXF &= \angle PXQ = -\angle QPX - \angle XQP = -\angle CPE - \angle FQP = \\ &= \angle PQF - \angle CPE = -\angle FAE = \angle EAF, \end{aligned}$$
from which we conclude that the points $A$, $E$, $F$ and $X$ are concyclic. The lines $QF$ and $PE$ thus intersect on the circumscribed circle of the triangle $AEF$.