Iranian Mathematical Olympiad

Let $M$ be the midpoint of $BC$. Since $BP = CQ$, it is clear that $MP = MQ$ and $BQ = CP$. Let $Z$ be the second intersection point of circumcircles of triangles $MPX$ and $MQY$. We claim that the circumcircles of triangles $APQ$ and $XYT$ are tangent to each other at $Z$.

We have
$$\left\{ \begin{array}{l} \widehat{PZQ} = \widehat{PZM} + \widehat{QZM}, \\ C_{\triangle MQY} : \widehat{QZM} = \widehat{MYQ} = \widehat{QFC}, \\ C_{\triangle APQ} : \widehat{QFC} = \widehat{APQ}, \\ C_{\triangle MPX} : \widehat{PZM} = \widehat{MXP} = \widehat{PEB}, \\ C_{\triangle APQ} : \widehat{PEB} = \widehat{AQP}, \end{array} \right.$$

And so $\widehat{PZQ} = \widehat{APQ} + \widehat{AQP} = 180^\circ - \widehat{PAQ}$. Which means $Z$ lies on the circumcircle of triangle $APQ$. Also
$$\begin{cases} C_{\triangle MQY} : \widehat{MZY} = 180^\circ - \widehat{MQT}, \\ C_{\triangle MPX} : \widehat{MZX} = 180^\circ - \widehat{MPT}. \end{cases}$$
These imply
$$\widehat{XYZ} = 360^\circ - (\widehat{MZY} + \widehat{MZX}) = \widehat{MQT} + \widehat{MPT} = 180^\circ - \widehat{PTQ}.$$
Which means $Z$ also lies on the circumcircle of triangle $XYZ$. We use a lemma to prove the problem.

Lemma. Given two circles $\Omega$ and $\Gamma$ with a common point $X$ and a line intersecting $\Omega$ at $Y$ and $P$, and intersecting $\Gamma$ at $Q$ and $Z$ ($P$ and $Q$ lie between $Y$ and $Z$). In this case, these circles are tangent to each other at $X$ if
$$\widehat{PXQ} = \widehat{XYZ} + \widehat{XZY}.$$

Proof. Let $l$ be the tangent line to $\Omega$ through $X$, therefore $\widehat{PXl} = \widehat{XYP}$ (with a fixed direction on $l$, and angles are considered with respect to this direction). $l$ is also tangent to $\Gamma$, if and only if $\widehat{XZQ} = \widehat{QXl}$. But $\widehat{PXl} = \widehat{PXQ} - \widehat{QXl}$. Therefore
$$\widehat{XZQ} = \widehat{PXQ} - \widehat{XYZ} = \widehat{PXQ} - \widehat{PXl} = \widehat{QXl}.$$

Now back to the problem, according to the lemma (for two circles $C_{\triangle MQY}$, $C_{\triangle APQ}$), it suffices to show that
$$\widehat{QZY} = \widehat{ZFT} + \widehat{ZTF}.$$
We have
$$\left\{ \begin{array}{l} \widehat{QZY} = \widehat{QMY} = \hat{C}, \\ C_{\triangle TXY} : \widehat{ZTF} = \widehat{ZXY}, \\ C_{\triangle APQ} : \widehat{TFZ} = \widehat{QPZ}, \\ C_{\triangle MPX} : \widehat{QPZ} = \widehat{MXZ}. \end{array} \right.$$
Note that $\widehat{XMY} = \hat{A}$ and the problem is equivalent to prove that
$$\begin{align*} \widehat{C} &= \widehat{MXZ} + \widehat{ZXY} = \widehat{MXY} &\iff MXY &\sim \widehat{ACB}, \\ & &\iff \frac{MY}{MX} &= \frac{AB}{AC}. \end{align*}$$
We also have:
$$\left. \begin{array}{l} MY \parallel AC \implies \frac{MY}{FC} = \frac{MQ}{QC} \\ MX \parallel AB \implies \frac{MX}{EB} = \frac{MP}{PB} \\ MP = MQ, \quad BP = CQ \end{array} \right\} \implies \frac{MY}{MX} = \frac{FC}{EB}.$$
So the problem is equivalent to show that $\frac{FC}{EB} = \frac{AB}{AC}$, or equivalently $FC \cdot AC = EB \cdot AB$. In order to prove this, we have
$$CF \cdot CA = \mathcal{P}_{C_{\triangle APQ}}(C) = CQ \cdot CP = BP \cdot BQ = \mathcal{P}_{C_{\triangle APQ}}(B) = BE \cdot BA,$$
(Where $\mathcal{P}_\lambda(S)$ is the power of point $S$ with respect to circle $\lambda$.) Hence the claim. ■