International Mathematical Olympiad

Let the reflections of the line $BC$ with respect to the lines $AB$ and $AC$ intersect at point $K$. We will prove that $P,Q$ and $K$ are collinear, so $K$ is the common point of the varying line $PQ$. Let lines $BE$ and $CF$ intersect at $I$. For every point $O$ and $d>0$, denote by ($O,d$) the circle centred at $O$ with radius $d$, and define ${\omega }_I=(I,IH)$ and ${\omega }_A=(A,AH)$. Let ${\omega }_K$ and ${\omega }_P$ be the incircle of triangle $KBC$ and the $P$-excircle of triangle $PBC$, respectively. Since $IH\perp BC$ and $AH\perp BC$, the circles ${\omega }_A$ and ${\omega }_I$ are tangent to each other at $H$. So, $H$ is the external homothetic centre of ${\omega }_A$ and ${\omega }_I$. From the complete quadrangle $BCEF$ we have $(A,I;Q,H)=-1$, therefore $Q$ is the internal homothetic centre of ${\omega }_A$ and ${\omega }_I$. Since $BA$ and $CA$ are the external bisectors of angles $\angle KBC$ and $\angle KCB$, circle ${\omega }_A$ is the $K$-excircle in triangle $BKC$. Hence, $K$ is the external homothetic centre of ${\omega }_A$ and ${\omega }_K$. Also it is clear that $P$ is the external homothetic centre of ${\omega }_I$ and ${\omega }_P$. Let point $T$ be the tangency point of ${\omega }_P$ and $BC$, and let $T^{\prime }$ be the tangency point of ${\omega }_K$ and $BC$. Since ${\omega }_I$ is the incircle and ${\omega }_P$ is the $P$-excircle of $PBC$, $TC=BH$ and since ${\omega }_K$ is the incircle and ${\omega }_A$ is the $K$-excircle of $KBC$, $T^{\prime }C=BH$. Therefore $TC=T^{\prime }C$ and $T\equiv T^{\prime }$. It yields that ${\omega }_K$ and ${\omega }_P$ are tangent to each other at $T$. Let point $S$ be the internal homothetic centre of ${\omega }_A$ and ${\omega }_P$, and let $S^{\prime }$ be the internal homothetic centre of ${\omega }_I$ and ${\omega }_K$. It's obvious that $S$ and $S^{\prime }$ lie on $BC$. We claim that $S\equiv S^{\prime }$. To prove our claim, let $r_A,r_I,r_P$, and $r_K$ be the radii of ${\omega }_A,{\omega }_I,{\omega }_P$ and ${\omega }_K$, respectively. It is well known that if the sides of a triangle are $a,b,c$, its semiperimeter is $s=(a+b+c)/2$, and the radii of the incircle and the $a$-excircle are $r$ and $r_a$, respectively, then $r\cdot r_a=(s-b)(s-c)$. Applying this fact to triangle $PBC$ we get $r_I\cdot r_P=BH\cdot CH$. The same fact in triangle $KCB$ yields $r_K\cdot r_A=CT\cdot BT$. Since $BH=CT$ and $BT=CH$, from these two we get $$\frac{HS}{ST}=\frac{r_A}{r_P}=\frac{r_I}{r_K}=\frac{H{S}^{\prime }}{S^{\prime }T},$$ so $S=S^{\prime }$ indeed. Finally, by applying the generalised Monge's theorem to the circles ${\omega }_A,{\omega }_I$, and ${\omega }_K$ (with two pairs of internal and one pair of external common tangents), we can see that points $Q$, $S$, and $K$ are collinear. Similarly one can show that $Q,S$ and $P$ are collinear, and the result follows.

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Alternative solution.

Again, let $BE$ and $CF$ meet at $I$, that is the incentre in triangle $BCP$; then $PI$ is the third angle bisector. From the tangent segments of the incircle we have $BP-CP=BH-CH$; hence, the possible points $P$ lie on a branch of a hyperbola $\mathcal{H}$ with foci $B,C$, and $H$ is a vertex of $\mathcal{H}$. Since $PI$ bisects the angle between the radii $BP$ and $CP$ of the hyperbola, line $PI$ is tangent to $\mathcal{H}$. Let $K$ be the second intersection of $PQ$ and $\mathcal{H}$, we will show that $AK$ is tangent to $\mathcal{H}$ at $K$; this property determines $K$. Let $G=KI\cap AP$ and $M=PI\cap AK$. From the complete quadrangle $BCEF$ we can see that ($H,Q;I,A$) is harmonic, so in the complete quadrangle $APIK$, point $H$ lies on line $GM$. Consider triangle $AIM$. Its side $AI$ is tangent to $\mathcal{H}$ at $H$, the side $IM$ is tangent to $\mathcal{H}$ at $P$, and $K$ is a common point of the third side $AM$ and the hyperbola such that the lines $AP$, $IK$ and $MH$ are concurrent at the generalised Gergonne-point $G$. It follows that the third side, $AM$ is also tangent to $\mathcal{H}$ at $K$. (Alternatively, in the last step we can apply the converse of Brianchon's theorem to the degenerate hexagon $AHIPMK$. By the theorem there is a conic section ${\mathcal{H}}^{\prime }$ such that lines $AI,IM$ and $MA$ are tangent to ${\mathcal{H}}^{\prime }$ at $H,P$ and $K$, respectively. But the three points $H,K$ and $P$, together with the tangents at $H$ and $P$ uniquely determine ${\mathcal{H}}^{\prime }$, so indeed ${\mathcal{H}}^{\prime }=\mathcal{H}$.)