Estonian Mathematical Olympiad

Let $S$ be a point such that $PESF$ is a parallelogram (Fig. 15). We first show that $PR\parallel AS$. For this note that $\angle PER=\angle SFA$ since $PE\parallel FS$ and $ER\parallel AF$. Additionally $PE=FS$ and $ER=AF$, thus triangles $PER$ and $SFA$ are congruent. From this we deduce that $PR\parallel AS$.

In order to prove the problem statement it now suffices to show that $A$, $S$ and $Q$ are collinear. Let $X$ be the point of intersection of $ES$ and $AF$ and $Y$ be the point of intersection of $FS$ and $AE$ (Fig. 16).

First we show that $XY\parallel BC$. Denote $\frac{AE}{AC}=\kappa$ and $\frac{AF}{AB}=\lambda$. As $EX\parallel CF$, we have $\frac{AX}{AF}=\frac{AE}{AC}=\kappa$. Thus $\frac{AX}{AB}=\frac{AX}{AF}\cdot \frac{AF}{AB}=\kappa \cdot \lambda$. Analogously, we get $\frac{AY}{AC}=\kappa \cdot \lambda$. Consequently, $\frac{AX}{AB}=\frac{AY}{AC}$, which implies $XY\parallel BC$.

The fact just proven implies $\frac{XY}{BC}=\frac{AX}{AB}$. It suffices to notice that the triangles $BCQ$ and $XYS$ are similar as their respective sides are parallel. Thus $\overrightarrow{XS}=\overrightarrow{XY}$ which implies $\overrightarrow{XS}=\frac{AX}{AB}$. Since $XS\parallel BQ$, this implies that the points $A$, $S$ and $Q$ are collinear.

Fig. 15 Fig. 16

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Alternative solution.

Denote $\overrightarrow{AC}={\vec{e}}_1$ and $\overrightarrow{AB}={\vec{e}}_2$. Take $\kappa ,\lambda$ such that $\overrightarrow{AE}=\kappa {\vec{e}}_1$ and $\overrightarrow{AF}=\lambda {\vec{e}}_2$. Then $$\begin{array}{rl}\overrightarrow{AR}& =\overrightarrow{AE}+\overrightarrow{AF}=\kappa {\vec{e}}_1+\lambda {\vec{e}}_2,\\ \overrightarrow{EB}& =\overrightarrow{AB}-\overrightarrow{AE}={\vec{e}}_2-\kappa {\vec{e}}_1,\\ \overrightarrow{FC}& =\overrightarrow{AC}-\overrightarrow{AF}={\vec{e}}_1-\lambda {\vec{e}}_2.\end{array}$$ Take also $\alpha ,\beta$ such that $\overrightarrow{EP}=\alpha \overrightarrow{EB}$ and $\overrightarrow{FP}=\beta \overrightarrow{FC}$. Then $$\begin{array}{rl}\overrightarrow{AP}& =\overrightarrow{AF}+\beta \overrightarrow{FC}=\lambda {\vec{e}}_2+\beta ({\vec{e}}_1-\lambda {\vec{e}}_2)=\beta {\vec{e}}_1+(1-\beta )\lambda {\vec{e}}_2,\\ \overrightarrow{AP}& =\overrightarrow{AE}+\alpha \overrightarrow{EB}=\kappa {\vec{e}}_1+\alpha ({\vec{e}}_2-\kappa {\vec{e}}_1)=(1-\alpha )\kappa {\vec{e}}_1+\alpha {\vec{e}}_2.\end{array}$$ Consequently, $\beta {\vec{e}}_1+(1-\beta )\lambda {\vec{e}}_2=(1-\alpha )\kappa {\vec{e}}_1+\alpha {\vec{e}}_2$. As vectors ${\vec{e}}_1$ and ${\vec{e}}_2$ are not parallel, the equality can hold only if the respective coefficients are equal, i.e., $\beta =(1-\alpha )\kappa$ and $(1-\beta )\lambda =\alpha$. This implies $\overrightarrow{AP}=\beta {\vec{e}}_1+\alpha {\vec{e}}_2$. Now we obtain $$\begin{array}{rl}\overrightarrow{PR}& =\overrightarrow{AR}-\overrightarrow{AP}=(\kappa {\vec{e}}_1+\lambda {\vec{e}}_2)-(\beta {\vec{e}}_1+\alpha {\vec{e}}_2)=(\kappa -\beta ){\vec{e}}_1+(\lambda -\alpha ){\vec{e}}_2,\\ \overrightarrow{AQ}& =\overrightarrow{AC}+\overrightarrow{CQ}=\overrightarrow{AC}+\overrightarrow{PB}=\overrightarrow{AC}+\overrightarrow{AB}-\overrightarrow{AP}\\ & ={\vec{e}}_1+{\vec{e}}_2-(\beta {\vec{e}}_1+\alpha {\vec{e}}_2)=(1-\beta ){\vec{e}}_1+(1-\alpha ){\vec{e}}_2.\end{array}$$ The lines $PR$ and $AQ$ being parallel is equivalent to the vectors $\overrightarrow{PR}$ and $\overrightarrow{AQ}$ being parallel, which in turn is equivalent to their coefficients being proportional. Hence it suffices to show that $\frac{\kappa -\beta }{1-\beta }=\frac{\lambda -\alpha }{1-\alpha }$, or equivalently, $$(\kappa -\beta )(1-\alpha )=(\lambda -\alpha )(1-\beta ).$$ As $\kappa (1-\alpha )=\beta$ and $\lambda (1-\beta )=\alpha$, we have $$\begin{array}{rl}(\kappa -\beta )(1-\alpha )& =\kappa (1-\alpha )-\beta (1-\alpha )=\beta -\beta +\alpha \beta =\alpha \beta ,\\ (\lambda -\alpha )(1-\beta )& =\lambda (1-\beta )-\alpha (1-\beta )=\alpha -\alpha +\alpha \beta =\alpha \beta .\end{array}$$ Hence the desired equality holds and the proof is complete.