IMO 2019 Shortlisted Problems

Solution 1. Throughout the solution we use oriented angles. Let rays $A{A}_1$ and $B{B}_1$ intersect the circumcircle of $△ACB$ at $A_2$ and $B_2$, respectively. By $$\angle QP{A}_2=\angle BA{A}_2=\angle B{B}_2{A}_2=\angle Q{B}_2{A}_2,$$ points $P,Q,A_2,B_2$ are concyclic; denote the circle passing through these points by $\omega$. We shall prove that $P_1$ and $Q_1$ also lie on $\omega$. By $$\angle C{A}_2{A}_1=\angle C{A}_{2}A=\angle CBA=\angle C{Q}_{1}Q=\angle C{Q}_1{A}_1,$$ points $C,Q_1,A_2,A_1$ are also concyclic. From that we get $$\angle Q{Q}_1{A}_2=\angle A_1{Q}_1{A}_2=\angle A_{1}C{A}_2=\angle BC{A}_2=\angle BA{A}_2=\angle QP{A}_2,$$ so $Q_1$ lies on $\omega$. It follows similarly that $P_1$ lies on $\omega$.

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Alternative solution.

Solution 2. First consider the case when lines $P{P}_1$ and $Q{Q}_1$ intersect each other at some point $R$. Let line $PQ$ meet the sides $AC$ and $BC$ at $E$ and $F$, respectively. Then $$\angle P{P}_{1}C=\angle BAC=\angle PEC,$$ so points $C,E,P,P_1$ lie on a circle; denote that circle by ${\omega }_P$. It follows analogously that points $C,F,Q,Q_1$ lie on another circle; denote it by ${\omega }_Q$. Let $AQ$ and $BP$ intersect at $T$. Applying Pappus' theorem to the lines $A{A}_{1}P$ and $B{B}_{1}Q$ provides that points $C=A{B}_1\cap B{A}_1,R=A_{1}Q\cap B_{1}P$ and $T=AQ\cap BP$ are collinear. Let line $RCT$ meet $PQ$ and $AB$ at $S$ and $U$, respectively. From $AB\Vert PQ$ we obtain $$\frac{SP}{SQ}=\frac{UB}{UA}=\frac{SF}{SE},$$ so $$SP\cdot SE=SQ\cdot SF.$$ So, point $S$ has equal powers with respect to ${\omega }_P$ and ${\omega }_Q$, hence line $RCS$ is their radical axis; then $R$ also has equal powers to the circles, so $RP\cdot R{P}_1=RQ\cdot R{Q}_1$, proving that points $P,P_1,Q,Q_1$ are indeed concyclic. Now consider the case when $P{P}_1$ and $Q{Q}_1$ are parallel. Like in the previous case, let $AQ$ and $BP$ intersect at $T$. Applying Pappus' theorem again to the lines $A{A}_{1}P$ and $B{B}_{1}Q$, in this limit case it shows that line $CT$ is parallel to $P{P}_1$ and $Q{Q}_1$. Let line $CT$ meet $PQ$ and $AB$ at $S$ and $U$, as before. The same calculation as in the previous case shows that $SP\cdot SE=SQ\cdot SF$, so $S$ lies on the radical axis between ${\omega }_P$ and ${\omega }_Q$. Line $CST$, that is the radical axis between ${\omega }_P$ and ${\omega }_Q$, is perpendicular to the line $\ell$ of centres of ${\omega }_P$ and ${\omega }_Q$. Hence, the chords $P{P}_1$ and $Q{Q}_1$ are perpendicular to $\ell$. So the quadrilateral $P{P}_1{Q}_{1}Q$ is an isosceles trapezium with symmetry axis $\ell$, and hence is cyclic.