THE Tenth ROMANIAN MASTER OF MATHEMATICS

Lemma 1. Let $PQRS$ be a convex quadrangle whose diagonals cross at $O$. Let ${\omega }_1$ and ${\omega }_2$ be the circles on diameters $PQ$ and $RS$, respectively, and let $\ell$ be their radical axis. Finally, let $A,B$, and $C$ be points outside this quadrangle so that: the point $P$ (respectively, $Q$) lies on the segment $AB$ (respectively, $BC$); and $AO\perp PS$, $BO\perp PQ$, and $CO\perp QR$. Then the three lines $AC,PQ$, and $\ell$ are concurrent or parallel.

Proof. Assume first that the lines $PR$ and $QS$ are not perpendicular. Let $H_1$ and $H_2$ be the orthocentres of the triangles $OSP$ and $OQR$, respectively; notice that $H_1$ and $H_2$ do not coincide. Since $H_1$ is the radical center of the circles on diameters $RS,SP$, and $PQ$, it lies on $\ell$. Similarly, $H_2$ lies on $\ell$, so the lines $H_1{H}_2$ and $\ell$ coincide. The corresponding sides of the triangles $AP{H}_1$ and $CQ{H}_2$ meet at $O,B$, and the orthocenter of the triangle $OPQ$ (which lies on $OB$). By Desargues' theorem, the lines $AC,PQ$ and $\ell$ are concurrent or parallel. The case when $PR\perp QS$ may be considered as a limit case, since the configuration in the statement of the lemma allows arbitrarily small perturbations. The lemma is proved.

Back to the problem, let the segments $PR$ and $QS$ cross at $O$, let ${\omega }_1$ and ${\omega }_2$ be the circles on diameters $PQ$ and $RS$, respectively, and let $\ell$ be their radical axis. By the Lemma, the three lines $AC,\ell$, and $PQ$ are concurrent or parallel, and similarly so are the three lines $AC,\ell$, and $RS$. Thus, if the lines $AC$ and $\ell$ are distinct, all four lines are concurrent or pairwise parallel. This is clearly the case if the lines $PS$ and $QR$ are not parallel, since $\ell$ crosses $OA$ and $OC$ at the orthocentres of $OSP$ and $OQR$, and these orthocentres are distinct from $A$ and $C$. In this case, denote the concurrency point by $T$. If $T$ is not ideal, then $$TP\cdot TQ=TR\cdot TS,$$ since $T$ lies on $\ell$, so $PQRS$ is cyclic. If $T$ is ideal (i.e., all four lines are parallel), then the segments $PQ$ and $RS$ have the same perpendicular bisector (namely, the line of centers of ${\omega }_1$ and ${\omega }_2$), and $PQRS$ is cyclic again.



Assume now $PS$ and $QR$ parallel. By symmetry, $PQ$ and $RS$ may also be assumed parallel: otherwise, the preceding argument goes through after relabeling. In this case, we need to prove that the parallelogram $PQRS$ is a rectangle. Suppose, by way of contradiction, that $OP>OQ$. Let the line through $O$ and parallel to $PQ$ meet $AB$ at $M$, and $CB$ at $N$. Since $OP>OQ$, the angle $SPQ$ is acute and the angle $PQR$ is obtuse, so the angle $AOB$ is obtuse, the angle $BOC$ is acute, $M$ lies on the segment $AB$, and $N$ lies on the extension of the segment $BC$ beyond $C$. Therefore: $OA>OM$, since the angle $OMA$ is obtuse; $OM>ON$, since $$OM:ON=KP:KQ,$$ where $K$ is the projection of $O$ onto $PQ$; and $ON>OC$, since the angle $OCN$ is obtuse. Consequently, $OA>OC$.



Similarly, $OR>OS$ yields $OC>OA$: a contradiction. Consequently, $OP=OQ$ and $PQRS$ is a rectangle. This ends the proof.



---

Alternative solution.

Second solution. (Ilya Bogdanov) To begin, we establish a simple, but useful lemma.

Lemma 2. If $P$ is a point on the side $AB$ of a triangle $OAB$, then $$\frac{\sin AOP}{OB}+\frac{\sin POB}{OA}=\frac{\sin AOB}{OP}.$$ Proof. Write $0=2(\text{area }AOB-\text{area }POB-\text{area }POA)=OA\cdot OB\cdot \sin AOB-OB\cdot OP\cdot \sin POB-OP\cdot OA\cdot \sin AOP$, and divide by $OA\cdot OB\cdot OP$ to get the required identity. A similar statement holds if the point $P$ lies on the line $AB$; the proof is obtained by using signed areas and directed lengths.

We now turn to the solution proper and prove the kind of converse below.

Claim. Let $PQRS$ be a cyclic quadrangle with $O=PR\cap QS$; assume that no its diagonal is perpendicular to a side. Let ${\ell }_A,{\ell }_B,{\ell }_C$, and ${\ell }_D$ be the lines through $O$ perpendicular to $SP,PQ,QR,\text{ and }RS$, respectively. Choose any point $A\in {\ell }_A$ and successively define $B=AP\cap {\ell }_B,C=BQ\cap {\ell }_C,D=CR\cap {\ell }_D$, and $A^{\prime }=DS\cap {\ell }_A$. Then $A^{\prime }=A$.



Proof. We restrict ourselves to the case when the points $A,B,C,D$, and $A^{\prime }$ lie on ${\ell }_A,{\ell }_B,{\ell }_C,{\ell }_D$, and ${\ell }_A$ on the same side of $O$ as their points of intersection with the respective sides of the quadrilateral $PQRS$. Again, a general case is obtained by suitable consideration of directed lengths. Write $$\begin{array}{rl}\alpha & =\angle QPR=\angle QSR=\pi /2-\angle POB=\pi /2-\angle DOS,\\ \beta & =\angle RPS=\angle RQS=\pi /2-\angle AOP=\pi /2-\angle QOC,\\ \gamma & =\angle SQP=\angle SRP=\pi /2-\angle BOQ=\pi /2-\angle ROD,\\ \delta & =\angle PRQ=\angle PSQ=\pi /2-\angle COR=\pi /2-\angle SOA,\end{array}$$ and apply Lemma 2 to the lines $APB,PQC,CRD$, and $DS{A}^{\prime }$ to get $$\begin{array}{rlrl}\frac{\sin (\alpha +\beta )}{OP}& =\frac{\cos \alpha }{OA}+\frac{\cos \beta }{OB},& \frac{\sin (\beta +\gamma )}{OQ}& =\frac{\cos \beta }{OB}+\frac{\cos \gamma }{OC},\\ \frac{\sin (\gamma +\delta )}{OR}& =\frac{\cos \gamma }{OC}+\frac{\cos \delta }{OD},& \frac{\sin (\delta +\alpha )}{OS}& =\frac{\cos \delta }{OD}+\frac{\cos \alpha }{O{A}^{\prime }}.\end{array}$$ Adding the two equalities on the left and subtracting the two on the right, the required equality $A=A^{\prime }$ (i.e., $\cos \alpha /OA=\cos \alpha /O{A}^{\prime }$, in view of $\cos \alpha \ne 0$) is equivalent to $$\frac{\sin QPS}{OP}+\frac{\sin SRQ}{OR}=\frac{\sin PQR}{OQ}+\frac{\sin RSP}{OS}.$$ Let $d$ denote the circumdiameter of $PQRS$, so $\sin QPS=\sin SRQ=QS/d$ and $\sin RSP=\sin PQR=PR/d$. Thus the required relation reads $$\frac{QS}{OP}+\frac{QS}{OR}=\frac{PR}{OS}+\frac{PR}{OQ},\text{or}\frac{QS\cdot PR}{OP\cdot OR}=\frac{PR\cdot QS}{OS\cdot OQ}.$$ The latter holds since $PQRS$ is cyclic.



Finally, we use the Claim to complete the proof. Assume that $PQRS$ is not cyclic, e.g., that $OP\cdot OR>OQ\cdot OS$, where $O=PR\cap QS$. Mark the point $S^{\prime }$ on the ray $OS$ so that $OP\cdot OR=OQ\cdot O{S}^{\prime }$. Notice that no diagonal of $PQRS$ is perpendicular to a side, so the quadrangle $PQR{S}^{\prime }$ satisfies the conditions of the claim. Let ${\ell }_A^{\prime }$ and ${\ell }_D^{\prime }$ be the lines through $O$ perpendicular to $P{S}^{\prime }$ and $R{S}^{\prime }$, respectively. Then ${\ell }_A^{\prime }$ and ${\ell }_D^{\prime }$ cross the segments $AP$ and $RD$, respectively, at some points $A^{\prime }$ and $D^{\prime }$. By the Claim, the line $A^{\prime }{D}^{\prime }$ passes through $S^{\prime }$. This is impossible, because the segment $A^{\prime }{D}^{\prime }$ crosses the segment $OS$ at some interior point, while $S^{\prime }$ lies on the extension of this segment. This contradiction completes the proof.