Subfield: Geometry


Question: Let $A B C D$ be any convex quadrilateral and let $P, Q, R, S$ be points on the segments $A B, B C, C D$, and $D A$, respectively. It is given that the segments $P R$ and $Q S$ dissect $A B C D$ into four quadrilaterals, each of which has perpendicular diagonals. Show that the points $P, Q, R, S$ are concyclic.

Solution:
We start with a lemma which holds even in a more general setup.



Lemma 1. Let $P Q R S$ be a convex quadrangle whose diagonals meet at $O$. Let $\omega_{1}$ and $\omega_{2}$ be the circles on diameters $P Q$ and $R S$, respectively, and let $\ell$ be their radical axis. Finally, choose the points $A, B$, and $C$ outside this quadrangle so that: the point $P$ (respectively, $Q$ ) lies on the segment $A B$ (respectively, $B C$ ); and $A O \perp P S, B O \perp P Q$, and $C O \perp Q R$. Then the three lines $A C, P Q$, and $\ell$ are concurrent or parallel.



Proof. Assume first that the lines $P R$ and $Q S$ are not perpendicular. Let $H_{1}$ and $H_{2}$ be the orthocentres of the triangles $O S P$ and $O Q R$, respectively; notice that $H_{1}$ and $H_{2}$ do not coincide.



Since $H_{1}$ is the radical centre of the circles on diameters $R S, S P$, and $P Q$, 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 $A P H_{1}$ and $C Q H_{2}$ meet at $O, B$, and the orthocentre of the triangle $O P Q$ (which lies on $O B$ ). By Desargues' theorem, the lines $A C, P Q$ and $\ell$ are concurrent or parallel.



<img_3558>



The case when $P R \perp Q S$ 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 $P R$ and $Q S$ cross at $O$, let $\omega_{1}$ and $\omega_{2}$ be the circles on diameters $P Q$ and $R S$, respectively, and let $\ell$ be their radical axis. By the Lemma, the three lines $A C, \ell$, and $P Q$ are concurrent or parallel, and similarly so are the three lines $A C, \ell$, and $R S$. Thus, if the lines $A C$ and $\ell$ are distinct, all four lines are concurrent or pairwise parallel.



This is clearly the case when the lines $P S$ and $Q R$ are not parallel (since $\ell$ crosses $O A$ and $O C$ at the orthocentres of $O S P$ and $O Q R$, these orthocentres being distinct from $A$ and $C$ ). In this case, denote the concurrency point by $T$. If $T$ is not ideal, then we have $T P \cdot T Q=T R \cdot T S$ (as $T \in \ell$ ), so $P Q R S$ is cyclic. If $T$ is ideal (i.e., all four lines are parallel), then the segments $P Q$ and $R S$ have the same perpendicular bisector (namely, the line of centers of $\omega_{1}$ and $\omega_{2}$ ), and $P Q R S$ is cyclic again.



Assume now $P S$ and $Q R$ parallel. By symmetry, $P Q$ and $R S$ may also be assumed parallel: otherwise, the preceding argument goes through after relabelling. In this case, we need to prove that the parallelogram $P Q R S$ is a rectangle.







<img_3452>



Suppose, by way of contradiction, that $O P>O Q$. Let the line through $O$ and parallel to $P Q$ meet $A B$ at $M$, and $C B$ at $N$. Since $O P>O Q$, the angle $S P Q$ is acute and the angle $P Q R$ is obtuse, so the angle $A O B$ is obtuse, the angle $B O C$ is acute, $M$ lies on the segment $A B$, and $N$ lies on the extension of the segment $B C$ beyond $C$. Therefore: $O A>O M$, since the angle $O M A$ is obtuse; $O M>O N$, since $O M: O N=K P: K Q$, where $K$ is the projection of $O$ onto $P Q$; and $O N>O C$, since the angle $O C N$ is obtuse. Consequently, $O A>O C$.



Similarly, $O R>O S$ yields $O C>O A$ : a contradiction. Consequently, $O P=O Q$ and $P Q R S$ is a rectangle. This ends the proof.
To begin, we establish a useful lemma.



Lemma 2. If $P$ is a point on the side $A B$ of a triangle $O A B$, then



$$

\frac{\sin A O P}{O B}+\frac{\sin P O B}{O A}=\frac{\sin A O B}{O P}

$$



Proof. Let $[X Y Z]$ denote the area of a triangle $X Y Z$, to write



$0=2([A O B]-[P O B]-[P O C])=O A \cdot O B \cdot \sin A O B-O B \cdot O P \cdot \sin P O B-O P \cdot O A \cdot \sin A O P$, and divide by $O A \cdot O B \cdot O P$ to get the required identity.



A similar statement remains valid if the point $C$ lies on the line $A B$; the proof is obtained by using signed areas and directed lengths.



We now turn to the solution. We first prove some sort of a converse statement, namely:



Claim. Let $P Q R S$ be a cyclic quadrangle with $O=P R \cap Q S$; 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 $S P$, $P Q, Q R$, and $R S$, respectively. Choose any point $A \in \ell_{A}$ and successively define $B=A P \cap \ell_{B}$, $C=B Q \cap \ell_{C}, D=C R \cap \ell_{D}$, and $A^{\prime}=D S \cap \ell_{A}$. 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 $P Q R S$. Again, a general case is obtained by suitable consideration of directed lengths.







<img_3651>



Denote



$$

\begin{gathered}

\alpha=\angle Q P R=\angle Q S R=\pi / 2-\angle P O B=\pi / 2-\angle D O S, \\

\beta=\angle R P S=\angle R Q S=\pi / 2-\angle A O P=\pi / 2-\angle Q O C, \\

\gamma=\angle S Q P=\angle S R P=\pi / 2-\angle B O Q=\pi / 2-\angle R O D, \\

\delta=\angle P R Q=\angle P S Q=\pi / 2-\angle C O R=\pi / 2-\angle S O A .

\end{gathered}

$$



By Lemma 2 applied to the lines $A P B, P Q C, C R D$, and $D S A^{\prime}$, we get



$$

\begin{aligned}

& \frac{\sin (\alpha+\beta)}{O P}=\frac{\cos \alpha}{O A}+\frac{\cos \beta}{O B}, \quad \frac{\sin (\beta+\gamma)}{O Q}=\frac{\cos \beta}{O B}+\frac{\cos \gamma}{O C} \\

& \frac{\sin (\gamma+\delta)}{O R}=\frac{\cos \gamma}{O C}+\frac{\cos \delta}{O D}, \quad \frac{\sin (\delta+\alpha)}{O S}=\frac{\cos \delta}{O D}+\frac{\cos \alpha}{O A^{\prime}}

\end{aligned}

$$



Adding the two equalities on the left and subtracting the two on the right, we see that the required equality $A=A^{\prime}$ (i.e., $\cos \alpha / O A=\cos \alpha / O A^{\prime}$, in view of $\cos \alpha \neq 0$ ) is equivalent to the relation



$$

\frac{\sin Q P S}{O P}+\frac{\sin S R Q}{O R}=\frac{\sin P Q R}{O Q}+\frac{\sin R S P}{O S}

$$



Let $d$ denote the circumdiameter of $P Q R S$, so $\sin Q P S=\sin S R Q=Q S / d$ and $\sin R S P=$ $\sin P Q R=P R / d$. Thus the required relation reads



$$

\frac{Q S}{O P}+\frac{Q S}{O R}=\frac{P R}{O S}+\frac{P R}{O Q}, \quad \text { or } \quad \frac{Q S \cdot P R}{O P \cdot O R}=\frac{P R \cdot Q S}{O S \cdot O Q}

$$



The last relation is trivial, due again to cyclicity.



Finally, it remains to derive the problem statement from our Claim. Assume that $P Q R S$ is not cyclic, e.g., that $O P \cdot O R>O Q \cdot O S$, where $O=P R \cap Q S$. Mark the point $S^{\prime}$ on the ray $O S$ so that $O P \cdot O R=O Q \cdot O S^{\prime}$. Notice that no diagonal of $P Q R S$ is perpendicular to a side, so the quadrangle $P Q R S^{\prime}$ 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 $A P$ and $R D$, 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 $O S$ at some interior point, while $S^{\prime}$ lies on the extension of this segment. This contradiction completes the proof.