@above I'm sorry, is that official solution?

Wow, yes, indeed, that is the second official solution. Actually, I came up with the solution in 15 minutes after the contest, after learning that the 'porism' Theorem is true, but Ilya Bogdanov did come up with it first.

Interestingly the problem is false if PQRS may lay outside the interiors of the sides. In particular, for arbitrary PQRS (not necessarily cyclic) with diagonals meeting at O, one may let ABCD be the orthocenters of POQ, QOR, ROS, SOP to get a valid construction. I have an algebraic solution (using degree counting of complex polynomials) which I'll post tonight once I write it up. Quite an impressive problem, from an algebraic perspective

Here is the brutal algebraic solution which I promised. It is quite involved, since I need to use complex numbers twice and Cartesian coordinates to finish. Despite this, the trickiest step is actually the one synthetic step in the middle. Suppose $PQRS$ is any quadrilateral (possibly concave or self-intersecting!). Let $O = {PR} \cap {QS}$. Let $W$, $X$, $Y$, $Z$ be the feet from $O$ to ${SP}$, ${PQ}$, ${QR}$, ${RS}$. We say that a quadruple of points \[ (A,B,C,D) \in \overline{OW} \times \overline{OX} \times \overline{OY} \times \overline{OZ} \]is good if $P$, $Q$, $R$, $S$ lies on lines $AB$, $BC$, $CD$, $DA$, and $A, B, C, D \neq O$. If additionally $ABCD$ is convex and $P$, $Q$, $R$, $S$ lie on the segments $AB$, $BC$, $CD$, $DA$, we say the quadruple is excellent. We say $PQRS$ is ordinary if there exists exactly one good quadruple, exceptional otherwise. Thus the problem asks us to show any quadrilateral with an excellent quadruple is cyclic. We show now that ``most quadrilaterals are ordinary''. Claim: A quadrilateral is exceptional if and only if $ \prod_{\text{cyc}} \cos \angle RPQ = \prod_{\text{cyc}} \cos \angle RPS. $

We now follow up by classifying all exceptional quadrilaterals. Claim: A quadrilateral $PQRS$ (possibly concave or self-intersecting) is exceptional if and only if it is cyclic, or $\overline{PR} \perp \overline{QS}$ or $\overline{PR} \parallel \overline{QS}$ (which means $PQRS$ is self-intersecting).

Thus nearly all quadrilaterals have a unique good quadruple. The surprise is that this quadruple is never excellent: Claim: The unique good quadruple of an ordinary quadrilateral $PQRS$ is not excellent.

Therefore all that remains is to check that if a quadrilateral has perpendicular diagonals but is not cyclic, then it has no good quadruples at all. This can be checked by Cartesian coordinates, which we outline before: set $O = (0,0)$, $P = (0,p)$, $Q = (-q,0)$, $R = (0,-r)$, $S=(s,0)$. We may pick $A$ and $B$ such that $\overline{AB} \parallel \overline{QS}$, hence $A = (p^2/s, s)$ and then $ D = \overline{AS} \cap \overline{OZ} = \frac{ps}{p^2+pr-s^2}(r,-s)$. Similarly, $C = \overline{BQ} \cap \overline{OX} = \frac{-pq}{p^2+pr-q^2}(r,q)$, (by replacing $s$ with $-q$ everywhere). Now the points $C$, $D$, $R = (0,-r)$ are collinear if and only if \[ 0 = \det \begin{bmatrix} prs & -ps^2 & p^2+pr-s^2 \\ -pqr & -pq^2 & p^2+pr-q^2 \\ 0 & -r & 1 \end{bmatrix} = pq(p+r)(q+s)(pr-qs) \]which amounts to $PQRS$ being cyclic, as desired.

randomusername wrote: Theorem. Fix a cyclic quadrilateral $PQRS$ whose diagonals meet at $O$. Let the perpendiculars from $O$ to $SP,PQ,QR,RS$ be $a,b,c,d$, and let $A\in a$ be arbitrary. Next, define $B=AP\cap b$, $C=BP\cap c$, $D=CP\cap d$, $A'=DP\cap a$. Then $A'=A$. Something is wrong with formulation of your lemma(theorem), I mean you define $B$ as intersection of $AP$ and $b$ and so $A$, $B$ and $P$ are collinear, so $BP\cap c$ is same as $AP\cap c$, similarly $DP\cap a$ is same as $AP\cap a$ which is obviously $A$... Am I missing something or am I right?

Oh, I got it... You have mistake in formulation, it should be $BQ$, $CR$ and $DS$... Anyway, now that I've got it, what is the motivation for such a lemma? I mean it's not known to me... So how would one come up with such an idea? And how do u conclude this: Quote: By the Theorem, however, $A',S',D'$ must be collinear

@artsolver: $A',S',D'$ collinear is a direct corollary of the Theorem applied to cyclic quad $PQRS'$ (yes, the notation is slightly different): we must have $D'S'\cap OA'=A'$. A memorable idea for motivation is to try to construct the diagram. If you start with $ABCD$, it's pretty arbitrary to find $P,Q,R,S$ which just happen to create four orthodiagonal quadrilaterals. The central part of the diagram is the cyclic quadrilateral $PQRS$. The conjecture that you will get back from not just one, but any point $A$ (i.e. that you always have $A'=A$ and not just sometimes) came for me from feeling like there should be enough freedom in the diagram, but you can also motivate/check things like this by, say, looking at the special case where $PQRS$ is a square or drawing a really good diagram (or both).

Thanks, that is pretty cool! When I tried constructing, I noticed, as probably everyone else, that it would be easier to start with $PQRS$, and eas afraid that I wouldn't get overlap of $A$ and $A'$, but now as you are saying motivation for your lemma, I am not surprised that my diagram was successfull at first take, so thanks.

IstekOlympiadTeam wrote: Let $ABCD$ be any convex quadrilateral and let $P, Q, R, S$ be points on the segments $AB, BC, CD$, and $DA$, respectively. It is given that the segments $PR$ and $QS$ dissect $ABCD$ into four quadrilaterals, each of which has perpendicular diagonals. Show that the points $P, Q, R, S$ are concyclic. That's very nice problem. I see the following problem, based on this problem, it can be considerd as an extension of Newton line in circumscribed quadrilateral. Let $ABCD$ be any convex quadrilateral and let $P, Q, R, S$ be points on the segments $AB, BC, CD$, and $DA$, respectively. It is given that the segments $PR$ and $QS$ dissect $ABCD$ into four quadrilaterals, each of which has perpendicular diagonals. a) Show that the points $P, Q, R, S$ lie on circle $(K)$. b) Let $M,N$ be midpoints of segment $AC,BD$. Prove that $M,K,N$ are collinear.

Here is an alternative solution, that I think is more elegant than both the official solutions. By applying inversion with center at the point of intersection of $PR$ and $QS$, it is not hard to reduce this problem to the following one: https://artofproblemsolving.com/community/c6h1389791_an_interesting_property_of_a_parallelogram. The latter one does not have too elegant solution (at least I am not aware of any such one), but I still think it is much easier than the original problem.

Here is my (almost) synthetic solution using dynamic geometry. However, I am not able to solve the case $PR\perp QS$ synthetically. If anyone did, please tell me. From now assume that $PR$ is not perpendicular to $QS$. Fix the quadrilateral $PQRS$. Let $T = PR\cap QS$ and let $W, X, Y, Z$ be the feet from $T$ to $SP, PQ, QR, RS$. Let $A$ move along $TW$ and redefine $B, D$ as $AP\cap TX, AS\cap TZ$ respectively. Let $C_1 = BQ\cap TY$ and $C_2 = DR\cap TY$. Observe that the problem's condition implies there exists $A$ on $\overrightarrow{TW}$ which satisfies $C_1=C_2$. Note that mapping $A\to C_1$ is an projective mapping as it's a sequence of two projections. Similarly $A\to C_2$ is also projective mapping. Note that $C_1=C_2$ for the following two cases. When $A=T$, then $B = D=T$ which implies $C_1=C_2=T$. When $A$ is the orthocenter of $\Delta PTS$, we get $B, D$ are orthocenters of $\Delta PTQ, \Delta RTS$. Thus $C_1, C_2$ are orthocenter of $\Delta QTR$. Thus the problem's condition implies there exists third point $A$ which makes $C_1=C_2$, which implies $C_1=C_2=C$ for every choice of $A$. Now we have to pick a good choices of $A$. Pick $A = A_1 = TW\cap PQ$. This makes $B=B_1 = X$ and $C=C_1 = TY\cap PQ$. Thus lines $A_1S, C_1R, TZ$ are concurrent at $E_1$. Pick $A = A_2 = TW\cap RS$. This makes $D=D_2=Z$ and $C=C_2 = TY\cap RS$. Thus lines $A_2P, C_2Q, TX$ are concurrent at $E_2$. Let $F_1 = A_2Q\cap C_2P$. We claim that $\{T, E_1, F_1\}$ are colinear. To prove that, take homography which sends $PQRS$ to square. Then it's easy to see that $\{A_1, A_2\}, \{C_1, C_2\}$ are symmetry across $T$. Thus $\{E_1, F_1\}=\{A_2Q\cap C_2P, A_1S\cap A_2R\}$ are symmetric across $T$ too. Hence $T, E_1, F_1$ are colinear as claimed. Now by Dual of Desargues Involution Theorem on point $T$ and quadrilateral $\{\overline{A_1P}, \overline{C_1Q}, \overline{A_1Q}, \overline{C_1P}\}$, we get and involution swapping $(TR, TS), (TA_1, TC_1) \equiv (TW, TY), (TF_1, TE_2)\equiv (TX, TZ)$. Rotate this involution by $90^{\circ}$ and project on line at infinity, we get an involution swapping $({\infty}_{PS}, {\infty}_{QR}), ({\infty}_{PQ}, {\infty}_{RS}), ({\infty}_{\perp PR}, {\infty}_{\perp QS})$. Now by Desargues Involution's Theorem on line at infinity and quadrilateral $PQRS$, we get that the current involution swaps $({\infty}_{PR}, {\infty}_{QS})$ too. Viewing this involution from $T$, it's easy to see that it must be reflection across $\angle(PR, QS)$. Thus $\measuredangle(PQ, PR) = \measuredangle(RS, QS)$ which implies that $PQRS$ is cyclic as desired. Now we have to eradicate the case $PR\perp QS$ which v_Enhance is already did for us. v_Enhance wrote: Set $T = (0,0)$, $P = (0,p)$, $Q = (-q,0)$, $R = (0,-r)$, $S=(s,0)$. We may pick $A$ and $B$ such that $\overline{AB} \parallel \overline{QS}$, hence $A = (p^2/s, s)$ and then $ D = \overline{AS} \cap \overline{OZ} = \frac{ps}{p^2+pr-s^2}(r,-s)$. Similarly, $C = \overline{BQ} \cap \overline{OX} = \frac{-pq}{p^2+pr-q^2}(r,q)$, (by replacing $s$ with $-q$ everywhere). Now the points $C$, $D$, $R = (0,-r)$ are collinear if and only if \[ 0 = \det \begin{bmatrix} prs & -ps^2 & p^2+pr-s^2 \\ -pqr & -pq^2 & p^2+pr-q^2 \\ 0 & -r & 1 \end{bmatrix} = pq(p+r)(q+s)(pr-qs) \]which amounts to $PQRS$ being cyclic, as desired.

IstekOlympiadTeam wrote: Let $ABCD$ be any convex quadrilateral and let $P, Q, R, S$ be points on the segments $AB, BC, CD$, and $DA$, respectively. It is given that the segments $PR$ and $QS$ dissect $ABCD$ into four quadrilaterals, each of which has perpendicular diagonals. Show that the points $P, Q, R, S$ are concyclic. First we prove the following result. Lemma: Let $POQ$ be a non-isosceles triangle with circumcenter $O_1$ and assume $\angle POQ \ne 90^{\circ}$. Let $\ell_P, \ell_Q$ be lines through $P$ and $Q$ respectively, perpendicular to $\overline{PQ}$. Suppose $R,S$ lie on $\overline{OQ}, \overline{OP}$ respectively; such that $\overline{RS} \perp \overline{OO_1}$. Let $X \in \ell_P, Y \in \ell_Q$ with $\overline{OX} \perp \overline{PS}$ and $\overline{OY} \perp \overline{QR}$. Let $Z_P=\overline{XS} \cap \overline{OO_1}, Z_Q=\overline{YR} \cap \overline{OO_1}$. Then $Z_P=Z_Q$. (Proof) Move $R$ with parameter $r$; $R \mapsto S$ is projective. Also $S \mapsto X$ and $R \mapsto Y$ are projective. Thus, $Z_P$ moves on $\overline{OO_1}$ with parameter $\tfrac{p(r)}{q(r)}$ where $p,q$ are polynomials of degree no more than $2$. Thus, $Z_P=Z_Q$ is a polynomial equation in $r$ of degree no more than $4$. In order to show it is an identity, we only need to check five cases: $R=R'$ for any of $R'=O, R'=\infty, R'=P, \angle OR'Q=\angle OQP, \angle OR'Q=90^{\circ}$ is clearly a solution. Thus, we are done! $\blacksquare$ Let $O=\overline{PR} \cap \overline{QS}$. Let $\ell_{PQ}$ be the line through $O$ perpendicular to $\overline{PQ}$. Define $\ell_{QR}, \ell_{RS}, \ell{SP}$ similarly. For now, suppose $\overline{PO} \perp \overline{OQ}$ is not the case. Suppose $A \in \ell_{PQ}$; define $B \overset{\text{def}}{:=} \overline{PA} \cap \ell_{PQ}$, $C \overset{\text{def}}{:=} \overline{QB} \cap \ell_{QR}$, $D \overset{\text{def}}{:=} \overline{RC} \cap \ell_{RS}$ and $E \overset{\text{def}}{:=} \overline{SD} \cap \ell_{SP}$. Define $f: \ell_{PQ} \mapsto \ell_{PQ}$ and $f(A)=E$ for all $A$ is a projective map. Claim: If $PQRS$ is cyclic then $f$ is the identity map. (Proof) If $PQRS$ is a rectangle then we are done by symmetry. WLOG $OP \ne OQ$ (else we could define $f$ with respect to other lines). Note that $A=E$ when $A=O$ and when $A$ coincides with the orthocenter of $\triangle POQ$. By the lemma, $A=E$ also when $A$ lies on the line at infinity. Thus, $f$ is indeed the identity map. $\blacksquare$ Onto the problem. Note that $f$ has three fixed points; the vertex $A$ of the convex quadrilateral $ABCD$, the orthocenter of triangle $POS$ and the point $O$. Thus, $f$ is an identity. We show that the only quadrilaterals for which $f$ is an identity are cyclic. Indeed, fix $P,Q,S,O$ and move point $R$ on line $\overline{PO}$ with parameter $r$; fix $\overline{AB} \parallel \overline{QS}$. Then $R \mapsto C$ and $R \mapsto D$ are projective maps (redefining $D$ to lie on $\overline{AS}$). Now $R,C,D$ collinear is a cubic (at best) equation in $r$. This fails at $r=\infty$ so it is not an identity and at most three values of $r$ work. Now $R=P, R=O$ and $PQRS$ cyclic work and so nor more roots exist. Thus, $PQRS$ is cyclic. $\blacksquare$ Remarks: If $\overline{PR} \perp \overline{QS}$ then $PQRS$ must be cyclic in order for the condition to hold. We handle this in the same manner as post #6. Let $\ell_1, \ell_2, \ell_3$ be lines concurrent at $O$ and $P_1(t), P_2(t), P_3(t)$ be rational-linear functions (projective maps on a point moving with parameter $t$) denoting positions of points $P_1, P_2, P_3$ on these lines. Then $P_1, P_2, P_3$ collinear is a polynomial equation in $t$ of degree $\le 3$. Indeed, take a homography mapping $O$ to a point at infinity. Then these lines become parallel. Now we just wish to solve $\overrightarrow{P_3}=\lambda \overrightarrow{P_1} +(1-\lambda)\overrightarrow{P_2}$ for fixed $\lambda$ (expressing $\ell_3$ by section formula in $\ell_1, \ell_2$). This is clearly of degree $\le 3$ in $t$. Let $\ell_1, \ell_2, \ell_3$ be lines concurrent at $O$ and $P_1(t), P_2(t)$ be rational-linear functions (projective maps on a point moving with parameter $t$) denoting positions of points $P_1, P_2$ on their locii. Then $P_3 =\ell_3 \cap \overline{P_1P_2}$ is parametrised by $\frac{p(r)}{q(r)}$ where $p,q$ have degree at most $2$. Indeed, take a homography mapping $O$ to a point at infinity. Then these lines become parallel. Now $\overrightarrow{P_3}=\lambda \overrightarrow{P_1} +(1-\lambda)\overrightarrow{P_2}$ for fixed $\lambda$ (expressing $\ell_3$ by section formula in $\ell_1, \ell_2$). Thus, $P_3$ has the desired parametric form.

Let us fix $PQRS$ and vary $ABCD$. Let $X = PR\cap QS$, and denote the line through $X$ perpendicular to $SP$ (resp. $PQ,QR,RS$) by $\ell_A$ (resp. $\ell_B, \ell_C, \ell_D$). We move $A$ linearly on $\ell_A$, and define $B = AP\cap \ell_B$, $C = BQ\cap \ell_C$, $D = CR\cap \ell_D$, $A' = DS\cap \ell_A$. By definition, the map $f: A\mapsto A'$ is a projective transformation on the real projective line $\ell_A \cup \{\infty\}$. We know that $f$ fixes $X$ and some given point $A$, and furthermore $f$ fixes the orthocenter $\tilde{A}$ of $PXS$. Clearly $A$ is distinct from $X$ and $\tilde{A}$, since $A$ and $X$ lie on different sides of line $PS$. If $X = \tilde{A}$, then $PR\perp QS$ and we can finish easily by using Cartesian coordinates and setting $X$ as the origin. So we can assume $A, \tilde{A}, X$ are three distinct fixed points of $f$, which implies that $f$ is the identity. Now, suppose for contradiction that $XP\cdot XR < XS\cdot XQ$. We choose $A\in\ell_A$ such that $AQ\perp SQ$, then we can easily derive a contradiction using power of a point at $X$.

Steiner inconic induced brain damage. Let $E = PR \cap QS$. Redefine the problems in terms of $P, Q, R, S, E$, and then define $A \mapsto B \mapsto C \mapsto D$ on the altitudes from $E$. It then remains to show that a solution $ABCD$ where the points lie strictly in the segments only exists when cyclic. By steiner inconic it follows that $AD$ has a fixed point. Claim: The fixed point of $AD$ is $Q$ when $PRQS$ is cyclic. Proof. It remains to show that when $A$ is on $PQ$, then $D$ also lies on $PQ$. Redefine this by defining $D'$ on $PQ$ and the perpendicular, and $C' = D'S \cap RD$. Then it remains to show that $EC' \perp RS$. This statement has degree $4$ when moving $S$, taking $S = P, Q, R, SP \parallel QR, SR \parallel QP$ finishes. $\blacksquare$ Claim: There's at most one value of $A$ such that $AD$ goes through $Q$ when not cyclic. Proof. The fixed point $F$ isn't $Q$, thus $AD$ must be the line $FQ$. $\blacksquare$ In fact, we claim the following: Claim: The fixed point moves quadratically on a fixed line through $Q$. Proof. Define $A_1$ as another point on $EA$. Define $B_1, C_1, D_1$ similarly. The fixed point is then $AD \cap A_1D_1$. Move $S$ along $QE$ fixing $A_1, B_1, P, Q, R, E$. Then $\deg EB = \deg EC = 1$, $\deg B = 1, \deg C = \deg D = \deg AD = 2$. As such, the fixed point has degree $4$. Note that $AD = A_1D_1$ when $\measuredangle PSQ = 90^\circ, \measuredangle SRP = 90^\circ$, so by Zack's lemma we can reduce to degree $2$. Then taking the two cases $S \in (PQR)$ finishes because the fixed point is $Q$ twice. $\blacksquare$ Now, by taking at $S$ at infinity and $A \in PQ$, $A_1$ approaching $E$, we get a fixed point at $PR \cap AD$, which evidently lies in the angle $\angle PQR$. This implies that there's no convex $ABCD$ when noncyclic.