Knock knock, who was there? Brazil 2016/6 Lemma: Consider a cyclic, convex quadrilateral $XYZW$ with circumcenter $O$ and Miquel Point $M$. Then, $MXOZ,MYOW$ are cyclic. Proof: Let $\omega= (XYZW), T= XZ \cap YW, U= XY \cap ZW, V=XW \cap YZ$. It's well known that $UV= \Pi_{\omega}(T)$ and $M \in UV$. Furthermore, $O,T,M$ are collinear. Now, invert about $\omega$. $\square$ Now, let $M=(OAC) \cap (OBD)$, with $O \neq M$. From the lemma, we know that $M$ is the center of the spiral similarity mapping $AB \mapsto DC$. Let $\Phi$ be the inversion centered at $M$ with radius $\sqrt{MA.MC}$ followed by a reflection WRT the internal angle bisector of $\angle AMC$. It is well known that $\Phi(A)=C, \Phi(B)=D, \Phi(C)=A, \Phi(D)=B$. Moreover, since $M \in (OAC),(OBD)$, then $\Phi(O)= AC \cap BD=P$. $\quad (\heartsuit)$ Observe that $\angle BYC= \angle XYZ= 180º- \frac{\angle ABC}{2}-\frac{\angle BCD}{2}= \frac{\angle BAD}{2} + \frac{\angle CDA}{2}= 180º- \angle XWZ \implies XYZW$ is cyclic. Now, observe that $\angle FZC= \angle AWE$, so $\Phi(W)$ lies on $(FZC)$. Similarly, $\Phi(W)$ lies on $(XBE)$, so $\Phi(W)$ is the Miquel Point of $BXZC$. Thus, $\Phi(Z) \in (XYZW)$. Similarly, $\Phi(X), \Phi(Y), \Phi(Z) \in (XYZW) \implies \Phi(XYZW)=(XYZW).$ $(\spadesuit)$ (here we could simply quote Brazil 2016/6) From $(\heartsuit)$, $(\spadesuit)$, $O \in (XYZW)$ if and only if its image, AKA $P$, lies on $(XYZW)$. Therefore, $O \in (XYZW)$ if and only if $P \in (XYZW)$, as desired.

Let's discard the case when $ABCD$ is trapezoid. Firstly, $XYZW$ is cyclic by easy angle chase. Let $AB$ and $CD$ intersect at $T$ and $BC$ and $AD$ intersect at $Q$. The key claim, is that line $TQ$ is the radical axis of the circle with diameter $OP$ and the circumcircle of $ABCD$, as well as its also radical axis of the circumcircle of $XYZW$ and the circumcircle of $ABCD$. To prove first claim, denote points of intersection of line $AC$ with the line $TQ$ as $K$, and line $BD$ with the line $TQ$ as $L$ also let $M, N$ be midpoints of $AC$ and $BD$ respectively. Then $(KP, AC) = -1 = (LP, BD)$, as a result $KA \cdot KC = KP \cdot KM$ and $LB \cdot LD = LP \cdot LN$, so points $K$ and $L$ belong to the desired radical axis. Since lines $KL$ and $TQ$ are the same, the first part follows. For the second part, note that lines $XZ$ and $YW$ are bisectors of angles between lines $AD$, $BC$ and $AB$, $CD$ respectively, then $AXZD, BXZC$ are cyclic by angle chase, $XZ$ passes through $Q$ and $QA \cdot QD = QX \cdot QZ = QB \cdot QC$, thus $Q$ belongs to the radical axis of circles $XYZT$ and $ABCD$, analogously for $T$. $Q.E.D.$

Pranjal's blog post makes this problem a lot more feasible to solve (obviously Brazil 2016/6 does too, but I didn't know about that.) $X,Y,Z,W$ are cyclic by an angle chase. Let $T=\overline{AD}\cap\overline{BC}$. Then $T-X-Z$ on the angle bisector of $\angle ATB$. Let $Z'$ be the incenter of $\triangle TAB$, then $TX\cdot TZ'=TA\cdot TB$ and $\frac{TZ}{TZ'}=\frac{TD}{TB}$, so $TX\cdot TZ=TA\cdot TD$ and $A,D,X,Z$ are cyclic. $T$ has the same power wrt $(ABCD)$ and $(XYZW)$. Let $T'$ be $\overline{AB}\cap\overline{CD}$. Then $T'$ also has the same power wrt $(ABCD)$ and $(XYZW)$. Let $M$ be the Miquel Point of $ABCD$, then $M$ is on the radical axis of $(ABCD)$ and $(XYZW)$. Because $\overline{OM}\perp\overline{TT'}$, $(XYZW)$ must be centered on $\overline{OM}$. Let $\varphi$ be the inversion with radius $\sqrt{MA\cdot MD}$ around $M$ plus the reflection across the angle bisector of $\angle AMC$. $\varphi$ takes $A$ to $C$, $B$ to $D$, so it takes $(ABCD)$ to itself. Thus it takes $(XYZW)$ to itself. $M,A,O,C$ and $M,B,O,D$ are cyclic by an angle chase, so $O$ goes to $P$ under $\varphi$, as desired. Some motivation I guess: $X,Y,Z,W$ being cyclic is motivated because this condition is necessary and you hope that $X,Y,Z,W$ cyclic is only true for some quadrilaterals $ABCD$, and even though it isn't, it at least helps. Constructing $T$ is motivated because $X,Z$ are in/ex centers of $\triangle TAB, \triangle TCD$, and $T-X-Z$. Then I saw that $TX\cdot TZ$ can be calculated in terms of $TA,TB,TC,TD$, and it turns out that it equals $TA\cdot TD$. At this point I remembered the blog post and tried to show that $(XYZW)$ goes to itself under Miquel Inversion. My original solution actually shows that $(DZC),(AZB),(XYZW)$ are coaxial (by PoP at $T'$) so then Clawson-Schmidt conjugate things finish, but this is not necessary as I showed above.

Does this problem have a normal solution which is not "complex bash for the win"? (Any idea if this calculation can be further optimized?) Sketch. It is immediate (trivial angle chasing) that $WXYZ$ is cyclic so we may focus only on $O$, $P$, $W$, $X$, $Y$. Set $A = a^2$, $B = b^2$, $C = c^2$, $D = d^2$ on the unit circle with center $O = 0$. Then $P = \frac{a^2c^2(b^2+d^2)- b^2d^2(a^2+c^2)}{a^2c^2-b^2d^2}$. The midpoint of arc $\widehat{BD}$ is can be taken to be $-bd$ and that of $\widehat{AC}$ to be $-ac$, so $X = \frac{bc(a^2-bd) - ad(b^2-ac)}{bc-ad}$. Arguing cyclically, we obtain $Y = \frac{cd(b^2-ac) - ab(c^2-bd)}{cd-ab}$ and $Z = \frac{ad(c^2-bd) - bc(d^2-ac)}{ad-bc}$. Now, $O$, $X$, $Y$ and $Z$ are concyclic if and only if $\frac{X-O}{Y-O}\frac{Y-Z}{X-Z} = \frac{XY-XZ}{XY-YZ}$ is real and $O$, $X$, $Y$ and $P$ are concyclic if and only if $\frac{XY-XP}{XY-YP}$ is real. Now it is just a matter of using $\bar{a} = \frac{1}{a}$ and similarly for $b$, $c$, $d$ to see that the latter two conditions are equivalent. Remark. I was initially hoping to prove that the ratio of the latter two is a real number, but this is equivalent to $X$, $Y$, $Z$ and $P$ being concyclic and hence does not help. To @Tafi_ak, I think it's alright since I rely only on the midpoints of $\widehat{AC}$ and $\widehat{BD}$ (even when arguing cyclically) and of no other arcs. Your link gives the whole picture when you care about at least (roughly) $4$ arcs.

Hello, projective overkill here (hope it's not a fakesolve). The problem dies to the following lemma: Lemma: Let $\omega$ be a circle (or a conic, anyways) and let $ABCD$ and $A_1B_1C_1D_1$ be quadrilaterals inscribed there. Let $X=AA_1\cap BB_1$, $Y=BB_1\cap CC_1$, $Z=CC_1\cap DD_1$ and $W=DD_1\cap AA_1$. Let $P=AC\cap BD$ and $P_1 = A_1C_1\cap B_1D_1$. Then $X,Y,Z,W,P,P_1$ lie on a conic. Proof: Define $M = DA_1\cap B_1C$, $N=CD_1\cap BA_1$, $L = DD_1\cap BB_1$. Begin by proving $M \in PX$. This follows from pascal with hexagon $AA_1DBB_1C$. Similarly one proves $M\in P_1Y$, $N\in P_1Z$ and $N\in WP$. Now, by Pascal's converse with hexagon $WYP_1ZXP$, we just need to prove $L$, $M$ and $N$ are collinear. But this follows from Pascal with $D_1CB_1BA_1D$. Notice that from the lemma, it follows that whenever five of $X,Y,Z,W,P,P_1$ lie on a circle, the sixth one also does. Okay so the original problem is basically this if we take $A_1,B_1,C_1$ and $D_1$ to be the second intersections of the bisectors of $\angle BAD$, $\angle ABC$, $\angle BCD$ and $\angle CDA$ with $\omega$, as we get $P=P$ and $P_1=O$.

Assassino9931 wrote: Does this problem have a normal solution which is not "complex bash for the win"? (Any idea if this calculation can be further optimized?) Sketch. It is immediate (trivial angle chasing) that $WXYZ$ is cyclic so we may focus only on $O$, $P$, $W$, $X$, $Y$. Set $A = a^2$, $B = b^2$, $C = c^2$, $D = d^2$ on the unit circle with center $O = 0$. Then $P = \frac{a^2c^2(b^2+d^2)- b^2d^2(a^2+c^2)}{a^2c^2-b^2d^2}$. The midpoint of arc $\widehat{BD}$ is $-bd$ and that of $\widehat{AC}$ is $-ac$, so $X = \frac{bc(a^2-bd) - ad(b^2-ac)}{bc-ad}$. Arguing cyclically, we obtain $Y = \frac{cd(b^2-ac) - ab(c^2-bd)}{cd-ab}$ and $Z = \frac{ad(c^2-bd) - bc(d^2-ac)}{ad-bc}$. Now, $O$, $X$, $Y$ and $Z$ are concyclic if and only if $\frac{X-O}{Y-O}\frac{Y-Z}{X-Z} = \frac{XY-XZ}{XY-YZ}$ is real and $O$, $X$, $Y$ and $P$ are concyclic if and only if $\frac{XY-XP}{XY-YP}$ is real. Now it is just a matter of using $\bar{a} = \frac{1}{a}$ and similarly for $b$, $c$, $d$ to see that the latter two conditions are equivalent. Remark. I was initially hoping to prove that the ratio of the latter two is a real number, but this is equivalent to $X$, $Y$, $Z$ and $P$ being concyclic and hence does not help. I think you are incorrect according to this.

Clearly XYZW is cyclic ( a simple angle chasing show either $\angle XYZ + \angle XWZ = 180^{\circ}$ or $ \angle WXY + \angle YZW = 180^{\circ}$ ); let $w$ the circle though $XYZW$. $\newline$ Let $E$ and $F$ be respectively AD $\cap$ BC and AB $\cap$ CD. $\newline$ By costruction $X$ is the $E$ excenter of $ABE$ , $Y$ is the $F$ excenter of $ACD$ , $Z$ Is the incenter of $CDE$ and $W$ is the incenter of $DFA$ $\newline$ Now consider the two inversion with center $E$ and $F$ which keep fixed the circumference $ABCD$ and call they rispectively $(1)$ and $(2)$ , it's easy to check that $(1)$ sends the excenter of $ABE$ in the incenter of $ECD$ so under $(1)$ $w$ stay fixed, similarly $w$ remain fixed also under $(2)$. $\newline$ Under $(1)$ $O$ is sent in $O'$ and $P$ in $P'$. Under $(2)$ $O$ is sent in $O''$ and $P$ in $P''$. An obvious fact, by the definition of inversion, is that $O,O',P,P'$ is cyclic. As a consecuence of the brocard's theorem, since $ABCD$ stay fixed under both $(1)$ and $(2)$ by definition, we have that $O'=P''$ and $P'=O''$. $\newline$ Now, since $w$ stay fixed under both $(1)$ and $(2)$, $O \in w \Leftrightarrow O'' \in w$ that is $O \in w \Leftrightarrow P' \in w $ and $O \in w \Leftrightarrow O' \in w$ so $ O\in w \Rightarrow w =(OO'P')$ but since $P \in (OO'P') $ we have $P \in w$. Similarly one can prove the other direction as well thus we are done.

Omitting most of angle chasing because I'm lazy. Let $K\neq P$ be the intersection of circumcircles of $DPA$, $BPC$ and $L\neq P$ be the intersection of the circumcircles of $ABP$, $CDP$. Easy angle chasing gives that $ABOK$ is cyclic. Analogously $CDOK$, $BCOL$, $DAOL$ are cyclic. Also $KLOP$ is cyclic. On the other hand angle chasing shows that $XYZW$ is cyclic. Let $BC$ intersect $AD$ at $E$. Then $E, X, Z$ are collinear (as $X$, $Z$ lie on the bisector of $CED$). $\angle EZD = \frac 12 \angle ECD = \frac 12 \angle BAE = \angle XAD$, hence $D, Z, X, A$ are concyclic. Similarly $B,C,X,Z$ are concyclic. Radical axes give that $AD, XZ, PK, OL, BC$ intersect at $E$ and $EX\cdot EZ = EA\cdot ED = EP\cdot EK = EO\cdot EL$. Now it is clear that $P\in (XYZW)\iff K\in (XYZW)$ and $O\in (XYZW)\iff L \in (XYZW)$. Mutatis mutandis, $P\in (XYZW)\iff L\in (XYZW)$. Therefore $P \in (XYZW) \iff L\in(XYZW) \iff O\in(XYZW)$.

Let $E=AB\cap CD$, $F=AD \cap BC$. Also let $Q$ be the circumcenter of cylcic (angle chasing) quadrilateral $XYZW$. Claim: $ABYW$ is cylcic and so on Pf: Easy angle chasing (I'll omit it) $\blacksquare$ Hence $EF$ is the radical axis off $(ABCD)$ and $(XYZW)$. In particular $OQ\perp EF$. But by Brokard's theorem, $OP\perp EF$, so $O-P-Q$ colinear. Let $\omega$ be the circle with diameter $OP$. I claim $\omega = (XYZW)$. Now let $M=OP\cap EF$ be the Miquel point of $ABCD$. $$Pow_\omega(E)=EM\cdot EF= EB\cdot EA=EY\cap EW$$and so on, which means that $(ABCD)$, $(XYZW)$ and $\omega$ are coaxial with radical axis $EF$. If either $O$ or $P$ lie on $(XYZW)$, then this point should also lie on the radical axis of $(XYZW)$ and $\omega$, i.e $EF$, unless they are equal. As this isn't possible, they are equal circles. $\square$

there are some ridiculous solutions here...

Solved with Pranjal and Abhay. Facts from here will be assumed without proof. First, since $\angle XYZ = 180 - \frac{B+C}{2} = \frac{C+D}{2} = 180 - \angle XWZ$ so $XYZW$ is cyclic. Let $K = AD \cap BC$ and $L = AB \cap CD$ and $M$ be the miquel point of $ABCD$. Note that $X$ is the incenter of $KAB$ and $Z$ is the excenter of $KCD$, so $K,X,Z$ are collinear on the $K$ angle bisector, similarly, $L,W,Y$ are also collinear. So since $\angle KXB = 90 + \frac{A}{2} = \angle ZCK$, we have $BCZX$ is cyclic. Let $\Phi_1$ be the inversion centered at $K$ that swaps $A,D$, $\Phi_2$ be the inversion centered at $L$ that swaps $A,B$ and let $\Phi_M$ be the inversion at $M$ with radius $\sqrt{MB.MD}$ followed by reflecting across the angle bisector of $\angle AMC$. Claim: $\Phi_1(\Phi_2) = \Phi_M$ Proof: Let $J$ be an arbitrary point, let $J' = JL \cap (DCJ)$ and $J'' = J'K \cap (BCJ')$, so $\Phi_1(\Phi_2(J)) = \Phi_1(J') = J''$. But $J''$ is just the clawson-schmidt conjugate of $J$, so $\Phi_1(\Phi_2)$ just sends every point to its clawson schmidt conjugate, which is also what $\Phi_M$, the miquel inversion does. $\square$ To finish, see that both $\Phi_1, \Phi_2$ fix $(WXYZ)$ and hence $\Phi_M$ does too by the above claim, but $\Phi_M$ also swaps $O,P$ so if one of them was on $(WXYZ)$, the other must be too, so we're done. $\blacksquare$

Solution. We can easily rule out the extreme case when at least two opposite sides of $ABCD$ are parallel to each other. Henceforth, let's assume that $AB\nparallel CD$ and $BC\nparallel DA$. Let $Q = \overline{BC}\cap \overline{DA}$ and $R = \overline{AB}\cap \overline{CD}$. Notice that $Z$ is the incenter of $\bigtriangleup CQD$ and $X$ is the $Q$-excenter of $\bigtriangleup AQB$. By using directed angles, we know that: $$\measuredangle WZY = \measuredangle DZC = 90^{\circ} + \measuredangle AQZ, \quad \measuredangle YXW = \measuredangle BXA = 90^{\circ} - \measuredangle AQZ$$implying that $WXYZ$ is cyclic, but: $$\measuredangle ZXA = \measuredangle DAX - \measuredangle AQX = 90^{\circ} - \measuredangle ZCD - \measuredangle AQX = \measuredangle ZDA$$thus $DAZX$ is cyclic. Similarly, we can prove that $CBZX$ is also inscribed. Hence, we can infer that $QA\cdot QD = QZ\cdot QX = QB\cdot QC$ $(1)$. Analogously, we can show that $ABYW$ and $CDWY$ are cyclic with $RA\cdot RB = RW\cdot RY = RD \cdot RC$ $(2)$. Let's define $P',\ Q'$ and $R'$ as the inverses of $P,\ Q$ and $R$ with respect to $(ABCD)$. It's known that $O,\ P,\ Q$ and $R$ form an orthocentric group, say, $P',\ Q'$ and $R'$ are the feet of altitudes of $\bigtriangleup PQR$. Therefore, $OR'PQ'$ is cyclic. Besides, observe that: \begin{align*} QP\cdot QR' &= QQ'\cdot QO = QP'\cdot QR = QA\cdot QD \tag{3}\\ RP\cdot RQ' &= RR'\cdot RO = RP'\cdot RQ = RA\cdot RB \tag{4} \end{align*}since $P'$ is the Miquel point of $ABCD$ and quadrilaterals $P'ADR$ and $P'ABQ$ are inscribed. By comparing $(1)$ and $(3)$ we conclude that $QP\cdot QR'= QZ\cdot QX = QQ'\cdot QO$ $(5)$; meanwhile, from $(2)$ and $(4)$ we infer that $RP\cdot RQ' = RW\cdot RY = RR'\cdot RO$ $(6)$. If $P$ lies on $(WXYZ)$, $(5)$ leads to discover that $R'$ also does so. On the other hand, $(6)$ implies that $PQ'YW$ and $YWR'O$ are cyclic, thus $Q'$ and $O$ also lie on $(WXYZ)$. If $O$ is the starting point lying on $(WXYZ)$, by $(5)$ we realize that $Q'$ also lies on such a circumference. By $(6)$, we deduce that also $P$ and $R'$ exhibit this property too. $\Box$

Sol Sketch:- Let $G$ be the miquel point of $ABCD$.Note that Pow of $G$ in (ABCD) is $GO \cdot GP$.Note that $AZXD$ and $CYWD$ are cyclic quads so $E-G-F$ is radax of $ABCD$ and $WXYZ$ which finishes the prob.

Actually,we can prove that $O,P,X,Y,Z,W$ are on the same conic section.(pretty hard)

Assassino9931 wrote: Does this problem have a normal solution which is not "complex bash for the win"? (Any idea if this calculation can be further optimized?) Sketch. It is immediate (trivial angle chasing) that $WXYZ$ is cyclic so we may focus only on $O$, $P$, $W$, $X$, $Y$. Set $A = a^2$, $B = b^2$, $C = c^2$, $D = d^2$ on the unit circle with center $O = 0$. Then $P = \frac{a^2c^2(b^2+d^2)- b^2d^2(a^2+c^2)}{a^2c^2-b^2d^2}$. The midpoint of arc $\widehat{BD}$ is can be taken to be $-bd$ and that of $\widehat{AC}$ to be $-ac$, so $X = \frac{bc(a^2-bd) - ad(b^2-ac)}{bc-ad}$. Arguing cyclically, we obtain $Y = \frac{cd(b^2-ac) - ab(c^2-bd)}{cd-ab}$ and $Z = \frac{ad(c^2-bd) - bc(d^2-ac)}{ad-bc}$. Now, $O$, $X$, $Y$ and $Z$ are concyclic if and only if $\frac{X-O}{Y-O}\frac{Y-Z}{X-Z} = \frac{XY-XZ}{XY-YZ}$ is real and $O$, $X$, $Y$ and $P$ are concyclic if and only if $\frac{XY-XP}{XY-YP}$ is real. Now it is just a matter of using $\bar{a} = \frac{1}{a}$ and similarly for $b$, $c$, $d$ to see that the latter two conditions are equivalent. Remark. I was initially hoping to prove that the ratio of the latter two is a real number, but this is equivalent to $X$, $Y$, $Z$ and $P$ being concyclic and hence does not help. To @Tafi_ak, I think it's alright since I rely only on the midpoints of $\widehat{AC}$ and $\widehat{BD}$ (even when arguing cyclically) and of no other arcs. Your link gives the whole picture when you care about at least (roughly) $4$ arcs. Your $Y$ seems wrong, the chord should go to $ac$ if the other one goes to $-ac$. The arguing cyclically also might need more care with signs since you have the squares.

I heard this question was also proposed by Ethan Tan of Australia

Pascal Spamming Time Let the internal angle bisectors of $A,B,C,D$ meet $(ABCD)$ again at $A',B',C',D'$, respectively. Clearly, $A'C'$ and $B'D'$ pass through $O$. Let $Q=AB'\cap DC', R=AD'\cap C'B, S=BB'\cap DD'$. Pascal on $AB'BDC'C$ and $AB'D'DC'A'$ give $Q,P,Y$ and $Q,W,O$ collinear. Pascal on $C'BDD'AC$ and $C'BB'D'AA'$ give $P,Z,R$ and $X,O,R$ collinear. Pascal again on $AB'BC'DD'$ gives $Q,R,S$ collinear. Finally, apply the converse of Pascal on $ZPYXOW$ using $Q,R,S$ collinear, we have $P,W,X,Y,Z,O$ all lie on the same conic, as desried.

It is trivial that $XYZW$ are concyclic. (simple angle-chasing) We prove that $X$, $Y$, $Z$, $W$, $O$, and $P$ lie on a conic section by Desargues Involution Theorem. It clearly implies the original proposition. Denote the two intersections of $OP$ and $\odot(ABCD)$ by $U$ and $V$. Let $\Phi: OP\rightarrow OP$ be the involution map that sends $O$ to $P$ and sends $U$ to $V$. It suffices to show $\Phi(XY\cap OP)=ZW\cap OP$ and $\Phi(YZ\cap OP)=WX\cap OP$. Suppose the internal angle bisectors of $A$, $B$, $C$, $D$ intersect $\odot (ABCD)$ again at $A'$, $B'$, $C'$, $D'$. Then $A'C'\cap B'D'=O$. By applying Desargues Involution Theorem on four points ($A$, $C$, $A'$, $C'$) and a line $OP$, we have an involution map $\Phi_1: OP\rightarrow OP$ satisfying \[\Phi_1(O)=\Phi_1(P), \quad \Phi_1(U)=\Phi_1(V), \quad \Phi_1(WX\cap OP)=\Phi_1(YZ\cap OP).\]Hence $\Phi\equiv \Phi_1$. It follows that $\Phi(YZ\cap OP)=WX\cap OP$. Similarly, we can consider applying Desargues Involution Theorem on four points ($B$, $D$, $B'$, $D'$) and a line $OP$. Then we conclude that $\Phi(XY\cap OP)=ZW\cap OP$. This completes the proof.

It's enough to prove, that $X,Y,Z,W,O,P$ lie on one conic. Let $AX,BY,CZ,DW$ meet $\odot (ABCD)$ again at $A',B',C',D'$ respectively, in particular $O=A'C'\cap B'D'.$ By Pascal on $A'C'CD'B'B,$ $CAA'BDD'$ we get $A'B\cap CD'=Q=YO\cap WP,$ and analogously $A'D\cap B'C=R=ZO\cap XP.$ By Pascal on $B'BA'DD'C$ we get $XY\cap ZW\in QR,$ so converse of Pascal for $XYOZWP$ finishes.

I solved this with complex numbers when I can't do geo at all. Now here's a synthetic solution. Claim 1. $WXYZ$ is cyclic. Proof: $\angle WXY+\angle YZW=\frac12(\angle ABC+\angle BCD+\angle CDA+\angle DAB)=180^\circ$ Claim 2. $WYAD$ is cyclic. Proof. Let $AB$ and $CD$ meets at $E$ and $AD$ and $BC$ meets at $F$. We know that $W$ is the $F$-excenter of $FAB$ and $Y$ is the incenter of $FCD$, so $WY$ is the angle bisector of $\angle AFB$. Now angle chasing gives $\angle DYW=180^\circ -\angle DYC=90^\circ-\frac12\angle BCD=\angle DAW$. Claim 3: $EF$ is the radical axis of $(ABCD)$ and $(WXYZ)$ Proof: $FA\times FD=FW\times FY$, so $F$ lies on the radical axis. Similar arguments for $E$. Claim 4: Let $(AOB)$ and $(COD)$ meet at $M$, then $PMCB, PMDA$ are cyclic. Proof: $M$ is the miquel point of $\{AB, CD, BC, AD\}$. As $ABCD$ is cyclic, this can be shown by the properties of the miquel point of a cyclic quadrilateral. Let $O$ lies on $XYZW$. As the radical center of $(ABCD), (CDO), (XYZWO), (ABO)$ is $E$, we know that the other intersection of $EO$ and $(XYZW)$ would be on both $(CDO)$ and $(ABO)$, which is $M$, so $M$ lies on the circle if $O$ lies on the circle. Similarly, $P$ lies on the circle if $M$ lies on the circle (for a similar radical axis argument). The converse can also be proven this way.

Set $(ABCD)$ unit circle; use $A=a^2$, etc. such that the internal angle bisectors of $A$, $B$, $C$, and $D$ hit the circle again at $bd$, $ac$, $-bd$, and $-ac$, respectively. It suffices to prove that $O,P,X,Y,Z,W$ lie on a common conic. The equations of the internal angle bisectors of $A$, $B$, $C$, and $D$ are \begin{align*} \ell_a(z)&=z-a^2-bd+a^2bd\overline z=0\\ \ell_b(z)&=z-b^2-ac+ab^2c\overline z=0\\ \ell_c(z)&=z-c^2+bd-bc^2d\overline z=0\\ \ell_d(z)&=z-d^2+ac-acd^2\overline z=0 \end{align*} By inspection, \[\Omega(z)=\ell_b(0)\ell_d(0)\ell_a(z)\ell_c(z)-\ell_a(0)\ell_c(0)\ell_b(z)\ell_d(z)=0\]is a conic passing through $W$, $X$, $Y$, $Z$, and $0$. So it suffices to confirm that \[\ell_b(0)\ell_d(0)\ell_a(p)\ell_c(p)\stackrel?=\ell_a(0)\ell_c(0)\ell_b(p)\ell_d(p)\] Indeed, the left hand side is \begin{align*} &(b^2+ac)(d^2-ac)\left(\frac{a^2c^2(b^2+d^2)-b^2d^2(a^2+c^2)}{a^2c^2-b^2d^2}-a^2-bd+\frac{a^2bd(a^2+c^2-b^2-d^2)}{a^2c^2-b^2d^2}\right)\left(\frac{a^2c^2(b^2+d^2)-b^2d^2(a^2+c^2)}{a^2c^2-b^2d^2}-c^2+bd-\frac{bc^2d(a^2+c^2-b^2-d^2)}{a^2c^2-b^2d^2}\right)\\ &=(b^2+ac)(d^2-ac)\left(\frac{(a^2-b^2)(d^2-a^2)(c^2-bd)}{a^2c^2-b^2d^2}\right)\left(\frac{(b^2-c^2)(c^2-d^2)(a^2+bd)}{a^2c^2-b^2d^2}\right) \end{align*} which is symmetrical under swapping $a$ and $b$ and swapping $c$ and $d$, so the right hand side equals the same thing, as desired.

Two MOPpers gave me hints toward this ridiculous solution. Let the internal angle bisectors of $A$, $B$, $C$, and $D$ intersect the circle again at $M_A$, $M_B$, $M_C$, and $M_D$, respectively. Let $S$ and $T$ be the midpoints of $AC$ and $BD$, respectively. Apply quartic Cayley Bacharach on $A,B,C,D,M_A,M_B,M_C,M_D,W,X,Y,Z,S,T,O,P$. We can verify that the quartic consisting of the $A$- and $C$- internal angle bisectors, the perpendicular bisector of $AC$, and the line $BD$ passes through all $16$ points; similarly, the quartic consisting of the $B$- and $D$- internal angle bisectors, the perpendicular bisector of $BD$, and the line $AC$ passes through all $16$ points. The quartic consisting of the circle $(ABCDM_AM_BM_CM_D)$ and the conic through $P,X,Y,Z,W$ must then pass through the remaining three points by quartic Cayley Bacharach. Namely, $X,Y,Z,W,O,P$ are coconic, which proves the problem statement.

Here goes a Pascal bash! We will show that $O,P,X,Y,Z,W$ lie on the same conic. Let the perpendicular bisector of $BD$ intersect arcs $BCD,BAD$ at $K,L$ and the perpendicular bisector of $AC$ intersect arcs $ABC,ADC$ at $M,N$ respectively. Let $R,S$ be the midpoints of $BD,AC$. Pascal at $LMDBAK$ gives $LM\cap AB,W,R$ are collinear. Pascal at $MNBACL$ implies $S,Y,AB\cap LM$ are collinear. Hence $AB,LM,SY,WR$ are concurrent. We can apply pascal at $ROSYZW$ since $L,M,SY\cap WR$ are collinear, $O,R,S,Y,Z,W$ lie on a conic. Let $\mathcal{C}$ be this conic. Since $Y,Z,W$ and $Z,W,X$ are cyclic, similarily we get that $O,R,S,Z,W,X$ are on the same conic. These must be the same hence $X\in \mathcal{C}$. Pascal at $BNKLCD$ yields $Y,NK\cap CD,R$ are collinear. Pascal at $DMNKAC$ gives $S,W,NK\cap CD$ are collinear. Combining these implies $NK,CD,RY,SW$ are concurrent. We can perform Pascal at $SPRYZW$ since $C,D,RY\cap SW$ are collinear. Thus, there is a conic passing through $S,P,R,Y,Z,W$. So $P\in \mathcal{C}$. We see that $X,Y,Z,W,P,R,S,O$ lie on a conic as desired.$\blacksquare$