Any solutions?

Solution for part b: Lemma: Let $ABCD$ be a cyclic quadrilateral with circumcenter $O$. $AC\cap BD=P,AB\cap CD=E,AD\cap BC=F$ and $Q$ is the center of spiral similarity that sends $AD$ to $CB$. (a) $E,Q,O$ are collinear. (b) $Q$ is second intersection of $\odot(AOB),\odot(DOC)$. Proof: Let $M=OP\cap EF$ be the miquel point of $ABCD$. Since $FP$ is polar of $E$ WRT $(ABCD)\Longrightarrow EO\perp FP$. Let $EO\cap FP=Q'$ then $P$ is orthocenter of $\triangle EFO\Longrightarrow FP.FQ=FM.FE=FD.FA=FC.FB\Longrightarrow Q'$ lies on $\odot(PAD),\odot(PBC)\Longrightarrow Q'=Q$. Also note that $ED.EC=EA.AB=EM.EF=EQ.EO\Longrightarrow Q\in\{\odot(OAB)\cap \odot(OCD)\}$. $\blacksquare$ [asy][asy] import graph; size(12.99302643151706cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -11.032651903971837, xmax = 16.96037452754522, ymin = -3.1577855361393623, ymax = 23.24165292451792; /* image dimensions */ draw((0.836267586914567,10.564691069034577)--(1.0402272703876467,11.169033971748888)--(0.43588436767333577,11.372993655221968)--(0.231924684200256,10.768650752507657)--cycle); draw((8.081476362499817,9.841603876523122)--(7.504824659110187,9.569018643024613)--(7.777409892608697,8.992366939634984)--(8.354061595998326,9.264952173133493)--cycle); /* draw figures */ draw(circle((9.05003483989132,7.792626796393265), 4.6731654617974066)); draw(circle((6.745106388532307,10.692155407325448), 2.1507314874990717)); draw(circle((9.131649361522589,5.643444393436535), 3.7040466157296907)); draw((3.319151558096868,19.91626080238579)--(8.354061595998326,9.264952173133493)); draw((8.354061595998326,9.264952173133493)--(9.05003483989132,7.792626796393265)); draw((9.05003483989132,7.792626796393265)--(6.826720910163575,8.542973004368717)); draw((6.826720910163575,8.542973004368717)--(0.231924684200256,10.768650752507657)); draw((4.809069939978866,9.75545438715276)--(12.773803745627207,4.969150115063894)); draw((8.094347362335453,12.367026962379956)--(5.551122860115973,4.694871094095113)); draw((8.094347362335453,12.367026962379956)--(4.809069939978866,9.75545438715276)); draw((4.809069939978866,9.75545438715276)--(-1.9135834263494587,4.411401235115413)); draw((-1.9135834263494587,4.411401235115413)--(5.551122860115973,4.694871094095113)); draw((5.551122860115973,4.694871094095113)--(12.773803745627207,4.969150115063894)); draw((3.319151558096868,19.91626080238579)--(0.231924684200256,10.768650752507657)); draw((0.231924684200256,10.768650752507657)--(-1.9135834263494587,4.411401235115413)); draw((3.319151558096868,19.91626080238579)--(4.809069939978866,9.75545438715276)); draw((4.809069939978866,9.75545438715276)--(5.551122860115973,4.694871094095113)); draw((3.319151558096868,19.91626080238579)--(8.094347362335453,12.367026962379956)); draw((8.094347362335453,12.367026962379956)--(12.773803745627207,4.969150115063894)); draw((6.826720910163575,8.542973004368717)--(-1.9135834263494587,4.411401235115413)); draw((8.354061595998326,9.264952173133493)--(6.826720910163575,8.542973004368717)); /* dots and labels */ dot((4.809069939978866,9.75545438715276),linewidth(2.pt)); label("$A$", (4.031264575287072,10.07200139175267), NE * labelscalefactor); dot((5.551122860115973,4.694871094095113),linewidth(2.pt)); label("$B$", (4.873160107062022,3.8479879961307364), NE * labelscalefactor); dot((12.773803745627207,4.969150115063894),linewidth(2.pt)); label("$C$", (13.231980029684731,4.599680435215511), NE * labelscalefactor); dot((8.094347362335453,12.367026962379956),linewidth(2.pt)); label("$D$", (8.240742234161818,12.838229567584639), NE * labelscalefactor); dot((3.319151558096868,19.91626080238579),linewidth(2.pt)); label("$E$", (2.948827463004995,20.776101724319858), NE * labelscalefactor); dot((-1.9135834263494587,4.411401235115413),linewidth(2.pt)); label("$F$", (-2.463358098405392,4.2088003668914284), NE * labelscalefactor); dot((6.826720910163575,8.542973004368717),linewidth(2.pt)); label("$P$", (5.474514058329842,8.057465655005474), NE * labelscalefactor); dot((8.354061595998326,9.264952173133493),linewidth(2.pt)); label("$Q$", (8.84209618542964,8.658819606273292), NE * labelscalefactor); dot((9.05003483989132,7.792626796393265),linewidth(2.pt)); label("$O$", (9.082637765936767,6.975028542723398), NE * labelscalefactor); dot((0.231924684200256,10.768650752507657),linewidth(2.pt)); label("$M$", (-0.6893639421653207,11.274709294288309), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Back to the problem: Assume that $\angle PBA+\angle QCA=90^{\circ}$. Since $\angle XPQ=\angle XQP=\frac{\angle A}{2}\Longrightarrow XP=XQ=XP'=XQ'\Longrightarrow PQQ'P'$ is cyclic quadrilateral with center $X$. Note that : $$\angle PXP'+\angle QXQ'=(180-2\angle XPB)+(180-2\angle XQC)=180^\circ\Longrightarrow PQ'\perp QP'$$(assume that $PQ'\cap QP'=V$). The tangents from $B,C$ to $\odot(ABC)$ intersect at $W$. Let $\omega(W,WB)$ cut $AB,AC$ at $R,L$ respectively. Since $\angle A+\angle PSQ=\angle PBA+\angle QCA=90^\circ\Longrightarrow \angle S=90^\circ-\angle A\Longrightarrow S\in \omega$ Let $U$ be the second intersection of $\odot(XPP'),\odot(XQQ')$. From the lemma $X,U,S$ are collinear and $U$ is spiral similarity sending $P'Q'$ to $QP$ so from $PQ'\perp QP'\Longrightarrow \angle PUQ=\angle P'UQ'=90^\circ$. since $\angle XUQ=\angle XQQ'=90-\angle QCA=\angle PBA=\angle RBS\Longrightarrow R,U,Q$ are collinear. similarly $L,U,P$ are collinear. Since $\angle PNB=\angle BRL=\angle C=\angle PUB\Longrightarrow PNUB$ is cyclic. similarly $QNUC$ is cyclic. Let $T'$ be the second intersection of $\odot(BUP')$ and $\odot(CUQ')$ then since $\angle UTP'+\angle UTQ'=\angle UCS+\angle UBS=180^\circ\Longrightarrow P',T',Q'$ are collinear. $$\angle T'BU=\angle UP'Q'=\angle UQP=\angle NCU=90^\circ-\frac{\angle A}{2}-\angle UCB\Longrightarrow \angle T'BC=90^\circ-\frac{\angle A}{2}-\angle UCB-\angle UBC=90^\circ-\frac{\angle A}{2}-(90^\circ-\angle A)=\frac{\angle A}{2}$$Similarly $\angle T'CB=\frac{\angle A}{2}$. 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[/asy][/asy]

$\textbf{Proof :}$ I think it's enough to prove that if $PBA + QCA = 90$ then $P'Q'T$ are collinear. $PQ$ is angle bisector of $\angle A$, so $XP =XQ = XP' = XQ'$ and $X$ is center of $PQP'Q'$. Let $A', B', C'$ be the second intersection points of $(ABC)$ with $PQ, PP', QQ'$ respectively. Easy to see that $A'C'\parallel PQ'$ and $A'B'\parallel QP'$. Also from simple angle computation from $PBA + QCA = 90$ we get that $QP'\perp Q'P$. Let $Y$ be on $P'Q'$, such that $P'Y/YQ' = QA/AP$. So we see that $YC'\parallel P'Q\parallel B'A'$ and $YB'\parallel Q'P\parallel C'A'$. So $A'B'C'Y$ is a rectangle. Easy to see that $A'$ is midpoint of bigger arc $BC$ of $(ABC)$ and $Y$ is symmetric to $A'$ wrt center of $(ABC)$, so $Y$ is midpoint of smaller arc $BC$ of $(ABC)$ and $Y\in P'Q'$. Done

Mosquitall wrote: $\textbf{Proof :}$ I think it's enough to prove that if $PBA + QCA = 90$ then $P'Q'T$ are collinear. $PQ$ is angle bisector of $\angle A$, so $XP =XQ = XP' = XQ'$ and $X$ is center of $PQP'Q'$. Let $A', B', C'$ be the second intersection points of $(ABC)$ with $PQ, PP', QQ'$ respectively. Easy to see that $A'C'\parallel PQ'$ and $A'B'\parallel QP'$. Also from simple angle computation from $PBA + QCA = 90$ we get that $QP'\perp Q'P$. Let $Y$ be on $P'Q'$, such that $P'Y/YQ' = QA/AP$. So we see that $YC'\parallel P'Q\parallel B'A'$ and $YB'\parallel Q'P\parallel C'A'$. So $A'B'C'Y$ is a rectangle. Easy to see that $A'$ is midpoint of bigger arc $BC$ of $(ABC)$ and $Y$ is symmetric to $A'$ wrt center of $(ABC)$, so $Y$ is midpoint of smaller arc $BC$ of $(ABC)$ and $Y\in P'Q'$. Done Why it's enough to show the "if" part? On the contrary, I found the real difficult part of this problem is the "only if" part. Still hoping someone post a solution for that ...

Let $D$ be so that $DT$ is a diameter. For the only if direction, we'll prove that for a fixed $Q$, there can be at most one $P$ on $AD$ such that $P', Q', T$ are collinear. Then, by part (a), we'd be done. Vary $P$ along the line $AD$. Claim. As $P$ varies, the line $P'Q'$ passes through a fixed point $Y$. If this were true, then $Q'$ would be uniquely defined as the intersection of $YT$ and $QC$. From here, $X$ is uniquely defined, and therefore so is $P$. Therefore it suffices to prove our claim. Proof of Claim. Let $P_1, Q_1$ be the intersections of $PB, QC$ respectively with the circumcircle of $ABC$. By the above posts, $DP_1 \parallel P'Q$, and $DQ_1 \parallel PQ'$. Thus, as $AP_1BD$ is cyclic, $\angle PAB = \angle PP_1D = \angle PP'Q$, and so $ABP'Q$ is cyclic, say on circle $\Omega$.. Let $Y'$ be the second intersection of $P'Q'$ with $\Omega$. Note that $\Omega$, $\angle AQC$ are both fixed since $Q$ is fixed. Therefore, as $\angle BQY$ = $180 - \angle BP'Y$ = $180 - \angle PP'Q'$ = $\angle APC$. Therefore, $Y$ is fixed since $B, Q, \Omega$, and $\angle BQY$ are all fixed.

andria wrote: since $\angle XUQ=\angle XQQ'=90-\angle QCA=\angle PBA=\angle RBS\Longrightarrow R,U,Q$ are collinear. I can't see the links between them. andria wrote: $\angle PNB=\angle BRL=\angle C=\angle PUB\Longrightarrow PNUB$ is concylic. Neither does this one make sense. Write it carefully or don't write down your solution please.

Mosquitall wrote: $P'Y/YQ' = QA'/AP'$. I think it should be like this. By the way beautiful solution.

liekkas wrote: andria wrote: since $\angle XUQ=\angle XQQ'=90-\angle QCA=\angle PBA=\angle RBS\Longrightarrow R,U,Q$ are collinear. I can't see the links between them. andria wrote: $\angle PNB=\angle BRL=\angle C=\angle PUB\Longrightarrow PNUB$ is concylic. Neither does this one make sense. Write it carefully or don't write down your solution please. Can you give a new solution?

Part $1 : \angle PBA+\angle QCA=90^{\circ}$. Let $PQ'$ and $P'Q$ meet at $K$ and Let $ABC$ meet $PQ$ and $PP'$ and $QQ'$ at $T'$ and $R$ and $S$ and Let $BS$ meet $AT$ at $Y$. Claim $: \angle PKQ = \angle 90$. Proof $:$ Note that $\angle PKQ = \angle QP'P + \angle Q'PP' = \frac{\angle PXQ + \angle P'XQ'}{2}= \angle 90$. Claim $: \angle PBS = \angle 90 = \angle QCR$. Proof $:$ Note that $\angle ACR = \angle ABR = \angle ABP = \angle 90 - \angle QCA$. we prove the other one with same approach. Note that $\angle TAT' = \angle 90$ so $PBYA$ is cyclic so $\angle BPY = \angle BAY = \angle BAT = \angle 90 - \angle PAB = \angle XPA$. Note that $PACQ'$ is cyclic so $\angle APQ' = \angle QCA = \angle ABS = \angle ABY = \angle APY$ so $P,Y,Q'$ are collinear. Claim $: T'S || PQ'$ and $T'R || P'Q$. Proof $: \angle APY = \angle ABY = \angle ABS = \angle ST'Q$ and we prove the other one with same approach. Note that $\angle T'ST = \angle 90 = \angle T'RT$ so $T'R = TS$ so $\frac{ST}{QP'} = \frac{RT'}{QP'} = \frac{PT'}{PQ} = \frac{Q'S}{Q'Q}$ which implies $Q',T,P'$ are collinear.