Let $M$ be the point on the perpendicular bisector of $PQ$ such that $M$ and $B$ are on the same side of $PQ$ and $\angle{PMQ}=180^\circ - \angle{PAQ}$. Since $AR=AS$, $\angle{ROA}=2\angle{RSA}=180^\circ - \angle{RAS}=\angle{PMQ}$. Since $OR=OA$, triangles $ORA$ and $MPQ$ are similar. Hence $OA/MP=OR/MP=RA/PQ=c$. Since $\angle{OAR}=\angle{PAQ}/2$, it follows by tangent-angle theorem that \[\angle{OAX} = \angle{OAP}-\angle{XAP}\\ = (180^\circ - \angle{PAQ}/2)-(180^\circ -\angle{XPQ})\\ = (180^\circ - \angle{MPQ})-(180^\circ -\angle{XPQ})\\ = \angle{XPM}\] This combined with the fact that $OA/MP=c=XP/XA$ implies that triangles $XPM$ and $XAO$ are similar which in turn implies that triangles $XMO$ and $XPA$ are similar. By the same argument, triangles $YQA$ and $YMO$ are similar. Hence $c=XM/XO=YM/YO$ which by the angle bisector theorem and the fact that $X$, $O$ and $Y$ are collinear implies that $\angle{XMO}=\angle{YMO}$ and hence that $\angle{XPA}=\angle{YQA}$.

Spiral similarity in the figure is obvious (could be proved by seg-angle-seg , on triangles) centered at $X$ : $A \sim P,R \sim Q$ , and assume $O \sim O_{1}$ , then $\angle XPA=\angle XO_{1}O$ centered at $Y$ : $A \sim Q,S \sim P$, and assume $O \sim O_{2}$ , then $\angle YQA=\angle YO_{2}O$ $\frac{dis<O,AR>}{dis<O_{1},PQ>}=\frac{dis<O,AS>}{dis<O_{2},QP>}=c$ , hence assume $O_{1}=O_{2}=Z$, then $\angle XPA=\angle XZO$ , $\angle YQA=\angle YZO$. then $\frac{XO}{XZ}=\frac{YO}{YZ}=c$, the given condition says that $X,O,Y$ are conlinear, so $ZO$ bisects $\angle XZY$, so $\angle XZO=\angle YZO$, hence $\angle XPA=\angle YQA$.

From the length condition we immediately obtain that there are spiral similarities $ X : RA \mapsto QP $ $ Y : SA \mapsto PQ $ We immediately obtain $PQRX$ and $PQSY$ are cyclic. Let $ O \in XY $ be the circumcenter of $ ASR $, and let $QPP'Q'$ be an isosceles trapezoid such that $\angle PP'Q = \angle PQ'Q = \angle ASR = \angle ARS $. Let $ O' $ be the circumcenter of the trapezoid. Then by the spiral similarities, $ X : O \mapsto O' \Rightarrow \angle XOO' = \angle XRQ = 180 - \angle XPQ $ $ Y : O \mapsto O' \Rightarrow \angle YOO' = \angle YSP = 180 - \angle YQP $ and we immediately get $PX || QY$. The problem follows quickly by angle chasing.

By the given length conditions, $AX / AR = PX / PQ.$ Moreover, by inscribed arcs, we have $\angle XAR = 180^{\circ} - \angle XAP = \angle XPQ.$ Therefore, $\triangle XAR \sim \triangle XPQ.$ Similarly, $\triangle YAS \sim \triangle YQP.$ Let $O$ be the circumcenter of $\triangle ARS.$ Define points $O'$ and $O''$ so that $\triangle XAR \cup O \sim \triangle XPQ \cup O'$ and $\triangle YAS \cup O \sim \triangle YQP \cup O''.$ Because $O$ lies on the perpendicular bisectors of $\overline{AR}$ and $\overline{AS}$, it follows that $O'$ and $O''$ lie on the perpendicular bisector of $\overline{PQ}.$ Therefore, \begin{align*} \angle O'QP = \angle O'PQ = \angle OAR = \angle OAS = \angle O''QP. \end{align*}It follows that $O' \equiv O''.$ Now, by spiral similarity, $\triangle XOO' \sim \triangle XAP.$ Therefore, $\angle XPA = \angle XO'O.$ Similarly, $\angle YQA = \angle YO'O.$ Thus, we need only show that $\angle XO'O = \angle YO'O$, which, by the Angle-Bisector Theorem in $\triangle O'XY$ is equivalent to $XO / XO' = YO / YO'.$ Indeed, \begin{align*} \frac{XO}{XO'} = \frac{XA}{XP} = \frac{YA}{YQ} = \frac{YO}{YO'}. \end{align*}

What a... strange problem. Anyways, here's a solution.

abacadaea wrote:

The only word which can describe this solution in this: WOW

Solution from Twitch Solves ISL: We begin as follows. Claim: There is a spiral similarity centered at $X$ mapping $AR$ to $PQ$. Similarly there is a spiral similarity centered at $Y$ mapping $SA$ to $PQ$. Proof. Since $\measuredangle XAR = \measuredangle XAP = \measuredangle XPQ$, and $AR/AX = PQ/PX$ is given. $\blacksquare$ Now the composition of the two spiral similarities \[ AR \xmapsto{X} PQ \xmapsto{Y} SA \]must be a rotation, since $AR = AS$. The center of this rotation must coincide with the circumcenter $O$ of $\triangle ARS$, which is known to lie on line $XY$. [asy][asy] import graph; size(10cm); pen uququq = rgb(0.25098,0.25098,0.25098); pen zzttqq = rgb(0.6,0.2,0.); draw((13.97096,-7.13004)--(20.23013,-5.53182)--(27.19232,-12.23004)--cycle, linewidth(0.6) + zzttqq); draw(circle((25.,-3.), 9.48683), linewidth(0.6) + uququq); draw(circle((13.,-3.), 4.24264), linewidth(0.6) + uququq); draw((13.97096,-7.13004)--(16.,0.), linewidth(0.6) + red); draw((13.97096,-7.13004)--(11.14589,0.81605), linewidth(0.6) + red); draw((11.14589,0.81605)--(25.35640,-1.57297), linewidth(0.6) + blue); draw((27.19232,-12.23004)--(20.85410,5.53296), linewidth(0.6) + red); draw((27.19232,-12.23004)--(16.,0.), linewidth(0.6) + red); draw((20.85410,5.53296)--(9.74299,-7.13207), linewidth(0.6) + blue); draw((13.97096,-7.13004)--(20.23013,-5.53182), linewidth(0.6) + zzttqq); draw((20.23013,-5.53182)--(27.19232,-12.23004), linewidth(0.6) + zzttqq); draw((27.19232,-12.23004)--(13.97096,-7.13004), linewidth(0.6) + zzttqq); draw((20.23013,-5.53182)--(19.26755,-9.17733), linewidth(0.6)); draw((11.14589,0.81605)--(20.85410,5.53296), linewidth(0.6)); dot("$A$", (16.,0.), dir(100)); dot("$Q$", (20.85410,5.53296), dir(120)); dot("$P$", (11.14589,0.81605), dir(120)); dot("$X$", (13.97096,-7.13004), dir(280)); dot("$Y$", (27.19232,-12.23004), dir(280)); dot("$R$", (25.35640,-1.57297), dir((8.043, 16.110))); dot("$S$", (9.74299,-7.13207), dir(225)); dot("$O$", (19.26755,-9.17733), dir(45)); dot("$O'$", (20.23013,-5.53182), dir(135)); [/asy][/asy] Thus, we may let $O'$ be the image under the rotation at $X$, so that \[ \triangle XPA \overset{+}{\sim} \triangle XO'O, \qquad \triangle YQA \overset{+}{\sim} \triangle YO'O. \]Because \[ \frac{XO}{XO'} = \frac{XA}{XP} = c \frac{YQ}{YA} = \frac{YO}{YO'} \]it follows $\overline{O'O}$ bisects $\angle XO'Y$. Finally, we have \[ \measuredangle XPA = \measuredangle XO'O = \measuredangle OO'Y = \measuredangle AQY. \] Remark: Indeed, this also shows $\overline{XP} \parallel \overline{YQ}$; so the positive homothety from $\omega_1$ to $\omega_2$ maps $P$ to $Q$ and $X$ to $Y$.

Claim $: XAR$ and $XPQ$, $YAS$ and $YQP$ are similar. Proof $:$ Note that $\frac{AX}{PX} = c = \frac{AR}{PQ} \implies \frac{AX}{AR} = \frac{PX}{PQ}$. we prove the other one with same approach. Let $T$ be point on perpendicular bisector of $PQ$ such that $\angle PTQ = \angle 180 - \angle PAQ$ so Now we have $\angle PTQ = \angle AOR$ so $\angle XPA = \angle XTO$ and with same approach $\angle YQA = \angle YTO$ also Note that $\frac{XO}{XT} = \frac{AX}{PX} = c = \frac{AR}{PQ} = \frac{YO}{YT}$ so $\angle YQA = \angle XPA$. we're Done.

oly kid tries to use isogonal conjugation in a quadrilateral, lives to regret it

Suppose the spiral similarity at $X$ sending $AR \mapsto PQ$ also sends $O \mapsto K$ and the spiral similarity at $Y$ sending $AS \mapsto QP$ also sends $O \mapsto K'$. Then \[\triangle OAR \sim \triangle KPQ, \quad \triangle OSA \sim \triangle K'PQ.\]\[\triangle XOK \sim \triangle XAP, \quad \triangle YOK' \sim \triangle YAQ.\] The first two similarities tells us $K = K'$. The last two similarities then finishes, as \[\frac{XO}{XK} = \frac{XA}{XP} = c = \frac{YA}{YQ} = \frac{YO}{YK}\]\[\implies \angle XOK = \angle YOK \implies \angle XPA = \angle AQY. \quad \blacksquare\]

The center of the spiral similarity sending $AR$ to $PQ$ is centered at $X$ and the center sending $AS$ to $PQ$ is centered at $Y$, since $\measuredangle{XAR} = \measuredangle XAP = \measuredangle XPQ$ and by our length conditions we have $\triangle XAR \sim \triangle XQP$ and analogously with $Y$. Let $O$ be the center of $(ASR)$. Then let the spiral similarity at $X$ send $O \to O_x$ and the spiral similarity at $Y$ send $O \to O_y$. Then since $\triangle OAR \cong \triangle OAS$ and $\triangle OAR \sim \triangle O_xPQ$ and $\triangle OAS \sim O_yPQ$ we have $\triangle O_yPQ \cong \triangle O_xPQ$ so $O_y = O_x = O'$. And since $\frac{XO}{XO'} = \frac{XA}{XP} = \frac{YA}{YQ} = \frac{YO}{YO'}$ and by angle bisector theorem we get that $\angle XO'O = \angle YO'O$ from which we get by our similarity that $\angle XO'O = \angle YO'O \implies \angle XPA = \angle AQY$ as desired.

Invert at $A.$ From now on, we take point names to mean the image of the point under inversion. First, notice that the length condition becomes \[\frac{AQ\cdot PX}{PQ}=AR=AS=\frac{AP\cdot QY}{PQ}.\]Then from $XP,YQ$ tangent to $(APQ)$ we have $\measuredangle QAR=\measuredangle QPX$ and $\measuredangle PAS=\measuredangle PQY.$ Thus we have $\triangle QRA\sim\triangle QXP$ and $\triangle PSA\sim\triangle PYQ,$ so $\measuredangle QRA=\measuredangle QXP$ and $\measuredangle PSA=\measuredangle PYQ$ and thus $PQRX,PQSY$ are cyclic. Let $X'$ be the point on $QY$ with $XX'\parallel PQ$ and let $Y'$ be the point on $PX$ with $YY'\parallel PQ.$ We see $X',Y'$ lie on $(PQRX),(PQSY)$ respectively. We have $\measuredangle SY'P+\measuredangle Y'PA=\measuredangle SY'P+\measuredangle PQS=0,$ so $AP\parallel SY'.$ Similarly, $AQ\parallel RX'.$ Now let $A'$ be the reflection of $A$ over $RS.$ We see that $ARA'S$ is a rhombus. Thus we have $\measuredangle Y'A'X'=\measuredangle PAQ=\measuredangle PQY=\measuredangle Y'YX,$ so $A'$ lies on $(XYX'Y').$ However, it is given that $A'$ lies on $(AXY),$ so we have that $A$ lies on $(XYX'Y').$ To finish, we have \[\measuredangle PXA=\measuredangle Y'A'A=\measuredangle SA'A=\measuredangle AA'R=\measuredangle AA'X=\measuredangle AYQ.\]