Interestingly enough, $AC, RP, OS$ are concurrent.

Let $R'$ be the second intersection of $QR$ with $\omega.$ As $RR'=AP=AC,$ we have that the parallel to $BC$ at $R'$ and the parallel to $AB$ at $R$ meet at a point $K$ on $\omega.$ Also, as $AP=AC,$ we have that the parallel to $AB$ at $P$ and the parallel to $BC$ at $A$ meet at a point $T$ on $\omega.$ Now by Pascal on $R'RTAPK$ we see that $PK$ and $TR$ meet at $S,$ hence $SK=SR$ and $\angle{SKR}=\angle{QR'P},$ hence $\angle{RSK}=\angle{R'QP},$ implying the conclusion.

Darn I overcomplicated this a lot. Let $O$ be the circumcenter of $ABC$. Let $D$ be the point on $\omega$ such that $AD\parallel BC$. Let $X = AS\cap BD$. A simple angle chasing gives $\triangle BXS\sim\triangle AQC$. Let $K$ be the circumcenter of $\triangle ABS$ and let $Y$ be the inverse of $X$ w.r.t. $\odot(ABS)$. The key claim is Claim: $Y,D,S$ are colinear. Proof: Let $L$ be the point such that $\triangle ADL\sim\triangle BSK$. By homothety, $K,L,X$ are colinear. As $KX\cdot KY = KA^2$, we get $$\angle XYA = \angle KAS = \angle KSA = \angle LDA\implies Y\in\odot(ALD).$$Therefore $$\angle KYD = \angle LAD = \angle KBS = \angle KSX = \angle KYS$$which implies $Y\in DS$ as desired. To finish, we note that $\triangle BXS\cup K\sim\triangle AQC\cup O$ thus \begin{align*} \measuredangle(AR,DS) &= 90^{\circ}+\measuredangle(OQ,DS) \\ &= 90^{\circ} + \measuredangle(OQ,KX) + \measuredangle(KX,DS) \\ &= 90^{\circ} + \measuredangle(AC,BS) + \measuredangle LAD \\ &= \measuredangle ACB + \measuredangle BAD & (\because\angle LAB = 90^{\circ}) \\ &= \measuredangle ACB + \measuredangle BCD \\ &= \measuredangle ARD \end{align*}thus $D\in RS$ therefore we can conclude by Reim's theorem.

Let $O$ be the circumcenter of $\omega$. $\angle OSQ\equiv \angle OSB=90^\circ-\beta$ $\angle ORQ=\angle OAQ\equiv\angle OAP=\frac{1}{2}\angle CAP=\frac{1}{2}(180^\circ-2\angle APC)=\frac{1}{2}(180^\circ-2\angle ABC)=90^\circ-\beta$ $\angle OPQ\equiv\angle OPA=\angle OCA=\frac{1}{2}(180^\circ-\angle AOC)=\frac{1}{2}(180^\circ-2\beta)=90^\circ-\beta$ Therefore, $O, R, S, P, Q$ are concyclic.

Let $O$ be the center of $\omega$. It's easy to see that $AOSC$ is cyclic, and that $AR \perp OQ$. We claim that in fact $OPQRS$ is cyclic. Fix $A,P,C$ on $\omega$ (such that $AP=AC$), and animate $B$ on $\omega$. Then $B \mapsto Q,S$ are projective maps by taking perspectivity from $C$ on $AP$ and $\odot (AOC)$ respectively. Also, using rotation about $O$ by angle $\angle 90^{\circ}$, we get that the point at infinity on $AR$ is projectively related to $Q$. Taking perspectivity from $A$ onto $\omega$, we get that $Q \mapsto R$ is also projective. Now, we claim that $\odot (OPQ) \cap \omega$ and $\odot (AOC) \cap \odot (OPQ)$ are projective maps from $Q$. This easily follows by inversion at $O$. Thus it suffices to show that $OPQRS$ is cyclic for three positions of $B$. The result is obvious when $B=A,P$. For the third position, take $B$ such that $BP \parallel AC$. This easily translates to $R=S=C$, which gives the result on some angle chasing. Hence, done. $\blacksquare$

math_pi_rate wrote: Thus it suffices to show that $\odot (QRS)$ passes through the fixed point $P$ for four positions of $B$. Can you elaborate on this part please? I think the condition is degree four by Pascal on $IRJSQP$ where $I,J$ are circle points.

MarkBcc168 wrote: Can you elaborate on this part please? I think the condition is degree four by Pascal on $IRJSQP$ where $I,J$ are circle points. I don't think we need circle points here. Here's how I thought about it: Vary $B$ linearly. Then $$\text{deg}(Q)=\text{deg}(R)=\text{deg}(S)=1$$Invert about $P$. Then it suffices to show that the inverse of $Q,R,S$ are collinear, given that the degree of each of these points is one (since inversion is also a projective map). This then requires checking $(1+1+1)+1=4$ positions of $B$.

math_pi_rate wrote: MarkBcc168 wrote: Can you elaborate on this part please? I think the condition is degree four by Pascal on $IRJSQP$ where $I,J$ are circle points. I don't think we need circle points here. Here's how I thought about it: Vary $B$ linearly. Then $$\text{deg}(Q)=\text{deg}(R)=\text{deg}(S)=1$$Invert about $P$. Then it suffices to show that the inverse of $Q,R,S$ are collinear, given that the degree of each of these points is one (since inversion is also a projective map). This then requires checking $(1+1+1)+1=4$ positions of $B$. Yeah this is a possible way to think about that. However, since image $S'$ moves along a fixed circle, $\deg S'$ should equal to two. Therefore we must check five cases which is as good as mine.

MarkBcc168 wrote: math_pi_rate wrote: Thus it suffices to show that $\odot (QRS)$ passes through the fixed point $P$ for four positions of $B$. Can you elaborate on this part please? I think the condition is degree four by Pascal on $IRJSQP$ where $I,J$ are circle points. $I$ and $J$ are concyclic points.

MarkBcc168 wrote: Yeah this is a possible way to think about that. However, since image $S'$ moves along a fixed circle, $\deg S'$ should equal to two. Therefore we must check five cases which is as good as mine. Yeah sorry about that. Anyway, I reduced the problem to checking just 3 positions, after some synthetic observations. I have edited my solution.

MEMO 2019 I3 wrote: Let $ABC$ be an acute-angled triangle with $AC>BC$ and circumcircle $\omega$. Suppose that $P$ is a point on $\omega$ such that $AP=AC$ and that $P$ is an interior point on the shorter arc $BC$ of $\omega$. Let $Q$ be the intersection point of the lines $AP$ and $BC$. Furthermore, suppose that $R$ is a point on $\omega$ such that $QA=QR$ and $R$ is an interior point of the shorter arc $AC$ of $\omega$. Finally, let $S$ be the point of intersection of the line $BC$ with the perpendicular bisector of the side $AB$. Prove that the points $P, Q, R$ and $S$ are concyclic. Solution: Note: $QO \perp AR$ $\implies$ $\angle ORQ=\angle OAP=\angle OPQ$. Now, $$\angle QPO=90^{\circ}-\frac{1}{2}\angle AOP=90^{\circ}-\angle ACP=90^{\circ}-\angle ABC=\angle OSQ \qquad \blacksquare$$

Axiomatical wrote: Interestingly enough, $AC, RP, OS$ are concurrent.

Nice observation. The proof is straightforward, using radical axes. The concurrence point of AC, RP, OS is the radical center of the three circles on the sketch.

JustKeepRunning wrote: $I$ and $J$ are concyclic points. wow such a genius! even albort instain didn't know 2 points are concyclic

$O,P,Q,R,S$ concyclic.

OK, this should have been quicker, struggled a wee bit. The entire solution is the following two key claims. Claim : $O$ lies on $(PQR)$. Proof : Note that since $AQ=QR$, $AO=OR$ (radii) and $OQ$ is a common side, we conclude that $\triangle AOQ \cong \triangle ROQ$. Thus, \[\measuredangle QRO = \measuredangle OAQ = \measuredangle OAP = \measuredangle APO = \measuredangle QPO \]which implies the claim. Claim : $S$ lies on $(PQO)$. Proof : Again, note that since $AP=AC$ , $PO=OC$ (radii) and $AO$ is a common side, we conclude that $\triangle AOP \cong \triangle AOC$. Thus, we have \[\measuredangle QPO = \measuredangle APO = \measuredangle OCA\]and \begin{align*} 2 \measuredangle OCA &= \measuredangle COA \\ &= 2\measuredangle CBA \\ &= 2\measuredangle SBA \\ &= 2(90+\measuredangle BSO)\\ &= 2\measuredangle BSO\\ &= 2\measuredangle QSO \end{align*}and thus we have that $\measuredangle OCA = \measuredangle QSO$. From this we conclude that $\measuredangle QPO = \measuredangle QSO$ which proves the claim. Now, combining these two claims together, we have that the points $P$ , $Q$ , $O$ , $R$ and $S$ are concyclic and we are done.