easy to see $AM,BN,OC$ concurrent on $(C)$ at $K$ $P'$ in $(C)$ such that $P'C$ is bisector of $\angle OCM$ from condition "Let $ l $be a line passing through $C$ such that the angle between $l$ and the segment $OC$ is $\frac{\pi}{3}$" we have $\angle P'CO=60^{\circ}$,so $OP'=OC=CP',\angle P'OC=60^{\circ}$ $A,P',O,K$ is concyclic so $ \angle MAP'=120^{\circ} $ and we have $\angle MCP'=60^{\circ}$ so $P',A,M,C$ is concyclic $(1)$ we have $\angle P'AK=P'BN=120^{\circ}$ and $\angle P'CN=120^{\circ}$ so $P',C,B,N$ is concyclic $(2)$ from$(1)(2)$ we have $P' \equiv P$ $O_{1},O_{2}$ is circumcenter of the triangle $ACM,BCN$ we have $OC=OP,O_{1}C=O_{1}P$ so $OO_{1}$ is perpendicular to $CP$ we have $OC=OP,O_{2}C=O_{2}P$ so $OO_{2}$ is perpendicular to $CP$ so $O_{1},O_{2},O$ is colinear but $O_{1}O$ is perpendicular to $AQ$ and $O_{2}O$ is perpendicular to $BR$ so $AQCP$ and $PCBR$ are trapeziums but $A,Q,C,P$ is concyclic and $P,C,B,R$ is concyclic so $PQ=AC=BC=PR$ and $OQ=OR$ so $OP\perp RQ$ P/S:i don't know how to use the condition $ \frac{\pi}{3}<\angle AOB <\frac{2\pi}{3} $

Attachments:

very nice problem. here's my solution in less detail: take $Q'$ on $\omega$ such that $OQ'B$ is equilateral and $Q',A$ are on the same side of $BO$ then angle chase to $POAB$ is cyclic and $POC$ is equilateral then to $PQ'O\cong OCB$ to get $PC=PQ'=PO$ now angle chase to $AMQ'C$ is cyclic to get $Q'=Q$ and $PQ=PO$(and similarly $PO=PR$) so $PO\perp RQ$. in more detail: .(hopefully not repeated). call $\angle BAO=x$ WLOG $\angle NCO=60$ Take $Q'$ on $\omega$ such that $Q',A$ are on the same side of $BO$ and $Q'B=OB=OQ'$ then $OQ'B$ is equilateral and $\angle Q'OB=60$ First of $S=BN\cap AM$ is on circle $k(C,CO)$ and $S\in CO$(since $\angle SBO=\angle SAO=90$). sbut since $\angle APB=\angle APC+\angle +\angle CPB=\angle CMS+\angle CNS=180-\angle ASB$ then $P\in k$ but then $\angle CPO=\angle BPO+\angle BNC=60-x+x=60$ but since $CP=CO$ then $CPO$ is equilateral. now $\angle POQ'=\angle COB$,$PO=OC$ and $OQ'=OB$ so $OPQ' \cong OCB$ so $PQ'=CB=CO=PO=PC$ and $P$ is the circumcenter of $CQ'O$ so $\angle CQ'O=\angle CPO/2=30$ and $\angle OQ'A=\angle Q'AO=\angle Q'AB+x=\angle Q'OB/2+x=30+x$ so $\angle CQ'A=30+30+x=60+x=60-x+2x=\angle MNS+\angle MSN=\angle AMC$ so $Q'$ is on circle $CAM$ therefore $Q'=Q$ and $PQ=PO$ similarly $PR=PO$ and since $OQ=OP$ we have that$OP$ is the perpendicular bisector of $RQ$ so $OP\perp RQ$.

Let cirlce $AOB=\gamma$. $AM\cap BN=E$. Note that $E\in \gamma$. (In fact $O, C, E$ are collinear.) (All angles are considered mod $\pi$) In $\triangle EMN$, $P$ is the Miquel point for $A,B,C$. So$P\in \gamma$. Take a point $P'$ on $\gamma$ such that $\angle OCP'=\frac \pi 3$. Since $OC=OP'$, $OCP'$ is an equilateral triangle. Hence $OP'||\ell$. Also note that $\angle CMA=\angle CEM+\angle MCE=\angle OEA+\angle P'OE$ $=\angle OBA+\angle CP'O=\angle OP'A+\angle CP'O=CP'A$. So $A,M,C,P'$ concyclic. Hence $P=P'$. Now take a point $Q'$ on $\omega$ such that $\angle BOQ'=-\frac \pi 3$. Since $OB=OQ'$, so $OBQ'$ is also an equilateral triangle. Note that $CB=CO$ and $Q'B=Q'O$ imply that $Q'C$ is the perpendicular bisector of $OB$. Since $OBQ'$ equilateral, we must have $\angle BQ'C=\frac \pi 6$. It is not hard to prove $\frac 1 2 \angle BOA=\angle ABO$. Therefore, $\angle AQ'C=\angle AQ'B-\frac \pi 6=\frac 1 2 \angle BOA-\frac \pi 3=\angle ABO-\frac \pi 3=\angle APO-\frac \pi 3=\angle APC$. So $A,P,C,Q'$ concyclic. So $Q=Q'$. Similarly we can prove $OAR$ is also an equilateral triangle. Since $OR=OQ$, $OBQ\cong OAR$. Now notice that a rotation $R(O,-\frac \pi 3)$ maps $OAB$ to $ORQ$. So this rotation maps $OE$, the perpendicular bisector of $AB$ to the perpendicular bisector of $RQ$. But since $\angle EOP = -\frac \pi 3$, $OE$ is mapped to $OP$. Therefore $OP$ is the perpendicular bisector of $RQ$. Hence $OP\perp RQ$. [asy][asy]/* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 0.94, xmax = 19.22, ymin = -1.58, ymax = 8.17; /* image dimensions */ /* draw figures */ draw(shift((5.98,4.7)) * scale(4, 4)*unitcircle); draw((2.2,3.38)--(7.68,1.08)); draw((5.98,4.7)--(7.68,1.08)); draw((5.98,4.7)--(2.2,3.38)); draw(shift((4.82,1.95)) * scale(2.99, 2.99)*unitcircle); draw((5.98,4.7)--(4.82,1.95)); draw((2.2,3.38)--(4.82,1.95)); draw((4.82,1.95)--(7.68,1.08)); draw(shift((3.67,2.95)) * scale(1.53, 1.53)*unitcircle); draw(shift((7.59,5.93)) * scale(4.85, 4.85)*unitcircle); draw((2.95,7.32)--(3.69,1.42)); draw((5.98,4.7)--(3.02,4.33)); draw((3.66,-0.8)--(4.82,1.95)); draw((4.82,1.95)--(3.69,1.42)); draw((7.68,1.08)--(3.69,1.42)); draw((2.2,3.38)--(3.69,1.42)); draw((2.95,7.32)--(2.2,3.38)); draw((2.2,3.38)--(3.66,-0.8)); draw((7.68,1.08)--(3.66,-0.8)); draw((11.26,2.76)--(7.68,1.08)); draw((2.79,1.69)--(11.26,2.76)); /* dots and labels */ dot((5.98,4.7),dotstyle); label("$O$", (6.05,4.8), NE * labelscalefactor); dot((7.68,1.08),dotstyle); label("$B$", (7.77,0.72), NE * labelscalefactor); dot((2.2,3.38),dotstyle); label("$A$", (1.75,3.5), NE * labelscalefactor); dot((4.82,1.95),dotstyle); label("$C$", (4.84,1.5), NE * labelscalefactor); dot((2.79,1.69),dotstyle); label("$M$", (2.43,1.3), NE * labelscalefactor); dot((11.26,2.76),dotstyle); label("$N$", (11.42,2.52), NE * labelscalefactor); dot((3.02,4.33),dotstyle); label("$P$", (2.97,4.62), NE * labelscalefactor); dot((3.69,1.42),dotstyle); label("$Q$", (3.6,1.05), NE * labelscalefactor); dot((2.95,7.32),dotstyle); label("$R$", (2.69,7.4), NE * labelscalefactor); dot((3.66,-0.8),dotstyle); label("$E$", (3.39,-1), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* re-scale y/x */ currentpicture = yscale(0.94) * currentpicture; /* end of picture */[/asy][/asy]

we want to prove OR^2 _ OQ^2 = PR^2 _ PQ^2 we have OR=OQ so we want to prove PR=PQ with complex number we can find P,Q,R in term of A,B and we can prove PQ=PR(with boring calculation)

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -2.4920984089472173, xmax = 3.577979310720535, ymin = -2.262823624143871, ymax = 1.3903171444812297; /* image dimensions */ /* draw figures */ draw(circle((0,0), 1), linewidth(0.4)); draw(circle((0,-0.8333333333333334), 0.8333333333333336), linewidth(0.4) + linetype("2 2")); draw(circle((-0.12687654216899405,0.2197566173253536), 1.0607054733200336), linewidth(0.4)); draw(circle((0.3859439706515187,-0.6684745660433021), 0.419679832294392), linewidth(0.4)); draw((0,-1.6666666666666667)--(-1.1023174629940151,-0.19691004934131331), linewidth(0.4)); draw((0,-1.6666666666666667)--(0.8,-0.6), linewidth(0.4)); draw((0.8,-0.6)--(0,0), linewidth(0.4)); draw((-0.8,-0.6)--(0,0), linewidth(0.4)); draw((0,-1.6666666666666667)--(0,0), linewidth(0.4)); draw((0,0)--(0.7216878364870327,-0.41666666666666685), linewidth(0.4)); draw((0.9196152422706645,0.3928203230275479)--(0,0), linewidth(0.4)); draw((-1.1023174629940151,-0.19691004934131331)--(0.4361440754675237,-1.0851412327099688), linewidth(0.4)); draw((0,0)--(0.11961524227066267,-0.992820323027551), linewidth(0.4)); draw((-0.12687654216899405,0.2197566173253536)--(0.3859439706515187,-0.6684745660433021), linewidth(0.4) + linetype("2 2")); /* dots and labels */ dot((0,0),dotstyle); label("$O$", (0.015968951724902265,0.03920976693116532), NE * labelscalefactor); dot((-0.8,-0.6),dotstyle); label("$A$", (-0.8755241859547611,-0.6304006342593358), NE * labelscalefactor); dot((0.8,-0.6),dotstyle); label("$B$", (0.8599157887283169,-0.6502115928744394), NE * labelscalefactor); dot((0,-1.6666666666666667),linewidth(4pt) + dotstyle); label("$D$", (-0.07912364962759516,-1.7437765084281571), NE * labelscalefactor); dot((0,-0.8333333333333334),linewidth(4pt) + dotstyle); label("$C$", (-0.07516145790457443,-0.9196406300398481), NE * labelscalefactor); dot((-1.1023174629940151,-0.19691004934131331),linewidth(4pt) + dotstyle); label("$M$", (-1.0855203472748596,-0.16682420266591194), NE * labelscalefactor); dot((0.4361440754675237,-1.0851412327099688),linewidth(4pt) + dotstyle); label("$N$", (0.45973442470322357,-1.1534099416980703), NE * labelscalefactor); dot((0.9196152422706645,0.3928203230275479),linewidth(4pt) + dotstyle); label("$Q$", (0.9351974314657107,0.42354236406417484), NE * labelscalefactor); dot((0.11961524227066267,-0.992820323027551),linewidth(4pt) + dotstyle); label("$R$", (0.07936401929323388,-1.0900148741297389), NE * labelscalefactor); dot((0.7216878364870327,-0.41666666666666685),linewidth(4pt) + dotstyle); label("$P$", (0.7648231873758194,-0.4402154315543414), NE * labelscalefactor); dot((0.3859439706515187,-0.6684745660433021),linewidth(4pt) + dotstyle); label("$O'$", (0.4003015488579127,-0.6383250177053772), NE * labelscalefactor); dot((-0.12687654216899405,0.2197566173253536),linewidth(4pt) + dotstyle); label("$O''$", (-0.11082118341176098,0.253168119974284), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We prove that $\angle ROP = \angle POQ$. We use $\measuredangle$ for directed angles modulo $180^\circ$, and $\angle$ for ordinary angles. Let $D = AM \cap BN$. First, we have $D \in \odot (AOB)$. Then we have \[\measuredangle APB = \measuredangle APC + \measuredangle CPB = \measuredangle AMC + \measuredangle CND = \measuredangle DMN + \measuredangle MND = \measuredangle MDN = \measuredangle ADB, \]i.e. $P \in \odot(AOBD)$. Because $P \in \odot(BCN)$, we have \[\measuredangle PCN = \measuredangle PBN = \measuredangle PBA + \measuredangle ABN = \measuredangle POA + \measuredangle ABD = \measuredangle POA + \measuredangle AOD = \measuredangle POD = \measuredangle POC.\]Therefore, $\ell$ is tangent to $\odot (PCO)$ at $P$. Thus, we have $ 60^\circ = \angle CPO = \angle POC$, i.e. triangle $PCO$ is equilateral. Consequently the perpendicular bisector of segment $CP$ goes through $O$. Let $O'$ be the centre of $\odot(CBN)$. Clearly, $OO'$ is perpendicular to $CP$. $OO'$ is also perpendicular to $BR$ as it is the radical axis of $\omega$ and $\odot(CBN)$. Let $O''$ be the centre of $\odot(CAM)$. $O'O''$ is perpendicular to $CP$, because it is the radical axis of $\odot(CAM)$ and $\odot(CBN)$. Hence, $O, O',$ and $O''$ all lie on a line. This line is perpendicular to the radical axis of $\omega$ and $\odot(CAM)$, $AQ$. Because $\angle COR = \angle BOP $, we have $ \angle ROP = \angle COB = \angle AOC$. But $\angle POQ = \angle AOC$ because $OO'$ bisects both $\angle AOQ$ and $\angle COP$. This concludes the proof. $\blacksquare$