Proof: Notice that $\angle ADB \equiv \angle ADQ=\angle APQ\equiv \angle CPQ=180-\angle CBQ=\angle CBD \implies AD \parallel BC \implies AD \parallel BC$. Now we know that $O$ is the external homothetic centre of $\omega_1,\omega_2$. So let $\phi:\omega_1\mapsto \omega_2$ be the homothety centred at $O$ that maps $\omega_1$ and $\omega_2$. Its easily visble that $\phi : A\mapsto B$. Let $OC \cap \omega_1=D'$. So we have that $\phi: D' \mapsto C$. So by the defination of homthety we have $D'A \parallel BC$ but we have already proved earlier in claim 1 that $DA \parallel BC$ and since both these points lie on the same circle $\implies D' \equiv D$ and we are done.$\blacksquare$.

Ok......

Attachments:
Sharygin_P6.pdf (125kb)

Ig this one also is not original. Anyways Nice and Easy! Sharygin 2020 CR P6 wrote: Circles $\omega_1$ and $\omega_2$ meet at point $P,Q$. Let $O$ be the common point of external tangents of $\omega_1$ and $\omega_2$. A line passing through $O$ meets $\omega_1$ and $\omega_2$ at points $A,B$ located on the same side with respect to line segment $PQ$.The line $PA$ meets $\omega_2$ for the second time at $C$ and the line $QB$ meets $\omega_1$ for the second time at $D$. Prove that $O-C-D$ are collinear. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); 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 = -9.86, xmax = 9.38, ymin = -5.93, ymax = 5.93; /* image dimensions */ pen qqttqq = rgb(0,0.2,0); /* draw figures */ draw(circle((-3.385429417571569,0.5257354392892398), 2.8914326054816426), linewidth(1) + qqttqq); draw(circle((0.10440993788819874,0.5688198757763976), 1.0801200625681069), linewidth(1) + qqttqq); draw((2.2,0.57)--(-1.95102497283907,-1.9848155761203277), linewidth(1)); draw((2.2,0.57)--(-1.9325429888198327,3.025636144755636), linewidth(1)); draw((2.2,0.57)--(-6.1,-0.47), linewidth(1)); draw((1.18,0.47)--(-4.911892812029705,-1.929929850896234), linewidth(1)); draw((2.2,0.57)--(-4.911892812029705,-1.929929850896234), linewidth(1) + red); draw((-0.62,1.37)--(-0.4464587973622666,-0.3602673031716728), linewidth(1)); draw((-0.4464587973622666,-0.3602673031716728)--(1.18,0.47), linewidth(1)); draw((-0.62,1.37)--(-0.6,-0.25), linewidth(1)); draw((-0.6,-0.25)--(-6.1,-0.47), linewidth(1)); draw((-0.5091043870530534,0.23054595632106312)--(-4.911892812029705,-1.929929850896234), linewidth(1)); /* dots and labels */ dot((-0.62,1.37),dotstyle); label("$P$", (-0.54,1.57), NE * labelscalefactor); dot((-0.6,-0.25),dotstyle); label("$Q$", (-0.52,-0.05), NE * labelscalefactor); dot((-6.1,-0.47),dotstyle); label("$T$", (-6.02,-0.27), NE * labelscalefactor); dot((1.18,0.47),dotstyle); label("$B$", (1.26,0.67), NE * labelscalefactor); dot((2.2,0.57),dotstyle); label("$O$", (2.28,0.77), NE * labelscalefactor); dot((-1.95102497283907,-1.9848155761203277),dotstyle); label("$Y$", (-1.88,-1.79), NE * labelscalefactor); dot((-1.9325429888198327,3.025636144755636),dotstyle); label("$X$", (-1.86,3.23), NE * labelscalefactor); dot((-0.5091043870530534,0.23054595632106312),dotstyle); label("$A$", (-0.42,0.43), NE * labelscalefactor); dot((-4.911892812029705,-1.929929850896234),dotstyle); label("D", (-4.84,-1.73), NE * labelscalefactor); dot((-0.4464587973622666,-0.3602673031716728),dotstyle); label("$C$", (-0.36,-0.17), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Let $\omega_1$ and $\omega_2$ be the two circles and let $\omega_2$ be closer to $O$. Let $OA\cap\omega_1=T\neq A$. Note that there exists a homothety $\mathcal H$ mapping $\omega_2$ to $\omega_1$ centered at $O$, also by this homothety $\mathcal H:B\mapsto A$. Now $$\angle ADB=\angle ATC=\angle QPC=\angle DBC\implies AD\|BC$$So, from this we get that $\mathcal H:C\mapsto D$. Hence, $\overline {O-C-D}$. $\blacksquare$

$\textbf{Diagram : }$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(7); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -14.9604349632152, xmax = 14.434708761976491, ymin = -8.838543117557512, ymax = 8.854005652888059; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw(circle((-3.2301087408037987,0.03546253533055041), 2.724588227721254), linewidth(1) + wrwrwr); draw(circle((2.926232680887291,-0.9351318329540811), 5.380513722851044), linewidth(1) + wrwrwr); draw((xmin, -0.6791606714703614*xmin-5.45185215716575)--(xmax, -0.6791606714703614*xmax-5.45185215716575), linewidth(1) + wrwrwr); /* line */ draw((xmin, 0.29173966449944705*xmin + 3.8159818082314376)--(xmax, 0.29173966449944705*xmax + 3.8159818082314376), linewidth(1) + wrwrwr); /* line */ draw((xmin, -0.11167637477303048*xmin-0.034869543283324374)--(xmax, -0.11167637477303048*xmax-0.034869543283324374), linewidth(1) + wrwrwr); /* line */ draw((-5.89060013024411,0.6229713244998791)--(4.524810009933331,4.202423854807213), linewidth(1) + wrwrwr); draw((-2.420620056977336,2.6370215066771823)--(-2.2228718914232912,-2.496109099027694), linewidth(1) + wrwrwr); draw((xmin, 0.22538594918519195*xmin + 3.182595255835732)--(xmax, 0.22538594918519195*xmax + 3.182595255835732), linewidth(1) + wrwrwr); /* line */ draw((-2.420620056977336,2.6370215066771823)--(-5.89060013024411,0.6229713244998791), linewidth(1) + wrwrwr); draw((-2.2228718914232912,-2.496109099027694)--(-1.4928456997159743,2.134342745515859), linewidth(1) + wrwrwr); draw((-2.3277026788727473,0.22507985344265558)--(4.524810009933331,4.202423854807213), linewidth(1) + wrwrwr); /* dots and labels */ dot((-3.2301087408037987,0.03546253533055041),linewidth(3pt) + dotstyle); label("$O_1$", (-3.895659164770403,-0.21411887365692625), NE * labelscalefactor); dot((2.926232680887291,-0.9351318329540811),linewidth(3pt) + dotstyle); label("$O_2$", (2.0942946509290357,-1.1015194389457321), NE * labelscalefactor); dot((-4.7608781438598875,-2.2184509601933016),linewidth(4pt) + dotstyle); dot((-3.9931694036763936,2.6510159061134293),linewidth(4pt) + dotstyle); dot((-1.4928456997159743,2.134342745515859),linewidth(4pt) + dotstyle); label("$P$", (-1.5385014132220127,2.392620286878941), NE * labelscalefactor); dot((-2.2228718914232912,-2.496109099027694),linewidth(4pt) + dotstyle); label("$Q$", (-2.4259019785108182,-3.042708175514995), NE * labelscalefactor); dot((-9.545607949697306,1.0311493475434912),linewidth(4pt) + dotstyle); label("$O$", (-9.441912697825439,1.2556383126026587), NE * labelscalefactor); dot((-5.89060013024411,0.6229713244998791),linewidth(4pt) + dotstyle); label("$A$", (-6.2805481839840684,0.03546253533055041), NE * labelscalefactor); dot((-2.3277026788727473,0.22507985344265558),linewidth(4pt) + dotstyle); label("$B$", (-2.841870993489946,-0.2973126766527518), NE * labelscalefactor); dot((4.524810009933331,4.202423854807213),linewidth(4pt) + dotstyle); label("$C$", (4.645571276134352,4.41700282644403), NE * labelscalefactor); dot((-2.420620056977336,2.6370215066771823),linewidth(4pt) + dotstyle); label("$D$", (-2.3149769078497178,2.864051837188619), NE * labelscalefactor); dot((-3.9931694036763936,2.6510159061134293),linewidth(4pt) + dotstyle); label("$X_1$", (-4.339359447414806,3.224558316837197), NE * labelscalefactor); dot((-4.7608781438598875,-2.2184509601933016),linewidth(4pt) + dotstyle); label("$X_2$", (-4.6444033917328325,-1.988920004234538), NE * labelscalefactor); dot((1.4193413546381013,4.230059978843752),linewidth(4pt) + dotstyle); label("$Y_1$", (1.5396692976235322,4.444734094109305), NE * labelscalefactor); dot((-0.09672898938034763,-5.386157631787548),linewidth(4pt) + dotstyle); label("$Y_2$", (0.014449576033397386,-5.178015785741184), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Since, $O$ is the intersection of the common external tangents to $\omega_1$ and $\omega_2$, $O$ is the center of homothety mapping $\omega_1$ to $\omega_2$. Call this map $\mathcal{H}.$ $\textbf{Claim : } AD \parallel BC. $ $\textbf{Proof : } \angle DAP = \angle DQP = \angle PCB$ We know that $\mathcal{H} : A \mapsto B$. Thus, by definition of homothety, $\mathcal{H} : D \mapsto D'$ such that $\frac{OA}{OB} = \frac{OD}{OD’}. $ This means that $AD \parallel BD’ \Longrightarrow BD' \parallel BC \Longrightarrow D' = C$ and hence, $\mathcal{H} : D \mapsto C$ or $O, D, C$ are collinear.

heres my solution-

Attachments:
p6.pdf (200kb)

Solution: Note: The diagram is in the second page. Let $OC\cap \omega_1=D'$. Now $O$ is the centre of homothety $\psi$ : $\omega_1 \rightarrow \omega_2$. $\implies \psi:A \rightarrow B $ and $D' \rightarrow C$. $\implies \triangle OD'A \sim \triangle OCB$. $\implies D'A \parallel CB$ . ---- (1) $BQ \cap \omega_1=D$ and $CP \cap \omega_1=A$. So , by Reim's theorem $AD\parallel CB$. (or rather we notice that $AQ$ is antiparallel to $PQ$ , and also $CB$, and as $B,Q,D$ and $C,A,P$ are collinear , we get that $AD\parallel CB$). But from (1) we know that $AD' \parallel CB \parallel AD$ and both $D$ and $D'$ lie on $\omega_1$ and also as $D,D'\neq A$ ; we get that $D=D'$. $\implies C,O,D$ are collinear . $\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(22.908cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -4.3, xmax = 33.88, ymin = -11.68, ymax = 6.3; /* image dimensions */ pen uuuuuu = rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666); pen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1.); /* draw figures */ draw(shift((10.84,0.02)) * scale(1.8601075237738267, 1.8601075237738267)*unitcircle, linewidth(2.)); draw(shift((15.04,-0.1)) * scale(3.388509997034094, 3.388509997034094)*unitcircle, linewidth(2.)); draw((xmin, -0.42380920264516764*xmin + 2.5938280928779434)--(xmax, -0.42380920264516764*xmax + 2.5938280928779434), linewidth(0.4)); /* line */ draw((xmin, 0.35794400857602104*xmin-1.8844337419592139)--(xmax, 0.35794400857602104*xmax-1.8844337419592139), linewidth(0.4)); /* line */ draw((xmin, 0.030981621938216172*xmin-0.011434492976192062)--(xmax, 0.030981621938216172*xmax-0.011434492976192062), linewidth(2.) + dotted); /* line */ draw((xmin, 0.31691054801142043*xmin-5.262263297882586)--(xmax, 0.31691054801142043*xmax-5.262263297882586), linewidth(1.6) + dotted); /* line */ draw((xmin, -1.6804761090198435*xmin + 21.663754350302092)--(xmax, -1.6804761090198435*xmax + 21.663754350302092), linewidth(2.) + dotted); /* line */ draw((xmin, -0.3954979646880618*xmin + 2.431647580383153)--(xmax, -0.3954979646880618*xmax + 2.431647580383153), linewidth(2.) + dotted + red); /* line */ draw((10.7998581447486,-1.8396743347847053)--(12.664752655705204,0.3809400857438858), linewidth(2.)); draw((xmin, 1.1907453249940225*xmin-21.309456846024904)--(xmax, 1.1907453249940225*xmax-21.309456846024904), linewidth(2.)); /* line */ /* dots and labels */ dot((10.114164126479789,-1.692647740988092),linewidth(2.pt) + uuuuuu); dot((10.213133958140643,1.7712963671417934),linewidth(2.pt) + uuuuuu); dot((5.728485371798595,0.1660432750914737),linewidth(3.pt) + uuuuuu); label("$O$", (5.8,0.28), NE * labelscalefactor,uuuuuu); dot((12.027617990536791,1.4516296687877603),linewidth(3.pt) + uuuuuu); label("$P$", (11.58,1.28), NE * labelscalefactor,uuuuuu); dot((11.94393992137185,-1.4771027519851505),linewidth(3.pt) + uuuuuu); label("$Q$", (12.14,-1.74), NE * labelscalefactor,uuuuuu); dot((12.664752655705204,0.3809400857438858),linewidth(3.pt) + xdxdff); label("$A$", (12.74,0.5), NE * labelscalefactor,xdxdff); label("$A_2$", (-4.2,-0.46), NE * labelscalefactor); dot((18.364104943905055,0.5575152636296009),linewidth(4.pt) + uuuuuu); label("$B$", (18.52,0.08), NE * labelscalefactor,uuuuuu); dot((14.966874615536696,-3.4877208678028797),linewidth(3.pt) + uuuuuu); label("$C$", (14.94,-3.04), NE * labelscalefactor,uuuuuu); dot((10.7998581447486,-1.8396743347847053),linewidth(3.pt) + uuuuuu); label("$D$", (10.78,-2.32), NE * labelscalefactor,uuuuuu); dot((11.68156758377308,0.3504794175499869),linewidth(4.pt) + uuuuuu); label("$M$", (11.76,0.52), NE * labelscalefactor,uuuuuu); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* re-scale y/x */ currentpicture = yscale(1.9166666666666665) * currentpicture; /* end of picture */ [/asy][/asy]

The solution I submitted

Nice prob

Attachments:

Notice $\measuredangle PAD=\measuredangle PQD=\measuredangle PCB$ so $\overline{AD}\parallel\overline{BC}.$ Consider the homothety $\psi:\omega_1\mapsto\omega_2,$ noting $A\mapsto B$ so $D\mapsto C.$ Hence, $O,D,$ and $C$ are collinear. $\square$

Spent more time than required for this Refer to lilavati_2005's diagram $$\angle DQP=\angle BQP \iff \angle DAP = \angle PCB \implies AD \parallel BC$$The homothety centred at $O$ that takes $\omega_1 \longrightarrow \omega_2$ takes $A\longrightarrow B$ but because $AD \parallel BC$ it must also take $D \longrightarrow C$ which implies $O-D-C$ as desired $\blacksquare$ Are 2020's problems easier than usual or are 2021's harder than usual?