Let $R$ be the intersection of $AA_1$, $BB_1$. Let $\Omega$ denote the circumcircle of triangle $ABC$, and let $A_0=AA_1 \cap \Omega, B_0=BB_1 \cap \Omega$. Step 1. The points $P,Q, A_0, B_0$ are concyclic. Since $PQ \parallel BC$, $\angle PQR=\angle ABB_0=\angle PA_0B_0$. The conclusion follows. Step 2. The points $P,Q, B_0, P_1$ are concyclic, hence $P_1$ and $Q_1$ lies in $\odot(PQA_0B_0)$. Note that $\angle B_1B_0C=\angle BAC=\angle B_1P_1C$, or $(B_0B_1P_1C)$ are concyclic. Hence $$\angle B_0P_1P=\angle B_1CB_0=\angle ABB_0=\angle PQB_0,$$or $P, Q, B_0, P_1$ are cyclic. This concludes the proof. $\square$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(13cm); 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 = -7.63900015685, xmax = 14.779588377, ymin = -2.38665690592, ymax = 8.57946862605; /* image dimensions */ /* draw figures */ draw((1.,4.96)--(-0.798898966671,-0.708008219337)); draw((1.,4.96)--(7.68847429913,-0.833514663193)); draw((7.68847429913,-0.833514663193)--(-0.798898966671,-0.708008219337)); draw(circle((5.34391751537,3.21734977311), 4.68043258616), green); draw(circle((0.279066234699,2.06933899636), 2.97920568808), green); draw(circle((1.75406921626,4.18630229154), 4.07948863865), blue); draw(circle((3.47180131167,1.0560363319), 4.62068545199), red); draw((3.88543194857,7.66473878747)--(2.88060830443,-0.762418678059)); draw((-2.28184506812,3.59163983096)--(1.27443840444,-0.738667551996)); draw((-1.14173984034,1.31288664799)--(7.68847429913,-0.833514663193)); draw((-0.798898966671,-0.708008219337)--(5.73321976419,5.08551776613)); draw((0.386107411678,0.343008384736)--(3.00779285736,0.304240392932)); /* dots and labels */ dot((1.,4.96),dotstyle); label("$C$", (1.0680093857,5.04959989258), NE * labelscalefactor); dot((7.68847429913,-0.833514663193),dotstyle); label("$B$", (7.75122752106,-0.739384830301), NE * labelscalefactor); dot((-0.798898966671,-0.708008219337),dotstyle); label("$A$", (-0.736145744742,-0.613878386444), NE * labelscalefactor); dot((3.3230046287,2.94782789959),dotstyle); label("$A_1$", (3.38987859704,3.04149679088), NE * labelscalefactor); dot((-0.22800366956,1.09078053776),dotstyle); label("$B_1$", (-0.171366747388,1.19027674399), NE * labelscalefactor); dot((0.386107411678,0.343008384736),dotstyle); label("$P$", (0.456165471895,0.437238080855), NE * labelscalefactor); dot((3.00779285736,0.304240392932),dotstyle); label("$Q$", (3.0761124874,0.405861469891), NE * labelscalefactor); dot((1.27443840444,-0.738667551996),dotstyle); label("$R$", (1.33471057889,-0.645254997408), NE * labelscalefactor); dot((2.88060830443,-0.762418678059),dotstyle); label("$S$", (2.95060604354,-0.66094330289), NE * labelscalefactor); dot((-2.28184506812,3.59163983096),dotstyle); label("$P_1$", (-2.22653476554,3.68471731564), NE * labelscalefactor); dot((3.88543194857,7.66473878747),dotstyle); label("$Q_1$", (3.9546575944,7.76367674098), NE * labelscalefactor); dot((-1.14173984034,1.31288664799),dotstyle); label("$E$", (-1.08128846535,1.40991302074), NE * labelscalefactor); dot((5.73321976419,5.08551776613),dotstyle); label("$D$", (5.7901893358,5.17510633644), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $AA_1,BB_1$ intersect the circumcircle of $\triangle ABC$ again at $D,E$ respectively, and $PB_1,QA_1$ meet $BC$ again at $R,S$ respectively. It is clear that $A,R,C,P_1$ and $B,S,C,Q_1$ are cyclic. By Reim's theorem, we know that $P,Q,D,E$ are cyclic. Taking power of a point at $B_1$, we get $B_1B\cdot B_1E=B_1A\cdot B_1C=B_1R\cdot B_1P_1$. Also, since $\triangle B_1PQ\sim \triangle B_1RB$, we get $B_1B\cdot B_1P=B_1R\cdot B_1Q$. Combining the two gives us $B_1P_1\cdot B_1P= B_1Q\cdot B_1E$, whence $P_1\in (PQDE)$. Similarly $Q_1\in (PQDE)$ are we are done.

$PB_1$ cuts $CB$,$AB$ at $E$,$D$,$QA_1$ cuts $CA$,$AB$ at $G$,$F$. Apply Pappus for ${B_1GA}$ and ${A_1BE}$ we have $B_1B$-$A_1G$=$Q$, $B_1E$-$A_1A$=$P$, $GE$-$AB$ collinear or parallel but $PQ$//$AB$ thus $GE//AB$. It's obvious that $P_1$,$Q_1$ also lies on $(AGE)$ thus by Reim we have the conclusion!

Nice problem Let $A_1Q \cap AC = X$, $B_1P \cap BC = Y$, and extension of $PQ$ meets $AC,BC$ at $M,N$ respectively. From the condition of the problem, $MN||AB$. Lemma $XY||AB||MN$.

Construct the circumcircle $\omega$ of $\bigtriangleup CXY$. Denote $P'$ as the second intersection of $YP$ and $\omega$. Since \[ \angle PP'C = \angle YP'C = \angle YXC = \angle BAC, \]therefore $P'=P_1$. Similarly, $Q'=Q_1$. So, $C,X,Y,P_1,Q_1$ are concyclic and \[ \angle Q_1P_1P = \angle Q_1P_1Y = \angle Q_1XY = \angle Q_1QP. \] Done. $\blacksquare$

mofumofu wrote: [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 = -7.63900015685, xmax = 14.779588377, ymin = -2.38665690592, ymax = 8.57946862605; /* image dimensions */ /* draw figures */ draw((1.,4.96)--(-0.798898966671,-0.708008219337)); draw((1.,4.96)--(7.68847429913,-0.833514663193)); draw((7.68847429913,-0.833514663193)--(-0.798898966671,-0.708008219337)); draw(circle((5.34391751537,3.21734977311), 4.68043258616), green); draw(circle((0.279066234699,2.06933899636), 2.97920568808), green); draw(circle((1.75406921626,4.18630229154), 4.07948863865), blue); draw(circle((3.47180131167,1.0560363319), 4.62068545199), red); draw((3.88543194857,7.66473878747)--(2.88060830443,-0.762418678059)); draw((-2.28184506812,3.59163983096)--(1.27443840444,-0.738667551996)); draw((-1.14173984034,1.31288664799)--(7.68847429913,-0.833514663193)); draw((-0.798898966671,-0.708008219337)--(5.73321976419,5.08551776613)); draw((0.386107411678,0.343008384736)--(3.00779285736,0.304240392932)); /* dots and labels */ dot((1.,4.96),dotstyle); label("$C$", (1.0680093857,5.04959989258), NE * labelscalefactor); dot((7.68847429913,-0.833514663193),dotstyle); label("$B$", (7.75122752106,-0.739384830301), NE * labelscalefactor); dot((-0.798898966671,-0.708008219337),dotstyle); label("$A$", (-0.736145744742,-0.613878386444), NE * labelscalefactor); dot((3.3230046287,2.94782789959),dotstyle); label("$A_1$", (3.38987859704,3.04149679088), NE * labelscalefactor); dot((-0.22800366956,1.09078053776),dotstyle); label("$B_1$", (-0.171366747388,1.19027674399), NE * labelscalefactor); dot((0.386107411678,0.343008384736),dotstyle); label("$P$", (0.456165471895,0.437238080855), NE * labelscalefactor); dot((3.00779285736,0.304240392932),dotstyle); label("$Q$", (3.0761124874,0.405861469891), NE * labelscalefactor); dot((1.27443840444,-0.738667551996),dotstyle); label("$R$", (1.33471057889,-0.645254997408), NE * labelscalefactor); dot((2.88060830443,-0.762418678059),dotstyle); label("$S$", (2.95060604354,-0.66094330289), NE * labelscalefactor); dot((-2.28184506812,3.59163983096),dotstyle); label("$P_1$", (-2.22653476554,3.68471731564), NE * labelscalefactor); dot((3.88543194857,7.66473878747),dotstyle); label("$Q_1$", (3.9546575944,7.76367674098), NE * labelscalefactor); dot((-1.14173984034,1.31288664799),dotstyle); label("$E$", (-1.08128846535,1.40991302074), NE * labelscalefactor); dot((5.73321976419,5.08551776613),dotstyle); label("$D$", (5.7901893358,5.17510633644), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $AA_1,BB_1$ intersect the circumcircle of $\triangle ABC$ again at $D,E$ respectively, and $PB_1,QA_1$ meet $BC$ again at $R,S$ respectively. It is clear that $A,R,C,P_1$ and $B,S,C,Q_1$ are cyclic. By Reim's theorem, we know that $P,Q,D,E$ are cyclic. Taking power of a point at $B_1$, we get $B_1B\cdot B_1E=B_1A\cdot B_1C=B_1R\cdot B_1P_1$. Also, since $\triangle B_1PQ\sim \triangle B_1RB$, we get $B_1B\cdot B_1P=B_1R\cdot B_1Q$. Combining the two gives us $B_1P_1\cdot B_1P= B_1Q\cdot B_1E$, whence $P_1\in (PQDE)$. Similarly $Q_1\in (BQDE)$ are we are done. How do you import a graph form geogebra?

Seicchi28 wrote: Nice problem Let $A_1Q$ meet $AC$ at $X$, $B_1P$ meet $BC$ at $Y$, extension of $PQ$ meets $AC,BC$ at $M,N$ respectively. From the question, $MN||AB$.

Now by the lemma, create the circumcircle $\omega$ of $\bigtriangleup CXY$. Denote $P'$ as the second intersection of $YP$ and $\omega$. Since \[ \angle PP'C = \angle YP'C = \angle YXC = \angle BAC, \]therefore $P'=P_1$. Similarly, $Q'=Q_1$. So, $C,X,Y,P_1,Q_1$ are concyclic and \[ \angle Q_1P_1P = \angle Q_1P_1Y = \angle Q_1XY = \angle Q_1QP.\]Done. Exactly how I did it. Though I proved the lemma with Pappus and then Desargues on perspective triangles.

Since Problem 2 was geometry and problem 2 is medium level, is there a chance that Problem 4 won't be geometry?

Seicchi28 wrote: Nice problem Let $A_1Q \cap AC = X$, $B_1P \cap BC = Y$, and extension of $PQ$ meets $AC,BC$ at $M,N$ respectively. From the condition of the problem, $MN||AB$. Lemma $XY||AB||MN$.

Construct the circumcircle $\omega$ of $\bigtriangleup CXY$. Denote $P'$ as the second intersection of $YP$ and $\omega$. Since \[ \angle PP'C = \angle YP'C = \angle YXC = \angle BAC, \]therefore $P'=P_1$. Similarly, $Q'=Q_1$. So, $C,X,Y,P_1,Q_1$ are concyclic and \[ \angle Q_1P_1P = \angle Q_1P_1Y = \angle Q_1XY = \angle Q_1QP. \] Done. $\blacksquare$ Just the same but my proof of the lemma as follows: Consider two lines $(P, B_1, Y)$ and $(Q, A_1, X)$. By Pappus's theorem points $S=PA_1\cap QB_1$, $T=PX\cap QY$ and $C=B_1Y\cap A_1X$ are collinear. Now, consider trinagles $APX$ and $BQY$. We claim that they are perspective. Indeed, $AP\cap BQ = S$, $AX\cap BY=C$ and $PX\cap QY=T$ are colinear. Hence, by Desargues's theorem lines $AB$, $PQ$ and $XY$ are concurrent. Since $AB\parallel PQ$ we have $AB\parallel PQ\parallel XY$. TUGUMEANDREW wrote: Since Problem 2 was geometry and problem 2 is medium level, is there a chance that Problem 4 won't be geometry? Usually (at least in recent years) problems 1,2,4,5 have different topics (ACGN).

Finally, the first medium geometry problem since the introduction of the new protocol! So surprised that they didn't vote for G1 this time.

TUGUMEANDREW wrote: Since Problem 2 was geometry and problem 2 is medium level, is there a chance that Problem 4 won't be geometry? P6 can be a Geo. P4,5 would be NT and Combi

Lol, this is my problem! Didn't expect to see this at P2

Beautiful problem!!! My solution is kinda long, especially because I used Moving Points during it. Let $B_1P_1$ cut $AB$ at $U$ and $Q_1A_1$ cut $AB$ at $V$. Moreover, let $L$ be the intersection point of lines $B_1P_1$ and $BC$ and $K$ the intersection point of lines $Q_1A_1$ and $AC$. Since $\angle PP_1C\equiv \angle BAC$, we conclude that $AP_1CU$ is cyclic. Similarly, since $\angle QQ_1C\equiv \angle ABC$, $BQ_1CV$ is cyclic, too. Keep these in mind as (1). The conclusion is equivalent to $\angle QPP_1\equiv \angle QQ_1P_1$. However, since $PQ\parallel AB$, we have $\angle QPP_1\equiv \angle AUP_1$, so it suffices to prove $\angle AUP_1\equiv \angle QQ_1P_1$, which is equivalent to $UVP_1Q_1$ being cyclic. $\textbf{Claim.}$ We have that $KL\parallel AB$. $\textbf{Proof:}$ We show this by the Moving Points method. Fix $\Delta ABC$, points $A_1$, $B_1$ and animate $P$ on $AA_1$. Consider $L'$ on $AB$ such that $KL'\parallel AB$. We prove that the map $$L\to P\to Q\to K\to L'~~by~~\overline{BC}\to \overline{AA_1}\to \overline{BB_1}\to \overline{AC}\to \overline{BC}$$is projective, thus it would ve enough to check $L=L'$ for $3$ distinct positions of $P$. $\textbf{(1)}$ $L\to P$ by $\overline{BC}\to \overline{AA_1}$ is projective since it is a perspectivity through $B_1$ from $BC$ to $AA_1$. $\textbf{(2)}$ $P\to Q$ by $\overline{AA_1}\to \overline{BB_1}$ is projective since it preserves linear motion. $\textbf{(3)}$ $Q\to K$ by $\overline{BB_1}\to \overline{AC}$ is projective since it is a perspectivity through $A_1$ from $BB_1$ to $AC$. $\textbf{(4)}$ $K\to L'$ by $ \overline{AC}\to \overline{BC}$ is projective since it preserves linear motion. Now it suffices to check for $3$ distinct positions of $P$ the desired parallelism. $\textbf{(1)}$ When $P\in AA_1\cap BB_1$, we actually have $Q=P$, $L=B$ and $K=A$ so there is nothing to prove. $\textbf{(2)}$ When $P=A$, $Q=B$ so $C=K=L$, again nothing to prove. $\textbf{(3)}$ When $P$ is the point at infinity on $\overline{AA_1}$, $Q$ is the point at infinity on $\overline{BB_1}$ so $B_1L\parallel AA_1$ and $A_1K\parallel BB_1$. Then apply Thales' Lemma to obtain $CB_1\cdot CA_1=CK\cdot CB$ and $CA_1\cdot CB_1=CL\cdot CA$, hence $CK\cdot CB=CL\cdot CA$ so indeed $KL\parallel BC$.$\blacksquare$ Now we are ready to finish. From (1) and Power of a Point Theorem we get $PB_1\cdot B_1U=AB_1\cdot CB_1$. By our lemma and Thales' Lemma, $B_1K\cdot B_1U=B_1L\cdot B_1A$. Divide these two last relations in order to get $PB_1\cdot B_1L=CB_1\cdot B_1K$, which again by Power of a Point leads to $CP_1KL$ being cyclic. In a similar manner one proves the cyclicity of $CQ_1KL$, so the pentagon $CP_1KLQ_1$ is cyclic. But then $\angle KQ_1P_1\equiv \angle KP_1C$, and since by (1) we have $\angle KCP_1\equiv \angle AUP_1$, we arrive at $\angle AUP_1\equiv \angle VQ_1P_1$, hence $UVQ_1P_1$ is cyclic and done.

-[]- wrote: Finally, the first medium geometry problem since the introduction of the new protocol! So surprised that they didn't vote for G1 this time. This probably just means that G1 and G2 got eliminated as known problems, and that G3 contained traces of combinatorial geometry.

MS_Kekas wrote: Lol, this is my problem! Didn't expect to see this at P2 Wowww. Where are you from?

Math-wiz wrote: MS_Kekas wrote: Lol, this is my problem! Didn't expect to see this at P2 Wowww. Where are you from? Ukraine

[quote=-[]-]Finally, the first medium geometry problem since the introduction of the new protocol! So surprised that they didn't vote for G1 this time.[/quote] What is the new protocol? When was it introduced?

See most solutions rely on angle chase. Any length chasing?

Okay, How long did it take to solve the Problem 2 of Day1 exam?

Solution. Let $X=\overline{BQ}\cap (ABC)$ and $Y=\overline{AP}\cap (ABC)$ so that $X\neq B$ and $Y\neq B$. By Reim's theorem, $PXYQ$ is cyclic. Define $P_2=\overline{PP_1}\cap \overline{AB}$ and $Q_2=\overline{QQ_1}\cap \overline{AB}$. We have $$\angle Q_2Q_1C=\angle QQ_1C=\angle CBA=\angle CBQ_2$$thus $Q_2CQ_1B$ is cyclic, therefore $$AA_1\cdot A_1Y=BA_1\cdot A_1C=Q_2A_1\cdot A_1Q_1$$so $AQ_1YQ_2$ is cyclic. Again, by Reim, we deduce that $PQ_1YQ$ is cyclic. Analogously, $PXP_1Q$ is cyclic. We conclude that $P,X,P_1,Q_1,Y,Q$ all lie on the same circle, which clearly implies the required result. $\blacksquare$

The Pappus step took me three times as long as it should have Solution. We begin by proving an important lemma. Lemma: Given a quadrilateral $BFEC$ such that points $P$ and $Q$ lie on diagonals $BE$ and $CF$ with $PQ \parallel BC$, suppose that $PF$ and $CE$ meet at $X$ and suppose that $QE$ and $BF$ meet at $Y$. Then $XY \parallel PQ \parallel BC$. Proof: By Pappus, we have $XY \cap PQ, XE \cap FQ,$ and $PE \cap FY$ collinear, which implies that $XY \cap PQ, B,$ and $C$ are collinear. But as $PQ \parallel BC$, this forces $XY \cap PQ$ to be the point at infinity. Back to the main problem: let $PP_1$ and $BC$ meet at $X$, and let $QQ_1$ and $AC$ meet at $Y$. Then $XY \parallel PQ \parallel AB$. Thus, as $\angle YXC= \angle CQ_1Q = \angle B$, we know that $XYQ_1C$ is cyclic, and similarly that $YXP_1C$ is cyclic, so $XYQ_1CP_1$ is a cyclic pentagon. Thus $XY$ and $P_1Q_1$ are antiparallel with respect to $PX \cap QY$, and as $XY$ and $PQ$ are parallel, we obtain that $P_1Q_1QP$ is cyclic, as desired.

Firstly. let $BB_1\cap (ABC)=B'$, similarly we define $A'$. As $\angle{CP_1P}=\angle{BAC}=\angle{CB'B}$, $CP_1B'B_1$ is a cyclic quadrilateral, similarly, $CQ_1A'A_1$ is a cyclic quadrilateral. As $\angle{BAA'}=\angle{QPA'}=\angle{QB'A'}$ and similar reason to get $\angle{B'A'P}=\angle{B'QP}$, $B'A'Q'P$ are concyclic Then, we use the given information $PQ||AB, \angle{A'PQ}=\angle{A'AB}=\angle{A'CA_1}=\angle{A'Q_1Q}$; thus $A'Q_1CA_1$ are concyclic. Same reason, $B'Q_1QP$ are also concyclic, $PQP_1Q_1$ are concyclic as desired

Let the intersection of $P_1P$ and $AB$ be denoted as point D, and the intersection of $Q_1Q$ and $AB$ be denoted as point E. Also let $AA_1\cap (ABC)=F$ and $BB_1\cap (ABC)=G$. Because $\angle PP_1C=\angle BAC$ and $\angle CB_1P_1=\angle B_1AD$, we get $\triangle CP_1B_1 \sim \triangle B_1AD$ and $CP_1AD$ are concyclic Because $\angle CQ_1Q=\angle CBA$ and $\angle CA_1Q_1=\angle BA_1E$, we get $\triangle CA_1Q_1 \sim \triangle BA_1E$ and $CQ_1BE$ are concyclic. Because $PQ \parallel AB$, $\angle B_1QP = \angle B_1BD$ and $\angle B_1PQ = \angle B_1DB$ so $\triangle B_1PQ\sim \triangle B_1DB$ Also because $PQ \parallel AB$, $\angle A_1PQ = \angle A_1AE$ and $\angle A_1QP = \angle A_1EA$ so $\triangle A_1PQ \sim \triangle A_1AE$ By Reim's, $FPQG$ are concyclic. Using Power of a point of $B_1$ with respect to $CP_1AD$, we get $CB_1 \cdot B_1A = P_1B_1 \cdot B_1D$. Using Power of a point of $B_1$ with respect to $(ABC)$, we get $EB_1 \cdot B_1B = CB_1 \cdot B_1A$. Using Power of a point of $A_1$ with respect to $CQ_1BE$, we get $CA_1 \cdot A_1B = Q_1A_1 \cdot A_1E$. Using Power of a point of $A_1$ with respect to $(ABC)$, we get $DA_1 \cdot A_1A = CA_1 \cdot A_1B$. Since $\triangle B_1PQ \sim \triangle B_1DB$, $B_1B\cdot B_1P=B_1D\cdot B_1Q$ Since $\triangle A_1PQ \sim \triangle A_1AE$, $A_1A\cdot A_1Q=A_1E\cdot A_1P$ We can eventually get $(B_1P_1)^2= B_1Q\cdot B_1G$ and $(A_1Q_1)^2= A_1P\cdot A_1F$ These prove that $P_1$ and $Q_1$ are both on circle $FGQP$

Let $AA_1$ and $BB_1$ intersect $(ABC)$ at $D$ and $E$, respectively. The following quadrilateral are cyclic: $PQDE$ because $\angle QPD=\angle BAD=\angle QED$. $B_1EP_1C$ because $\angle B_1P_1C=\angle BAC=\angle BEC$. $PDP_1E$ because $\angle PP_1E=\angle B_1CE=\angle ABE=\angle PDE$. $QDQ_1E$ similarly to $(3)$. Thus, $PQP_1Q_1$ is cyclic.

[asy][asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions (otherwise throws error due to the missing xmin) are done using bubu-asy.py. This adds back the dps, real xmin coordinates, changes the linewidth, the fontsize and adds the directions to the labellings. */ pair A = (-80,-52.9); pair B = (66.99396,-52.9); pair C = (-42.42817,70.47100); pair T = (-39.70091,-9.14504); pair P = (-57.87643,-28.87921); pair Q = (8.42014,-28.87921); pair S = (-72.68462,-28.87921); pair R = (45.68907,-28.87921); pair B_1 = (-63.67993,0.68861); pair A_1 = (-5.11751,28.40407); pair P_1 = (-78.09177,74.11437); pair Q_1 = (-21.94016,99.58748); pair X = (-90.44944,11.66663); pair Y = (32.33519,69.06855); import graph; size(13cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); pen ffxfqq = rgb(1,0.49803,0); draw(A--B, linewidth(0.5) + ffxfqq); draw((-2.75727,-52.9)--(-4.63014,-55.30797), linewidth(0.5) + ffxfqq); draw((-2.75727,-52.9)--(-4.63014,-50.49202), linewidth(0.5) + ffxfqq); draw((-6.50302,-52.9)--(-8.37589,-55.30797), linewidth(0.5) + ffxfqq); draw((-6.50302,-52.9)--(-8.37589,-50.49202), linewidth(0.5) + ffxfqq); draw(B--C, linewidth(0.5)); draw(C--A, linewidth(0.5)); draw(circle((-6.50302,-7.87639), 86.19124), linewidth(0.5)); draw(circle((-65.28053,23.14823), 52.55164), linewidth(0.5) + red); draw(circle((27.05460,43.34548), 74.58988), linewidth(0.5) + red); draw(circle((-24.72814,31.10775), 68.53645), linewidth(0.5) + linetype("4 4") + blue); draw(A--Y, linewidth(0.5) + blue); draw(X--B, linewidth(0.5) + blue); draw(P--P_1, linewidth(0.5)); draw(Q--Q_1, linewidth(0.5)); draw(S--R, linewidth(0.5) + ffxfqq); draw((-9.75203,-28.87921)--(-11.62490,-31.28719), linewidth(0.5) + ffxfqq); draw((-9.75203,-28.87921)--(-11.62490,-26.47123), linewidth(0.5) + ffxfqq); draw((-13.49777,-28.87921)--(-15.37064,-31.28719), linewidth(0.5) + ffxfqq); draw((-13.49777,-28.87921)--(-15.37064,-26.47123), linewidth(0.5) + ffxfqq); draw(P_1--C, linewidth(0.5)); draw(C--Q_1, linewidth(0.5)); dot("$A$", A, W); dot("$B$", B, dir(0)); dot("$C$", C, 1.5*dir(105)); dot("$T$", T, N); dot("$Q$", Q, NE); dot("$S$", S, NW); dot("$R$", R, NE); dot("$B_1$", B_1, 2*dir(125)); dot("$A_1$", A_1, dir(0)); dot("$P_1$", P_1, NW); dot("$Q_1$", Q_1, N); dot("$X$", X, W); dot("$Y$", Y, NE); [/asy][/asy] Let $R=PQ\cap CB$ and $S=PQ\cap CA$. Also let $X=BB_1\cap\odot(ABC)$ and $Y=AA_1\cap\odot(ABC)$. Finally let $T=AA_1\cap BB_1$. Firstly, as we have that $PQ\parallel AB$, we get $\measuredangle PP_1C=\measuredangle BAC=\measuredangle PSC\implies PSP_1C$ is cyclic. Similarly, $QCQ_1R$ is cyclic. Also, we have that $\measuredangle XYP=\measuredangle XYA=\measuredangle XBA=\measuredangle XQP\implies XPQY$ is cyclic. Now also note that $\measuredangle XCS=\measuredangle XCA=\measuredangle XBA=\measuredangle XQS\implies XSQC$ is cyclic. Similarly, $YRPC$ is also cyclic. Now then, we have that $B_1P_1\cdot B_1P=B_1C\cdot B_1S=B_1X\cdot B_1Q\implies P_1XPQ$ is cyclic. Similarly, we also get that $Q_1YQP$ is also cyclic. Combining this with the fact that $XYPQ$ is cyclic, we get that $P_1XPQYQ_1$ is cyclic which finishes.

Define $X=QA_1\cap AB$, $Y=PB_1\cap AB$, $S= AQ\cap PB$, and $T=QX\cap PY.$ Claim: $P_1Q_1XY$ is cyclic Proof. Notice that by radical axis this is equivalent to showing $T$ is on the radical axis of $(BXCP_1),(AYCQ_1)$. First, pappus gives $C$, $S$, $T$ are collinear. Then, DDIT on the complete quad formed by lines $AB$, $PQ$, $AQ$, $BP$ gives an involution swapping pairs $(TA,TY)$ $(TX,TB)$ and $(TS,T\infty)$. Projecting onto $AB$ gives an involution swpaping $(A,Y)$ $(B,X)$ $(\infty, F'=AB\cap TS)$ But if the radical axis from above hits $AB$ at $F$, then $FY\cdot FA=FX\cdot FB$ by powers so the involution swapping $(A,Y)$ and $(B,X)$ is an inversion at $F$ so $(F,\infty)$ is an involute pairing and $F=F'$. Thus line $CST$ is the desired radical axis. $\blacksquare$ Reim's theorem now finishes.

Let $\overline{AA_1}$ and $\overline{BB_1}$ intersect $(ABC)$ again at $D$ and $E$ respectively, so the angle condition implies $CA_1DQ_1$ and $CB_1EP_1$ are cyclic. By Reim's, $PQDE$ is cyclic as well, since $ABDE$ is. I claim that $PQDQ_1$ is cyclic. Indeed, note that $$\measuredangle DQ_1Q=\measuredangle DAB=\measuredangle DBC=\measuredangle DPQ;$$$PQEP_1$ cyclic follows similarly. Combining these, it follows that $PQDEP_1Q_1$ is cyclic as desired. $\blacksquare$ Remark: wow. The author of this problem seems to always come up with these really clean, fresh geo problems (JMO 2023/6 is another example) Remark (motivation): The construction of $D$ and $E$ is motivated by trying to make sense of the strange angle condition: I personally noticed that as $\overline{PQ}$ varies, $P_1$ and $Q_1$ lie on fixed circles, which naturally led me to figure out what these circles were by introducing these points. Then drawing an accurate diagram makes it clear how to proceed.

pretty straightforward $U=PQ\cap BC$ $V=PQ\cap AC$ $R=PB_1\cap QA_1$ $X$ is the center of homothety sending $PQ$ to $UV$ $S=AQ\cap BP$ Notice that $PVP_1C$ and $QUQ_1C$ are cyclic. By Monge, we have $C,S,X$ collinear. By Pappus on $A_1PA$ and $B_1QB$ we have that $R,C,S$ are collinear. Notice that $CX$ is the radical axis of $(PVC)$ and $(QUC)$. Thus $R$ lies on the radical axis. Put another way, $RP\cdot RP_1=RQ\cdot RQ_1$. Thus $P,Q,P_1,Q_1$ are concyclic. Done.

A Plain figure with no additions [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15.29453946345512cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(13); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -3.545414608519562, xmax = 32.74912485493556, ymin = -5.711343740006539, ymax = 10.520050302010716; /* image dimensions */ pen qqttzz = rgb(0.,0.2,0.6); pen qqzzcc = rgb(0.,0.6,0.8); pen qqttqq = rgb(0.,0.2,0.); pen qqqqcc = rgb(0.,0.,0.8); pen qqwwtt = rgb(0.,0.4,0.2); pen qqttcc = rgb(0.,0.2,0.8); /* draw figures */ draw(circle((10.,0.), 5.604319762468949), linewidth(1.) + qqttzz); draw((4.583966606974815,-1.4404798803302028)--(15.506441446309776,-1.0428339265491575), linewidth(1.) + qqzzcc); draw((15.506441446309776,-1.0428339265491575)--(8.202344722423005,5.30818570728263), linewidth(1.) + qqttqq); draw((4.583966606974815,-1.4404798803302028)--(8.202344722423005,5.30818570728263), linewidth(1.) + qqttqq); draw((6.171195380420182,1.5198730350512988)--(15.506441446309776,-1.0428339265491575), linewidth(1.) + qqqqcc); draw((4.583966606974815,-1.4404798803302028)--(10.151186194117859,3.6136393563995135), linewidth(1.) + qqqqcc); draw(circle((13.7340954217603,4.2944591119622375), 5.623869433911009), linewidth(1.) + qqwwtt); draw((8.689119314947433,6.779639888357758)--(8.202344722423005,5.30818570728263), linewidth(1.)); draw(circle((4.999461201237492,2.681098940226504), 4.142468796735001), linewidth(1.) + qqwwtt); draw((6.927967288402725,6.347283878781809)--(8.202344722423005,5.30818570728263), linewidth(1.)); draw((6.927967288402725,6.347283878781809)--(8.689119314947433,6.779639888357758), linewidth(1.)); draw(circle((8.762115488952197,2.679196248573975), 4.101093327938611), linewidth(1.) + linetype("2 2") + qqttcc); draw((12.354548218091669,-1.1575824251620528)--(8.689119314947433,6.779639888357758), linewidth(1.) + qqttzz); draw((5.713560776514278,-1.3993556335327275)--(6.927967288402725,6.347283878781809), linewidth(1.) + qqttzz); draw((5.893480461757436,-0.2516567953731126)--(11.836276385567048,-0.03530209664319856), linewidth(1.) + qqzzcc); /* locus construction */ draw((6.9741,7.6879)--(6.9741,7.6879)^^(7.2834,7.8069)--(7.2834,7.8069), linewidth(1.6)); /* dots and labels */ dot((10.,0.),dotstyle); dot((15.506441446309776,-1.0428339265491575),dotstyle); label("$B$", (15.613342288582977,-0.8086059501890119), NE * labelscalefactor); dot((4.583966606974815,-1.4404798803302028),dotstyle); label("$A$", (4.6892809025332065,-1.2132008163390022), NE * labelscalefactor); dot((8.202344722423005,5.30818570728263),dotstyle); label("$C$", (8.306834999874306,5.545913418166718), NE * labelscalefactor); dot((6.171195380420182,1.5198730350512988),dotstyle); label("$B_1$", (6.260060971115526,1.7617614347638675), NE * labelscalefactor); dot((10.151186194117859,3.6136393563995135),dotstyle); label("$A_1$", (10.234610538588973,3.856134859540288), NE * labelscalefactor); dot((5.893480461757436,-0.2516567953731126),dotstyle); label("$P$", (5.998264293018473,-0.023215915897854326), NE * labelscalefactor); dot((11.836276385567048,-0.03530209664319856),linewidth(4.pt) + dotstyle); label("$Q$", (11.924389097215407,0.14338197016390639), NE * labelscalefactor); dot((12.354548218091669,-1.1575824251620528),linewidth(1.pt) + dotstyle); label("$J$", (12.447982453409514,-0.9752038362507727), NE * labelscalefactor); dot((8.689119314947433,6.779639888357758),linewidth(4.pt) + dotstyle); label("$Q'$", (8.782828960050766,6.973895298696095), NE * labelscalefactor); dot((5.713560776514278,-1.3993556335327275),linewidth(4.pt) + dotstyle); label("$L$", (5.807866708947889,-1.2132008163390022), NE * labelscalefactor); dot((6.927967288402725,6.347283878781809),linewidth(4.pt) + dotstyle); label("$P'$", (7.021651307397863,6.545500734537282), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(14.473791467295925cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(12); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -3.722705371673187, xmax = 36.75108609562274, ymin = -7.619324939355466, ymax = 10.565718542968266; /* image dimensions */ pen qqttzz = rgb(0.,0.2,0.6); pen qqzzcc = rgb(0.,0.6,0.8); pen qqttqq = rgb(0.,0.2,0.); pen qqqqcc = rgb(0.,0.,0.8); pen qqwwtt = rgb(0.,0.4,0.2); pen wwzzff = rgb(0.4,0.6,1.); pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882); pen qqttcc = rgb(0.,0.2,0.8); pen eqeqeq = rgb(0.8784313725490196,0.8784313725490196,0.8784313725490196); pen zzccff = rgb(0.6,0.8,1.); draw(circle((8.506812245675057,2.9604181885520444), 4.140885015934622), linewidth(1.) + linetype("2 2") + wwzzff); /* draw figures */ draw(circle((10.,0.), 5.604319762468949), linewidth(1.) + qqttzz); draw((4.583966606974815,-1.4404798803302028)--(15.506441446309776,-1.0428339265491575), linewidth(1.) + qqzzcc); draw((15.506441446309776,-1.0428339265491575)--(8.202344722423005,5.30818570728263), linewidth(1.) + qqttqq); draw((4.583966606974815,-1.4404798803302028)--(8.202344722423005,5.30818570728263), linewidth(1.) + qqttqq); draw((6.008784688561027,1.2169595715492947)--(15.506441446309776,-1.0428339265491575), linewidth(1.) + qqqqcc); draw((4.583966606974815,-1.4404798803302028)--(10.151186194117859,3.6136393563995135), linewidth(1.) + qqqqcc); draw(circle((13.426534840451424,3.940743903404536), 5.4001906888830495), linewidth(1.) + qqwwtt); draw((9.024571191585899,7.068806484358715)--(8.202344722423005,5.30818570728263), linewidth(1.)); draw(circle((5.045453049041576,2.6564398586017166), 4.122829257131737), linewidth(1.) + qqwwtt); draw((6.426784885318526,6.540978884219707)--(8.202344722423005,5.30818570728263), linewidth(1.)); draw((11.714520493171745,-1.1808834113273448)--(9.024571191585899,7.068806484358715), linewidth(1.) + qqttzz); draw((5.803629606185249,-1.3960765684666794)--(6.426784885318526,6.540978884219707), linewidth(1.) + qqttzz); draw((5.893480461757436,-0.2516567953731126)--(11.346795115678864,-0.053122256385905266), linewidth(1.) + qqzzcc); draw((5.803629606185249,-1.3960765684666794)--(8.851538800117314,1.071689929142913), linewidth(1.) + cqcqcq); draw((6.008784688561027,1.2169595715492947)--(4.613952109782934,1.5488344405675873), linewidth(1.) + qqttcc); draw((10.151186194117859,3.6136393563995135)--(11.9558341730438,5.251962746207756), linewidth(1.) + qqttcc); draw((11.9558341730438,5.251962746207756)--(9.024571191585899,7.068806484358715), linewidth(1.) + cqcqcq); draw((4.613952109782934,1.5488344405675873)--(6.426784885318526,6.540978884219707), linewidth(1.) + cqcqcq); draw((4.613952109782934,1.5488344405675873)--(11.9558341730438,5.251962746207756), linewidth(1.) + eqeqeq); draw((11.346795115678864,-0.053122256385905266)--(11.9558341730438,5.251962746207756), linewidth(1.) + eqeqeq); draw((5.893480461757436,-0.2516567953731126)--(4.613952109782934,1.5488344405675873), linewidth(1.) + eqeqeq); draw((11.346795115678864,-0.053122256385905266)--(6.426784885318526,6.540978884219707), linewidth(1.) + cqcqcq); draw((4.613952109782934,1.5488344405675873)--(8.202344722423005,5.30818570728263), linewidth(1.) + qqttzz); draw(circle((5.910994086898991,3.903056024991948), 2.6878759567678383), linewidth(1.) + zzccff); draw((4.583966606974815,-1.4404798803302028)--(4.613952109782934,1.5488344405675873), linewidth(1.) + qqttzz); draw((4.613952109782934,1.5488344405675873)--(5.803629606185249,-1.3960765684666794), linewidth(1.) + qqttzz); draw(circle((10.082420780384243,5.502476635340945), 1.890088608622575), linewidth(1.) + zzccff); /* dots and labels */ dot((15.506441446309776,-1.0428339265491575),dotstyle); label("$B$", (15.615666827081196,-0.779817434941671), NE * labelscalefactor); dot((4.583966606974815,-1.4404798803302028),dotstyle); label("$A$", (4.699276418075602,-1.1821414057895412), NE * labelscalefactor); dot((8.202344722423005,5.30818570728263),dotstyle); label("$C$", (8.320192155706449,5.576901304454677), NE * labelscalefactor); dot((6.008784688561027,1.2169595715492947),dotstyle); label("$B_1$", (6.120821115071417,1.4731968018064017), NE * labelscalefactor); dot((10.151186194117859,3.6136393563995135),dotstyle); label("$A_1$", (10.251347215776235,3.8871406268936224), NE * labelscalefactor); dot((5.893480461757436,-0.2516567953731126),dotstyle); label("$P$", (6.013534722845317,0.024830506754069278), NE * labelscalefactor); dot((11.346795115678864,-0.053122256385905266),linewidth(4.pt) + dotstyle); label("$Q$", (11.458319128319852,0.15893849703669266), NE * labelscalefactor); dot((11.714520493171745,-1.1808834113273448),linewidth(4.pt) + dotstyle); label("$J$", (11.8338215011112,-0.9675686213373438), NE * labelscalefactor); dot((9.024571191585899,7.068806484358715),linewidth(4.pt) + dotstyle); label("$Q'$", (9.124840097402194,7.293483580072256), NE * labelscalefactor); dot((5.803629606185249,-1.3960765684666794),linewidth(4.pt) + dotstyle); label("$L$", (5.906248330619218,-1.1821414057895412), NE * labelscalefactor); dot((6.426784885318526,6.540978884219707),linewidth(4.pt) + dotstyle); label("$P'$", (6.523145085919288,6.757051618941763), NE * labelscalefactor); dot((8.851538800117314,1.071689929142913),linewidth(4.pt) + dotstyle); label("$N$", (8.963910509063044,1.285445615410729), NE * labelscalefactor); dot((4.613952109782934,1.5488344405675873),linewidth(4.pt) + dotstyle); label("$E$", (4.726098016132126,1.7682343804281733), NE * labelscalefactor); dot((11.9558341730438,5.251962746207756),linewidth(4.pt) + dotstyle); label("$F$", (12.075215883619922,5.469614912228579), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Sketch of solution: 1) Let $AA_1$ meets circle $(ABC) $at $F$, and let $BB_1$ meets circle $(ABC)$ at $E$. 2) By Reim theorem you have that $EFPQ$ cyclic. 3) By some angle chase you show $EB_1P'C $ cyclic, and similarly $FA_1Q'C$. By simple angle chase you will se that $EPQP'$ cyclic, and similarly $FPQQ'$ as desired.

WHAT IN THE WORLD IS THIS Define $Q_2$ and $P_2$ as the intersection of side $AB$ with lines $QQ_1$ and $PP_1$ respectively, and define $X$ as the intersection of lines $QQ_1$ and $PP_1$. Claim: The problem is affine. Proof: The angle conditions imply that $CQ_1 BQ_2$ and $CP_1 A P_2$ are cyclic – it suffices to show that $X$ lies on the radical axis of the two circles. Let $Y$ be the intersection of lines $CX$ and $AB$ – it is equivalent to the original problem statement to show that $YQ_2 \cdot YB = YP_2 \cdot YA$. Now, points $P_1$ and $Q_1$ don't matter anymore; the definition of all points is affine, and what we're asked to prove is affine. Now we will solve the original problem (adding points $P_1$ and $Q_1$ back) with the additional assumption that $\overline{AA_1}$ and $\overline{BB_1}$ are altitudes. Since $BA_1 B_1A$ is cyclic, by Reim's theorem, it follows that $A_1 QPB_1$ too is cyclic.

Since $PQA_1 B_1$ is cyclic, by Reim once more, $PQP_1 Q_1$ is cyclic.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(14.473791467295925cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.4) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -33.90294294377877, xmax = 61.659999263181504, ymin = -41.602313770778665, ymax = 14.038772683926569; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); pen ffdxqq = rgb(1,0.8431372549019608,0); pen ffqqtt = rgb(1,0,0.2); pen ttttff = rgb(0.2,0.2,1); pen ttffqq = rgb(0.2,1,0); pen ffwwqq = rgb(1,0.4,0); pen qqzzff = rgb(0,0.6,1); pen ffqqff = rgb(1,0,1); pen xfqqff = rgb(0.4980392156862745,0,1); draw((-5,10)--(-16.24999735332854,-20.518583701619058)--(20.270971501626907,-20.662633307357392)--cycle, linewidth(1.) + zzttqq); /* draw figures */ draw((-5,10)--(-16.24999735332854,-20.518583701619058), linewidth(1.) + ffdxqq); draw((-16.24999735332854,-20.518583701619058)--(20.270971501626907,-20.662633307357392), linewidth(1.) + ffdxqq); draw((20.270971501626907,-20.662633307357392)--(-5,10), linewidth(1.) + ffdxqq); draw(circle((2.0525255237098183,-9.932579284308634), 21.143458401780634), linewidth(1.) + ffqqtt); draw((-5,10)--(-4.913765920945284,-29.895460256151182), linewidth(1.) + ttttff); draw((-16.24999735332854,-20.518583701619058)--(16.300614749971015,5.689129539899033), linewidth(1.) + ttffqq); draw((xmin, 2.712763633903391*xmin + 12.457690285813076)--(xmax, 2.712763633903391*xmax + 12.457690285813076), linewidth(0.4) + linetype("0 3 4 3") + white); /* line */ draw((-4.976134034432313,-1.041385160223868)--(-10.428035306194664,-15.831104665892425), linewidth(1.) + ffwwqq); draw((-4.976134034432313,-1.041385160223868)--(31.423730105365635,-7.674407099508636), linewidth(1.) + ttffqq); draw((31.423730105365635,-7.674407099508636)--(16.300614749971015,5.689129539899033), linewidth(1.) + qqzzff); draw((31.423730105365635,-7.674407099508636)--(20.270971501626907,-20.662633307357392), linewidth(1.) + qqzzff); draw((-10.428035306194664,-15.831104665892425)--(14.190417657765865,-36.92397129981129), linewidth(1.) + ffqqtt); draw((14.190417657765865,-36.92397129981129)--(20.270971501626907,-20.662633307357392), linewidth(1.) + ffqqtt); draw((16.300614749971015,5.689129539899033)--(20.270971501626907,-20.662633307357392), linewidth(1.) + ttffqq); draw((-4.913765920945284,-29.895460256151182)--(20.270971501626907,-20.662633307357392), linewidth(1.) + red); draw(circle((18.120084891832406,-7.490787774969996), 13.30491232680282), linewidth(1.) + ffwwqq); draw(circle((7.6504156543471025,-25.202159646033778), 13.412148570376223), linewidth(1.) + ffqqff); draw(circle((11.275454012939369,-15.42082374137463), 21.699824146074572), linewidth(1.) + linetype("4 4") + blue); draw((-5,10)--(16.300614749971015,5.689129539899033), linewidth(1.) + xfqqff); /* dots and labels */ dot((-5,10),dotstyle); label("$A$", (-5.708132318748717,10.857499803668759), NE * labelscalefactor); dot((-16.24999735332854,-20.518583701619058),dotstyle); label("$B$", (-17.7470669440381,-21.329496396586734), NE * labelscalefactor); dot((20.270971501626907,-20.662633307357392),dotstyle); label("$C$", (21.176742414410473,-21.89089749310282), NE * labelscalefactor); dot((-4.913765920945284,-29.895460256151182),dotstyle); label("$S$", (-4.647708025329445,-29.251489647424812), NE * labelscalefactor); dot((16.300614749971015,5.689129539899033),dotstyle); label("$T$", (16.560777843055995,6.303913131927187), NE * labelscalefactor); dot((-4.933937573159862,-20.56321762018564),dotstyle); label("$A_1$", (-6.706178712555091,-21.641385894651226), NE * labelscalefactor); dot((5.632094760585414,-2.9004942652053733),dotstyle); label("$B_1$", (5.0208664146698,-1.3061906208463983), NE * labelscalefactor); dot((-4.976134034432313,-1.041385160223868),dotstyle); label("$P$", (-6.331911314877701,-0.9319232231690088), NE * labelscalefactor); label("$h$", (0.8415471406056099,12.541703093217011), NE * labelscalefactor,white); dot((-10.428035306194664,-15.831104665892425),dotstyle); label("$Q$", (-11.821166480812755,-15.902619130264586), NE * labelscalefactor); dot((31.423730105365635,-7.674407099508636),dotstyle); label("$P_1$", (31.656229549377397,-7.044957385233036), NE * labelscalefactor); dot((14.190417657765865,-36.92397129981129),dotstyle); label("$Q_1$", (14.190417657765856,-38.35866299090795), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let, $T=BB_1\cap (ABC)$ and $S=AA_1\cap (ABC).$ $\color{red}\textbf{Claim:-}$ $P_1,T,B_1,C$ and $Q_1,C,A,S$ are concyclic points. $\color{blue}\textbf{Proof:-}$ Since, the points $A,B,C,T$ are cyclic and $PQ||AB$ we get some angle conditions$$\angle TCB_1=\angle ABT=\angle PQB\implies\angle BAC=\angle BTC=\angle PP_1C$$Therefore the points $P_1,T,B_1,C$ are concyclic. Now From second Intersection of points we get,$$\angle CBA=\angle ASC=\angle A_1SC=\angle CQ_1A_1$$Therefore the points $Q_1,C,A,S$ are concyclic also. Now By Simple Angle chase we get,$$\angle PST=\angle ABT=\angle PQT$$Therefore the points $P,Q,S,Q_1,P_1,T$ are Concyclic points.

One small construction and it all comes crashing down. Let $AA_{1}$ meet $\odot (ABC)$ again in $X$ and $BB_{1}$ meet $\odot (ABC)$ again in $Y$. CLAIM 1: $Q,X,Q_{1},C$ and $P,Y,P_{1},C$ are concyclic PROOF: Indeed, note that $\angle QQ_{1}C=\angle ABC=\angle QXC$. Similarly, $\angle PP_{1}C=\angle BAC=\angle PYC$ CLAIM 2: $P,Q,X,Q_{1},P_{1},Y$ are concyclic PROOF: Note that $\angle QQ_{1}X=\angle QCX=\angle BCX=\angle BAX$$=\angle QPX=\angle BYX=\angle QYX$ and $\angle PP_{1}Y=\angle PCY=\angle ACY=\angle ABY$$=\angle PQY=\angle AXY=\angle PXY$, and we are done.$\blacksquare$