soltion

I will say trivial but the configuration was nice. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(14cm); 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 = -15.950602401055523, xmax = 18.004246773231156, ymin = -11.782602124577654, ymax = 10.640827492467496; /* image dimensions */ pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen qqqqcc = rgb(0,0,0.8); pen xfqqff = rgb(0.4980392156862745,0,1); pen ttffqq = rgb(0.2,1,0); pen ttffcc = rgb(0.2,1,0.8); /* draw figures */ draw(circle((-1.86,0.39), 10), linewidth(1.2) + dtsfsf); draw((-7.164140632588331,8.867387106280182)--(-11.172434198964863,-3.2539771253329275), linewidth(1.2) + qqqqcc); draw((-11.172434198964863,-3.2539771253329275)--(6.810423011789218,-4.59234531085873), linewidth(1.2) + blue); draw((6.810423011789218,-4.59234531085873)--(5.50353913515792,7.155965666850353), linewidth(1.2) + qqqqcc); draw((5.50353913515792,7.155965666850353)--(-7.164140632588331,8.867387106280182), linewidth(1.2) + qqqqcc); draw((-7.164140632588331,8.867387106280182)--(6.810423011789218,-4.59234531085873), linewidth(1.2) + xfqqff); draw((5.50353913515792,7.155965666850353)--(-2.7750231063269384,4.639967277352538), linewidth(1.2) + red); draw((-2.7750231063269384,4.639967277352538)--(-11.172434198964863,-3.2539771253329275), linewidth(1.2) + xfqqff); draw((-11.172434198964863,-3.2539771253329275)--(2.8781732821735044,-0.8049619975136506), linewidth(1.2) + xfqqff); draw(circle((-4.086274011214042,-2.378618099679658), 7.140022383036164), linewidth(1.2) + ttffqq); draw((xmin, 1.2436106632578734*xmin + 2.7031158336596466)--(xmax, 1.2436106632578734*xmax + 2.7031158336596466), linewidth(1.2) + linetype("4 4") + ttffcc); /* line */ draw((-3.4203371749014244,-9.487517294372136)--(-11.172434198964863,-3.2539771253329275), linewidth(1.2) + qqqqcc); draw((-2.7750231063269384,4.639967277352538)--(-3.4203371749014244,-9.487517294372136), linewidth(1.2) + xfqqff); draw((-3.4203371749014244,-9.487517294372136)--(6.810423011789218,-4.59234531085873), linewidth(2) + qqqqcc); draw((-3.4203371749014244,-9.487517294372136)--(2.8781732821735044,-0.8049619975136506), linewidth(1.2) + xfqqff); draw((-7.164140632588331,8.867387106280182)--(-3.4203371749014244,-9.487517294372136), linewidth(1.2) + blue); draw((5.50353913515792,7.155965666850353)--(2.8781732821735044,-0.8049619975136506), linewidth(1.2) + red); draw(circle((1.9689708559065824,3.9082382920653767), 4.80009437632402), linewidth(1.2) + green); /* dots and labels */ dot((-7.164140632588331,8.867387106280182),dotstyle); label("A", (-7.3957442974040735,9.053328050552794), NE * labelscalefactor); dot((-11.172434198964863,-3.2539771253329275),dotstyle); label("D", (-11.607027539150018,-3.4261814511655477), NE * labelscalefactor); dot((6.810423011789218,-4.59234531085873),dotstyle); label("C", (7.024042299987802,-4.815243462840911), NE * labelscalefactor); dot((-2.7750231063269384,4.639967277352538),dotstyle); label("$P$", (-2.6773431783798203,4.864093412166779), NE * labelscalefactor); dot((5.50353913515792,7.155965666850354),dotstyle); label("B", (5.723174701752144,7.135099558239198), NE * labelscalefactor); dot((2.8781732821735044,-0.8049619975136506),linewidth(4pt) + dotstyle); label("$Q$", (2.967099281761342,-0.6260088244548949), NE * labelscalefactor); dot((-4.086274011214042,-2.378618099679658),linewidth(4pt) + dotstyle); label("$O_1$", (-4.000259379975406,-2.1914596630096694), NE * labelscalefactor); dot((-1.86,0.39),linewidth(4pt) + dotstyle); label("$O$", (-1.7733504406228373,0.5646157569811308), NE * labelscalefactor); label("l", (5.833417718551776,10.288049838708673), NE * labelscalefactor,ttffcc); dot((-3.4203371749014244,-9.487517294372136),linewidth(4pt) + dotstyle); label("$J$", (-3.515190106057024,-9.97461664906369), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ /* end of picture */ [/asy][/asy] Let $O_1$ be the circumcentre of $\Delta PDQ$ . We claim that the locus of $O_1$ is a fixed line which passes through $O$ the centre of $\odot(ABCD)$. Lemma : If $AP,AQ$ are isogonal cevians w.r.t $\angle BAC$ in a triangle $\Delta ABC$ and $P,Q \in BC$ then the circumcircle of $\Delta APQ$ is tangent to the circumcircle of $\Delta ABC$. Proof: Let $l$ be the tangent to $\Delta ABC$ at $A$ . We claim that this line is also tangent to $\odot(APQ)$. Note that $\angle(l,AP)=\angle(l,AB)+\angle BAP=\angle BCA+\angle CAQ=\angle AQP \implies l $ is tangent to $\Delta APQ$ where the last step follows from the assumption that $l$ is tangent to $BC$ and as $AP,AQ$ were isogonal as stated in the statement of our lemma. Done. $\square$. Now back to the main problem . By our lemma we have that $\odot(BPQ)$ is tangent to $\odot(ABC)=\Omega$ so their radical axis is nothing but the tangent at $B$ to both these circles. Now let $J=\odot(DPQ) \cap \Omega$ . We claim that $J$ is a fixed point. Observe that the radical axis of $\odot(DPQ)$ and $\odot(BPQ)$ is $PQ \equiv AC$ and the radical axis of $\odot(DPQ)$ and $\Omega$ is $DJ$ . By radical axis theorem (given as example 2.7 in egmo on page 26. Proved on page 28) we have that the radical axis of these three circles are concurrent or that the tangent at $B$ ,$AC$ and $DJ$ are concurrent. Note that the intersection of $AC$ and the tangent at $B$ is fixed and $J$ could only move about $\Omega \implies J$ is a fixed point.Now the proof smoothly follows. Now we have that $DJ$ is a fixed line and also since $DJ$ is the radical axis of $\Omega$ and $\odot(PDQ)$ we have that $DJ \perp OO_1$ .Let $l$ denote the perpendicular line from $O$ onto $DJ$. Note that $l$ is also a fixed line since $D,J,O$ are fixed and so $O_1$ varies on $l$.We are done . $\blacksquare$

Here's an Useless Inversive Solution. Sharygin 2020 CR P22 wrote: Let $\Omega$ be the circumcircle of cyclic quadrilateral $ABCD$. Consider such pairs of points $P$, $Q$ of diagonal $AC$ that the rays $BP$ and $BQ$ are symmetric with respect the bisector of angle $B$. Find the locus of circumcenters of triangles $PDQ$.

Answer:- . The Locus will be a line. Let $\{P_n,Q_n\}$ be the pair of points on $AC$ such that $BP_n,BQ_n$ are Isogonal where $n=\{1,2,3,\cdots\}$ and let $O_n$ be the center of $\odot(DP_nQ_n)$. Claim 1:- $\odot(BP_nQ_n)$ and $\odot(ABC)$ are tangent to each other.

Proof

Now we will Invert around $D$ with any arbitary radius (Let this map be $\Psi$) and let the inverse of a point $X$ be $X'$. This is what happens after this Inversion $(\Psi)$. $$\begin{cases} \Psi:\odot(ABC)\leftrightarrow A'B'C' \\ \Psi:AC\leftrightarrow \odot(DA'C') \\ \Psi:\odot(BP_nQ_n)\leftrightarrow\text{Circle tangent to } A'C' \text{ at} B' \\ \Psi: (P_n,Q_n)\leftrightarrow \{\odot(DA'C')\cap\odot(BP_nQ_n)'\}\\ \Psi:O_n\leftrightarrow \text{Reflection of } D \text{ on } P_n'Q_n' \end{cases}$$

Claim 2:- $(P_1'Q_1'),(P_2'Q_2'),\cdots (P_n'Q_n'), A'C'$ are concurrent.

Proof

Claim 3:- Reflection of $D$ on $P_n'Q_n'$ and $A'C'$ are concyclic.

Proof

Now note that the Reflection of $D$ on $P_n'Q_n'$ is $O_n'$. Hence we get that $D,O_1',O_2',\cdots, O_n'$ are all concyclic. Hence, Inverting back we get that $O_1,O_3,O_3,\cdots O_n$ are collinear. Hence, the resulting locus of the circumcenters is a line. $\blacksquare$

A short solution: Let tangent at $A$ meet $BC$ in $X$.It is well known that $XA^2=XP\times XQ$ for all isogonal points $P$ and $Q$ on $BC$.Let $P$ be centre of $(DPQ)$. Therefore:$XP^2-PD^2=XA^2$ and the RHS is a constant.Take any point $Y$ on $XD$ such that $YP^2-PD^2=XA^2$ and by perpendicularity criterion the locus is a line $\perp$ to $XD$ and passing through $Y$

By ddit the locus of circles are coaxial hence the locus of centers is a line

This was proposed by KV Sudharshan and me Here's our original proposal: Original proposal wrote: Let $ABC$ be a triangle, with a fixed point $D$ on its circumcircle $\Omega$. Let $P$ be an arbitrary point on the segment $BC$, and $Q$ be its isogonal with respect to $\angle BAC$, $Q \in BC$. Let $M$ be the intersection of $CD$ and the $A$-symmedian. Define $E$ to be the second intersection of $BM$ with $\Omega$. Prove that $P, Q, D, E$ are concyclic. In particular, this implies that $E$ is independent of $P$ and so the locus is the perpendicular bisector of $ED.$ Here's our proof to this version: Redefine $E$ to be the second intersection of $\Omega$ and $\odot(PQD)$. We need to show that $M=BE \cap CD$ lies on the $A$-symmedian of $\triangle ABC$. Define $X$ to be the intersection of $BC$ and tangent to $\Omega$ at $A$. Lemma 1: The circles $\Omega$ and $\odot(APQ)$ are tangent at $A$. Proof: Let $AP,AQ$ meet $\odot(ABC)$ again at $W,Y$ respectively. Clearly, $PQ \parallel WY$, and so there exists a homothety $\mathcal{H}$ centered at $A$ under which $\triangle APQ \mapsto \triangle AXY$. Hence, $\mathcal{H}: \odot(APQ) \mapsto \odot(AWY)=\Omega$ and so these circles are tangent at $A$, as desired. $\square$ Lemma 2: $X$ lies on $ED$. Proof: Simply note that $X$ is the radical center of $\Omega, \odot(APQ)$ and $\odot(PQD)$, since $BC$ is the radical axis of $\odot(APQ)$ and $\odot(PQD)$, and the tangent at $A$ is the radical axis of $\Omega, \odot(APQ)$ by lemma 2. Hence $X$ lies on $ED$, the radical axis of $\odot(APQ)$ and $\odot(PQD)$. $\square$ Now, let $XA'$ be the other tangent from $X$ to the circumcircle. From \textbf{Lemma 2}, we have $X,D,E$ are collinear. Notice by Brokard's theorem on $DEBC$, we get that $M$ lies on the polar of $X$, but since $AA'$ is the polar of $X$, hence $A, A', M$ are collinear. . Now clearly, $ABA'C$ is harmonic, and so, $AA'$ is a symmedian $\Rightarrow AM$ is a symmedian, as desired. $\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(14cm); 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 = -23.16939756138639, xmax = 27.41782896932673, ymin = -21.052579667701877, ymax = 12.721833339568468; /* image dimensions */ pen ccqqqq = rgb(0.8,0,0); pen zzttff = rgb(0.6,0.2,1); draw((-4.323175912689353,6.770394924190434)--(-7.679337856199352,-5.916869398291695)--(13.386374534257135,-5.717194399330021)--cycle, linewidth(0.5) + ccqqqq); /* draw figures */ draw(circle((2.816444897604075,-1.905783828492775), 11.236114233278629), linewidth(0.3) + blue); draw((-4.323175912689353,6.770394924190434)--(-7.679337856199352,-5.916869398291695), linewidth(0.5) + ccqqqq); draw((-7.679337856199352,-5.916869398291695)--(13.386374534257135,-5.717194399330021), linewidth(0.5) + ccqqqq); draw((13.386374534257135,-5.717194399330021)--(-4.323175912689353,6.770394924190434), linewidth(0.5) + ccqqqq); draw((-4.323175912689353,6.770394924190434)--(-3.0464313221985004,-11.491030069177468), linewidth(0.3)); draw((-7.679337856199352,-5.916869398291695)--(0.13143773754642263,-17.730879273067618), linewidth(0.3) + linetype("4 4")); draw((8.859962213810459,-11.378173258409609)--(-4.323175912689353,6.770394924190434), linewidth(0.3)); draw(circle((0.7214116448818275,-9.78790224671698), 5.710198272996486), linewidth(0.5) + blue); draw((0.13143773754642263,-17.730879273067618)--(13.386374534257135,-5.717194399330021), linewidth(0.5)); draw(circle((0.6214158917097529,0.76164971293657), 7.78164552815772), linewidth(0.3) + linetype("4 4") + zzttff); draw((-19.88148982559575,-6.032529606532421)--(-7.679337856199352,-5.916869398291695), linewidth(0.5)); draw((-19.88148982559575,-6.032529606532421)--(-4.323175912689353,6.770394924190434), linewidth(0.5)); draw((-4.323175912689353,6.770394924190434)--(0.13143773754642263,-17.730879273067618), linewidth(0.5)); draw((-19.88148982559575,-6.032529606532421)--(5.573567853948765,-12.798374707933887), linewidth(0.5) + linetype("4 4")); /* dots and labels */ dot((-4.323175912689353,6.770394924190434),linewidth(4pt) + dotstyle); label("$A$", (-4.918319754227145,7.613515366368989), NE * labelscalefactor); dot((-7.679337856199352,-5.916869398291695),linewidth(4pt) + dotstyle); label("$B$", (-8.687564083966555,-6.7691274707945945), NE * labelscalefactor); dot((13.386374534257135,-5.717194399330021),linewidth(4pt) + dotstyle); label("$C$", (13.878306574341753,-6.719532150666445), NE * labelscalefactor); dot((-3.0464313221985004,-11.491030069177468),linewidth(4pt) + dotstyle); label("$W$", (-2.93450694910114,-12.819756526428929), NE * labelscalefactor); dot((-3.4389579680402016,-5.876676223996348),linewidth(4pt) + dotstyle); label("$P$", (-4.868724434098995,-6.7691274707945945), NE * labelscalefactor); dot((5.573567853948765,-12.798374707933887),linewidth(4pt) + dotstyle); label("$D$", (5.645483433068832,-14.307616130273438), NE * labelscalefactor); dot((0.13143773754642263,-17.730879273067618),linewidth(4pt) + dotstyle); label("$M$", (-0.008383061540282555,-19.019171542447715), NE * labelscalefactor); dot((8.859962213810459,-11.378173258409609),linewidth(4pt) + dotstyle); label("$Y$", (9.06756052191119,-10.98472968168737), NE * labelscalefactor); dot((4.806893942361303,-5.798516490248464),linewidth(4pt) + dotstyle); label("$Q$", (5.7446740733251325,-6.620341510410143), NE * labelscalefactor); dot((-4.985140994855612,-9.991915204241046),linewidth(4pt) + dotstyle); label("$E$", (-6.009416797046448,-10.736753081046617), NE * labelscalefactor); dot((-19.88148982559575,-6.032529606532421),linewidth(4pt) + dotstyle); label("$X$", (-20.888012835491487,-6.917913431179046), NE * labelscalefactor); dot((-0.8123626230734491,-12.539786792624431),linewidth(4pt) + dotstyle); label("$A'$", (-0.2563596621810332,-13.811662928991936), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Remark: This also gives us that $T=BD \cap CE$ lies on the $A$-symmedian of $\triangle ABC$.

Wizard_32 wrote: This was proposed by KV Sudharshan and me Here's our original proposal: Original proposal wrote: Let $ABC$ be a triangle, with a fixed point $D$ on its circumcircle $\Omega$. Let $P$ be an arbitrary point on the segment $BC$, and $Q$ be its isogonal with respect to $\angle BAC$, $Q \in BC$. Let $M$ be the intersection of $CD$ and the $A$-symmedian. Define $E$ to be the second intersection of $BM$ with $\Omega$. Prove that $P, Q, D, E$ are concyclic. In particular, this implies that $E$ is independent of $P$ and so the locus is the perpendicular bisector of $ED.$ Here's our proof to this version: Redefine $E$ to be the second intersection of $\Omega$ and $\odot(PQD)$. We need to show that $M=BE \cap CD$ lies on the $A$-symmedian of $\triangle ABC$. Define $X$ to be the intersection of $BC$ and tangent to $\Omega$ at $A$. Lemma 1: The circles $\Omega$ and $\odot(APQ)$ are tangent at $A$. Proof: Let $AP,AQ$ meet $\odot(ABC)$ again at $W,Y$ respectively. Clearly, $PQ \parallel WY$, and so there exists a homothety $\mathcal{H}$ centered at $A$ under which $\triangle APQ \mapsto \triangle AXY$. Hence, $\mathcal{H}: \odot(APQ) \mapsto \odot(AWY)=\Omega$ and so these circles are tangent at $A$, as desired. $\square$ Lemma 2: $X$ lies on $ED$. Proof: Simply note that $X$ is the radical center of $\Omega, \odot(APQ)$ and $\odot(PQD)$, since $BC$ is the radical axis of $\odot(APQ)$ and $\odot(PQD)$, and the tangent at $A$ is the radical axis of $\Omega, \odot(APQ)$ by lemma 2. Hence $X$ lies on $ED$, the radical axis of $\odot(APQ)$ and $\odot(PQD)$. $\square$ Now, let $XA'$ be the other tangent from $X$ to the circumcircle. From \textbf{Lemma 2}, we have $X,D,E$ are collinear. Notice by Brokard's theorem on $DEBC$, we get that $M$ lies on the polar of $X$, but since $AA'$ is the polar of $X$, hence $A, A', M$ are collinear. . Now clearly, $ABA'C$ is harmonic, and so, $AA'$ is a symmedian $\Rightarrow AM$ is a symmedian, as desired. $\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(14cm); 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 = -23.16939756138639, xmax = 27.41782896932673, ymin = -21.052579667701877, ymax = 12.721833339568468; /* image dimensions */ pen ccqqqq = rgb(0.8,0,0); pen zzttff = rgb(0.6,0.2,1); draw((-4.323175912689353,6.770394924190434)--(-7.679337856199352,-5.916869398291695)--(13.386374534257135,-5.717194399330021)--cycle, linewidth(0.5) + ccqqqq); /* draw figures */ draw(circle((2.816444897604075,-1.905783828492775), 11.236114233278629), linewidth(0.3) + blue); draw((-4.323175912689353,6.770394924190434)--(-7.679337856199352,-5.916869398291695), linewidth(0.5) + ccqqqq); draw((-7.679337856199352,-5.916869398291695)--(13.386374534257135,-5.717194399330021), linewidth(0.5) + ccqqqq); draw((13.386374534257135,-5.717194399330021)--(-4.323175912689353,6.770394924190434), linewidth(0.5) + ccqqqq); draw((-4.323175912689353,6.770394924190434)--(-3.0464313221985004,-11.491030069177468), linewidth(0.3)); draw((-7.679337856199352,-5.916869398291695)--(0.13143773754642263,-17.730879273067618), linewidth(0.3) + linetype("4 4")); draw((8.859962213810459,-11.378173258409609)--(-4.323175912689353,6.770394924190434), linewidth(0.3)); draw(circle((0.7214116448818275,-9.78790224671698), 5.710198272996486), linewidth(0.5) + blue); draw((0.13143773754642263,-17.730879273067618)--(13.386374534257135,-5.717194399330021), linewidth(0.5)); draw(circle((0.6214158917097529,0.76164971293657), 7.78164552815772), linewidth(0.3) + linetype("4 4") + zzttff); draw((-19.88148982559575,-6.032529606532421)--(-7.679337856199352,-5.916869398291695), linewidth(0.5)); draw((-19.88148982559575,-6.032529606532421)--(-4.323175912689353,6.770394924190434), linewidth(0.5)); draw((-4.323175912689353,6.770394924190434)--(0.13143773754642263,-17.730879273067618), linewidth(0.5)); draw((-19.88148982559575,-6.032529606532421)--(5.573567853948765,-12.798374707933887), linewidth(0.5) + linetype("4 4")); /* dots and labels */ dot((-4.323175912689353,6.770394924190434),linewidth(4pt) + dotstyle); label("$A$", (-4.918319754227145,7.613515366368989), NE * labelscalefactor); dot((-7.679337856199352,-5.916869398291695),linewidth(4pt) + dotstyle); label("$B$", (-8.687564083966555,-6.7691274707945945), NE * labelscalefactor); dot((13.386374534257135,-5.717194399330021),linewidth(4pt) + dotstyle); label("$C$", (13.878306574341753,-6.719532150666445), NE * labelscalefactor); dot((-3.0464313221985004,-11.491030069177468),linewidth(4pt) + dotstyle); label("$W$", (-2.93450694910114,-12.819756526428929), NE * labelscalefactor); dot((-3.4389579680402016,-5.876676223996348),linewidth(4pt) + dotstyle); label("$P$", (-4.868724434098995,-6.7691274707945945), NE * labelscalefactor); dot((5.573567853948765,-12.798374707933887),linewidth(4pt) + dotstyle); label("$D$", (5.645483433068832,-14.307616130273438), NE * labelscalefactor); dot((0.13143773754642263,-17.730879273067618),linewidth(4pt) + dotstyle); label("$M$", (-0.008383061540282555,-19.019171542447715), NE * labelscalefactor); dot((8.859962213810459,-11.378173258409609),linewidth(4pt) + dotstyle); label("$Y$", (9.06756052191119,-10.98472968168737), NE * labelscalefactor); dot((4.806893942361303,-5.798516490248464),linewidth(4pt) + dotstyle); label("$Q$", (5.7446740733251325,-6.620341510410143), NE * labelscalefactor); dot((-4.985140994855612,-9.991915204241046),linewidth(4pt) + dotstyle); label("$E$", (-6.009416797046448,-10.736753081046617), NE * labelscalefactor); dot((-19.88148982559575,-6.032529606532421),linewidth(4pt) + dotstyle); label("$X$", (-20.888012835491487,-6.917913431179046), NE * labelscalefactor); dot((-0.8123626230734491,-12.539786792624431),linewidth(4pt) + dotstyle); label("$A'$", (-0.2563596621810332,-13.811662928991936), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Remark: This also gives us that $T=BD \cap CE$ lies on the $A$-symmedian of $\triangle ABC$. Why did they change the problem;this would have been a little interesting instead of the original one.

Solution : Let $\odot (ABCD) = \Omega ; \odot (BPQ)=\omega_1 ; \odot (PDQ)=\omega_2$. Let $P$ be closer to $A$ than $Q$. We claim that the locus of circumcentre of $\triangle PQD$ is a line segment passing through $O$ . Let $O,X,Y$ be the circumcentres of $ABCD, \triangle BPQ ; \triangle PDQ$ respectively. Let $l$ be a line tangent to $\Omega$ at $B$ . And let $S$ be a point on it closer to $C$ than $A$ . $\angle SBC=\angle BAC ; \angle BPQ=\angle BAP+\angle ABP=\angle BAC+\angle CBQ=\angle SBC+\angle CBQ=\angle SBQ$. So , by converse of alternate segment theorem , $SB$ is also tangent to $\omega_1$. $\implies O,X,B$ are collinear . As $O$ and $B$ are fixed , $X$ moves linearly on $OB$. $PQ$ or $AC$ is the radical axis of $\omega_1$ and $\omega_2 \implies XY\perp PQ$. As $X$ moves linearly on $OB$ , and $XY$ is always perpendicular to $AC$ locus of $Y$ also moves linearly on a line as locus of $XY$ is always parallel to each other . that is for some isogonal points $P_1,Q_1$ and $P_2,Q_2$ ,$X_1Y_1\parallel X_2Y_2$. When $P,Q$ are at extreme points on $AC$ that is on $A$ and $C$ , then $\odot (PDQ)=\odot (ABCD) \implies Y=O$. And $O$ is one extreme end of the locus of $Y$ as it is the centre when $P,Q$ are at the extreme ends. So , the locus line of $Y$ passes through $O$. Let $M$ be second point of intersection of $\Omega$ and $\omega_2$. Thus $OY$ which is the line joining the centres of $\Omega$ and $\omega_2$ is the perpendicular bisector of $DM$ , which is their radical axis . And we have just proved that locus $OY$ is a line and is fixed, and even $D$ is fixed $\implies M$ is also a fixed point . So , $P,Q,M,D$ are concyclic . When $P,Q$ are the same , then we get the second extreme end as , if one point moves further on $AC$, we just flip $P,Q$ , that is, basically the feet of angle bisector is the other extreme end of $P,Q$ . Let $E$ be the feet of angle bisector of $\angle ABC$ on $AC$. So , $Y=$ the circumcentre of $\odot DEM$ which is fixed $\implies $ the second extreme point is also a fixed point . $\implies$ the locus of circumcentre of $\odot PDQ$ is a line segment (as both ends of it are fixed). $\blacksquare$ [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 = -6.08, xmax = 32.1, ymin = -10.04, ymax = 7.84; /* image dimensions */ pen ududff = rgb(0.30196078431372547,0.30196078431372547,1.); pen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1.); pen zzttqq = rgb(0.6,0.2,0.); pen uuuuuu = rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666); pen yqqqyq = rgb(0.5019607843137255,0.,0.5019607843137255); draw((6.58,1.98)--(13.492053304128861,1.8720081524477756)--(14.707352734844084,-3.228992071478425)--(7.555990338236781,-6.011296032181511)--cycle, linewidth(2.) + zzttqq); draw((8.89755920049959,0.494627346143377)--(13.492053304128861,1.8720081524477756)--(13.190134493526493,-2.2565748468829203)--cycle, linewidth(2.) + green); draw((7.555990338236781,-6.011296032181511)--(8.89755920049959,0.494627346143377)--(13.190134493526493,-2.2565748468829203)--cycle, linewidth(2.) + yqqqyq); /* draw figures */ draw(shift((9.98,-1.66)) * scale(4.980923609131142, 4.980923609131142)*unitcircle, linewidth(2.)); draw((6.58,1.98)--(13.492053304128861,1.8720081524477756), linewidth(2.) + zzttqq); draw((13.492053304128861,1.8720081524477756)--(14.707352734844084,-3.228992071478425), linewidth(2.) + zzttqq); draw((14.707352734844084,-3.228992071478425)--(7.555990338236781,-6.011296032181511), linewidth(2.) + zzttqq); draw((7.555990338236781,-6.011296032181511)--(6.58,1.98), linewidth(2.) + zzttqq); draw((6.58,1.98)--(14.707352734844084,-3.228992071478425), linewidth(2.)); draw((8.89755920049959,0.494627346143377)--(13.492053304128861,1.8720081524477756), linewidth(2.)); draw((8.89755920049959,0.494627346143377)--(13.492053304128861,1.8720081524477756), linewidth(2.) + green); draw((13.492053304128861,1.8720081524477756)--(13.190134493526493,-2.2565748468829203), linewidth(2.) + green); draw((13.190134493526493,-2.2565748468829203)--(8.89755920049959,0.494627346143377), linewidth(2.) + green); draw((7.555990338236781,-6.011296032181511)--(8.89755920049959,0.494627346143377), linewidth(2.) + yqqqyq); draw((8.89755920049959,0.494627346143377)--(13.190134493526493,-2.2565748468829203), linewidth(2.) + yqqqyq); draw((13.190134493526493,-2.2565748468829203)--(7.555990338236781,-6.011296032181511), linewidth(2.) + yqqqyq); draw(shift((11.568332185354123,-0.06264319995636441)) * scale(2.7282922860608423, 2.7282922860608423)*unitcircle, linewidth(2.) + green); draw((xmin, -0.9943502824859157*xmin + 15.287835166724546)--(xmax, -0.9943502824859157*xmax + 15.287835166724546), linewidth(2.)); /* line */ draw(shift((9.6522166577321,-3.0522708592044903)) * scale(3.6262921499593186, 3.6262921499593186)*unitcircle, linewidth(2.) + yqqqyq); draw((9.522980354324158,-3.6012064394607846)--(9.98,-1.66), linewidth(2.8) + red); draw((9.98,-1.66)--(13.492053304128861,1.8720081524477756), linewidth(2.)); /* dots and labels */ dot((9.98,-1.66),linewidth(3.pt) + ududff); label("$O$", (10.06,-1.54), NE * labelscalefactor,ududff); dot((6.58,1.98),linewidth(4.pt) + ududff); label("$A$", (6.66,2.14), NE * labelscalefactor,ududff); dot((13.492053304128861,1.8720081524477756),linewidth(4.pt) + xdxdff); label("$B$", (13.58,2.04), NE * labelscalefactor,xdxdff); dot((14.707352734844084,-3.228992071478425),linewidth(3.pt) + xdxdff); label("$C$", (14.82,-3.56), NE * labelscalefactor,xdxdff); dot((7.555990338236781,-6.011296032181511),linewidth(3.pt) + xdxdff); label("$D$", (7.2,-6.68), NE * labelscalefactor,xdxdff); dot((8.89755920049959,0.494627346143377),linewidth(4.pt) + xdxdff); label("$P$", (8.5,-0.02), NE * labelscalefactor,xdxdff); dot((13.190134493526493,-2.2565748468829203),linewidth(4.pt) + uuuuuu); label("$Q$", (12.86,-2.86), NE * labelscalefactor,uuuuuu); dot((13.492053304128298,1.8720081524471315),linewidth(4.pt) + uuuuuu); dot((16.229793453396322,-0.8502645383505616),linewidth(3.pt) + xdxdff); label("$S$", (16.3,-0.74), NE * labelscalefactor,xdxdff); dot((10.208149200356374,-6.635695724456704),linewidth(3.pt) + uuuuuu); label("$M$", (10.18,-7.12), NE * labelscalefactor,uuuuuu); dot((11.568332185354123,-0.06264319995636441),linewidth(3.pt) + uuuuuu); label("$X$", (11.64,0.06), NE * labelscalefactor,uuuuuu); dot((9.6522166577321,-3.0522708592044903),linewidth(3.pt) + uuuuuu); label("$Y$", (9.74,-2.94), NE * labelscalefactor,uuuuuu); dot((11.201623484341527,-0.9820961305250085),linewidth(3.pt) + uuuuuu); label("$E$", (11.28,-0.86), NE * labelscalefactor,uuuuuu); dot((9.522980354324158,-3.6012064394607846),linewidth(3.pt) + uuuuuu); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* re-scale y/x */ currentpicture = yscale(1.6666666666666667) * currentpicture; /* end of picture */ [/asy][/asy]

A neat and easy way to do Problem 22. Let $\Omega$ be the circumcircle of cyclic quadrilateral $ABCD$. Consider such pairs of points $P, Q$ of diagonal $AC$ that the rays $BP$ and $BQ$ are symmetric with respect the bisector of angle $B$. Find the locus of circumcenters of triangles $PDQ$. [asy][asy] /* 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 = -8.257793872625976, xmax = 34.72772903207368, ymin = -16.47152058245142, ymax = 8.991838217226268; /* image dimensions */ /* draw figures */ draw(circle((12.12,-2.66), 7.367849075544366), linewidth(0.8) + linetype("2 2") + red); draw(circle((12.999546880066505,1.193318133955158), 3.415424078697192), linewidth(0.8)); draw(circle((14.022543448505224,-5.670491731962898), 7.039082366028591), linewidth(0.8)); draw((5.00491186027269,-0.7466989876363526)--(24.26845358090226,2.124379533059874), linewidth(0.8) + blue); draw((7.551334874711067,-8.440354571561215)--(24.26845358090226,2.124379533059874), linewidth(0.8) + linetype("4 4") + blue); draw((5.591429390005569,6.3875412510798695)--(24.26845358090226,2.124379533059874), linewidth(0.8) + blue); draw((5.00491186027269,-0.7466989876363526)--(13.75959307826595,4.523100621437956), linewidth(0.8)); draw((13.75959307826595,4.523100621437956)--(18.367923520227076,1.244952200141924), linewidth(0.8)); /* dots and labels */ dot((13.75959307826595,4.523100621437956),dotstyle); label("$B$", (13.85573943638573,4.75184930946025), NE * labelscalefactor); dot((5.00491186027269,-0.7466989876363526),dotstyle); label("$A$", (4.602724969161551,-0.589131193140037), NE * labelscalefactor); dot((18.367923520227076,1.244952200141924),dotstyle); label("$C$", (18.47053399345703,1.4722998780390213), NE * labelscalefactor); dot((9.810876906360894,-0.030407912238523904),dotstyle); label("$P$", (9.522049116293391,-0.6125565462216173), NE * labelscalefactor); dot((16.406478537892422,0.9526143403662415),linewidth(4pt) + dotstyle); label("$Q$", (16.45595362844113,0.4181589893679119), NE * labelscalefactor); dot((7.551334874711067,-8.440354571561215),dotstyle); label("$D$", (7.366916632788015,-9.115959714835233), NE * labelscalefactor); dot((19.30173640074807,-1.0144416539757532),linewidth(4pt) + dotstyle); label("$I$", (19.384122763638654,-0.8233847239558391), NE * labelscalefactor); dot((24.26845358090226,2.124379533059874),linewidth(4pt) + dotstyle); label("$E$", (24.373722970015237,2.3156125889759087), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Solution. Notations and some terminologies. Let $I\neq D$ be the intersection point of $\odot(PQD)$ and $\Omega$. Let tangent to $\Omega$ at $B$ intersect $AC$ at $E$. Obviously, $BP, BQ$ are isogonal lies with respect to $\angle ABC$. Thus, $\angle ABP=\angle CBQ$. Lemma. $\odot(BPQ)$ is internally tangent to $\odot(ABC)$ at $B$. Proof of lemma. It is enough to show that $BE$ is tangent to $\odot(BPQ)$ too. For that, it is enough to show $\angle EBQ=\angle BPQ$ but this is true because, $\angle EBC=\angle BAC$ and $\angle QBC=\angle PBA\implies \angle EBQ=\angle EBC+\angle CBQ=\angle ABP+\angle PAB=\angle BPQ$. This proves the lemma. Proof. From our lemma, we know that the radical axis of $\odot(BPQ)$ and $\Omega$ is $BE$. Also, the radical axis of $\odot(BPQ),\odot(DPQ)$ is $AE\implies E$ is the radical center of $\Omega,\odot(BPQ),\odot(DPQ)\implies E$ lies on radical axis of $\Omega, \odot(DPQ)\implies E,I,D$ are collinear. Now note that, as $A,B,C,D$ are fixed $\implies E$ is fixed $\implies I$ is fixed. Now, the circumcenter $\odot(DPQ)$ will lie on perpendicular bisector of $DI$. Thus, the locus for circumcenter of $\triangle DPQ$ is perpendicular bisector of $DI$ where $I=ED\cap\Omega$ where $E=BB\cap AC$. \textbf{Answer: The locus for circumcenter of $\triangle DPQ$ is perpendicular bisector of $DI$ where $I=ED\cap\Omega$ where $E=BB\cap AC$.} \textbf{Feedback:} Nice problem. This results holds true for any $P,Q$ for which $\odot(BPQ)$ is internally tangent to $\odot(ABC)$.

Jupiter_is_BIG wrote: A neat and easy way to do Hmm but the isogonality condition was confusing T_T

Rays $BP$ and $BQ$ are symmetric wrt bisector of $\angle B.$ We can conclude that $\widehat {PBA}=\widehat{CBQ}. $ Let $AC$ intersects the tangent of $\odot(BPQ),$ which is drawn from point $B$ at $\mathcal{H}. $ $\mathcal{H}D$ intersects $\Omega$ at $E$(twice). $\widehat{HBP}=\widehat {BQP}\implies\widehat{HBA}=\widehat {HBP}-\widehat {PBA}=\widehat{BQP}-\widehat{CBQ}=\widehat{BCA}+\widehat{CBQ}=\widehat {BCA}. $ Note, $BH $ is tangent to $\Omega. $ Cause $\Omega$ is a fixed circle, and $H, B, A, C,D $ are the fixed points and $E $ is also a fixed point. We know that $HB $ is tangent to $\Omega, $ and $\odot (HED) $ is secant of $\Omega, $ so $HE\cdot HD=HB^2\hdots (\clubsuit)$ $HB $ is tangent to $\odot(BPQ),$ $(HPQ)$ is a secant of $\odot (BPQ)\implies HP\cdot HQ=HB^2\hdots (\heartsuit)$ By $(\clubsuit),$ and $(\heartsuit)$ $PH\cdot QH= EH\cdot DH.$ Therefore, $PQDE $ is cyclic. $E $ is a fixed point when $P, Q $ move on diagonal $AC,\odot (PQD)$ passes through $E. $ If $J $ is the center of $\odot(PDQ)\implies JE=JD. $ Now, locus of the circumcircles of $\triangle PDQ$ is perpendicular bisector of the segment $DE. $

Dunno why this was P22, but oh well...

Attachments:
P22_diagram1.pdf (2kb)
P22.pdf (85kb)
P22_diagram2.pdf (3kb)

[asy][asy] import graph; size(11.398123948425162cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -4.990537470988406, xmax = 6.407586477436757, ymin = -3.8268462744485827, ymax = 3.498021723082299; /* image dimensions */ pen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1.); pen zzttqq = rgb(0.6,0.2,0.); pen uuuuuu = rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666); pair A = (-0.9596865640705633,1.7547084369610915), B = (0.47685591378396835,1.9423203745750226), C = (1.8819164476711245,-0.677045407624109), D = (-1.799991034786933,-0.8717982993139333), P = (-0.5682032702609432,1.4196894350531077), Q = (0.739215981770902,0.30084145401001905), I = (-0.028539867352876638,0.9578625896489479), F = (-1.8225588686621645,0.8235770578767301), G = (1.0585210277197676,-1.6969187469867517), H = (-0.6581791975097588,-0.008761292138955597); draw(A--B--C--D--cycle, linewidth(1.2) + zzttqq); draw(circle(H, 1.431281645064946), linewidth(1.2) + linetype("4 4") + green); /* draw figures */ draw(circle((0.,0.), 2.), linewidth(1.2)); draw(A--B, linewidth(1.2) + zzttqq); draw(B--C, linewidth(1.2) + zzttqq); draw(C--D, linewidth(1.2) + zzttqq); draw(D--A, linewidth(1.2) + zzttqq); draw(A--C, linewidth(1.2)); draw(B--P, linewidth(1.2)); draw(B--Q, linewidth(1.2)); draw(P--D, linewidth(1.2)); draw(Q--D, linewidth(1.2)); draw(B--I, linewidth(1.2)); draw(I--D, linewidth(1.2)); draw(circle((0.26177003883128575,1.0662367083605924), 0.9021000630816286), linewidth(1.2)); draw((-1.8450390150224578,2.5123650193616003)--A, linewidth(1.2)); draw((-1.8450390150224578,2.5123650193616003)--B, linewidth(1.2)); draw((-1.8450390150224578,2.5123650193616003)--D, linewidth(1.2)); draw(Q--G, linewidth(1.2)); draw(G--C, linewidth(1.2)); draw(G--A, linewidth(1.2)); /* dots and labels */ dot(A,linewidth(4.pt) + xdxdff); label("$A$", (-0.9609859830865658,1.828511236771561), NE * labelscalefactor,xdxdff); dot(B,linewidth(4.pt) + xdxdff); label("$B$", (0.5074839734590074,2.0120699813397573), NE * labelscalefactor,xdxdff); dot(C,linewidth(4.pt) + xdxdff); label("$C$", (1.984694822603066,-0.6801249389937887), NE * labelscalefactor,xdxdff); dot(D,linewidth(4.pt) + xdxdff); label("$D$", (-1.97492952451089,-0.977315287342297), NE * labelscalefactor,xdxdff); dot(P,linewidth(4.pt) + xdxdff); label("$P$", (-0.6375729569426002,1.5488026736200238), NE * labelscalefactor,xdxdff); dot(Q,linewidth(4.pt) + xdxdff); label("$Q$", (0.7609698588150885,0.40374574321841816), NE * labelscalefactor,xdxdff); dot(I,linewidth(4.pt) + xdxdff); label("$I$", (-0.025710475048611407,0.7271587693623831), NE * labelscalefactor,xdxdff); dot((-1.8450390150224578,2.5123650193616003),linewidth(4.pt) + xdxdff); label("$E$", (-1.8088525651396645,2.5802280002413176), NE * labelscalefactor,xdxdff); dot(F,linewidth(4.pt) + xdxdff); label("$F$", (-1.957447739313919,0.8670130509381517), NE * labelscalefactor,xdxdff); dot(G,linewidth(4.pt) + xdxdff); label("$G$", (1.0144557441711697,-1.965036150971163), NE * labelscalefactor,xdxdff); dot(H,linewidth(4.pt) + uuuuuu); label("$H$", (-0.6200911717456291,0.06285093187748214), NE * labelscalefactor,uuuuuu); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] \[\color{yellow} \rule{10mm}{4pt} \color{cyan} \rule{10mm}{4pt} \color{yellow} \rule{10mm}{4pt} \color{cyan} \rule{10mm}{4pt} \color{yellow} \rule{10mm}{4pt} \color{cyan} \rule{10mm}{4pt} \color{yellow} \rule{10mm}{4pt} \color{cyan} \rule{10mm}{4pt} \color{yellow} \rule{10mm}{4pt} \color{cyan} \rule{10mm}{4pt}\color{yellow} \rule{10mm}{4pt} \color{cyan} \rule{10mm}{4pt} \color{yellow} \rule{10mm}{4pt} \color{cyan} \rule{10mm}{4pt} \color{yellow} \rule{10mm}{4pt} \color{cyan} \rule{10mm}{4pt} \color{yellow} \rule{10mm}{4pt} \color{cyan} \rule{10mm}{4pt} \]$E=\overline{BB}\cap \overline{AC} \thickspace \text{and} \thickspace F=\overline{ED} \cap \Omega $ $\color{yellow} \textbf{claim:}$ circumcircle of $BPQ$ is tangent to $\Omega$ $\color{green} \textbf{proof:}$ extend $AQ$ to intersect $\Omega$ at $G$ then by spiral similarity we have: $ \triangle{ABQ}\sim \triangle{BPC} \\ \implies \angle{BQP}=\angle{EBP} \\ \color{green} \text{which obviously proves our claim} \blacksquare$ now by our $\color{green} \text{claim}$ we have: \[EP\times EQ=EB^2=EF\times ED \\ \implies \color{cyan} \text{PQDF is cyclic}\]now by $PQDF$ cyclic it is obvious that the circumcenter of $PQD$ moves along the perpendicular bisector of $DF$ so we are done. \[\color{magenta} \huge{\boxed{\mathbb{Q.E.D}}}\]\[\color{yellow} \rule{10mm}{4pt} \color{cyan} \rule{10mm}{4pt} \color{yellow} \rule{10mm}{4pt} \color{cyan} \rule{10mm}{4pt} \color{yellow} \rule{10mm}{4pt} \color{cyan} \rule{10mm}{4pt} \color{yellow} \rule{10mm}{4pt} \color{cyan} \rule{10mm}{4pt} \color{yellow} \rule{10mm}{4pt} \color{cyan} \rule{10mm}{4pt}\color{yellow} \rule{10mm}{4pt} \color{cyan} \rule{10mm}{4pt} \color{yellow} \rule{10mm}{4pt} \color{cyan} \rule{10mm}{4pt} \color{yellow} \rule{10mm}{4pt} \color{cyan} \rule{10mm}{4pt} \color{yellow} \rule{10mm}{4pt} \color{cyan} \rule{10mm}{4pt} \]

Eliti wrote: Easy by inversin I am curious about the solution by inversion can you write the full solution by inversion if you have one?

NJOY wrote: Let $\Omega$ be the circumcircle of cyclic quadrilateral $ABCD$. Consider such pairs of points $P$, $Q$ of diagonal $AC$ that the rays $BP$ and $BQ$ are symmetric with respect the bisector of angle $B$. Find the locus of circumcenters of triangles $PDQ$. a similar problem to this one about locus of the circumcenter purposed by me

Let $O$ be the circumcenter of triangle $DPQ$. We claim that line $DO$ is fixed. Equivalently, we will prove that the second point of intersection of circles $(DPQ)$ and $(DAC)$ is fixed. We have the following Claim: Claim: Circles $(BPQ)$ and $(BAC)$ are tangent at point $B$. Proof: Indeed, if the tangent to $(BAC)$ at $B$ intersects $AC$ at $K$, then $\angle KBP=\angle KBA+\angle ABP=\angle ACB+\angle QBC=\angle PQB,$ and so $KB$ is tangent to $(BPQ)$, as desired $\blacksquare$ To the problem, if $KD$ intersects $(ABCD)$ at point $S$, then $S$ is obviously fixed, and $KS \cdot KD=KB^2=KP \cdot KQ,$ and so $S \in (PDQ)$, as desired.