nice problem! We have that some useful lemmas and theorems: appalonious's circle, coaxality theory, Klamkin's batterfly theorem ... .?

First invert in $P$ then denote $\angle O'PA',\angle O'PB',\angle O'PC'$ by $90-x,90-y,90-z$ then you need a concurrence of lines which you get via Ceva theorem in trigo form on triangle $A'B'C'$ . You get that you need to prove some equality $A$ with only $5$ points and sines of angles in terms of $x,y,z$ and interior angles of $A'B'C'$ To make a fast finish extend $PO'$ to meet circle $P'B'C'$ and the same for $B',C'$ to reduce $A$ to Ceva theorem in trigo form for point $P$ wrt triangle $A'B'C'$. It's kinda long so i didn't write the entire solution just the main steps. Remember here sines are used only by sine threorems for ratios so this can be done without trigonometry by the same method.

solution without invert lemma1: let $W_{1},W_{2},W_{3}$ be three coaxial circles then $W_{3}$ is locus of points like $S$ that $ \frac{P{}_W{}_{1}(S)}{P{}_W{}_{2}(S)}=constant $ so now let $W_{1}$ be the circumcircle of triangle $BYP$ and $W_{2}$ be the circumcircle of triangle $CZP$ now let $E$ be the intersection of $W_{1},AB$ and $F$ be the intersection of $W_{2},AC$ by lemma 1 we need to prove that $ \frac{P{}_W{}_{1}(X)}{P{}_W{}_{2}(X)}=\frac{P{}_W{}_{1}(A)}{P{}_W{}_{2}(A)} \Longleftrightarrow \frac{AE}{AF}=\frac{CY}{BZ} $ [rest]we have this: $AZ.ZB-AY.YC=OY^2-OZ^2=YP.YZ-ZP.ZY=YF.YC-ZE.ZB$ so $ZB.(AZ+ZE)=CY.(AY+YF) \Longrightarrow ZB.AE=CY.AF \Longrightarrow \frac{AE}{AF}=\frac{CY}{ZB}$ and we are done[/rest]

Two solutions. First is by me, the second is by Hyungwoo Jo from Korean Minjok Leadership Academy. Solution 1 Invert in P with any power. Put an apostrophe to denote the transformed image. We need to show that $A'X'$, $B'Y'$, and $C'Z'$ are concurrent. Let $A'Y'$, $B'X'$ meet in R', and $A'Z'$, $C'X'$ meet in Q'. A simple angle chase gives that $Q'$, $R'$ lies on $\omega '$. Now we try to use desargue's theorem on triangle $X'B'C'$ and $A'Y'Z'$. We need that $Q'R'$, $B'C'$, and $Y'Z' = l$ is concurrent. But since $P, X' , B', C'$ is concyclic, it is sufficient to prove that quadrilateral $Q'R'PX'$ is concyclic, which is again a simple angle chase. Solution 2 Denote by $A_1$ the intersection of $l$ and the polar of $X$ wrt $\omega$. Then it's not difficult to see that A_1 is the radical center of $(APX)$, $(OPX)$, $\omega$. Therefore $AA_1$ is the radical axis of $(APX)$ and $\omega$. Now if $AA_1$, $BB_1$, $CC_1$ meet at a point $Q$, power of $Q$ wrt $(APX)$, $(BPY)$, $(CPZ)$, and $\omega$ are all equal thus implying the problem. Also, if the pole of $l$ wrt $\omega$ is $H$, it can be easily shown that $H$ is the orthocenter of triangle $OXA_1$. Therefore $XP * PA_1 = YP*PB_1 = ZP*PC_1 = PI*OP$. Now it's just ceva (non-trigonometric); and menelaus, preservation of cross ratio, the identity above can help to establish the statement of ceva's theorem.

and for proof of lemma 1 we can use this let $ \ell $ be the radical axis of $W_{1},W_{2}$ and point $P$ be an arbitrary point now let $H$ be the foot of $P$ to $ \ell$ $ P{}_W{}_{1}(P)-P{}_W{}_{2}(P)=-2.PH.O_{1}O_{2} $

Hmm, looks projective Lemma 1: (Generalised) Suppose we have a triangle $\triangle ABC$ with circumscribed conic $\Gamma(ABC)$. Let $\ell$ be a line intersecting $\triangle ABC$ at $D, E, F$, such that $D$ is on $BC$ etc... Let the polar of $D$ wrt $\Gamma(ABC)$ be $\ell_D$ and let it intersect $\ell$ at $D'$ and define others similarly. Then $AD', BE', CF'$ are concurrent. Proof: Under a projective transformation take the conic to a circle. Note that if $\ell \cap \Gamma(ABC) = X, Y$ then $(D, D'), (E, E'), (F, F')$ with $(X, Y)$ form involutions. Suppose $BE' \cap CF' = P$, and consider the intersections of $\ell$ with the opposite sides of complete quadrangle $ABCP$. We get if $AP \cap \ell$ is $D_1$ then $(E, E'), (F, F'), (D, D_1)$ are involutions, hence $D_1 = D'$ so done. Main Proof: There are a couple of crucial things to note. We want to prove that three circle are co-axial, and since they are already concurrent, we must prove that there exists a point other than $P$ such that it is the same power wrt to all circles. Let $w_A, w_B, w_C$ be the circle passing through $OPX, OPY, OPZ$. Note that the radical axis of $w_C$ and $w$ is the polar of $X$ wrt $w$. Hence, by the radical axis theorem on $\odot APX, w, w_C$ we get that $AD$, where $D$ is the intersection of $\ell$ and the polar of $X$, is the radical axis of $\odot APZ, w$. If we similarly define $E, F$ as the intersection of the polars of $Y, Z$ with $\ell$ such that $BE, CF$ are the radical axis of $\odot BPY, \odot CZP$ respectively. However, by lemma $1$ these lines concur, hence the point of concurrence $Q$ is the radical centre of $\odot APX, \odot BPY, \odot CZP$, hence they are co-axial with radical axis $PQ$. So, done.

After inverting through $P$, the problem becomes the following: Let $l\cap (BCP)=A'$, define $B',C'$ similarly, prove that $AA',BB',CC'$ concur. We apply Desargues' theorem on $AB'C',A'BC$ and by simple angle chasing we can prove $A'B\cap AB'=B_a\in (O), A"C\cap AC'=C_a\in (O)$ and we just have to prove $B_aC_a,BC,A'P\iff $by radical axis theorem $B_a,C_a,P,A'$ are concyclic. Let the midpoints of $B_aB,C_aC$ be $E,F$. Hence $O,E,P,A',E$ are concyclic$\Rightarrow PBE\sim PCF\Rightarrow PB_aB\sim PC_a,C$ and the result follows.

XmL I understood your proof (I think) but where did you use the fact that in the original diagram P was the perpendicular from O to l?

well it's because of that fact that $O,E,P,A',E$ are concyclic

Let circles $(PBC),(PAC),(PAB)$ intersect $\ell$ at $D,E,F$. Lemma: \[ \frac{PE}{PF}=\frac{PZ}{PY} \] Proof: Consider an inversion about $P$ which preserves the circumcircle of $ABC$. Suppose $A,B,C$ map to $A',B',C'$, and $E,F$ map to $E',F'$. Then of course $B',A',F'$ are collinear and $C',A',E'$ are collinear. We claim that $P$ is the midpoint of $ZF'$. Let $Q$ be the intersection of $A'B'$ and $AB$; then $(QP,\ell_1,QZ,QF')$ is harmonic, where $\ell_1$ is the polar of $P$ passing through $Q$. Since $OP\perp \ell$, $\ell_1\parallel \ell$, so $P$ is the midpoint of $ZF'$. Therefore \[ \frac{PE}{PF}=\frac{PF'}{PE'}=\frac{PZ}{PY} \] as desired. Now consider the three circles $\omega_A,\omega_B,\omega_C$ with diameters $XD,YE,ZF$. By the lemma, $P$ has equal power to all three. So $(X,D), (Y,E), (Z,F)$ are in involution. By the converse of Desargues' Involution Theorem, $AD,BE,CF$ concur. So inverting the diagram yields that the original statement is true.

Sorry I meant to post this on another problem

Here is a powers bash using the Coaxial Lemma requiring no synthetic insight. We use directed lengths. Let $\omega_A = (APX), \omega_B=(BPY), \omega_C = (CPZ)$. We want to show that $\omega_A, \omega_B, \omega_C$ are coaxial. Note that $p(Z, \omega_A)=ZP\cdot ZX$ and $p(Z, \omega_B)=ZP\cdot ZY$. Then we want to show that $\frac{p(Z, \omega_A)}{p(Z, \omega_B)}=\frac{ZX}{ZY}=\frac{p(C, \omega_A)}{p(C, \omega_B)}$. Let $f(G)=p(G, \omega_A)-p(G, \omega)$ for all points $G$ in the plane. Then $f$ is linear. Note that $f(A)=0$ and $f(Y)=YP\cdot YX-YA\cdot YC$. On the other hand, $f$ is linear, so $p(C, \omega_A)=f(C)=\frac{AC}{AY}f(Y)=\frac{YP\cdot YX\cdot AC}{AY}+AC\cdot YC$, where the first line is true since $p(C, \omega)=0$ and the second line true since $f$ is linear and $f(A)=0$. Similarly, we get that $p(C, \omega_B)=\frac{XP\cdot XY\cdot BC}{BX}+BC\cdot XC$. It remains to prove that $$\displaystyle\frac{\frac{YP\cdot YX\cdot AC}{AY}+AC\cdot YC}{\frac{XP\cdot XY\cdot BC}{BX}+BC\cdot XC}=\frac{ZX}{ZY}$$By Menelaus on $XYC$, however, the RHS equals $\frac{AC\cdot BX}{AY\cdot BC}$. Thus we want to show $$\displaystyle\frac{\frac{YP\cdot YX\cdot AC}{AY}+AC\cdot YC}{\frac{XP\cdot XY\cdot BC}{BX}+BC\cdot XC}=\frac{AC\cdot BX}{AY\cdot BC}$$, or after cross multiplying, $YC\cdot AY - XC\cdot BX=XY(XP+YP)$. But note that $XY(XP+YP)=(XP-YP)(XP+YP)=XP^2-YP^2=XO^2-YO^2$ where $O$ is the circumcenter. Thus we want to show $YC\cdot AY-XC\cdot BX=XO^2-YO^2$, or $XO^2-XC\cdot XB=YO^2-YC\cdot YA$. But both sides are equal to $R^2$, where $R$ is the radius of $\omega$, so we are done. $\square$

This nice problem can be regarded as a property of isogonal center. Her is a complicated solution,which showes the connect of G8 and isogonal center: Let $Q$ be the image of $P$ under the inversion WRT $\odot (ABC),$ $R$ be the isogonal conjugate of $Q$ WRT $\triangle ABC$. $AR\cap BC=X,BR\cap CA=Y,CR\cap AB=Z.$ Let $N$ be the Miquel point of $X,Y,Z$ WRT $\triangle ABC.$ Then it’s known that $\{N,P \}=\odot (APDX)\cap \odot (BPEY)\cap \odot (CPFZ).$ Here showes more details.

Note that $\Delta ABC$ is a cevian triangle of the triangle formed by line $AX, BY, CZ$ and $XYZ$ is the perspective axis, so I restate the problem as the following stronger version: $\textbf{Theorem:}$ Given a triangle $\Delta ABC$ and a point $P$, let $\Delta DEF$ be the cevian triangle of $P$ with respect to $\Delta ABC$, $M$ is the Miquel point of $D, E, F$ with respect to $\Delta ABC$. $D_1E_1F_1$ is the trilinear polar of $P$ with respect to $\Delta ABC$. Let $U$ be the pedal point of the circumcenter $O$ of $\Delta DEF$ on line $D_1E_1F_1$. Then $(MDD_1), (MEE_1), (MFF_1)$ are coaxial and the second intersection point of these three circles is $U$. $\textbf{Lemma1:}$ Given $\Delta ABC$ and two points $P, Q$. $\Delta DEF$, $\Delta XYZ$ are the cevian triangles of $P, Q$ with respect to $\Delta ABC$, respectively. $D_1E_1F_1$ and $X_1Y_1Z_1$ are the trilinear polar of $P, Q$ with respect to $\Delta ABC$, respectively. $EF\cap YZ=A_1$, similarly define $B_1, C_1$. Then the intersection point of $D_1E_1F_1$ and $X_1Y_1Z_1$ is the perspective center of $\Delta ABC$ and $\Delta A_1B_1C_1$ $Proof:$ Let $T=D_1E_1F_1\cap X_1Y_1Z_1$, so it's suffice to prove $A, A_1, T$ are collinear. Apply Desargue's theorem to $\Delta A_1D_1X_1$ and $\Delta AF_1Y_1$ we then have to prove $FY, BC, F_1Y_1$ are concurrent. By Carnot's theorem we know that $D, E, F, X, Y, Z$ lie on one conic $\Gamma $, apply Pascal's theorem to hexagon $ZXDEYF$ with conic $\Gamma $ we have $FY, BC, F_1Y_1$ are concurrent. $\textbf{Lemma2:}$ Given $\Delta ABC$ and two cyclocevian conjugate points $P, Q$. $\Delta DEF$, $\Delta XYZ$ are the cevian triangles of $P, Q$ with respect to $\Delta ABC$, respectively. $D_1E_1F_1$ and $X_1Y_1Z_1$ are the trilinear polar of $P, Q$ with respect to $\Delta ABC$, respectively. Then the pole of $PQ$ with respect to $(DEFXYZ)$ is the intersection point of $D_1E_1F_1$ and $X_1Y_1Z_1$. $Proof:$ Let $T=D_1E_1F_1\cap X_1Y_1Z_1$, $EF\cap YZ=A_1$, similarly define $B_1, C_1$, $EZ\cap FY=A_2$, similarly define $B_2, C_2$. It's well known that $A_2, B_2, C_2$ lie on $PQ$, and by lemma$1$ we have $T$ lies on the polar of $A_2$ with respect to $(DEFXYZ)$, hence the polar of $T$ is $A_2B_2C_2$. $\textbf{Lemma3:}$ Given $\Delta ABC$ and two cyclocevian conjugate points $P, Q$. $\Delta DEF$, $\Delta XYZ$ are the cevian triangles of $P, Q$ with respect to $\Delta ABC$. $M, N$ are the Miquel points of $D, E, F$; $X, Y, Z$ with respect to $\Delta ABC$. $O$ is the center of $(DEFXYZ)$. Then $PQ$ is the radical axis of $(MON)$ and $(DEFXYZ)$. $Proof:(TelvCohl)$ See here https://artofproblemsolving.com/community/u300541h1409250p7909467 $\textbf{Corollary4:}$ Given $\Delta ABC$ and two cyclocevian conjugate points $P, Q$. $\Delta DEF$, $\Delta XYZ$ are the cevian triangles of $P, Q$ with respect to $\Delta ABC$, respectively. $M, N$ are the Miquel points of $D, E, F$; $X, Y, Z$ with respect to $\Delta ABC$, respectively. $D_1E_1F_1$ and $X_1Y_1Z_1$ are the trilinear polar of $P, Q$ with respect to $\Delta ABC$, respectively. $T=D_1E_1F_1\cap X_1Y_1Z_1$. Let $U, V$ be the pedal point of the center $O$ of $(DEFXYZ)$ on $D_1E_1F_1$ and $X_1Y_1Z_1$, respectively. Then $O, M, N, U, V, T$ are concyclic. $Proof:$ It's suffice to prove $T$ is the antipode of $O$ with respect to $(OMN)$. Since by lemma$3$ we have $PQ$ is the radical axis of $(MON)$ and $(DEFXYZ)$, so we only need to prove $T$ is the pole of $PQ$ with respect to $(DEFXYZ)$, but this is just lemma$2$. $\textbf{Back to the main problem:}$ Let $Q$ be the cyclocevian conjugate of $P$ with respect to $\Delta ABC$ and $\Delta XYZ$ be its cevian triangle with respect to $\Delta ABC$. $N$ is the Miquel point of $X, Y, Z$ with respect to $\Delta ABC$. Since by corollary$4$ and $OM=ON$(well known) we have $O, U, M, N$ are concyclic, so $M, D, D_1, U$ are concyclic $\Leftrightarrow $ $\angle MDD_1+\angle ONM=90^{\circ }$. Let $MD\cap NX=A'$, since it's well known $A'$ lies on $(OMN)$ and the angle bisector of $\angle MA'N$ is perpendicular to $BC$, so $\angle MDD_1+\angle ONM=90^{\circ }$.$\Box $

Invert the problem at $P$ so it reads: ISL 2012/G8 Inverted wrote: Let $ABC$ be a triangle with circumcircle $\omega$ and $\ell$ a line without common points with $\omega$. Denote by $P$ the foot of the perpendicular from the center of $\omega$ to $\ell$. The circles $(PBC),(PCA),(PAB)$ intersect $\ell$ at $X,Y,Z$ respectively. Prove that lines $AX,BY,CZ$ concur (possibly at infinity). [asy][asy] usepackage("amssymb"); import cse5; import olympiad; pathpen = black; pointpen = black; size(12cm); pen n_purple = rgb(0.7,0.4,1), n_blue = rgb(0,0.6,1), n_green = rgb(0,0.4,0), n_orange = rgb(1,0.4,0.1), n_red = rgb(1,0.2,0.4); pair O=origin, A=dir(60), B=dir(250), C=dir(10), P=(1.4,0), Q=(1.4,2), D=extension(B,C,P,Q), E=extension(C,A,P,Q), F=extension(A,B,P,Q); pair X=(1.4,-1.836), Y=(1.4,2.29), I=extension(A,X,B,Y), Z=extension(P,Q,I,C); DPA(A--E^^B--F^^B--D, n_orange); D((F+(0,0.5))--(2*P-F-(0,0.5))); DPA(circumcircle(P,B,C)^^circumcircle(P,C,A)^^circumcircle(P,A,B), n_blue); D(circumcircle(A,B,C), green+dotted); DPA(X--A+(A-X)*0.1^^Z--I-(Z-I)*0.4^^B--Y, blue+dashed); D("A",A,dir(115)); D("B",B,dir(240)); D("C",C,dir(85)); D("D",D,D); D("E",E,dir(-45)); D("F",F,F); D("P",P,dir(30)); D("X",X,X); D("Y",Y,Y); D("Z",Z,dir(45)); D("I",I,dir(-95)); D("O",O,dir(90)); [/asy][/asy] Let $D,E,F$ be the intersections of lines $BC,CA,AB$ with $\ell$. Let $O$ be the center of $\omega$, and let $\omega$ have radius $R$. Claim. $PD\cdot PX=PE\cdot PY=PF\cdot PZ$ (with all lengths directed). Proof. $PD\cdot PX = PD\cdot (DX-DP) = -DP\cdot DX +DP^2=-DC\cdot DB+DP^2=-\text{Pow}(D,\omega)+DP^2=-(DO^2-DP^2-R^2)=R^2-OP^2$ by the Pythagorean theorem. This expression is independent of $D,X$, so the same argument gives this as the value for $PE\cdot PY,PF\cdot PZ$.$\square$ This means that there is a negative (since $P$ lies outside $\omega$) inversion at $P$ swapping $(D,X),(E,Y),(F,Z)$, thus these are pairs of an involution on line $\ell$. Let $AX\cap BY=I$, and let $CI\cap \ell=Z'$, so we wish to show $Z'=Z$. Consider the complete quadrangle induced by $ABCI$ (we must take care of the case $I=A,B,C$, but this case is trivial). By Desargues Involution Theorem with respect to line $\ell$, we have that $(BC\cap \ell=D,AI\cap \ell =X), (CA\cap \ell=E,BI\cap\ell=Y), (AB\cap\ell = F,CI\cap\ell=Z')$ are pairs of an involution on $\ell$. Thus $Z'=Z$, so $AX,BY,CZ$ concur at $I$. $\blacksquare$

Can anybody (thoughtlessly) barybash this problem ?

which I'm pretty sure is undoable by hand.

lyukhson wrote: Let $ABC$ be a triangle with circumcircle $\omega$ and $\ell$ a line without common points with $\omega$. Denote by $P$ the foot of the perpendicular from the center of $\omega$ to $\ell$. The side-lines $BC,CA,AB$ intersect $\ell$ at the points $X,Y,Z$ different from $P$. Prove that the circumcircles of the triangles $AXP$, $BYP$ and $CZP$ have a common point different from $P$ or are mutually tangent at $P$. 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/* line */ draw((xmin, 0.9961735888211992*xmin + 3.280268679470318)--(xmax, 0.9961735888211992*xmax + 3.280268679470318), linewidth(0.6) + dbwrru); /* line */ draw((xmin, 0.18916960911360972*xmin + 1.2385281123334804)--(xmax, 0.18916960911360972*xmax + 1.2385281123334804), linewidth(0.6) + dbwrru); /* line */ draw((xmin, -1.9958163055240987*xmin-3.0289550465836204)--(xmax, -1.9958163055240987*xmax-3.0289550465836204), linewidth(0.6) + ffqqff); /* line */ draw((xmin, 0.5210374952339091*xmin-0.4022592653468127)--(xmax, 0.5210374952339091*xmax-0.4022592653468127), linewidth(0.6) + ffqqff); /* line */ draw((xmin, -0.25618070728372083*xmin-1.2133972714712349)--(xmax, -0.25618070728372083*xmax-1.2133972714712349), linewidth(0.6) + ffqqff); /* line */ /* dots and labels */ dot((-2.64,2.24),dotstyle); label("$A$", (-2.5284761904761894,2.4836380952381094), NE * labelscalefactor); dot((-3.22,-2.08),dotstyle); label("$B$", (-3.1115809523809514,-1.826266666666652), NE * labelscalefactor); dot((2.491295238095238,-1.8516190476190386),dotstyle); label("$C$", (2.5927047619047627,-1.5980952380952234), NE * labelscalefactor); dot((-4.72,1.82),dotstyle); label("$P$", (-4.607371428571428,2.078), NE * labelscalefactor); dot((-0.4327523956737076,-0.255278613543807),linewidth(4pt) + dotstyle); label("$O$", (-0.3228190476190466,-0.0516), NE * labelscalefactor); dot((-6.674698419170175,-2.2181450760759867),linewidth(4pt) + dotstyle); label("$X$", (-6.584857142857142,-2.0037333333333187), NE * labelscalefactor); dot((-3.9940555793004893,3.319703923225688),linewidth(4pt) + dotstyle); label("$Y$", (-3.897504761904761,3.5230857142857284), NE * labelscalefactor); dot((-1.919690420831666,7.605064451736557),linewidth(4pt) + dotstyle); label("$Z$", (-1.8186095238095228,7.807638095238109), NE * labelscalefactor); dot((0.7363065967824595,-2.28524323465199),linewidth(4pt) + dotstyle); label("$Q$", (0.8433904761904772,-2.0797904761904613), NE * labelscalefactor); dot((-2.0507782376869756,-0.7640892711244693),linewidth(4pt) + dotstyle); label("$X'$", (-1.9453714285714276,-0.5586476190476045), NE * labelscalefactor); dot((-2.530025400713196,0.7599241962330618),dotstyle); label("$P'$", (-2.427066666666666,1.0132), NE * labelscalefactor); dot((-1.9849663816880367,1.3028975953347164),linewidth(4pt) + dotstyle); label("$Y'$", (-1.8946666666666656,1.4948952380952525), NE * labelscalefactor); dot((-0.6906212512376372,1.1078835601913044),linewidth(4pt) + dotstyle); label("$Z'$", (-0.6016952380952371,1.3174285714285858), NE * labelscalefactor); dot((-3.594530913455832,4.145068361201962),linewidth(4pt) + dotstyle); label("$Ta$", (-3.491866666666666,4.3597142857143), NE * labelscalefactor); dot((-7.750469801218311,-4.440544637459655),linewidth(4pt) + dotstyle); label("$Tb$", (-7.649657142857142,-4.234742857142842), NE * labelscalefactor); dot((-5.505610512729866,0.19703392370859119),linewidth(4pt) + dotstyle); label("$Tc$", (-5.393295238095237,0.40474285714287156), NE * labelscalefactor); dot((-1.0436425748868363,-0.9460361784853173),linewidth(4pt) + dotstyle); label("$T$", (-0.9312761904761895,-0.7361142857142712), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $\ell_a, \ell_b, \ell_c$ denote the polars of $X, Y, Z$ w.r.t $\triangle ABC$. Let $T_a, T_b, T_c$ be the intersections of $\ell_a, \ell_b, \ell_c$ with $\ell$ respectively. Claim 1: $T_a$ is the radical centre of $(APX), (OPX), \omega$. Proof: Observe that $PX$ is the radical axis of $(APX), (OPX)$ and $\ell_a$ is the radical axis of $\omega, (OPX)$. So, $T_a = \ell_a \cap PX$ is the radical centre of $(OPX), (APX), \omega$. $\square$ Similar observations hold true for $T_b, T_c$. Claim 2: $AT_a, BT_b, CT_c$ concur. Proof: Note that there exists an inversion centered at $P$ swapping $(X, T_a), (Y, T_b), (Z,T_c)$. So, $(X,T_a), (Y,T_b), (Z,T_c)$ are involution pairs. Let $AT_a \cap BT_b = T$. Note that there is an involution swapping $(X,T_a), (Y,T_b), (C,CT \cap \ell )$.So, $AT_a, BT_b, CT_c$ are concurrent. $\square$ By Claim 1, $AT_a$ is the radical axis of $\omega, (APX)$. So, $T$ lies on the radical axis of $\omega, (APX), (BPY), (CPZ)$. So, the circles $(APX), (BPY), (CPZ)$ are coaxial. $\blacksquare$

If you mess up your geogebra diagram like me by putting $Y$ on $AB$ and $Z$ on $AC$ instead, then you will notice that $(BPY)$ and $(CPZ)$ instead meet on $(ABC)$

Nice problem. Main Lemma: Let $\triangle DEF$ be the antipedal triangle of $P$ WRT $\triangle ABC$, $O$ is the circumcenter of $\triangle ABC$, a line $l$ passing through $P$ perpendicular to $OP$ intersect $BC, CA, AB$ at $X, Y, Z$ , respectively. $U$ is chosen on $EF$ such that $UX$ is perpendicular to $l$, define $V, W$ similarly. Then $U, V, W$ are collinear. Proof: Let $Q$ be the infinite point of direction $OP$, we wish to use Desargues's Involution with point $Q$ and quadrilateral $VW, EF, FD, DE$, that means we have to show that $(QD, QU), (QE, QV), (QF, QW)$ are reciprocal pairs of some involution. So if we project them on to $l$, let $D', E', F'$ be the projection of $D, E, F$ on $l$, we need to show that $(D', X), (E', Y), (F,' Z)$ are reciprocal pairs of some involution. Notice that $X$ lies on the radical axis of $\odot(PD)$ and $\odot (O)$, we get $$ PX\times PD' = PX^2 + PX\times XD' =PX^2 -(OX^2- r^2) = -Power (P, \odot(O))$$hence $PX\times PD' = PY\times PE' = PZ\times PF' $, $(D', X), (E', Y), (F,' Z)$ are reciprocal pairs of some involution, we're done. $\square$ Back to the original problem. We're enough to show that the center of $\odot(APX), \odot(BPY), \odot(CPZ)$ are collinear. Take homothety with center P and ratio $2$, use the Lemma directly then we're done.

lyukhson wrote: Let $ABC$ be a triangle with circumcircle $\omega$ and $\ell$ a line without common points with $\omega$. Denote by $P$ the foot of the perpendicular from the center of $\omega$ to $\ell$. The side-lines $BC,CA,AB$ intersect $\ell$ at the points $X,Y,Z$ different from $P$. Prove that the circumcircles of the triangles $AXP$, $BYP$ and $CZP$ have a common point different from $P$ or are mutually tangent at $P$. Proposed by Cosmin Pohoata, Romania We define an inversion $\Psi$ about a circle with center $P$ and an arbitrary radius. We now tell the inverted form of the problem - Inverted Problem : Let $\triangle ABC$ have circumcircle $\omega$ and let $\ell$ be a line such that $\ell$ and $\omega$ do not intersect with each other. Let $O$ be the circumcenter of $\triangle ABC$ and let $P$ be the foot of the perpendicular from $O$ to $\ell$. If circumcircles of the triangles $\triangle PBC, \triangle PAC, \triangle PAB$ intersect the line $\ell$ at points $X, Y, Z$ respectively ($X, Y$ or $Z$ can equal $P$ if and only if there is just one intersection of the respective circle and otherwise distinct from $P$), then show that the lines $AX, BY, CZ$ concur at a point $K$ ($K$ can possibly lie on the line of infinity) (Let us call circumcircle of $\triangle BCP$ as $\Omega$) Leet $D, E, F$ be the intersections of line $\ell$ with lines $\overline{BC}, \overline{AC}, \overline{AC}$ respectively. Now, we see that $PD \cdot PX = PD \cdot (DX - DP) = PD \cdot DX - PD^2 \overset{{\huge \mathrm{Pow}_{\Omega}(D)}}{=} DC \cdot DB - PD^2 = \overset{{\huge \mathrm{Pow}_{\omega}(D)}}{=} DO^2 - OA^2 - PD^2 = OP^2 - OA^2$ and symmetrically we can also achieve that $PD \cdot DX = PE \cdot EY = PF \cdot FZ = OP^2 - OA^2$. This means there is some inversion (negative inversion?) centered at $P$ taking points $D, E, F$ to $X, Y, Z$ respectively. This is an involution (since inverting a point twice about the same circle returns the original reference point). Now, we are motivated to consider $AX \cap CZ = K'$, it suffices to show that $K' = K$ by showing that $K' \in \overline{BY}$. But let us say that $\overline{BK'} \cap \ell = Y'$, so we can alternatively prove that $Y = Y'$. We see that using the inversion involution swapping the pairs of points $(D, X), (E, Y), (C, Z)$, we use DIT on complete quadrilateral $ABCK'$, we know that DIT should swap the intersection of line $\ell$ with the opposite sides (including diagonals) of quadrilateral $ABCK'$, we can see that this DIT swaps $(D, X), (F, Z), (E, Y')$ which means that $Y = Y'$ since there is a unique involution swapping two known pairs of points, which completes our proof.

We need two lemmas first: Lemma1: Let $T_{A}$ be the second intersection point of the circle $\Omega$ with (circle) $(AXP)$ then the common chord $AT_{A}$ intersects $\ell$ at a point $K_{A}$ where $K_{A}BCP$ is cyclic. Proof: Let $XP_1,XP_2$ be the tangents to $\Omega$. Then $O,P_1,P_2,P,X$ lie on the same circle through $(OXP)$. The three radical axes of the circles $\Omega , (AXP) ,(OXP)$ are $P_1P_2 , \ell ,AT_A$ and concur on $K_A$ . Next verify that$\angle XK_AP_1=\angle PP_1X$ and $XP_1^2=XP\cdot XK_A=XC\cdot XB$ and thus $K_ABCP$ is cyclic. Later on we will need this following special lemma. Lemma2:Let a triangle $ABC$ inscribed in a circle and let $S_AS_BS_C$ be a circumcevian triangle wrt some point. Let $ S_AB\cap S_BA=K_C, S_AC\cap S_CA=K_B,S_BC\cap S_CB=K_A$ then $AS_A,BS_B,CS_C$ are concurrent iff $AK_A,BK_B,CK_C$ are concurrent. Proof: We are going to compute $\frac{\sin{BCK_C}}{\sin{ACK_C}}$. From triangle $BCK_C$ we get :$\frac{\sin{BCK_C}}{BK_C}=\frac{\sin{CBK_C}}{CK_C}$ and from $ACK_C$ : $\frac{\sin{ACK_C}}{AK_C}=\frac{\sin{CAK_C}}{CK_C}$ and then divide to get $\frac{\sin{BCK_C}}{\sin{ACK_C}}=\frac{BK_C}{AK_C}\frac{\sin{CBK_C}}{\sin{CAK_C}}$ . Since $K_CBS_B$ , $K_CAS_A$ are similar :$\frac{BK_C}{AK_C}=\frac{BS_B}{AS_A}$ and we also have that $\frac{\sin{CBK_C}}{\sin{CAK_C}}=\frac{\sin{S_BBC}}{\sin{S_AAC}}$. Finally get that $\frac{\sin{BCK_C}}{\sin{ACK_C}}=\frac{BS_B}{AS_A}\frac{\sin{S_BBC}}{\sin{S_AAC}}$ and multiply the similar relations from all three sides to get the desired result.$$\frac{\sin{BCK_C}}{\sin{ACK_C}}\frac{\sin{CAK_A}}{\sin{BAK_A}}\frac{\sin{ABK_B}}{\sin{CBK_B}}=\frac{\sin{S_BBC}}{\sin{S_BBA}}\frac{\sin{S_AAB}}{\sin{S_AAC}}\frac{\sin{S_CCA}}{\sin{S_CCB}}$$Now we proceed to the solution. It suffices to show that the three radical axes $AT_A,BT_B,CT_C$ pass through a common point. We will show that $BK_C , CK_B$ intersect on a point $S_A$ on $\Omega$ and that $PS_AOA$ is cyclic. Then let $BK_C$ intersect $\Omega$ at $S_A$ :$\angle OS_AA=90-\angle ABK_C=90-APK_C=OPA$ (the second to last equality due to $BAK_CP$ being cyclic) thus $PS_AOA$ is cyclic. Next define $S_B , S_C$ similarly , $AS_A, BS_B,CS_C$ concur on the pole of $\ell$ and conclude using lemma 2 that $AK_A,BK_B,CK_C$ are concurrent.

We use coaxial circles lemma and linearity of power of a point, pow. Define $f(\bullet)=P(\bullet,(APX))-P(\bullet,(ABC))$. As $f$ is linear, we have \begin{align*}f(A)=\frac{AZ}{BZ}f(B)+\frac{AB}{BZ}f(Z)\implies P(B,(APX))=\frac{ZA\cdot ZB-ZP\cdot ZX}{AZ}\cdot AB. \end{align*}Similarly, define $g(\bullet)=P(\bullet,(CPZ))-P(\bullet,(ABC))$ and we obtain \begin{align*}g(C)=\frac{CX}{BX}g(B)+\frac{BC}{BX}g(X)\implies P(B,(CPZ))=\frac{XB\cdot XC-XP\cdot XZ}{CX}\cdot BC. \end{align*} Also, $P(Y,(APX))=YX\cdot YP$ and $P(Y,(CPZ))=YP\cdot YZ$. By the coaxial circles lemma we need, \begin{align*}\frac{P(B,(APX))}{P(B,(CPZ))}&=\frac{P(Y,(CPZ))}{P(Y,(CPZ))}\Longleftrightarrow\\ \frac{AB\cdot CX\cdot (ZA\cdot ZB-ZP\cdot ZX)}{BC\cdot AZ\cdot (XB\cdot XC-XP\cdot XZ)}&=\frac{YX\cdot YP}{YP\cdot YZ}\Longleftrightarrow\\ \frac{ZA\cdot ZB-ZP\cdot ZX}{XB\cdot XC-XP\cdot XZ}&=\frac{YX\cdot AZ\cdot BC}{YZ\cdot CX\cdot AB}. \end{align*}Claim. $\dfrac{YX\cdot AZ\cdot BC}{YZ\cdot CX\cdot AB}=1$. Proof. This is true by Law of Sines, \begin{align*} \frac{YX\cdot AZ\cdot BC}{YZ\cdot CX\cdot AB}=\frac{\sin{\angle YCX} \cdot \sin{\angle AYZ}\cdot \sin{\angle BAC}}{\sin{\angle ZAY}\cdot \sin{\angle CYX}\cdot \sin{\angle ACB}}=1.\quad \square \end{align*} Now, we have \begin{align*}ZO^2-XO^2&=(ZO^2-r^2)-(XO^2-r^2)\\&=ZA\cdot ZB-XB\cdot XC\\&=XZ(ZP-XP)=(ZP+XP)(ZP-XP)\\&=ZP^2-XP^2, \end{align*}which is true as $OP\perp XZ$. We are done. [asy][asy] size(12cm); defaultpen(fontsize(10pt));pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;pen org=magenta;pen heavy=heavymagenta; pair O,A,B,C,a,b,P,X,Y,Z; O=(0,0);B=dir(195);A=dir(118);C=dir(345);path w=circumcircle(A,B,C); a=(1.6,0);b=(0,3);P=foot(O,a,b);X=extension(B,C,a,b);Y=extension(a,b,A,C);Z=extension(a,b,A,B); draw(w,heavyblue);draw(A--B--C--cycle,heavygreen); draw(circumcircle(P,C,Z),olive);draw(circumcircle(P,A,X),olive);draw(circumcircle(P,B,Y),olive); draw(Z--Y,heavygreen);draw(Z--A,heavygreen);draw(C--Y,heavygreen);draw(X--C,heavygreen); draw(O--P,olive); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$A$",A,dir(A)); dot("$O$",O,dir(90)); dot("$P$",P,dir(P)); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); dot("$Z$",Z,dir(Z)); [/asy][/asy]

This was cool. Define $P'$ as the inverse of $P$ wrt $\omega$. Next, let $A_1 = (APX)\cap (ABC), B_1 = (BPY)\cap (ABC), C_1 = (CPZ)\cap (ABC)$. If we show that $AA_1, BB_1, CC_1$ are concurrent at a point $K$, then $K$ is the radical center of $(APX), (BPY), (CPZ)$, so $PK$ is the radical axis and the three circles concur. Now, define $D = AA_1\cap \ell$. If $A_2 = A_1X\cap (ABC), A_3 = AX\cap (ABC)$, then since $P\in (XAA_1)$, $X = AA_3\cap A_2A_1$, and $\angle OPX = 90$, this means $P$ is the miquel point of $AA_1A_2A_3$. Therefore, since $P'$ is the polar of $\ell$, this means $P' \in AA_2$. Furthermore, $D$ lies on the pole of $X$ (by brocard). If we define $B_2, C_2$ similarly, then $AA_2, BB_2, CC_2$ are all collinear at $P'$ Now, observe that if the following expression is equal to $1$, we're done by ceva: \[\frac{\sin \angle BAA_1}{\sin \angle CAA_1} \cdot \frac{\sin \angle ACC_1}{\sin \angle BCC_1}\cdot \frac{\sin\angle CBB_1}{\sin\angle ABB_1} = \frac{BA_1}{CA_1}\cdot \frac{CB_1}{AB_1}\cdot \frac{AC_1}{BC_1}\]We have \[\frac{BA_1}{CA_1} = \frac{\frac{BA_1}{\sin \angle BXA_1}}{\frac{CA_1}{\sin\angle CXA_1}} = \frac{\frac{BX}{\sin \angle BA_1 A_2}}{\frac{CX}{\sin \angle A_2A_1 C}} = \frac{BX}{CX} \cdot \frac{\sin \angle A_2A C}{\sin \angle A_2 AB} \]Now, when we get similar results for $\frac{CB_1}{AB_1}$ and $\frac{AC_1}{BC_1}$, plugging it into our expression gives \[\frac{BX}{CX} \cdot \frac{CY}{AY} \cdot \frac{AZ}{ZB} \cdot \frac{\sin \angle A_2AB}{\sin \angle A_2 AC}\cdot \frac{\sin \angle C_2 CA}{\sin\angle C_2 CB}\cdot \frac{\sin \angle B_2 BC}{\sin \angle B_2 BA}\]However, since $X,Y,Z$ are collinear, $\frac{BX}{CX} \cdot \frac{CY}{AY} \cdot \frac{AZ}{ZB} =1$ by menelaus. Furthermore, since $AA_2, BB_2, CC_2$ are concurrent at $P'$, then $\frac{\sin \angle A_2AB}{\sin \angle A_2 AC}\cdot \frac{\sin \angle C_2 CA}{\sin\angle C_2 CB}\cdot \frac{\sin \angle B_2 BC}{\sin \angle B_2 BA} = 1$ by ceva. Thus, our entire expression is $1$, so we conclude that $AA_1, BB_1, CC_1$ are collinear.

We invert at $P,$ images are marked by primes. Let $Q=X'B'\cap A'Y',R=X'C'\cap A'Z',E,F$ are midpoints of $B'Q,C'R$ respectively. By Miquel theorem in $QX'Y',RX'Z'$ we get $Q,R\in \omega '.$ Clearly $P,X',E,F$ are concyclic with center of $\omega ',$ hence $P$ is center of spiral similarity $B'EQ\mapsto C'FR$ and $P\in \odot (X'QR).$ By radical center theorem for $\odot (PB'C'X'),\odot (B'C'QR),\odot (PX'QR)$ we conclude that $PX',B'C',QR$ concur, and by Desargues' theorem for $A'Y'Z',Y'B'C'$ lines $A'X',B'Y',C'Z'$ concur. The conclusion follows.

Radical axis theorem on $\omega$, $(AXP)$ and $(OXP)$ where $O$ is the center of $\omega$ gives us that the radical axis of $\omega$ and $(AXP)$ is $AD$, where $D$ is the intersection of the polar of $X$ wrt. $\omega$ and $\ell$. By radical axis theorem, it is sufficient to have $AD$, $BE$, $CF$ concur. Take a homography fixing $\omega$ and sending $\ell$ to infinity. The problem is now trivial since $AD$, $BE$, $CF$ are the altitudes of $\triangle ABC$.

wowowowow very cool problem Invert at $P$ such that $(ABC)$ stays fixed; then $X^* = (PA^*B^*) \cap \ell$, etc, and it suffices to show that $A^*X^*, B^*Y^*, C^*Z^*$ concur. If $B^*C^*, A^*C^*, A^*B^* \cap \ell = Q, R, S$ respectively then by Trig Ceva on $\triangle{A^*B^*C^*}$, it suffices to show that $\prod_{\text{cyc}} \frac{\sin \angle{SA^*X^*}}{\sin \angle{RA^*X^*}} = 1$. By the Ratio Lemma on $\triangle{SA^*R}$ this is equal to $\prod_{\text{cyc}} \frac{SX^*}{RX^*} \div \prod_{\text{cyc}} \frac{SA^*}{RA^*}$, the latter of which is equal to $1$ by Law of Sines, so it suffices to show that $\prod_{\text{cyc}} \frac{SX^*}{RX^*} = 1$. However, by Power of a Point at $Q$, $QP \cdot QX^* = \text{Pow}_{(A^*B^*C^*)}(Q) = QO^2 - R^2$ where $R$ is the radius of $(A^*B^*C^*)$, which is equal to $QP^2 + OP^2 - R^2$, so $PX^* = \frac{T^2}{QP}$ if $T$ is the length of the tangent from $P$ to $(A^*B^*C^*)$. Thus, the desired product is $\prod_{\text{cyc}} \frac{|SP - \frac{T^2}{QP}|}{|RP - \frac{T^2}{QP}|} = \prod_{\text{cyc}} \frac{|QP \cdot SP - T^2|}{|QP \cdot RP - T^2|}$, which clearly equals $1$. $\square$

We use moving points. Consider the diagram upon applying an inversion at $P$ that fixes $(ABC)$. We have $X = (PBC) \cap \ell$ and $Y, Z$ are defined similarly. It suffices to show that $\overline{AX}$, $\overline{BY}$, and $\overline{CZ}$ concur (or that they are parallel). Animate $B$ on $(ABC)$. The key idea is that $X \mapsto B \mapsto Y$ is projective. Now setting $\deg X = \deg Y = 1$, we have $\deg B = 2$. Thus, we must check that $\deg X + \deg B + \deg Y + 1 =5$ distinct positions for $B$ satisfy the hypotheses of the problem. If $B=A$, then the conclusion is immediate as lines $\overline{BC}$ and $\overline{AC}$ coincide. Likewise $B=C$ implies the conclusion. Suppose that $B=\overline{AY} \cap (ABC) \backslash \{A\}$. Let $O$ be the center of $(ABC)$. Then, $ACPY$ is cyclic, so $\angle CPY = 180^{\circ}-\angle A$, from which $\angle CPO = 90^{\circ}-\angle A$. As $\angle CBO = 90^{\circ} - \angle A$, we have that $OCPB$ is cyclic. Thus, $\angle OPB = \angle OCB = 90^{\circ} - \angle A$, so $\angle BPZ = 180^{\circ}-\angle A$. This yields that $ABPZ$ is cyclic, so $\overline{CZ}$ contains $A$; hence the problem holds for this position of $B$. Likewise $B=\overline{CY} \cap (ABC) \backslash \{C\}$ implies the conclusion. By geometric continuity, we can find $B$ such that $\overline{AX}$, $\overline{BY}$, and $\overline{CZ}$ are pairwise parallel. The above observations complete the proof.

Projective geo (oly proj geo) is now certified by me to be way beyond my aukaat. Also thanks to solutions of other problems of Arjun bro, Tafi bhai and Arnab da from where I gathered more than half of the motivation of the solution! Anyways, here is a solution without any of Involutions and stuffs; this uses Inversion, projective geo and spiral sim. [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 = (-47.92316,44.41909); pair B = (-92.70353,-47.20610); pair C = (48.72624,-47.73503); pair O = (-21.90479,-25.05241); pair E = (-74.70683,-10.38303); pair D = (-6.24812,4.68241); pair P = (-27.42295,0.02255); pair Z = (-125.13669,-21.48094); pair Y = (190.77414,48.04036); pair X = (-49.20486,-4.77091); pair K = (1.99417,45.17638); pair L = (-86.42531,11.55835); pair T = (-27.35926,-99.03548); pair Q = (-239.53862,-46.65694); pair R = (-67.97211,184.28143); pair F = (-95.39563,-14.93594); pair G = (40.54973,14.98104); import graph; size(15cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); pen wwzzqq = rgb(0.4,0.6,0); draw(A--B, linewidth(0.5)); draw(C--A, linewidth(0.5)); draw(circle(O, 74.18386), linewidth(0.5)); draw(circle((-79.30094,2.99904), 51.96331), linewidth(0.5) + red); draw(circle((71.02825,72.41390), 122.20125), linewidth(0.5) + red); draw(circle((-22.09560,-76.07155), 76.28036), linewidth(0.5) + blue); draw(circle((-49.01405,46.24822), 51.01949), linewidth(0.5) + blue); draw(Q--K, linewidth(0.5) + linetype("4 4") + blue); draw(Q--B, linewidth(0.5)); draw(Z--T, linewidth(0.5)); draw(T--Y, linewidth(0.5)); draw(B--R, linewidth(0.5)); draw(R--C, linewidth(0.5)); draw(T--L, linewidth(0.5)); draw(T--R, linewidth(0.5)); draw(T--K, linewidth(0.5)); draw(Q--Z, linewidth(0.5)); draw(K--X, linewidth(0.5)); draw(X--L, linewidth(0.5)); draw(O--P, linewidth(0.5)); draw(P--R, linewidth(0.5)); draw(X--B--C--cycle, wwzzqq+linewidth(1.6)); draw(A--Z--Y--cycle, wwzzqq+linewidth(1.6)); dot("$A$", A, NW); dot("$B$", B, SW); dot("$C$", C, SE); dot("$O$", O, SW); dot("$E$", E, NW); dot("$D$", D, N); dot("$P$", P, NW); dot("$Z$", Z, NW); dot("$Y$", Y, NE); dot("$X$", X, NW); dot("$K$", K, NE); dot("$L$", L, NW); dot("$T$", T, dir(270)); dot("$Q$", Q, NW); dot("$R$", R, NW); dot("$F$", F, NW); dot("$G$", G, NE); [/asy][/asy] Firstly, note that it suffices to prove that $\left\{\odot(AXP),\odot(BYP),\odot(CZP)\right\}$ are coaxial. Now perform an Inversion centered at $P$ with any arbitrary radius, but later drop off the superscripts of $^*$ of the images. Thus we get the following current problem. $\textbf{PROBLEM: }$ Let $\triangle ABC$ be a triangle with circumcircle and $\ell$ be an arbitrary line. Let $O$ denote the circumcenter of $\odot(ABC)$ and $P$ denote the foot of the perpendicular from $O$ onto $\ell$. Now let $X$ be the second intersection of $\ell$ with $\odot(PBC)$. Similarly define $Y$ and $Z$ too. Prove that $AX$, $BY$ and $CZ$ are concurrent. Now note that proving $AX$, $BY$ and $CZ$ are concurrent is equivalent to proving that $\triangle AZY$ and $\triangle XCB$ are perspective by using Desargues' Theorem. This is further equivalent to proving that $\overline{AZ\cap XC-ZY\cap CB-YA\cap BX}$ are collinear. Let $D=\ell\cap AC$ and $E=\ell\cap AB$. Now let $K=BX\cap\odot(ABC)$ and $L=CX\cap\odot(ABC)$. Also, let $T=ZB\cap YC$. Now let $Q=\ell\cap BC$. Now by using $\operatorname{Pow}_{\odot(PXBC)}(Q)$, we get that $QX\cdot QP=QB\cdot QC$. Now temporarily add in the intersections of $\ell$ with $\odot(ABC)$ as $F$ and $G$. Now this gives that $QF\cdot QG=QB\cdot QC=QX\cdot QP$. Moreover notice that $P$ is the midpoint of $FG$. Now by using EGMO Lemma 9.17, we get that $(Q,X;F,G)=-1$, that is $Q$ lies on the polar of $X$, $Q\in\mathcal P_X$. Now let $Q'=KL\cap BC$. By using Brokard Theorem on $BCKL$, we get that $Q'\in\mathcal P_X$. Thus this finally gives that $Q\equiv Q'\equiv\mathcal P_X\cap\ell$. Now let $R=BL\cap CK$. Now note that as we have $X=BK\cap CL$ and that $P=\odot(LKX)\cap\odot(CBX)$, it follows from the definition of spiral similarity that $P$ is the center of the spiral similarity mapping $\overline{CB}\mapsto\overline{LK}$. However, note that $CBKL$ (the concave quadrilateral as in the diagram) is cyclic, so from the properties of miquel point of cyclic quadrilaterals, we have that $P$ is the inverse of $R$ w.r.t. $\odot(ABC)$. This means that $P$, and thus it also follows that $\mathcal P_R\equiv\ell$. Now let $T=LE\cap KD$. Now by the converse of Pascal's Theorem on $LTKBAC$, we get that as $\overline{E-D-X}$ are collinear, thus $\left\{A,L,B,T,C,K\right\}$ lie on a conic. But we also have that $ALBCK$ are cyclic, so this gives us that $T\in\odot(ABC)$. So now we have that $\overline{Q-L-K}$, $\overline{Q-E-D}$ and $\overline{Q-B-C}$ are collinear. This gives us that $\triangle BEL$ and $\triangle CDK$ are perspective from $Q$, which on applying Desargues' Theorem means that $\overline{BL\cap CK-BE\cap CD-LE\cap KD}$ are collinear; that is $\overline{R-A-T}$ are collinear. This gives us that $R=BL\cap AT$. Now let $Z'=AL\cap BT$. Then by applying Brokard's Theorem on $ALBT$, we get that $Z'E$ is the polar of $R$, that is $\mathcal P_R\equiv Z'E$. But from before, we had that $\mathcal P_R\equiv\ell$, and thus $Z'\in\ell$. Now finally note that as $E$ lies on the polar of $Z'$, we get that $(Z,E;F,G)=-1$, which on combining with with the fact that $P$ is the midpoint of $FG$ and EGMO Lemma 9.17 gives us that $EZ'\cdot EP=EF\cdot EG=EA\cdot EB$ which finally implies that $PABZ'$ is cyclic. This means that $Z\equiv Z'=\ell\cap\odot(APB)$, that is $\overline{A-L-Z}$ and $\overline{T-B-Z}$ are collinear. Similarly we also get that $\overline{A-K-Y}$ and $\overline{T-C-Y}$ are collinear. So we finally conclude that $\overline{K-L-Q}$, that is $\overline{AZ\cap XC-ZY\cap CB-YA\cap BX}$ are collinear and we are done!!!!!

Solved with Malay Mahajan and Kanav Talwar.

Claim: The intersection of the polar of $X$ with respect to $\omega$ and $\ell$ is on the radical axis of $\omega$ and $(APX).$ Proof: Let $A_1$ be the intersection of the polar and the radical axis. Note that the inversion of $(OPX)$ with respect to $\omega$ is the polar of $X$ as $OX$ is the diameter of $(OPX).$ Thus, $A_1$ must be the radical center of $\omega, (OPX),(APX),$ so it must lie on the radical axis of $\omega$ and $(APX)$ as well. Same goes for the other points. Call them $B_1,C_1.$ Thus, we need to show that $\overline{AA_1}, \overline{BB_1}, \overline{CC_1}$ pass through a a common point. Take a homography fixing $\omega$ and sending $\ell$ to the line at infinity. Then $\overline{AA_1}, \overline{BB_1}, \overline{CC_1}$ turn into altitudes of $\triangle ABC,$ which concur at the orthocenter, so we're done.

By the Coaxiality Lemma, $(AXP)$ is coaxial with $(BYP)$, $(CZP)$ if and only if \[\frac{\operatorname{pow}(A,(BYP))}{\operatorname{pow}(A,(CZP))} = \frac{\operatorname{pow}(X,(BYP))}{\operatorname{pow}(X,(CZP))} \iff \frac{AB \cdot AK}{AC \cdot AL} = \frac{XY}{XZ},\] where $K = AB \cap (BYP)$ and $L = AC \cap (CZP)$. Notice that \[\frac{AB}{AC} = \frac{\sin \angle C/\sin \angle X}{\sin \angle B/\sin \angle X} = \frac{XY/CY}{XZ/BZ},\] so the desired is now equivalent to \begin{align*} \frac{AK}{AL} = \frac{CY}{BZ} &\iff BZ(AZ-KZ) = CY(AY+YL) \\ &\iff ZA \cdot ZB - ZK \cdot ZB = YA \cdot YC + YC \cdot YL \\ &\iff \operatorname{pow}(Z,(ABC)) - \operatorname{pow}(Z,(BYP)) = \operatorname{pow}(Y,(ABC)) - \operatorname{pow}(Y,(CZP)) \\ &\iff \operatorname{pow}(Z,(ABC)) - \operatorname{pow}(Y,(ABC)) = \operatorname{pow}(Z,(BYP)) - \operatorname{pow}(Y,(CZP)) \\ &\iff ZP^2 - YP^2 = ZY \cdot ZP + ZY \cdot YP = (ZP-YP)(ZP+YP). \quad \blacksquare \end{align*}

Let $D,E,F=\ell\cap(BCP),(ACP),(ABP)$ respectively. Let $I$ be a point circle coaxial to $\omega,\ell$ so we have $IP\perp\ell$. Then $XI^2=XB\cdot XC=XP\cdot XD$ so $\triangle XPI\sim\triangle XID$ so $\angle XID=90^\circ$ and similarly $\angle YIE=\angle ZIF=90^\circ$. Thus $(D,X),(E,Y),(F,Z)$ are pairs of an involution on $\ell$ by projecting the perpendicular involution from $I$. Now call $Q=BE\cap CF$, and projecting DDIT from $Q$ to complete quadrilateral $\triangle ABC\cup\ell$ onto line $\ell$ gives $(AQ\cap\ell,X),(E,Y),(F,Z)$ are pairs of an involution, so $AQ\cap\ell=D$ and $AD,BE,CF$ concur. Inversion at $P$ finishes.