Inverting a G2 is a bit overkill, but I hope that someone finds this solution informative. We invert about $P$. The new problem is as follows: Quote: Let $PBC$ be a triangle and $\Gamma=(PBC)$. The $P$-Apollonius circle intersects $\Gamma$ again at $M$ and the tangent to $\Gamma$ at $P$ again at $A$. Points $X$ and $Y$ lie on segments $PB$ and $PC$ respectively so that $\angle PMX=\angle PMY$. Show that $A, X, Y$ are collinear. [asy][asy] size(12cm); pair P=dir(125); pair B=dir(200); pair C=dir(-20); pair A= 2*extension(B, C, P, P+dir(90)*P)-P; pair M=2*foot(origin, A, -P)+P; pair X=P+0.6*(B-P); pair Y=extension(A, X, P, C); pair Z=extension(A, X, P, M); draw(anglemark(P, M, X, t=10),green); draw(anglemark(Y, M, P, t=10),green); draw(unitcircle); draw(P--B--C--cycle); draw(M--A); draw(circumcircle(P, A, M), dotted); draw(A--P--M); draw(X--M--Y, red); draw(A--Y, dashed+blue); string[] names = {"$P$", "$B$", "$C$", "$A$", "$M$", "$X$", "$Y$", "$Z$"}; pair[] pts = {P, B, C, A, M, X, Y, Z}; pair[] labels = {dir(90), B, C, A, M, dir(120), dir(45), Z}; for(int i=0; i<names.length; ++i){ dot(names[i], pts[i], dir(labels[i])); } [/asy][/asy] Let $Z=PM\cap AX$ and $Y'=AX\cap PC$. Note that $\angle PMA=90^{\circ}$ by the inversion. Since $MP$ bisects $\angle XMY$, we have \[(MA, MZ; MX, MY)=-1=(P, M; B, C)\stackrel{P}{=}(A, Z; X, Y').\]Therefore $Y=Y'$, which solves the problem.

Let $S = \overline{XM} \cap \overline{PY}$, $T = \overline{YM} \cap \overline{PX}$, so that quadrilateral $XYST$ is cyclic by the given angles. We denote by $O$ the center of this circle. [asy][asy]size(250); defaultpen(fontsize(11pt)); pair D = dir(130); pair E = -conj(D); pair M = midpoint(D--E); pair X = dir(230); pair Y = dir(25); pair S = -X+2*foot(origin, X, M); pair T = -Y+2*foot(origin, Y, M); pair B = extension(X, T, D, E); pair C = extension(Y, S, D, E); pair P = extension(X, T, S, Y); pair A = 1/conj(M); pair O = origin; markscalefactor *= 0.6; draw(anglemark(M, X, P), lightblue); draw(anglemark(P, Y, M), lightblue); filldraw(A--B--C--cycle, invisible, red); draw(P--X--S, lightblue); draw(P--Y--T, lightblue); draw(D--E, red); filldraw(unitcircle, invisible, heavycyan); draw(O--A, red); draw(P--A, dotted+red); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$M$", M, dir(295)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$S$", S, dir(S)); dot("$T$", T, dir(T)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$P$", P, dir(P)); dot("$A$", A, dir(A)); dot("$O$", O, dir(-90)); /* TSQ Source: D = dir 130 E = -conj(D) M = midpoint D--E R295 X = dir 230 Y = dir 25 S = -X+2*foot origin X M T = -Y+2*foot origin Y M B = extension X T D E C = extension Y S D E P = extension X T S Y A = 1/conj(M) O = origin R-90 !markscalefactor *= 0.6; anglemark M X P lightblue anglemark P Y M lightblue A--B--C--cycle 0.1 lightred / red P--X--S lightblue P--Y--T lightblue D--E red unitcircle 0.1 lightcyan / heavycyan O--A red P--A dotted red */ [/asy][/asy] Claim: We have $\overline{OM} \perp \overline{BC}$. Hence $\overline{PA} \perp \overline{AOM}$. Proof. Follows by butterfly theorem if one extends $\overline{BC}$ to a chord of the circle, since then $M$ is the midpoint of that chord. $\blacksquare$ Thus, since $M \in \overline{OA}$ and $\overline{PA} \perp \overline{AOM}$, we conclude $A$ coincides with the Miquel point of quadrilateral $STXY$. Therefore $PAXY$ is cyclic.

Take $X$ and $Y$ as the intersection points of a circle $\Omega$ passing through $A$ and $P$. I shall prove that $\angle{PXM}=\angle{PYM}$. Let $Q$ be the antipodal of $P$ in $\Omega$. Then $Q \in AM$ and $\angle{CYQ}=\angle{CMQ}=\angle{BMQ}=\angle{BXQ}=90^{\circ}$, implying that both $CMQY$ and $BXQM$ are cyclic. Therefore, $=\angle{PYM}=\angle{CYM}=\angle{CQM}=\angle{BQM}=\angle{BXM}=\angle{PXM}.$ The conclusion follows.

Let $X_1, Y_1$ be reflections of $B,C$ across $X,Y$ respectively. The angle condition gives $$\angle BX_1C = \angle BXM = \angle CYM = \angle BY_1C\implies B,C,X_1,Y_1\text{ concyclic.}$$Let $P_1$ be the reflection of $P$ across $A$. Observe that $\triangle PX_1Y_1\stackrel{-}{\sim}\triangle PCB\stackrel{-}{\sim}\triangle P_1BC$. Thus Gliding Principle gives $\triangle AXY\stackrel{-}{\sim}\triangle PCB$, hence we can conclude that $$\angle AXY = \angle PCB = \angle YPA\implies A,P,X,Y\text{ concyclic}$$as desired.

Perform inversion around the circle with center $M$ and radius $MC.$ We obtain the following equivalent problem: "$M$- midpoint of $BC$ $P$ a random point (not on $BC$). Let $\omega_1=(PBM)$ and $\omega_2=(PCM)$. Let $D$ be the antipode of $M$ wrt $\omega_2$. $A$ is a point on $PD$ such that $AB=AC$.Let $X \in \omega_1$ and $Y \in \omega_2$ such that $\angle XPM=\angle YPM$. Prove that $APXY$ is cyclic" $\angle XBC=\angle YCB$ $\Rightarrow$ $BX,CY$, and $AM$ are concurrent at $L.$ $\angle PMB=\angle PAM=180-\angle PMC=180-\angle PYC=\angle PYL$ $\Rightarrow$ $APLY$ is cyclic. Symmetrically, $APXL$ is also cyclic. This implies that $APXY$ is cyclic.

Easy for a G2 [asy][asy] unitsize(2.7cm); defaultpen(fontsize(10pt)); pair P = dir(110); pair B = dir(230); pair C = dir(310); pair P1 = dir(70); pair A = (P+P1)/2; pair M = (B+C)/2; pair Q = 1.4*M-0.4*A; pair X = foot(Q,P,B); pair Y = foot(Q,P,C); dot(P);dot(B);dot(C);dot(A);dot(Q);dot(X);dot(Y);dot(M); draw(X--P--Y); draw(P--B--C--cycle,linewidth(1.5)); draw(X--Q--A--P); draw(Y--Q,dashed); draw(X--M--Y); draw(B--Q--C); draw(circumcircle(A,X,Y),dotted); markscalefactor = 0.02; draw(anglemark(M,X,P)); draw(anglemark(P,Y,M)); draw(anglemark(M,Q,B)); draw(anglemark(C,Q,M)); label("$A$",A,N); label("$B$",B,W); label("$C$",C,NE); label("$P$",P,dir(P)); label("$Q$",Q,S); label("$X$",X,SW); label("$Y$",Y,SE); label("$M$",M,2*NE); [/asy][/asy] Let $Q$ be the point on $AM$ such that $\angle QXB=90^{\circ}$. Notice the cylic quadrilateral $BMQX$, thus $$\angle CYM = \angle BXM = \angle BQM = \angle CQM\implies CYQM\text{ concyclic.}$$Thus $\angle CYQ=90^{\circ}$ or $A,X,Y,P,Q$ lies on circle with diameter $PQ$.

Daniel Zhu notes that this problem and USA TSTST 2019/5 can be simultaneously generalized as follows: Quote: Let $ABC$ be a triangle with point $M$ on segment $BC$. Points $K$, $X$, and $Y$ are chosen so that $X$ and $Y$ lie on the extensions of $AB$ and $AC$ respectively, $K$ lies inside $ABC$, and \[\angle BXM=\angle CYM=\angle KBC=\angle KCB.\]Line $MK$ meets the parallel from $A$ to $BC$ at $P$. Show that $APXY$ is cyclic. If you found the angle chasing solution to either problem, then this should be a breeze.

The next solution drew the following reaction from graders when I submitted a version of it. So consider yourself warned

liberator wrote:

The next solution drew the following reaction from graders when I submitted a version of it. So consider yourself warned

What does $\text{[D]DIT}$ mean?

I only used MPT!!!!

Nineth problem in this pdf

A slightly different solution : Let XM intersect PC at Z and YM intersect PB at W. Clearly XYZW is cyclic. CLAIM : A is the Miguel point wrt. XYZW. Proof: We prove XY and ZW intersect on PA. Let PM intersect ZW at Q and let XY intersect ZW at R. Then -1=(Z,Y;Q,R)=(PC,PB;PM,PR). So PR parallel to BC. So R lies on PA. So A is the foot of altitude from M to PR. Thus A is the Miquel point of (XYZW) [Since (XYZW) is cyclic]. XW intersects YZ at P so A lies on (PXY) proving the result.

Generalisation of G2 wrote: Let $ABC$ be a triangle with point $M$ on segment $BC$. Points $K$, $X$, and $Y$ are chosen so that $X$ and $Y$ lie on the extensions of $AB$ and $AC$ respectively, $K$ lies inside $ABC$, and \[\angle BXM=\angle CYM=\angle KBC=\angle KCB.\]Line $MK$ meets the parallel from $A$ to $BC$ at $P$. Show that $APXY$ is cyclic. $\widehat{AYQ}+\widehat{AXQ}=\widehat{BMQ}+\widehat{CMQ}=180^{\circ}\implies Q\in\odot(AXY)$. Note, $K'\cap\odot(BQC)$. $\widehat{K'QC}=\widehat{MYC}=\widehat{MXB}=\widehat{K'QB}\implies K'\equiv K$. $M$,$P$,$Q$ are collinear. Now, $\widehat{APQ}=\widehat{PMB}=\widehat{AXQ}$. Hence, $APXY$ is cyclic.

aops29 wrote:

After a quick glance, this is the roughly the same as my solution (and the official shortlist solution too). However, you might need to be a bit more careful about the case that $\odot(BXM)$ and $\odot(CYM)$ are tangent.

@above that case can be handled separately, I guess

$$\mathbf{SOLUTION}$$ Consider an inversion around the circle centred at $M$ and radius $MB$. Then we have the following inverted problem: Inverted Problem wrote: Let $A^*BC$ be an isosceles triangle with $A^*B=A^*C$ and $M$ be the midpoint of $BC$. Let $\omega$ be the circle with diameter $MA^*$. Let $P^*$ be any point on $\omega$ such that $P^*B<P^*C$. Let $X^*$ be any point on $\odot (P^*BM)$ not lying in the region of $\triangle ABC$. Let $Y^*$ be a point on $\odot (P^*CM)$ such that $P^*M$ lies on angle bisector of $\angle X^*P^*Y^*$. Then prove that $A^*P^*X^*Y^*$ is concyclic. Let $T=BX^*\cap CY^*$. Then $TB=TC$. Clearly $T-M-A^*$. Also $\measuredangle P^*X^*B=\measuredangle P^*Y^*C=\measuredangle P^*Y^* T$ which implies $P^*X^*TY^*$ is concyclic. Now, $\measuredangle A^*TC=90^\circ -\measuredangle TCM=90^\circ -\measuredangle Y^*P^*M=\measuredangle A^*P^*Y^*$. Hence $P^*A^*Y^*T$ is concyclic. $\therefore A^*P^*X^*Y^*$ is concyclic. [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 = 7.667462578894224, xmax = 54.00025977272606, ymin = -62.97441228572777, ymax = -11.513636235807901; /* image dimensions */ pen qqccqq = rgb(0,0.8,0); pen ccqqqq = rgb(0.8,0,0); pen ubqqys = rgb(0.29411764705882354,0,0.5098039215686274); /* draw figures */ draw((19.06967438831378,-40.83360946002954)--(40.788173072922454,-40.893938623042345), linewidth(1)); draw((29.956308872027144,-31.00512313428571)--(19.06967438831378,-40.83360946002954), linewidth(1)); draw((29.956308872027144,-31.00512313428571)--(40.788173072922454,-40.893938623042345), linewidth(1)); draw(circle((29.94261630132263,-35.93444858791083), 4.929344471047745), linewidth(1) + qqccqq); draw(circle((24.506816321799704,-38.142477310631804), 6.066688112173986), linewidth(1) + ccqqqq); draw(circle((35.378416280845556,-33.72641986518985), 8.979910561558464), linewidth(1) + blue); draw(circle((30.828114623730084,-38.11439838352692), 7.16253026787154), linewidth(1) + dotted + ubqqys); draw((29.91682665685946,-45.21872059466098)--(40.788173072922454,-40.893938623042345), linewidth(1)); draw((29.91682665685946,-45.21872059466098)--(19.06967438831378,-40.83360946002954), linewidth(1)); draw((29.956308872027144,-31.00512313428571)--(29.91682665685946,-45.21872059466098), linewidth(1) + linetype("2 2")); draw((26.49537898620796,-32.41095716497108)-- (26.546217827874187,-43.85610541783429),linewidth(1)); draw((26.49537898620796,-32.41095716497108)-- (36.37047814841509,-42.65136283108624),linewidth(1)); draw((29.92892373061812,-40.86377404153595)-- (26.49537898620796,-32.41095716497108),linewidth(1)); draw((26.49537898620796,-32.41095716497108)--(29.956308872027144,-31.00512313428571), linewidth(1) + linetype("2 2")); /* dots and labels */ dot((19.06967438831378,-40.83360946002954),dotstyle); label("$B$", (18.310991040364986,-40.23031782990152), W* labelscalefactor); dot((40.788173072922454,-40.893938623042345),dotstyle); label("$C$", (41.52948972497366,-40.290646992914326), SE * labelscalefactor); dot((29.956308872027144,-31.00512313428571),dotstyle); label("$A^*$", (30.17024038266932,-30.39666425881484), NE * labelscalefactor); dot((29.92892373061812,-40.86377404153595),linewidth(4pt) + dotstyle); label("$M$", (30.17024038266932,-40.41130531893993), NE * labelscalefactor); dot((29.92892373061812,-40.86377404153595),linewidth(4pt) + dotstyle); dot((26.49537898620796,-32.41095716497108),dotstyle); label("$P^*$", (26.731478090939614,-31.78423500810928), NE * labelscalefactor); dot((26.546217827874187,-43.85610541783429),dotstyle); label("$X^*$", (26.791807253952417,-43.246775980541614), NE * labelscalefactor); dot((36.37047814841509,-42.65136283108624),linewidth(4pt) + dotstyle); label("$Y^*$", (36.62546082503913,-43.16085104631118), SE * labelscalefactor); dot((29.91682665685946,-45.21872059466098),linewidth(4pt) + dotstyle); label("$T$", (30.17024038266932,-44.75500505586165), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Somebody please check my solution.

Me: ihatemath123, you should do 2018 g2 ihatemath123: I think i've already done that, isnt it a one liner? Me: What????? Is this the problem ur thinking of [proceeds to read problem] ihatemath123: yeah Me: AAAAAAAAAAAAA Projective > elem Let $D=YM\cap PX$ and $E=XM\cap PY$. Note $DXYE$ is cyclic. Claim: $AP$, $DE$, and $XY$ concur at $T$. Proof. Note that \[-1=(BC;M\infty)\overset{P}{=}(X,E;M,XM\cap AP)\]and \[-1=(CB;M\infty)\overset{P}{=}(Y,D;M,DY\cap AP).\]The prism lemma finishes. $\blacksquare$ Then, we can say the following which finishes. Claim: $A$ is the miquel point of $DXYE$. Proof. It suffices to show $M$ inverts to $A$. By Brokard, the polar of $M$ is line $PT$ or $AP$, so the inverse of $M$ lies on $AP.$ Moreover, if $BC$ hits $(DXYE)$ at $P,Q$ the butterfly theorem gives $M$ is the midpoint of $PQ$. Thus $OM$ is the perpendicular bisector $BC$ so $M$ is sent to $OM\cap AP$, which is $A$ as $AB=AC$. $\blacksquare$ This finishes as $APDE$ is cyclic so $TY\cdot TX=TE\cdot TD=TA\cdot TP$.

The angle condition tells us that $(BMX)$ and $(CMY)$ congruent by LOS, so if we suppose they intersect for a second time at $N$, the two circles are symmetric about $MN$. From here, we can see \[\angle PAN = \angle PYN = \angle PXN = 90 \implies APXYN \text{ cyclic.} \quad \blacksquare\]

Solved with cursed_tangent1434(who also did the asy). Let $Z = (BXM) \cap AM$. Note that $Z \in (PAX)$ since $\measuredangle XZA = \measuredangle XZM = \measuredangle XBM = \measuredangle XPA$. We also have $Z \in (CYM)$ since $\angle CYM = \angle BXM = \angle BZM = \angle MZC$. Since $\angle ZMC = \angle ZYC = \angle ZYP = ZAP = 90^\circ$ we have $Z \in (PAY)$ and it follows that $X$, $Y \in (PAZ)$ as desired. [asy][asy] import geometry; size(10cm); defaultpen(fontsize(9pt)); pen pri; pri=RGB(24, 105, 174); pen sec; sec=RGB(217, 165, 179); pen tri; tri=RGB(126, 123, 235); pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair A,B,C,X,Y,P,Z,M; pair A = dir(90); pair B = dir(210); pair C = dir(330); pair M = midpoint(B--C); P=A-1; Z=(0, -1); pair X = intersectionpoints(circle(B,M,Z),line(P,B))[0]; pair Y = intersectionpoints(circle(C,M,Z),line(P,C))[1]; filldraw((path)(A--B--C--cycle), white+0.1*pri, pri); draw(A--P); draw(circumcircle(B,M,Z), sec+dashed); draw(circumcircle(Z,M,C), sec+dashed); draw(P--X); draw(P--C); draw(A--Z); filldraw(circumcircle(A,X,Y), tfil, tri); dot("$A$",A,dir(90)); dot("$B$",B,dir(210)); dot("$C$",C,dir(260)); dot("$M$",M,dir(210)); dot("$P$",P,dir(180)); dot("$X$",X,dir(260)); dot("$Y$",Y,dir(260)); dot("$Z$",Z,dir(260)); [/asy][/asy]

Hmmmm By the Law of Sines, note that $(BMX)$ and $(CMY)$ have the same radius, so their radical axis by symmetry is just the perpendicular bisector of $BC,$ which $A$ lies on. Let the two circles intersect again at point $N.$ Note that $\measuredangle PXN = \measuredangle BXN = \measuredangle BMN = 90^\circ$ and similarly $\measuredangle PYN = 90^\circ.$ We also know that $\measuredangle PAN = 90^\circ$ since $AP \parallel BC.$ Thus $$\measuredangle PXN = \measuredangle PAN = \measuredangle PYN = 90^\circ,$$which implies the result.

Let $(BMX)$ meet line $AM$ at $G$. Then $\angle BXM = \angle BGM = \angle MGC = \angle MYC$, so $MGYC$ is cyclic. We have that $\angle PXG = \angle PAG = 90^\circ$, so $PAGX$ is cyclic. Similarly, $\angle PXG = \angle PAG = 90^\circ$, so $PAYG$ is cyclic. Therefore, $X$ and $Y$ both lie on the circumcircle of $PAG$, so $PAXY$ is cyclic.

Claim: The circumcircles of $\Delta{BMX}$ and $\Delta{CMY}$ intersect on $\overrightarrow{\rm AM}$. Proof: Let $D'$ be the second intersection of these circles. We have to show that $A$ is on $MD'$, the radical axis, so it would suffice to show that $A$ has equal powers with respect to $(BMX)$ and $(CMY)$. However, by the extended law of sines in $\Delta{BMX}$ and $\Delta{CMY}$, they have equal circumradii. Hence it is only required to show that $AE^2 = AF^2$ where $E$ and $F$ are the circumcenters of $\Delta{BMX}$ and $\Delta{CMY}$. Notice that $ME = MF$ (equal circumradii) and $\angle{AME} = 90^{\circ} + \angle{BME} = 90^{\circ} + \angle{CMF} = \angle{AMF}$, so $\Delta{AME} \cong \Delta{AMF}$by SAS, hence $AE = AF$ as desired. Now let us call this intersection $D$. Let $\measuredangle$ denote directed angles modulo $180^{\circ}$. Notice that $\measuredangle{ADX} = \measuredangle{MDX} = \measuredangle{MBX} = \measuredangle{MBP} = \measuredangle{APX}$, hence $A, P, X, D$ are concyclic. Similarly, $A, P, Y, D$ are concyclic. Hence, $A, P, X, Y$ are concyclic as desired.

Neothehero wrote: Let $ABC$ be a triangle with $AB=AC$, and let $M$ be the midpoint of $BC$. Let $P$ be a point such that $PB<PC$ and $PA$ is parallel to $BC$. Let $X$ and $Y$ be points on the lines $PB$ and $PC$, respectively, so that $B$ lies on the segment $PX$, $C$ lies on the segment $PY$, and $\angle PXM=\angle PYM$. Prove that the quadrilateral $APXY$ is cyclic. Make the circles $(BXM),(CYM)$ then since $\angle PXM=\angle PYM$ and $MB=MC$ we get that the two circle are equal. Let $F=(MBX)\cap (MCY)$ then we have that $\angle MBF=\angle MCF$ so $A,M,F$ are collinear. Now $\angle PXF=180-\angle BMF=180-90=90=\angle PYF=\angle PAF$ so $A,P,X,Y,F$ lie on the same circle.

construct circles $BXM$ and $CYM$ they intersect on $AM$ since $BXM=CYM$ and $BM=CM$ let that point be $O$ we also have that $PAM=AMB=AMC=CYO=BXO$ since cyclic quadrilaterals exist and $BMOX$ and $CMOY$ are cyclic then, $APXO$ and $APYO$ are cyclic, and $X$ and $Y$ both lie on circle $APO$ thus, $APXY$ are concyclic

Let $K= \overline{MY} \cap \overline{PB}, L=\overline{MX} \cap \overline {PC}.$ Then $A$ is the Miquel point of cyclic quadrilateral $YXKL$ by butterfly. Therefore $APXY$ is also cyclic$.\blacksquare$

Let $P = XM \cap PC \text{ } Q = YM \cap PB \text{ and } Z = (BXM) \cap AM$ By the given angle We get that $XPQY$ is cyclic. Observe that by reim's we have $PAZX$ cyclic. Also, as $ABC$ is isosceles so is $BZC$ then we have \[ \measuredangle MYC = \measuredangle BXM = \measuredangle BZM = \measuredangle MZC\]Where the last equality is due to the isosceles condition. So $YCMZ$ is cyclic hence \[ \measuredangle ZYP = \measuredangle ZYC = \measuredangle ZMC = 90^{\circ} = \measuredangle ZAP\]Hence we get that $ZYAP$ is cyclic. Combining this with the above para we get that $PAYZX$ is cyclic proving the required result.

Inversion + Harmonics! What else can one ask for? Invert about $P$ with radius $\sqrt{PB\cdot PC}$ then reflect about the angle bisector of $BPC$. Observe that $M*$ is the intersection of the symmedian and the circumcircle of $(PB^*C^*)$, lines $X^*M^*$ and $Y^*M^*$ are reflections over the symmedian and $X^*\in PB^*$ and $Y^* \in PC^*$. Let $\ell$ be the line through $M^*$ perpendicular to $PM^*$. Observe that we need to prove that we need to prove that $X^*Y^*\cap PP$ is the same as $X^*Y^* \cap \ell$. Observe that $(X^*Y^*\cap PP, X^*Y^* \cap PM^*; X^*, Y^*)\overset{P}{=} (B^*C^* \cap PP, BC \cap PM^*; B^*, C^* )=-1$ (harmonic quadrilateral) and $(X^*Y^*\cap \ell, X^*Y^* \cap PM^*; X^*, Y^*)=-1$ (angle bisector). $\blacksquare$ Edit: As it turns out this question could have been angle chased +Pop. But I think I have forgotten to do both of those things.

Here are few proofs: Proof-1: (Angle-Chase) Let $Z = (BMX) \cap (CMY)$ with $Z \neq M$. Note that: $(BMZX)$ is cyclic and $AP \parallel BM$ which implies $APXZ$ to be cyclic. Similarly, $APYZ$ is also cyclic. Therefore, $APXYZ$ is cyclic and we are done. Proof-2: (Config-Geo) Let $S=MX \cap CY$ and $T=MY \cap BX$. Note that: $STXY$ is cyclic. Extend $BC$ into chord $B'C'$ of circle $(STXY)$ and $O$ denote the center of $(STXY)$. As $M$ is midpoint of $BC$, by Butter-fly theorem $M$ is the midpoint of $B'C'$. Therefore, $OM \perp BC$ and $OM \perp AP$. Notice that: $A$ is the Miquel point of quadrilateral $STXY$ which implies $APXY$ to be cyclic and we are done.

Let $(BXM), (CYM)$ meet at $K$. I claim $(PK)$ contains $A,X,Y$, it is easy to see that $\angle PXM = \angle PYM$ implies $\angle BXM = \angle CYM$ implies $\angle BKM = \angle CKM$, so $K$ lies on the perpendicular bisector of $BC$, and so does $A$, so $AK \perp BC$ and thus $AK \perp PA$, as desired. Likewise, this implies $KM \perp BC$, so $\angle PXK = \angle BXK = 180 - \angle BMK = 90$, likewise $\angle PYK = 90$, so we are done.

Solution: Let $PB \cap \odot (APY)=X’ \neq P$. Let $AM \cap \odot (APY)= N \neq A$. Note that $\angle PAN=90=\angle PX’N=\angle BMN$, so $BMX’N$ must be cyclic. Also, $\angle PYN=\angle CYN=\angle CMN$, which gives us that $YCMN$ is also cyclic. Now finally note that $ \angle PX’M=\angle BX’M=\angle BNM=\angle CNM=\angle CYM=\angle PYM=\angle PXM$, so $X’ \equiv X$, and we are done. $\blacksquare$

Nice problem, took me some time