[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -10.169, xmax = 8.2, ymin = -3, ymax = 7.3; /* image dimensions */ draw((-3.142967880809327,3.741704410289035)--(-4.27,-1.08)--(0.99,-1.08)--cycle, linewidth(1) + blue); /* draw figures */ draw((-3.142967880809327,3.741704410289035)--(-4.27,-1.08), linewidth(1) + blue); draw((-4.27,-1.08)--(0.99,-1.08), linewidth(1) + blue); draw((0.99,-1.08)--(-3.142967880809327,3.741704410289035), linewidth(1) + blue); draw(circle((-1.64,0.8478292787166926), 3.2608933941294866), linewidth(1)); draw(circle((-3.7362180134430876,0.22962892231873314), 1.4142316370900383), linewidth(1)); draw(circle((-0.16375861470797104,0.7966499870857753), 2.20294668910573), linewidth(1)); draw((-1.1069556946477261,-2.3692018831480532)--(-4.27,-1.08), linewidth(1)); draw((-1.1069556946477261,-2.3692018831480532)--(0.99,-1.08), linewidth(1)); draw(circle((-5.097700151769647,3.090432973082939), 2.060371989727628), linewidth(1)); draw(circle((-0.6303772705960715,4.578841302197107), 2.648378702211095), linewidth(1)); draw((-2.605842442388212,2.1295710231108576)--(-3.142967880809327,3.741704410289035), linewidth(1)); draw((-2.605842442388212,2.1295710231108567)--(xmax, -6.300399792383099*xmax-14.288278159894903), linewidth(1)); /* ray */ draw((-4.815305376023,1.5900297212366055)--(-1.64,-2.4130641154127943), linewidth(1)); /* dots and labels */ dot((-3.142967880809327,3.741704410289035),dotstyle); label("$A$", (-3.051170840703382,3.956205358613116), NE * labelscalefactor); dot((-4.27,-1.08),dotstyle); label("$B$", (-4.173819337640587,-0.8582295417137585), NE * labelscalefactor); dot((0.99,-1.08),dotstyle); label("$C$", (1.0724034461236613,-0.8582295417137585), N * labelscalefactor); dot((-1.64,-2.4130641154127943),linewidth(4pt) + dotstyle); label("$M_{A}$", (-1.5615026428443977,-2.2399507687133995), S *6* labelscalefactor); dot((-2.339470425047464,0.4513208203605414),linewidth(4pt) + dotstyle); label("$I$", (-2.2523632563442164,0.6314386561452294), NE * labelscalefactor); dot((0.8358349509755364,2.970013771970114),linewidth(4pt) + dotstyle); label("$M_{B}$", (0.921277686920576,3.135808380082079), N * labelscalefactor); dot((-4.815305376023,1.5900297212366055),linewidth(4pt) + dotstyle); label("$M_{C}$", (-4.73514358610919,1.7540871530824378), W * 2*labelscalefactor); dot((-2.605842442388212,2.1295710231108576),dotstyle); label("$X$", (-2.5114359864066484,2.3370007957229113), SW * 4*labelscalefactor); dot((-3.6341824169645403,1.640174866411884),linewidth(4pt) + dotstyle); label("$M$", (-3.5477269066563766,1.818855335598046), NW * labelscalefactor); dot((-1.841899916754811,2.2238205744391006),linewidth(4pt) + dotstyle); label("$N$", (-1.7558071903912218,2.4017689782385196), N * labelscalefactor); dot((-4.895223969272359,1.040033955434055),linewidth(4pt) + dotstyle); label("$P$", (-4.799911768624798,1.214352298785703), W * 2*labelscalefactor); dot((1.1171955142980874,2.5888917772536733),linewidth(4pt) + dotstyle); label("$Q$", (1.2019398111548774,2.7687886791602994), N * labelscalefactor); dot((-1.1069556946477261,-2.3692018831480532),linewidth(4pt) + dotstyle); label("$Y$", (-1.0217677885476644,-2.196771980369661), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] First note $Y \in (ABC)$ by a simple angle chase:$$\angle ABY + \angle ACY = (180^\circ - \angle MIB)+(180^\circ - \angle NIC) = \angle BIC + \angle MIN = (90^\circ + \frac{\angle A}{2})+\angle MBI + \angle NCI = 180^\circ.$$Key Claims: $MNPQ$ is cyclic and $(MPA),(NQA)$ are tangent at $A$. Proof: Invert about the incircle: $I$ is now the orthocenter of $\triangle ABC$, and $\overline{BPM} \parallel \overline{CQN}$. Since $$\angle APM = 180^\circ - \angle APB = \angle ACB = 180^\circ - \angle AIB = \angle AMP$$we have $AM=AP$, and similarly $AN=AQ$. The claim follows (since $MP \parallel NQ$). Return to the uninverted picture. Since $X$ is the radical center of the circles $\omega_B, \omega_C, (MPA), (NQA)$ we have $XA=XI$. Let $M_A$ be the midpoint of $\widehat{BC}$ not containing $A$, and define $M_B, M_C$ similarly. Then the desired result follows from Pascal on $AYBM_BM_CM_A$.

[asy][asy] size(6cm); 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 = -4.68, xmax = 13.16, ymin = -3.5, ymax = 6.98; /* image dimensions */ pen zzttff = rgb(0.6,0.2,1); pen ffvvqq = rgb(1,0.3333333333333333,0); /* draw figures */ draw(circle((-1.0964533667921372,2.808556043496445), 3.1646044581100514), linewidth(0.8) + zzttff); draw(circle((4.863546633207864,2.4885560434964464), 3.164604458110053), linewidth(0.8) + zzttff); draw(circle((3.276453366792137,0.8314439565035552), 3.1646044581100523), linewidth(0.8) + ffvvqq); draw((-0.35870469944593764,5.885965389053015)--(0.24,-0.06), linewidth(0.8)); draw((0.24,-0.06)--(6.2,-0.38), linewidth(0.8)); draw((6.2,-0.38)--(5.601295300554062,5.565965389053019), linewidth(0.8)); draw((5.601295300554062,5.565965389053019)--(-0.35870469944593764,5.885965389053015), linewidth(0.8)); draw((1.94,3.7)--(1.827093266415726,1.5971120869928908), linewidth(0.8)); draw((0.24,-0.06)--(1.94,3.7), linewidth(0.8)); draw((1.94,3.7)--(6.2,-0.38), linewidth(0.8)); draw((0.24,-0.06)--(1.827093266415726,1.5971120869928908), linewidth(0.8)); draw((1.827093266415726,1.5971120869928908)--(6.2,-0.38), linewidth(0.8)); draw((-0.35870469944593764,5.885965389053015)--(1.827093266415726,1.5971120869928908), linewidth(0.8)); draw((1.827093266415726,1.5971120869928908)--(5.601295300554062,5.565965389053019), linewidth(0.8)); /* dots and labels */ dot((1.94,3.7),dotstyle); label("$A'$", (1.76,4.26), NE * labelscalefactor); dot((0.24,-0.06),dotstyle); label("$B'$", (-0.4,-0.48), NE * labelscalefactor); dot((6.2,-0.38),dotstyle); label("$C'$", (6.28,-0.18), NE * labelscalefactor); dot((1.827093266415726,1.5971120869928908),linewidth(4pt) + dotstyle); label("$I$", (1.5,1.54), NE * labelscalefactor); dot((-0.35870469944593764,5.885965389053015),dotstyle); label("$M'$", (-0.28,6.08), NE * labelscalefactor); dot((0.12323437020220891,1.0996441351657311),linewidth(4pt) + dotstyle); label("$P'$", (0.2,1.26), NE * labelscalefactor); dot((5.601295300554062,5.565965389053019),linewidth(4pt) + dotstyle); label("$N'$", (5.68,5.72), NE * labelscalefactor); dot((5.899799138628696,2.6014095882821118),linewidth(4pt) + dotstyle); label("$Q'$", (5.98,2.76), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Claim. $M,N,P,Q$ are concyclic. Proof. Consider the inversion at $I$ which preserves $(ABC)$, let $X'$ be the image of an object $X$. Then $I$ is the orthocenter of $A'B'C'$. Let $A=\angle B'A'C'$, $B=\angle A'B'C'$ and $C=\angle A'C'B'$. Notice that $B'M'\|C'N'$. Hence $$\angle M'IB'+\angle N'IC'=360^{\circ}-\angle B'M'I-\angle M'B'I-\angle IN'C'-\angle IC'N'=360^{\circ}-A-(180^{\circ}-A)=180^{\circ}$$hence $$M'B'=\frac{B'I\sin\angle M'IB'}{\sin\angle B'M'I}=\frac{C'I\sin\angle C'IN'}{\sin\angle C'N'I}=N'C'$$which implies $M'N'C'B'$ is a parallelogram, hence $M'N'=BC=P'Q'$, which implies $M'N'Q'P'$ is an isoceles trapezoid, which is indeed cyclic, invert back yields the result. $\square$ [asy][asy] 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 = -9.871875000000001, xmax = 15.215624999999994, ymin = -9.28312499999999, ymax = 5.454374999999993; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); pen ffvvqq = rgb(1,0.3333333333333333,0); pen zzttff = rgb(0.6,0.2,1); pen fuqqzz = rgb(0.9568627450980393,0,0.6); pen zzttqq = rgb(0.6,0.2,0); /* draw figures */ draw(circle((-4.26,-1.6), 2.1883326986543885), linewidth(0.8) + qqwuqq); draw(circle((2.1093749999999964,-0.6768749999999999), 4.24758974154406), linewidth(0.8) + qqwuqq); draw((-5.38,-3.48)--(4.663783792716652,-4.070547690782389), linewidth(0.8)); draw((-5.38,-3.48)--(-2.09429466218645,-1.286120421553271), linewidth(0.8)); draw((-2.09429466218645,-1.286120421553271)--(4.663783792716652,-4.070547690782389), linewidth(0.8)); draw(circle((-0.23732540859718557,-1.7210531721695757), 5.435162913491629), linewidth(0.8) + blue); draw(circle((-6.5227070845296256,11.380965686923354), 11.108458046184419), linewidth(0.8) + linetype("4 4") + ffvvqq); draw((-3.1794087991084052,2.8489732193814428)--(-5.38,-3.48), linewidth(0.8)); draw((-3.1794087991084052,2.8489732193814428)--(4.663783792716652,-4.070547690782389), linewidth(0.8)); draw((-5.264927145060398,0.34394481226210705)--(-2.402847685913124,0.8428327705919746), linewidth(0.8) + zzttff); draw((-2.402847685913124,0.8428327705919746)--(1.2985163948385723,3.492600618696674), linewidth(0.8) + zzttff); draw((-5.264927145060398,0.34394481226210705)--(-1.7609218532430773,-6.938298591934416), linewidth(0.8) + linetype("4 4") + fuqqzz); draw((-1.7609218532430773,-6.938298591934416)--(1.2985163948385723,3.492600618696674), linewidth(0.8) + linetype("4 4") + fuqqzz); draw((-1.7609218532430773,-6.938298591934416)--(-3.1794087991084052,2.8489732193814428), linewidth(0.8)); draw((0.6643662859899618,-7.08089906399402)--(-3.1794087991084052,2.8489732193814428), linewidth(0.8) + linetype("4 4") + zzttqq); /* dots and labels */ dot((-5.38,-3.48),dotstyle); label("$B$", (-5.990625000000002,-3.995624999999996), NE * labelscalefactor); dot((-4.26,-1.6),dotstyle); label("$O$", (-4.134375000000002,-1.32375), NE * labelscalefactor); dot((2.1093749999999964,-0.6768749999999999),dotstyle); label("$O_{1}$", (2.2218749999999963,-0.3956250000000002), NE * labelscalefactor); dot((-2.09429466218645,-1.286120421553271),linewidth(4pt) + dotstyle); label("$I$", (-1.96875,-1.0706249999999995), NE * labelscalefactor); dot((4.663783792716652,-4.070547690782389),dotstyle); label("$C$", (4.78125,-3.79875), NE * labelscalefactor); dot((-3.1794087991084052,2.8489732193814428),linewidth(4pt) + dotstyle); label("$A$", (-3.0656250000000025,3.06375), NE * labelscalefactor); dot((-3.3680335904423444,-3.5982982570729587),linewidth(4pt) + dotstyle); label("$D$", (-3.2625,-3.3768749999999965), NE * labelscalefactor); dot((-0.825135705191316,-3.747813868422477),linewidth(4pt) + dotstyle); label("$E$", (-0.703125000000003,-3.5175), NE * labelscalefactor); dot((-5.264927145060398,0.34394481226210705),linewidth(4pt) + dotstyle); label("$P$", (-5.146875000000002,0.5606249999999986), NE * labelscalefactor); dot((-3.9720518999683647,0.5693053938272891),linewidth(4pt) + dotstyle); label("$M$", (-3.853125000000002,0.7856249999999984), NE * labelscalefactor); dot((-1.5762320381717025,1.4345982706453033),linewidth(4pt) + dotstyle); label("$N$", (-1.4625,1.6575), NE * labelscalefactor); dot((1.2985163948385723,3.492600618696674),linewidth(4pt) + dotstyle); label("$Q$", (1.40625,3.710624999999995), NE * labelscalefactor); dot((-2.402847685913124,0.8428327705919746),linewidth(4pt) + dotstyle); label("$X$", (-2.278125000000003,1.066874999999998), NE * labelscalefactor); dot((0.6643662859899618,-7.08089906399402),linewidth(4pt) + dotstyle); label("$Y$", (0.7875,-6.864374999999993), NE * labelscalefactor); dot((-1.7609218532430773,-6.938298591934416),linewidth(4pt) + dotstyle); label("$Z$", (-1.659375000000003,-6.72375), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Now we return to the original problem, let $\omega_B$ and $\omega_C$ intersect $BC$ at $D,E$ respectively. Let the isogonal line of $AY$ w.r.t. $\angle BAC$ meet $(ABC)$ at $Z$. Firstly, we have $$\angle BYC=180^{\circ}-\angle YBC-\angle YCB=180^{\circ}-\angle BMD-\angle CNE=180^{\circ}-\angle BAC$$The last equality follows from the fact that $MD,NE$ are both perpendicular to the radical axis of $\omega_B$ and $\omega C$, so $Y$ lies on $(BAC)$. Notice that $$\angle BPD=\angle YBC=\angle BPZ$$hence $P,D,Z$ are collinear. Now by spiral sim. lemma $$\triangle PAM\sim\triangle PCD$$hence $$\angle APM=\angle CPD=\angle CPZ=\angle BAY$$so $AY$ is tangent to $(APM)$, and by symmetry tangent to $(AQN)$ as well. By radical axis on $(APM),(AQN)$ and $(MNPQ)$, $MN\cap PQ=X$ lies on $AY$ as desired.

Let \(\overline{PM}\) and \(\overline{QN}\) intersect \(\Gamma\) again at \(P'\) and \(Q'\). Claim: \(BCNM\) is tangential, and \(Y\in\Gamma\). Proof. If \(\ell\) is the common tangent to \(\omega_B\) and \(\omega_C\) at \(I\), then \begin{align*} \measuredangle MIN&=\measuredangle(\overline{MI},\ell)+\measuredangle(\ell,\overline{NI})\\ &=\measuredangle MBI+\measuredangle ICN=\measuredangle IBC+\measuredangle BCI=\measuredangle BIC, \end{align*}implying that \(BCNM\) is tangential. Moreover \[\measuredangle YBM=\measuredangle BIM=\measuredangle CIN=\measuredangle YCN,\]so \(Y\) lies on \(\Gamma\). \(\blacksquare\) [asy][asy] size(8cm); defaultpen(fontsize(10pt)); pen pri=blue; pen pri2=lightblue; pen sec=heavygreen; pen tri=orange; pen qua=lightred; pen fil=invisible; pen fil2=invisible; pen sfil=invisible; pen tfil=invisible; pen qfil=invisible; pair A,B,C,I,P,M,T,NN,Q,X,Pp,Qp,Y,D; A=dir(115); B=dir(225); C=dir(315); I=incenter(A,B,C); P=dir(200); M=2*foot(circumcenter(B,I,P),A,B)-B; T=reflect(M,I)*foot(I,A,B); NN=extension(A,C,M,T); Q=reflect(origin,circumcenter(C,I,NN))*C; X=extension(P,M,Q,NN); Pp=2*foot(origin,P,M)-P; Qp=2*foot(origin,Q,NN)-Q; Y=2*foot(origin,A,X)-A; D=foot(I,B,C); draw(circumcircle(M,P,Q),tri+linewidth(.3)+dashed); draw(circumcircle(A,M,P),qua+dashed); draw(circumcircle(A,NN,Q),qua+dashed); draw(A--Y,qua+linewidth(1)); draw(Pp--Qp,pri2+linewidth(1)); draw(P--Pp,tri+linewidth(.3)); draw(Q--Qp,tri+linewidth(.3)); filldraw(circumcircle(B,I,M),sfil,sec); filldraw(circumcircle(C,I,NN),sfil,sec); draw(M--NN,pri); filldraw(unitcircle,fil,pri); filldraw(incircle(A,B,C),fil2,pri2); filldraw(A--B--C--cycle,fil,pri); dot("\(A\)",A,NW); dot("\(B\)",B,SW); dot("\(C\)",C,SE); dot("\(M\)",M,NW); dot("\(N\)",NN,dir(80)); dot("\(I\)",I,E); dot("\(T\)",T,dir(75)); dot("\(P\)",P,dir(160)); dot("\(Q\)",Q,dir(-10)); dot("\(P'\)",Pp,Pp); dot("\(Q'\)",Qp,Qp); dot("\(X\)",X,NW); dot("\(Y\)",Y,S); dot("\(D\)",D,S); real t=.2; clip( (-1-t,-1-t)--(1+t,-1-t)--(1+t,1+t)--(-1-t,1+t)--cycle); [/asy][/asy] Claim: \(\overline{YA}\) is tangent to \((AMP)\) and \((ANQ)\). Proof. Note that \[\measuredangle APM=\measuredangle APB+\measuredangle BPM=\measuredangle AYB+\measuredangle YBM=\measuredangle YAM,\]and analogously \(\measuredangle AQN=\measuredangle YAN\). \(\blacksquare\) Claim: \(M\), \(N\), \(P\), \(Q\) are concyclic. Proof. If \(\overline{PM}\) and \(\overline{QN}\) intersect \(\Gamma\) again at \(P'\) and \(Q'\), then By Reim's theorem it will suffice to show \(\overline{MN}\parallel\overline{P'Q'}\). If the incircle touches \(\overline{BC}\) and \(\overline{MN}\) at \(D\) and \(T\), then \begin{align*} \measuredangle(\overline{P'Q'},\overline{BC}) &=\measuredangle Q'P'C+\measuredangle P'CB =\measuredangle NQC+\measuredangle MPB\\ &=\measuredangle NIC+\measuredangle MIB =2\measuredangle MIB=\measuredangle TID=\measuredangle(\overline{MN},\overline{BC}), \end{align*}as required. \(\blacksquare\) By radical axis theorem on \((AMP)\), \((ANQ)\), \((MNPQ)\), the desired conclusion follows, with points \(A\), \(P\), \(Q\) on the common tangent of \((AMP)\) and \((ANQ)\).

EGMO 2019 P4 vibe

Solution by L567 We will ignore $\omega_C$ for now. Let $R,S$ be midpoints of minor arcs $AC$ and $AB$. Define $X$ instead as $RS \cap PM$ and redefine $Y$ as tangent to $\omega_B$ meeting $(ABC)$. Let $M_A$ be the third arc midpoint First, see that $\angle XSI = \angle RSI = \angle RBC = \angle MBI = \angle MPI = \angle XPI$, which means $PXIS$ is cyclic. Next, since $\angle XIM + \angle MIP = \angle XIP = \angle XSP = \angle RSP = \angle RBP = \angle IBP = \angle IMX = \angle MPI + \angle MIP$, we have $\angle XIM = \angle MPI$, which means $XI$ is tangent to $\omega_B$ Now, we have $\angle YAI = \angle YAM_A = \angle YBM_A = \angle IBM_A - \angle IBY = \angle IBM_A - \angle IMB$ Let $AI \cap \omega_B = Z$ Since $X$ lies on $RS$, the perpendicular bisector of $AI$, $\angle XAI = \angle XIZ = \angle ZMI = \angle BMZ - \angle IMB$ But since $\angle BMZ = 180 - \angle AIB = \angle IBM_A$, this means $\angle YAI = \angle XAI \implies A,X,Y$ are collinear. Since we have $\angle BYI = 360 - (\angle IBY - \angle ICY) - \angle BIC = 360 - 2 \angle BIC = 180 - \angle BAC$ (The middle thing follows because the circles are tangent) which means $Y \in (ABC)$ and so $Y$ is indeed the $Y$ in the problem. So, $PM$ and $QN$ both meet $RS$ at a point on $AY$. So, $X = PM \cap QN$. So, $X$ is also the $X$ in the problem. So, we indeed have $A,X,Y$ collinear, as desired. $\blacksquare$

Easy for a G8? Here is a solution using only angle chasing. Let $E,F$ be the intersections of $\omega_B,\omega_C$ such that $E \neq B, F \neq C$. Let $U = \overline{EM} \cap \overline{AC}$ and $V = \overline{FN} \cap \overline{AB}$. Claim 1: $(MNEF)$ is an iscosceles trapezoid and $Y \in \odot(ABC)$ Proof: Since $\overline{BI}$ is an angle bisector so $\overline{IM} = \overline{IE}$. Similiarly $\overline{IN} = \overline{IF}$ . Thus $\overline{O_1O_2}$ is a perpendicular bisector of $\overline{ME},\overline{NF}$ implying the first result. Now we have $$\angle YBE = \angle BME = \angle BAC - \angle FNC = \angle BAC - \angle FCY$$implying the second part $\qquad \square$ Claim 2: $\overline{YA}$ tangent to $\odot(AMP),\odot(ANQ)$ Proof: Notice that $\angle MPE = \angle ABC$ and $\angle APB = 180^\circ - \angle ACB$ and so $$\angle APM = \angle BAC - \angle BPE = \angle BAC - \angle YBE = \angle FCY = \angle BAY$$ Implying $\overline{YA}$ tangent to $\odot(AMP)$. Similiarly we can prove the other part $\qquad \square$ Claim 3: $U \in \odot(AMP), V \in \odot(ANQ)$ Proof: $$\angle AUM = \angle CNF = \angle BCY = \angle MAY = \angle MPA$$similiarly the other part $\qquad \square$ Claim 4: $(MNQP)$ cyclic Proof: Notice that $$\angle QPM = \angle APM - \angle APQ = \angle FNC - \angle APQ = \angle FNC - \angle NCQ$$also note that $$\angle MNQ = \angle MNV + \angle VNQ = \angle NFC + \angle QCF$$and so $$\angle QPM + \angle MNQ = \angle FNC + \angle NFC + \angle QCF - \angle NCQ = 180^\circ$$hence the result $\qquad \square$ Finish by applying radical axis on $\odot(MNQP), \omega_B,\omega_C$ $\qquad \blacksquare$

Let $\triangle DEF$ be the contact triangle of $\triangle ABC$. Invert about $(DEF)$. Let the image of $\bullet$ after the inversion be $\bullet'$. Then $A',B',C'$ are the midpoints of $EF,FD,DE$ respectively. Note that $I$ is the orthocenter of $\triangle A'B'C'$. Also, $(BPMI)$ and $(CQNI)$ map to two parallel lines, $\overline{B'P'M'}$ and $\overline{C'Q'N'}$, where $P'$ and $Q'$ lie on $(A'B'C')$ and $M'$ and $N'$ lie on $(A'IB'F)$ and $(A'IC'E)$, respectively. Now $X'=(M'P'I) \cap (N'Q'I)$. Finally, $Y'$ lies on the intersection of the two circles through $I$ tangent to $\overline{B'P'M'}$ and $\overline{C'Q'N'}$ at $B'$ and $C'$, respectively. From here on out, we omit the $'$ in $\bullet '$. Note that the circumradii of $\triangle ABC, \triangle AIBF,$ and $\triangle AICE$ are all equal to half the circumradius of $\triangle DEF$. Then $$\measuredangle APM = \measuredangle APB = \measuredangle BIA = \measuredangle BMA = \measuredangle PMA,$$so $AP=AM$. Similarly, $AQ=AN$. Thus, $PMNQ$ is an isosceles trapezoid. We delete points $D,E,$ and $F$, and restate the problem, and relabel $I$ as $H$. New Problem wrote: Let $\triangle ABC$ have circumcircle $\Gamma$. Two parallel lines, $\ell_B$ and $\ell_C$, pass through $B$ and $C$, respectively. Let $\ell$ be a line through $A$ perpendicular to $\ell_B$ and $\ell_C$. Let $P = \ell_B \cap \Gamma \neq B$ and $Q = \ell_C \cap \Gamma \neq C$, and let $M$ and $N$ be the reflections of $P$ and $Q$ over $\ell$, respectively. Let $X = (PMH) \cap (QNH) \neq H$, and let $Y$ be the intersection of the two circles through $H$ and tangent to $\ell_B$ and $\ell_C$ at $B$ and $C$, respectively. Prove that $A,X,H,$ and $Y$ lie on a circle. The first observation we make is that by symmetry, $X$ is just the reflection of $H$ over $\ell$. We can now delete $M,P,N,$ and $Q$. [asy][asy] defaultpen(fontsize(12)); size(12cm); pair A,B,C,H,Y,OB,X,TB,TC; A = dir(110); B = dir(210); C = dir(330); H = orthocenter(A,B,C); Y = dir(260); OB = circumcenter(B,H,Y); X = 2*foot(H,A,A+B-OB)-H; TB = B + (B-OB)*dir(90); TC = C + (B-OB)*dir(90); draw(circumcircle(A,B,C),red); filldraw(A--B--C--cycle,blue+white+white+white); draw(circumcircle(B,Y,H),green); draw(circumcircle(C,Y,H),green); draw(TB--B+(B-TB)*5,magenta); draw(TC--C+(C-TC)*5,magenta); draw(A+2*(B-OB)--A-4*(B-OB),blue); draw(X--H,dashed); label("$\ell_B$",B+(B-TB)*5,dir(0)); label("$\ell_C$",C+(C-TC)*5,dir(0)); label("$\ell$",A-4*(B-OB),dir(270)); dot("$A$",A,dir(A)); dot("$B$",B,dir(180)); dot("$C$",C,dir(350)); dot("$Y$",Y,dir(270)); dot("$X$",X,dir(0)); dot("$H$",H,dir(300)*1.5); real r = circumradius(X,A,H); draw(arc(circumcenter(X,A,H),circumcenter(X,A,H)+r*dir(160),circumcenter(X,A,H)+r*dir(200)),dotted+purple); [/asy][/asy] Since $AH=AX$, we prove instead that the tangent at $A$ to $(AHY)$ is parallel to $\ell_B$ and $\ell_C$. We now finish the problem with the following lemma: Lemma: If $\triangle ABC$ is a triangle, $P$ is a point on $(ABC)$, and $H$ is the orthocenter of $\triangle ABC$, then the tangent to $(BPH)$ at $B$ and the tangent to $(CPH)$ at $C$ are parallel if and only if $P$ lies on $(ABC)$. Proof: Suppose the two tangent lines are parallel, and call them $\ell_B$ and $\ell_C$. Then $$\measuredangle BPC = \measuredangle BPH + \measuredangle HPC = \measuredangle (\ell_B, BH) + \measuredangle (CH, \ell_C) = \measuredangle (\ell_B,BC) + \measuredangle CBH + \measuredangle(BC,\ell_C) + \measuredangle HCB$$$$= \measuredangle CBH + \measuredangle HCB = \measuredangle CHB = \measuredangle BAC,$$so $P$ lies on $\Gamma$. If $P$ lies on $\Gamma$, then using the above equation, $\measuredangle (\ell_B, BC) + \measuredangle (BC, \ell_C) = 0^\circ$, so $\ell_B \parallel \ell_C$. $\blacksquare$ Now, using this lemma, $Y$ lies on $\Gamma$, and again using this lemma, the tangent to $(AHY)$ at $A$ is parallel to $\ell_B$ and $\ell_C$, as desired.

Claim-1-$Y \in (ABC)$ Proof: Let $l$ be the common tangent of $w_b$ and $w_c$ If $l \cap BY=J$ and $l \cap CY=J_1$ then by tangent secant theorem $$\angle YBI= \angle JIB=\angle BMI \; \; \; \; \angle YCI= \angle J_1IC=\angle CNI$$$$\implies \angle YBI +\angle YCI=\angle BIC$$But notice that if $CI \cap (ABC)=H$ and $BI \cap (ABC)=K$ then $\angle HBI=\frac{B}{2}+\frac{C}{2}=90-\frac{A}{2}$ If we take the angles to be directed it is clear that $Y \in (ABC)$ Claim-2- $X = HK \cap l$ Proof: Assume a fantom point $Y= l \cap HK$ Now, because of symmetry it suffice to show that $Y \in PM$ Notice that $\angle YIP=\angle PBI=\angle PHK \implies Y \in (HPI)$ Further $\angle YPI=\angle YHI=\frac{B}{2}=\angle MBI=\angle MPI \implies P-M-X$ To finish, observe that $HK$ is the perpendicular bisector of $AI$ (well known, also trivial using angle chasing). Hence $\angle KAX=\angle KIX=\angle IBY$ Hence we are done!

Let $\omega_P$ and $\omega_Q$ denote the circumcircles of $\triangle APM$ and $\triangle AQN$. By the same trick used in G3 and G6, \[ \measuredangle MIN = \measuredangle MBI + \measuredangle ICN = \measuredangle IBC + \measuredangle BCI = \measuredangle BIC, \]so $I$ has an isogonal conjugate with respect to $BCNM$. Hence it must also be the incenter of $BCNM$. Claim 1: $\omega_P$ and $\omega_Q$ are tangent to each other. Proof. This is just angle chasing. \begin{align*} \measuredangle APM + \measuredangle NQA &= \measuredangle APB + \measuredangle BPM + \measuredangle NQC + \measuredangle CQA \\ &= \measuredangle ACB + \measuredangle BIM + \measuredangle NIC + \measuredangle CBA\\ &= \measuredangle CAB. \qquad\square \end{align*} Let their common tangent at $A$ be $\ell$. Claim 2: $P$, $Q$, $M$, $N$ are concyclic. By angle chasing again, one has \begin{align*} \measuredangle QPM &= \measuredangle QPB + \measuredangle BPM\\ &= \measuredangle QCB + \measuredangle BIM\\ &= \measuredangle QCI + \measuredangle ICB + \measuredangle CIN\\ &= \measuredangle QNI + \measuredangle NCI + \measuredangle CIN\\ &= \measuredangle QNI + \measuredangle CNI\\ &= \measuredangle QNI + \measuredangle INM = \measuredangle QNM.\qquad\square \end{align*} Claim 3: $Y$ lies on $(ABC)$. Proof. Let $M_A$, $M_B$ and $M_C$ be midpoints of arcs $BC$, $CA$ and $AB$. Let $O_B$ and $O_C$ be the centers of $\omega_B$ and $\omega_C$. Since $O_B$, $I$, $C$ are collinear, by Pascal’s theorem $\overline{BO_B}$ and $\overline{CO_C}$ intersect on $(ABC)$. Thus $Y$ also lies on $(ABC)$. $\square$ By radical axes on $(PQNM)$, $\omega_P$, and $\omega_Q$, it follows that $X$ lies on $\ell$. Therefore, it suffices to show that $Y$ also lies on $\ell$. Just notice that \[ \measuredangle APM = \measuredangle APB + \measuredangle BPM = \measuredangle ACB + \measuredangle YBA = \measuredangle ACB + \measuredangle YCA = \measuredangle YCB = \measuredangle YAM, \]so $Y$ lies on $\ell$ as desired. $\blacksquare$

This is really a cute problem imo so here goes the solution Claim 1:- $Y$ lies on $\odot(ABC)$ This can be proved by clever angle chase,Join $MI$ and $NI$ as $YB$ and $YC$ are tangent $\angle YBA$=$\angle BPM$=180-$\angle BIM$ $\angle YCN$=$\angle NQC$=180-$\angle NIC$ Now $\angle BIC$=90+$\frac{A}{2}$,$\angle XIN$=$\frac{C}{2}$, $\angle MIX$=$\frac{B}{2}$ By some computation one can easily get $\angle A$+$\angle BCY$=180$^\circ$ This concludes our claim. Claim 2:-M,N,P,Q are concyclic This is also simple angle chase with the use of dangles $\angle QPM$=$\angle QPB$-$\angle BPM$ =$\angle QCB$+$\angle BIM$-180 =$\angle QCI$+$\angle ICB$-$\angle CIN$ =180-$\angle QNI$+$\angle NCI$-$\angle CIN$ =-$\angle QNI$ - $\angle QCI$ =-$\angle QNM$ This concludes our claim and hence M,N,P,Q are concyclic Claim 3:- $YA$ is the radical axis $\odot (AMP)$ and $\odot (ANQ)$ This is again angle chase, BYPA is Concyclic so $\angle BYA$=180-$\angle BPA$ We know that $\angle ABY$=$\angle BPI$ From here, $\angle APM$=$\angle MAP$ similarly do the same angle chasing to prove $\angle XAN$=$\angle AQN$ and this concludes our claim. Claim 4:- $XA$ is the radical axis of $\odot (ANQ) $ and $\odot (AMP)$ As $XM \cdot XP=XN \cdot XQ$ it follows that X lies on the radical axis of $\odot (AMP)$ and $\odot (ANQ)$ Applying radical axis lemma we hence showed that $X-Y-A$ are collinear as desired.

[asy][asy] size(11cm); defaultpen(fontsize(10pt)); pair A = dir(110), B = dir(220), C = dir(320), I = incenter(A,B,C), Y = dir(285), P1 = intersectionpoint(Y+dir(B--A)*0.001--Y+dir(B--A)*100, unitcircle), Q1 = intersectionpoint(Y+dir(C--A)*0.001--Y+dir(C--A)*100, unitcircle), X = extension(A,Y,circumcenter(A,I,B),circumcenter(A,I,C)), M = extension(P1,X,A,B), N = extension(Q1,X,A,C), P = X+dir(P1--X)*abs(A-X)*abs(X-Y)/abs(P1-X), Q = X+dir(Q1--X)*abs(A-X)*abs(X-Y)/abs(Q1-X), X1 = 2*foot(X,A,I)-X, M1 = extension(M,M+dir(Q1--B),B,C), N1 = extension(N,N+dir(P1--C),B,C); draw(A--B--C--A^^unitcircle, heavycyan); draw(circumcircle(B,I,M)^^circumcircle(C,I,N), heavygreen); draw(B--Q1^^M--M1^^N--N1^^C--P1, magenta); draw(P--P1^^Q--Q1, heavygreen); draw(arc(circumcenter(M,N,P),circumradius(M,N,P),60,120), heavygreen+dashed); draw(B--Y--C, heavygreen); draw(X--I^^A--X1, magenta); draw(X--X1^^A--I, magenta+dotted); draw(A--Y, orange+dashed); draw(extension(M,N,B,Q1)--extension(M,N,C,P1)^^Q1--P1, heavycyan); dot("$A$", A, dir(110)); dot("$B$", B, dir(225)); dot("$C$", C, dir(315)); dot("$Y$", Y, dir(285)); dot("$P_1$", P1, dir(45)); dot("$Q_1$", Q1, dir(135)); dot("$P$", P, dir(135)); dot("$Q$", Q, dir(30)); dot("$X$", X, dir(60)); dot("$M$", M, dir(120)); dot("$N$", N, dir(80)); dot("$M_1$", M1, dir(280)); dot("$N_1$", N1, dir(230)); dot("$I$", I, dir(180)); dot("$X'$", X1, dir(170)); [/asy][/asy] Define $M_1 = \omega_B \cap \overline{BC}$, $N_1 = \omega_C \cap \overline{BC}$, $P_1 = \overline{PM} \cap \Gamma$, and $Q_1 = \overline{QN} \cap \Gamma$. Note that $MNN_1M_1$ is an isosceles trapezoid because $\overline{MM_1}$ and $\overline{NN_1}$ share the same perpendicular bisector. By Reim's we have $\overline{MM_1} \parallel \overline{CP_1}$, so in fact $\overline{BQ_1} \parallel \overline{MM_1} \parallel \overline{II} \parallel \overline{NN_1} \parallel \overline{CP_1}$. It follows by symmetry that $\overline{MN} \parallel \overline{P_1Q_1}$, so $PMNQ$ is cyclic by Reim's. Then $XM \cdot XP = XN \cdot XQ$, so $X$ lies on the radical axis of $\omega_B$ and $\omega_C$, which is the common tangent at $I$. From $\angle BCY = \angle N_1NC = \angle ACP_1 = \angle AQ_1P_1$ (similarly $\angle CBY = \angle AP_1Q_1$) and $BC = P_1Q_1$, we have $Y \in \Gamma$ with $\triangle BCY \cong \triangle P_1Q_1A$. I now claim that $XA = XI$. Let $X'$ be the reflection of $X$ over $\overline{AI}$. We have $\angle BAX' = \angle CAY = \angle ABQ_1$, so $\overline{AX'} \parallel \overline{BQ_1} \parallel \overline{XI}$. But by construction $AX'IX$ is a kite, so it must be a rhombus. To finish, note that from $$XA^2 = XI^2 = XN \cdot XQ,$$$\overline{XA}$ must be tangent to $(ANQ)$. Therefore $$\angle CAX = \angle NAX = \angle AQN = \angle AQQ_1 = \angle CAY$$which implies $A$, $X$, $Y$ are collinear, as required. $\blacksquare$

Didn't want to use directed angles since I'm not much used to them but was forced to use it in some places. Also extremely long proof because I wrote each and every small detail (which are equally important) as claims :< Claim: $Y$ lies on $\Gamma$. Proof: Let $\ell$ be the common tangent of $\omega_B , \omega_C$, $G=\omega_B \cap BC$ and $H=\omega_C \cap BC$. \begin{align*} \angle YBA + \angle YCA &= \angle YBC + \frac{\angle B}{2} + \frac{\angle B}{2}+ \angle YCB + \frac{\angle C}{2} + \frac{\angle C}{2}\\ &= \angle BIG + \angle(MI , \ell ) + \angle (\ell , GI) + \angle CIH + \angle (\ell , NI) + \angle(HI , \ell)\\ &= \angle(MI , \ell ) + \angle (\ell , NI) + \angle BIG + \angle (\ell , GI) + \angle(HI , \ell) + \angle CIH\\ &= \angle MIN + \angle BIC \\ &= \frac{\angle B + \angle C}{2} + 90^\circ + \frac{\angle A}{2} \\ &=180^\circ \end{align*} Claim: $\angle IMN = \angle IMB$ and $\angle INM = \angle INC$. Proof: It's enough to show that $MI$ and $NI$ are angle bisectors of angle $MBN$ and $CNM$ respectively. But this is true because $I$ is midpoint of arc smaller $MG$ and smaller arc $NH$ so $IM = IG$ and $IN = IH$ and $$\angle MIN = \angle GIH = \dfrac{\angle B + \angle C}{2} \implies \triangle GIH \cong \triangle MIN$$So $\angle IMN = \angle IGH = \angle IMB$ and $\angle INM = \angle IHG = \angle INC$. Claim: $QPMN$ is cyclic. [Let's call the circumcircle $\omega$] Proof: (Sorry if I mess up dangles, it was really hard to write "$\pm 180^\circ , \pm 360^\circ$" in vanilla so many times ) \begin{align*} \measuredangle QPM &= \measuredangle QPB - \measuredangle MPB \\ &= \measuredangle QCH - \measuredangle MIB \\ &= \measuredangle QCI + \measuredangle ICH - \measuredangle MIB \\ &= \measuredangle QNI + \measuredangle NCI - \measuredangle NIC \\ &= \measuredangle QNI - \measuredangle INC \\ &= \measuredangle QNI + \measuredangle CNI \\ &= \measuredangle QNI + \measuredangle INM \\ &= \measuredangle QNM \end{align*}Now by Radical Axis theorem on $(\omega_B,\omega,\omega_C)$ we see that $PM$ and $QN$ concur on $\ell$ so $X \in \ell$. Claim: $XA=XI$. Proof: Let the perpendicular bisector of $AI$ intersect minor arc $AB$, $AC$ at $M_C$ and $M_B$ respectively. By Fact 5 $M_C$ and $M_B$ are the midpoints of the respective arcs. $$\measuredangle XIP = \measuredangle IBP = \measuredangle M_BBP = \measuredangle M_CM_BP = \measuredangle XM_CP$$So $XM_CPI$ is cyclic, similarly $XIM_BQ$ is cyclic. $$\measuredangle XM_CI = \measuredangle XPI = \measuredangle MPI = \measuredangle IBC = \measuredangle M_BBC = \measuredangle M_BM_CC = \measuredangle M_BM_CI$$So $X \in M_BM_C \implies X \in$ perpendicular bisector of $AI$. $XM \cdot XP = XI^2 = XA^2 \implies (MAP)$ is tangent to $AX$ thus $$\angle MPA = \angle MAX$$Let $F = PM \cap (ABC)$ Claim: $FABY$ is an isosceles trapezium. $$\angle FAB = \angle FPB = \angle MPB = \angle MBY = \angle YBA$$Finally to finish off $$\angle MPA = \angle FPA = \angle FYA = \angle BAY = \angle MAY$$$\implies \angle MAX = \angle MAY \implies \overline{A-X-Y}$ are collinear.

[asy][asy] size(350); //stole geogebra points because i am too lazy to do asymptote correctly pair A=(2.999429319069,8.7145022896488), B = origin, C = (11.707076895692, 0); pair I = incenter(A, B, C); pair P = (-0.6582634654393, 3.0392879500232), Q = (5.4004931584466, 9.3572483063131), M = (1.6101594341007, 4.6781359327132), NN = (4.2178365368594, 7.4951359343011); pair X = extension(P, M, Q, NN); pair Y = (3.4938614778841, -3.2148956397776); draw(A--B--C--cycle, red); draw(circumcircle(A, B, C), blue); draw(circumcircle(B, P, I), orange); draw(circumcircle(C, Q, I), orange); draw(circumcircle(A, M, P), magenta); draw(circumcircle(A, NN, Q), magenta); pair OB = circumcenter(B, P, I), OC = circumcenter(C, Q, I); pair B1 = (3.6565170629108, 0), C1 = (7.4924761733508, 0); draw(M--B1, purple+dashed); draw(NN--C1, purple+dashed); draw(OB--OC, gray+dashed); pair Z = (8.2164182887114, 8.9293807829784); draw(B--Z--C--Y--cycle, green); draw(X--P, brown); draw(X--NN, brown); draw(X--Q, brown); draw(X--M, brown); draw(X--A, brown); draw(X--Y, brown); pair MB = (-0.3061221855607, 4.9787989737885), MC = (10.4616597135134, 7.4632127191535); draw(MB--MC, cyan); draw(A--I, gray+dashed); dot("$A$", A, N); dot("$B$", B, SW); dot("$C$", C, SE); dot("$I$", I, NE); dot("$O_B$", OB, S); dot("$O_C$", OC, NE); dot("$Y$", Y, S); dot("$X$", X, NW); dot("$M$", M, NW); dot("$N$", NN, NE); dot("$B_1$", B1, S); dot("$C_1$", C1, S); dot("$Z$", Z, NE); dot("$P$", P, W); dot("$Q$", Q, NE); dot("$M_C$", MB, W); dot("$M_B$", MC, NE); [/asy][/asy] Neat problem! Denote by $O_B$, $O_C$ the centers of $\omega_B$ and $\omega_C$, and let $\omega_B$ intersect $\overline{BC}$ at $B_1$, while $\omega_C$ intersects $\overline{BC}$ at $C_1$. We begin with the following claim. Claim. $\overline{MB_1}$ and $\overline{NC_1}$ are parallel. Proof. Observe that because $\overline{BI}$ and $\overline{CI}$ are angle bisectors, $I$ is the arc midpoint of $\widehat{MB_1}$ and $\widehat{NC_1}$, and hence $\overline{O_BO_C}$ is perpendicular to both $\overline{MB_1}$ and $\overline{NC_1}$. It follows that $\overline{MB_1}$ is parallel to $\overline{NC_1}$, as desired. $\blacksquare$ In fact we realize the stronger fact that $MNC_1B_1$ is an isosceles trapezoid, because the perpendicular bisectors of $\overline{MB_1}$ and $\overline{NC_1}$ coincide. Claim. $Y$ lies on the circumcircle $\Gamma$. Proof. Let $Z = \overline{BO_B} \cap \overline{CO_C}$. It suffices to show that $Z$ lies on $\Gamma$ by definition of $Y$. Observe that \begin{align*} \measuredangle BZC &= \measuredangle BO_BI + \measuredangle IO_CC \\ &= 2\measuredangle BMB_1 + B + 2\measuredangle C_1NC + C \\ &= 2(\measuredangle BB_1M + \measuredangle NC_1C) - B - C \\ &= \measuredangle BAC, \end{align*}because $\overline{MB_1} \parallel \overline{NC_1}$ by the first claim. This proves the result. $\blacksquare$ Claim. The triangles $XMN$ and $XQP$ are similar. Proof. We will show that $\measuredangle XMN = \measuredangle PQX$. Observe that \begin{align*} \measuredangle XMN &= \measuredangle B_1MN + \measuredangle XMB_1 \\ &= \measuredangle C_1B_1M + \measuredangle PBB_1 \\ &= \measuredangle B_1C_1N + \measuredangle PBB_1 \\ &=\measuredangle CQN + \measuredangle PQC \\ &= \measuredangle PQX, \end{align*}as required. This proves the claim. $\blacksquare$ From this claim, we obtain that $XM \cdot XP = XN \cdot XQ$, so $X$ lies on the radical axis of $(AMP)$ and $(ANQ)$. However, $X$ also lies on $\overline{PM}$ and $\overline{QN}$, so it has equal power to all of $\omega_B, \omega_C$, $(ANQ)$ and $(AMP)$. It follows that $X$ is the radical center of these four circles. Now, we prove the following: Claim. $\overline{XA}$ is the common tangent to $(AMP)$ and $(ANQ)$. Proof. First observe that $\overline{XI}$ is the common tangent of $\omega_B$ and $\omega_C$ because $X$ is the radical center. Hence it suffices to show that $XA = XI$. We will show that $X$ lies on the perpendicular bisector of $\overline{AI}$ -- in other words, $M_B, X, M_C$ are collinear where $M_B, M_C$ are the arc midpoints of minor $\widehat{AB}$ and $\widehat{AC}$. This is just angle chasing. First, we claim that $M_BXIP$ is cyclic. Indeed, this is just true by angle chasing: $$\measuredangle PXI = \measuredangle PMB_1 = \measuredangle PBC = \measuredangle PM_CC,$$in particular because $\overline{XI} \parallel \overline{MB_1}$, and $C, I, M_C$ are collinear. Then, we have \begin{align*} \measuredangle IXM_C &= \measuredangle IPM_C \\ &= \measuredangle BPM_C + \measuredangle IPB \\ &= \measuredangle BCI + \measuredangle IB_1C_1. \end{align*}For the next step, observe that $$\measuredangle IBC + \measuredangle IB_1C_1 = \measuredangle IPM + \measuredangle IPB = \measuredangle BB_1M = \measuredangle CC_1N = \measuredangle ICB + \measuredangle IC_1B_1,$$so this implies that $$\measuredangle IB_1C_1 + \measuredangle BCI = \measuredangle IC_1B_1 + \measuredangle CBI.$$Now we can just angle chase symmetrically in reverse around $C$, obtaining $$\measuredangle IXM_C = \measuredangle IXM_B,$$as desired. This proves the claim. $\blacksquare$ With this information, we can now find \begin{align*} \measuredangle XAB &= \measuredangle XAM \\ &= \measuredangle MPA \\ &= \measuredangle BPA + \measuredangle MPB \\ &= \measuredangle BCA + \measuredangle MB_1B \\ &= \measuredangle BCA + \measuredangle CC_1N \\ &= \measuredangle CNC_1 \\ &= \measuredangle YCB = \measuredangle YAB. \end{align*}Finally, this implies that $X, A, B$ are collinear, so we are done. $\square$

Let $\omega_b, \omega_c$ meet $BC$ again at $D, E$ respectively, $M_b, M_c$ denote the midpoints of arcs $CA, AB$ respectively, $K$ be the second intersection between $PM$ and $\Gamma$, $T$ be the second intersection between $QN$ and $\Gamma$, and $l$ be the tangent at $I$. Claim: $Y \in (ABC)$. Proof. Notice $$180^{\circ} - \angle BYC = \angle YBC + \angle YCB = \angle BID + \angle CIE$$$$= \angle BIC - \angle DIE = \left(90^{\circ} + \frac{\angle A}{2} \right) - (\angle DBI + \angle ECI)$$$$= \left(90^{\circ} + \frac{\angle A}{2} \right) - \left(\frac{\angle B}{2} + \frac{\angle C}{2} \right) = \angle A$$which suffices. $\square$ Claim: $ABY \sim NEC$ and $ACY \sim MDB$. Proof. Observe $$\angle AYB = \angle ACB = \angle NCE$$and $$\angle BAY = \angle BCY = \angle ECY = \angle ENC$$implying the first similarity. The second result follows analogously. $\square$ Claim: $AB \parallel KY$ and $AC \parallel TY$. Proof. Because $P$ is the Miquel Point of $ACDM$, it's also the center of the spiral similarity taking $DM$ to $CA$. But the previous claim implies $B$ goes to $Y$ under the same transformation, so $PBDM \sim PYCA$. Now, the Spiral Similarity Lemma yields $PBY \sim PMA$. Hence, $$\angle ABK = \angle APK = \angle APM = \angle YPB = \angle YKB$$so $AB \parallel KY$. The second result follows in an analogous fashion. $\square$ Claim: $BT \parallel DM \parallel EN \parallel CK$. Proof. Notice $$\angle DMB = \angle CAY = \angle TYA = \angle TBA = \angle TBM$$so $BT \parallel DM$. Similarly, we conclude $CK \parallel EN$. Now, observe $$\angle BTC = \angle BTY + \angle YTC = \angle BAY + \angle ACT = \angle KYA + \angle AYT$$$$= \angle KYT = \angle KCT$$so $BT \parallel KC$. At this point, Transitivity of Parallelism finishes. $\square$ Claim: $MN \parallel KT$. Proof. Let $F = DM \cap KT$. Since $\angle DBI = \frac{\angle B}{2} = \angle MBI$, $I$ is the midpoint of arc $DM$, so $ID = IM$ and $DM \parallel l$. Similarly, we conclude $IE = IN$ and $EN \parallel l$, so $DENM$ is an isosceles trapezoid. As a result, $$\angle DMN = \angle MDE = \angle TBE = \angle TBC = \angle BTK = \angle DFK$$as desired. $\square$ Now, notice $$180^{\circ} - \angle PMN = \angle XMN = \angle XKT = \angle PKT = \angle PQT = \angle PQN$$so $MNQP$ is cyclic. Thus, $$Pow_{\omega_b}(X) = XM \cdot XP = Pow_{(MNQP)}(X) = XN \cdot XQ = Pow_{\omega_c}(X)$$so $X$ lies on the Radical Axis of $\omega_b$ and $\omega_c$, i.e. $l$. Claim: $XIPM_c$ is cyclic. Proof. Observe $$\angle IXP = 180^{\circ} - \angle XPI - \angle PIX = 180^{\circ} - \angle MPI - \angle PBI$$$$= 180^{\circ} - \angle MBI - \angle PBM_b = 180^{\circ} - \angle CBM_b - \angle PBM_b$$$$= 180^{\circ} - \angle CBP = \angle CM_cP = \angle IM_cP$$as required. $\square$ Claim: $X \in M_bM_c$. Proof. Notice $$\angle IM_cX = \angle IPX = \angle IPM = \angle IBM = \angle CBM_b = \angle CM_cM_b = \angle IM_cM_b$$which suffices. $\square$ It's well-known that $M_bM_c$ is the perpendicular bisector of $AI$, so $XA = XI$, which implies $$XA^2 = XI^2 = Pow_{\omega_b}(X) = XM \cdot XP.$$As a result, we know $XMA \sim XAP$, so $$\angle BAX = \angle MAX = \angle APX = \angle APK = \angle ABK = \angle YKB = \angle BAY$$which finishes the problem. $\blacksquare$

We have $90^{\circ}+\frac{\angle BAC}{2}=\angle BIC = \angle BMI + \angle CNI = \angle IBY + \angle ICY = \frac{\angle ABC}{2} + \frac{\angle ACB}{2} + 180^{\circ} - \angle BAC$, which implies that $\angle BAC + \angle BAC = 180^{\circ}$ and $Y$ lies on $\Gamma$. We also have $\angle PAY = 180^{\circ}-\angle PBY = \angle PMB$, so $AY$ is tangent to $(AMP)$. Similarly, $AY$ is tangent to $(ANQ)$ and hence $(AMP)$ is tangent to $(ANQ)$. Now invert about $A$, we have $AY \parallel MP \parallel NQ$, so we need to prove that $AX \parallel MP \parallel NQ$. But this is easy. Notice that $I$ is now the midpoint of both the major arc $MP$ of $BMP$ and the major arc $NQ$ of $CNQ$, so $MP$ and $NQ$ share a common perpendicular bisector. This is also the line connecting the centers of $(AMP)$ and $(ANQ)$, which is perpendicular to the radical axis $AX$ of the two circles. Thus we are done.

Claim: Consider triangle with incenter $I$ and circumcircle $\Gamma$. Take any point $D$ on $AI$ and let $(IDB)$ intersect $(ABC),AB$ at $E,F$, respectively. If $P$ is the intersection of $EF$ and the tangent from $I$ to $(IDB)$ and $S$ is the intersection of $(ABC)$ and the tangent from $B$ to $(IDB)$, prove that $A,P,S$ are collinear. Proof. [asy][asy]import olympiad; size(6cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta; pair O,A,B,C,D,E,F,I,P,Q,R,S; O=(0,0);A=dir(120);B=dir(205);C=dir(335);I=incenter(A,B,C); D=1.2I-0.2A; E=intersectionpoints(circumcircle(I,D,B),circumcircle(A,B,C))[0];F=intersectionpoints(circumcircle(I,D,B),A--B)[0]; Q=extension(E,D,I,B);P=extension(A,Q,E,F);R=extension(B,D,A,Q);S=intersectionpoints(circumcircle(A,B,C),A--10Q-9A)[1]; draw(circumcircle(A,B,C),heavyblue); draw(A--B--C--cycle,red+1);draw(circumcircle(I,D,B),heavyblue);draw(A--D,red+1);draw(P--S,heavygreen);draw(P--E,heavygreen);draw(P--I,heavygreen);draw(D--E,red);draw(B--I,red);draw(B--D,red);draw(E--I,red);draw(P--A,heavygreen); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$I$",I,dir(I)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$P$",P,dir(P)); dot("$Q$",Q,dir(Q)); dot("$R$",R,dir(R)); dot("$S$",S,dir(S)); [/asy][/asy] Let $Q=IB \cap DE$ and let $R=BD\cap AP$. Let $S'$ be the intersection of $AP$ and the tangent from $B$ to $(IDB)$, we claim that $S\equiv S'$, i.e. $S'$ lies on $(ABC)$. Firstly, by Pascal's theorem on $IIBFED$, we obtain that $A,P,Q$ are collinear. Note that $AIQE$ is cyclic, because $$\measuredangle QEA=\measuredangle BEA-\measuredangle BED=\measuredangle BCA-\measuredangle BID=\measuredangle QIA.$$Furthermore, we get that $AERD$ is cyclic, because $\measuredangle EDR=\measuredangle EIQ=\measuredangle EAR$. Hence, $$\measuredangle BRS'=\measuredangle DRA=\measuredangle QEA=\measuredangle BID=\measuredangle S'BR\implies \measuredangle AS'B=2\measuredangle DIB=\measuredangle ACB.$$The claim follows. $\square$ By the claim, all we need is to show that $Y$ lies on $(ABC)$. Indeed as $(BIP)$ and $(CIQ)$ are tangent to each other, \begin{align*} \measuredangle BYC=\measuredangle BO_1I+\measuredangle IO_2C=2\measuredangle BPI+2\measuredangle IQC=2\measuredangle BIC=\measuredangle BAC, \end{align*}we are done. [asy][asy]import geometry; size(7cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta; pair O,A,B,C,K,P,M,I,O1,O2,Q,N,X,Y; O=(0,0);A=dir(120);B=dir(205);C=dir(335);I=incenter(A,B,C); K=1.1I-0.1A; P=intersectionpoints(circumcircle(I,K,B),circumcircle(A,B,C))[0];M=intersectionpoints(circumcircle(I,K,B),A--B)[0]; O1=circumcenter(I,K,B);O2=intersectionpoint(line(O1,I),perpendicular(midpoint(I--C),line(I,C))); Q=intersectionpoints(circumcircle(A,B,C),circle(O2,abs(O2-I)))[0];N=intersectionpoints(A--C,circle(O2,abs(O2-I)))[0];X=extension(P,M,Q,N);Y=intersectionpoints(circumcircle(A,B,C),A--100X-99A)[1]; draw(circumcircle(A,B,C),heavyblue); draw(A--B--C--cycle,red+1);draw(circumcircle(I,P,B),heavyblue);draw(A--I,red+1); draw(circumcircle(I,Q,C),heavyblue);draw(B--Y--C,heavygreen+1);draw(A--Y,dashed+orange+0.5); draw(P--X--Q,red+0.5);draw(I--X,red+0.5);draw(B--O1--O2--C,lightblue+0.5); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$I$",I,dir(I)); dot("$P$",P,dir(P)); dot("$M$",M,dir(M)); dot("$O_1$",O1,dir(O1)); dot("$O_2$",O2,dir(O2)); dot("$Q$",Q,dir(Q)); dot("$N$",N,dir(N)); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); [/asy][/asy]

Let $T_1=\omega_B\cap BC$ and $T_2=\omega_C\cap BC$. We make the following claims: Claim: $Y$ lies on $\Gamma$ Proof: We angle-chase this, getting: \begin{align*} \angle T_1IT_2&=\angle T_1MI+\angle T_2NI\\ &=\angle T_2CI+\angle T_1BI\\ &=90^{\circ}-\frac{\angle A}{2}.\\ \end{align*} Therefore: \begin{align*} \angle A&=\angle BIT_1+\angle CIT_2\\ &=\angle BMT_1+\angle CNT_2\\ &=\angle YBC+\angle YCB\\ &=180^{\circ}-\angle BCY.\\ \end{align*} As a result $ABYC$ is cyclic $\blacksquare$ Claim: $T_1T_2NM$ is an isosceles trapezoid Proof: We can angle chase to see: \[\angle IT_1M=\angle IBM=\angle IBT_1=\angle IMT_1,\]so $\triangle{IT_1M}$ is isosceles. Similarly, $\triangle{IT_2M}$ is isosceles. As the centers of the two circles are collinear with $I$, we have our desired result $\blacksquare$ Claim: $PMNQ$ is cyclic Proof: We angle chase: \begin{align*} \angle MNQ&=\angle MNT_2+T_2NQ\\ &=MNT_2+180^{\circ}-\angle BCQ\\ &=180^{\circ}-\angle MT_1T_2+\angle BPQ\\ &=\angle BT_1M+\angle BPQ\\ &=180^{\circ}-\angle BPM+\angle BPQ\\ &=180^{\circ}-\angle MPQ,\\ \end{align*}as desired $\blacksquare$ Claim: $AY$ is tangent to both $(AMP)$ and $(ANQ)$ Proof: For $(AMP)$, we angle chase, getting: \begin{align*} \angle BAY&=\angle BCY\\ &=\angle NCB+\angle BCY-\angle BCA\\ &=\angle NQT_2+\angle T_2QC-\angle BCA\\ &=\angle NQC-\angle BCA\\ &=180^{\circ}-\angle NT_2C-\angle BCA\\ &=\angle BT_1M-\angle BCA\\ &=180^{\circ}-\angle BCA-\angle BPM\\ &=\angle BPA-\angle BPM\\ &=MPA,\\ \end{align*}as desired. We use the exact same approach for $(ANQ)$ and we get the same result $\blacksquare$ Using the radical center on $(PMNQ),(AMP),(ANQ)$, it is $X$. This means that $XA$ is tangent to $(ANQ)$ and $(AMP)$, so $A,X,Y$ are collinear $\square$

okay this is doable with nothing but radical axis and angle chase Angle chase: Y lies on ABC. Draw PM cap ABC and QN cap ABC = U, V. Angle chase. UAV = BYC. Find angles UV, BC and MN, BC. UV and MN are parallel (this angle chase takes a while, but it does work). Now PMQN is cyclic by Reim's. Angle chase: APM and AQN are tangent. So MAX = APM, but angle chasing gives APM = BAY. Ta-da!!!!

First denote $S = \omega_B \cap BC$ and $T = \omega_C \cap BC$. We claim $Y$ lies on $\Gamma$, as \[90 + \frac{\angle A}{2} = \angle BIC = \angle IBY + \angle ICY \implies \angle BYC = 180 - \angle A.\] We also note $MNTS$ is an isosceles trapezoid by symmetry about $O_BO_C$. Thus $\triangle XMN \sim \triangle XQP$ since \[\measuredangle NMX = \measuredangle NMS + \measuredangle SMP = \measuredangle NQC + \measuredangle CQP = \measuredangle XQP.\] As a result, $AX$ forms the radical axis of $(APM)$ and $(AQN)$, so we must show $Y$ lies on this radical axis. In fact, we claim $AY$ is tangent to $(APM)$, as \[\measuredangle APM = \measuredangle APB + \measuredangle BPM = \measuredangle ACB + \measuredangle YCN = \measuredangle YCB = \measuredangle YAM,\] and analogously $AY$ is also tangent to $(AQN)$. $\blacksquare$

WHAT A PROBLEM. My first recent-year G8 solve!!! Solved with a few hints (specifically to arrive at Claim 2 and construct $(AMP)$ and $(ANQ)$), but still very awesome. Claim: $Y$ lies on $\Gamma.$ Proof: Let $Z$ be an arbitrary point on the tangent at $I$ to both circles closer to $Y$ than $A.$ Then we have $\measuredangle ZIB = \measuredangle IMB = \measuredangle IBY$ and $\measuredangle CIZ = \measuredangle CNI = \measuredangle YCI.$ Therefore, adding these two gives $$\measuredangle CIB = \measuredangle IBY + \measuredangle YCI.$$Consequently, because the sum of the angles in a quadrilateral is $360$ degrees, we get $$\measuredangle BYC = 2 \measuredangle BIC = \measuredangle BAC,$$as claimed. Claim 2: Let $\omega_B$ and $\omega_C$ intersect side $BC$ again at $F \ne B$ and $G \ne C,$ respectively. Then $MFGN$ is an isosceles trapezoid. Proof: Note that $I$ is the arc midpoint of minor arc $MF$ of $\omega_B$ since $\measuredangle MBI = \measuredangle IBF.$ Thus $M$ and $F$ are symmetric about the line connecting the centers of $\omega_B$ and $\omega_C,$ as are $N$ and $G.$ Furthermore, we also have that $MF$ is parallel to the tangent at $I$ on both circles, as is $NG,$ so $MF \parallel NG.$ This is enough to imply that $MFGN$ is an isosceles trapezoid, as claimed. Claim 3: $P,M,N,Q$ are concyclic. Proof: First, we can write $$\measuredangle PMN = \measuredangle PMB + \measuredangle BMF + \measuredangle FMN.$$We now note that $\measuredangle PMB = \measuredangle PBY$ since $BY$ is tangent to $\omega_B.$ We similarly note that $\measuredangle BMF = \measuredangle YBF = \measuredangle YBC.$ Finally, by Claim 2, we can write $\measuredangle FMN = \measuredangle GFM = \measuredangle BFM = \measuredangle YBM = \measuredangle YBA$. Therefore, $$\measuredangle PMN = \measuredangle PBY + \measuredangle YBC + \measuredangle YBA = \measuredangle PBC + \measuredangle YBA.$$Now, $$\measuredangle PQN = \measuredangle PQC + \measuredangle CQN = \measuredangle PBC + \measuredangle YCN = \measuredangle PBC + \measuredangle YCA.$$However, by Claim 1, $Y$ lies on $\Gamma,$ so $\measuredangle YBA = \measuredangle YCA.$ This readily implies $\measuredangle PMN = \measuredangle PQN,$ as claimed. Let $\omega$ be the common circle of $P,M,N,Q$ as proven to exist by Claim 3. Now we will finish the problem. First, since $Y$ lies on $\Gamma$ and $YB$ is tangent to $\omega_B,$ we note that $$\measuredangle YAP = \measuredangle YBP = \measuredangle BMP = \measuredangle AMP,$$so $YA$ is tangent to $(AMP).$ Similarly, $YA$ is tangent to $(ANQ).$ This means that $(PMA)$ and $(QNA)$ are tangent, with $Y$ lying on the common external tangent. On the other hand, we can apply the radical center theorem on $(PMA), (QNA),$ and $\omega$ to get that $PM \cap QN = X$ lies on this external tangent. Therefore, $A,X,Y$ are all collinear on the common external tangent of $(PMA)$ and $(QNA),$ as desired.

Let $K$ be the intersection point of the tangent at $I$ and $\Gamma$. Then $\angle BIC = \angle BIK + \angle CIK = \angle IMB + \angle CNI = \angle YCI + \angle YBI$. Note that $\angle BIC = 90^{\circ} + \frac{\angle A}{2}$ so $\angle BYC = 360^{\circ} - (180^{\circ} - \angle A) = 180^{\circ} - \angle A \implies ABYC$ is cyclic. Notice that $MNC_1B_1$ is an isosceles trapezoid due to the fact that the perpendicular bisectors of $MB_1$ and $NC_1$ coincide. Note that $PBCQ$ is cyclic. We can then prove that $\triangle XMN \sim \triangle XQP$. $\angle XNM = 180^{\circ} - (\angle MNC_1 + \angle C_1NQ) = 180^{\circ} - (\angle MBB_1 + 180^{\circ} - \angle C_1CQ)$ $= \angle BPM + \angle C_1CQ - 180^{\circ} = \angle BPM - \angle BPQ = \angle XPQ$. By using similar triangle ratios, we find that $XM \cdot XP = XN \cdot XQ$, so $AX$ is the radical axis of $\omega_B$ and $\omega_C$. Similarly, we have $XN \cdot XQ = XM \cdot XP \implies AX$ is the radical axis of $(AMP)$ and $(ANP)$. All that is left is to show that $AY$ is the radical axis of $(AMP)$ and $(ANP)$ which is equivalent to $AY$ tangent to both circle. Notice that $PAYB$ is cyclic. Then $\angle PMB = 180^{\circ} - PBY = \angle PAY$, which proves the tangency, so we are done.

Angle chase We use directed angles mod $180^\circ$ throughout this solution.

mfw g8 is just a massive chase

Take an inversion about I. In the inverted image, we have $I$ is the orthocenter of $\triangle ABC$, $B,P,M$ collinear and $C,Q,N$ collinear. Note that \[\angle APM=\angle ACB=180^\circ-\angle AIB=\angle AMP\]so $AP=AM$ and similarly $AQ=AN$. This implies that $PMNQ$ is an isosceles trapezoid and $(MAP)$ is tangent to $(NAQ)$. In our uninverted image, note that since $PMNQ$ is cyclic, $X$ is on the radical axis of $\omega_B$ and $\omega_C$. Furthermore, it is on the radical axis of $(MAP)$ and $(NAQ)$. It is also on the common tangents from $I$ to $\omega_B$ and $\omega_C$ and from $A$ to $(MAP)$ and $(NAQ)$. Note that \[\angle BIC+\angle MIN=180^\circ-\angle IBC-\angle ICB+\angle MIX+\angle XIN=180^\circ-\angle IBC-\angle ICB+\angle IBM+\angle ICN=180^\circ\]so $\angle BIM+\angle CIN = 180^\circ\implies \angle BPM+\angle CQN=180^\circ\implies \angle ABY+\angle ACY=180^\circ$, so $Y$ is on $(ABC)$. Furthermore, \[\angle AQN=\angle AQC-\angle NQC=180^\circ-\angle AYC-\angle ACY=\angle YAC\]so $YA$ is tangent to $(NAQ)$. Similarly, it is tangent to $(MAP)$ and so $Y$ is also on the common tangent from $A$ to $(MAP)$ and $(NAQ)$. We are done.

Invert about $I$ and rename $I$ to $H$ to get the following equivalent statement: Let $\triangle ABC$ have orthocenter $H$. Let $\ell_B$, $\ell_C$ be paralell lines through $B$ and $C$, respectively. Let $P$ and $Q$ be the intersections of $\ell_B$ and $\ell_C$ with $(ABC)$, respectively. Let $M$ and $N$ be the intersections of $\ell_B$ and $\ell_C$ with $(AHB)$ and $(AHC)$, respectively. Let $X$ be the intersection of $(HPM)$ and $(HQN)$. Let $Y$ be the second intersection of the circle through $H$ tangent to $\ell_B$ at $B$ and the circle through $H$ tangent to $\ell_C$ at $C$. Then, $AHYX$ is cyclic. We will prove this statement instead of the original. Let $H'$ be the reflection of $H$ over the angle bisector of $BC$ and $PQ$. Then, $\triangle BHC \cong PH'Q$. Claim: $X$ is the intersection of $HH'$ with $(H'PQ)$. Proof: We have \[\angle PXQ = \angle PXH + \angle HXQ = \angle BMH + \angle HNC\]\[= \angle BAH + \angle HAC = \angle BAC = \angle CHB = \angle PH'Q.\]Thus, $X$ lies on $(PH'Q)$. We have \[\angle PXH = \angle PMH = \angle BMH = \angle BAH = \angle HCB = \angle PQH' = \angle PXH'.\]Thus, $X$ is on $HH'$. Claim: $A$ is the orthocenter of $\triangle PXQ$. Proof: Let $D$ be the intersection of $PQ$ and $HH'$. We have \[\angle APQ = \angle ACQ = \angle BCQ + \angle ACB\]\[\angle PQX = \angle DXQ + \angle QDX = \angle HCB + \angle QCB.\]Thus, \[\angle APQ + \angle PQX = \angle BCQ + \angle QCB + \angle ACB + \angle HCB = 90.\]Repeating the same logic on the other side, $A$ is the orthocenter of $\triangle PXQ$. Claim: $AHYX$ is cyclic Proof: We have \[\angle AXH = \angle AXP + \angle PXH' = \angle PQA + \angle PQH'\]\[= \angle PBA + \angle HCB = \angle PBC + \angle CBA + \angle HCB = \angle PBC + 90\]\[= \angle PBC + \angle CBH + \angle ACB = \angle PBH + \angle ACB = \angle BYH + \angle AYB\]\[= \angle AYH.\]

Eyed wrote: Let $ABC$ be a triangle with incenter $I$ and circumcircle $\Gamma$. Circles $\omega_{B}$ passing through $B$ and $\omega_{C}$ passing through $C$ are tangent at $I$. Let $\omega_{B}$ meet minor arc $AB$ of $\Gamma$ at $P$ and $AB$ at $M\neq B$, and let $\omega_{C}$ meet minor arc $AC$ of $\Gamma$ at $Q$ and $AC$ at $N\neq C$. Rays $PM$ and $QN$ meet at $X$. Let $Y$ be a point such that $YB$ is tangent to $\omega_{B}$ and $YC$ is tangent to $\omega_{C}$. Show that $A,X,Y$ are collinear. We'll prove this problem follows these claims : $\textbf{1/}$ Let $B'$ be the intersection of line $BI$ with $\omega_c$ ($B' \neq I$); $C'$ be the intersection of line $CI$ with $\omega_b$ ($C' \neq I$) $\hspace{1cm}$$M'$ be the intersection of lines $C'M$ and $AC$; $N'$ be the intersection of lines $B'N$ and $AB$ Then $N'MM'N$ is a isosceles trapezoid and $I$ lie on circumcircle of $N'MM'N$ $\textit{Prove :}$ $\, \,$Cuz $M$, $I$, $B$, $C$ are concyclic and $I$ is incenter of $\triangle ABC$, we see that : $$ \angle AM'M = \angle MC'C + \angle ACC' = \angle IBC + \angle ICB = \angle C'IB = \angle C'MB $$$\, \,$So $\triangle AMM'$ is a isosceles triangle at $A$. Similar, $\triangle ANN'$ is a isosceles triangle at $A$ $\, \,$Therefore, $N'MM'N$ is a isosceles trapezoid $\, \,$Next, let $\ell$ be the line which tangent both $\omega_b$, $\omega_c$ at $I$, we see that : $$ \angle MIN = \angle MI\ell + \angle NI\ell = \angle MBI + \angle NCI = \angle IBC + \angle ICB = \angle AM'M $$$\, \,$Therefore, $M'$ lies on circle $(IMN)$. Similar, $N'$ lies on circle $(IMN)$. $\textbf{2/}$ $M$, $N$, $P$, $Q$ are concyclic $\textit{Prove :}$ $\, \,$Let $\omega_b$, $\omega_c$ meet line $BC$ at $T$, $L$ respectively ($T \neq B$, $L \neq C$) $\hspace{1cm}$$O_b$, $O_c$ be center of $\omega_b$, $\omega_c$ respectively $\, \,$Cuz $CI$ is the internal bisector of $\angle NCL$ so $O_cI \perp NL$. Similar, $O_bI \perp MT$ $\, \,$But $O_b$, $I$, $O_c$ are colibnear, so $NL$ $\|$ $MT$ $\, \,$Combine with claim $\textbf{1/}$, we see that : $$\angle N'IN = 2\angle NIA = 2(\angle AIC - \angle NIC) = 180^\circ + \angle BAC - 2\angle NIC = \angle BAC + \angle NQC - \angle NLC = \angle NQC - \angle BMT$$$\, \,$So $$ \angle PMN = \angle PMA + \angle AMN = \angle PC'B + \angle N'IN = \angle PBT + \angle NQC = 180^\circ - (\angle PQC - \angle NQC) = 180^\circ - \angle PQN$$$\, \,$Therefore, $M$, $N$, $P$, $Q$ are concyclic $\textbf{3/}$ Let $V$ be the midpoint of arc $BAC$ of $\Gamma$ Then $V$ is the intersection of lines $B'Q$, $C'P$ $\textit{Prove :}$ $\, \,$Cuz $Q$ is the second intersection of $\omega_c$ and $\Gamma$ so $\angle B'QC = \angle BIC = 90^\circ + \frac12 \angle BAC = \angle VQC$ $\, \,$Lead to $V$, $B'$, $Q$ are colinear. Similar, $P$, $C'$, $P$ are colinear $\textbf{4/}$$\triangle AXI$ is a isosceles triangle at $X$ $\textit{Prove :}$ $\, \,$Let $V_b$, $V_c$ be the intersections of lines $BI$, $CI$ with $\Gamma$ respectively ($V_b \neq B$, $V_c \neq C$) $\hspace{0.7cm}$$U_b$ be the intersection of lines $V_bQ$ and $B'N$ $\, \,$We ez to see that $V_bV_c$ is the perpendicular bisector of $AI$, so we just need to prove that $V_b$, $X$, $V_c$ are colinear $\, \,$Cuz $M$, $N$, $P$, $Q$ are concyclic from $\textbf{2/}$, so $X$ lie on radical axis of $\omega_b$ and $\omega_c$ $\, \,$Which means $XI$ tangent with both $\omega_b$, $\omega_c$ at $I$ $\, \,$Lead to $\triangle XNI \sim \triangle XIQ$. As a result : $\dfrac{XN}{XQ} = \dfrac{XN}{XI} .\dfrac{XI}{XQ} = \left( \dfrac{IN}{IQ} \right)^2 $ $\, \,$In the order hand : $\hspace{0.5cm}$From claim $\textbf{3/}$ and $NB' \perp AI$ at claim $\textbf{1/}$, we see that : $$\angle V_bQB' = \angle V_bQV = 90^\circ - \frac12(\angle BAC + \angle ABC) = 90^\circ - \angle AIB' = \angle NB'I = \angle V_bB'U_b$$$\hspace{0.5cm}$Lead to $\triangle V_bU_bB' \sim \triangle V_bB'Q$. As a result : $\dfrac{V_bU_b}{V_bQ} = \dfrac{V_bU_b}{V_bB'} .\dfrac{V_bB'}{V_bQ} = \left( \dfrac{B'U_b}{B'Q} \right)^2 $ $\, \,$Therefore $$ \dfrac{XN}{XQ} = \left( \dfrac{IN}{IQ} \right)^2 = \left( \dfrac{\sin \angle NQI}{\sin \angle INQ} \right)^2 = \left( \dfrac{\sin \angle V_bB'U_b}{\sin \angle V_bB'Q} \right)^2 = \left( \dfrac{V_bU_b}{B'V_b} .\dfrac{V_bB'}{V_bQ} \right)^2 = \dfrac{V_bU_b}{V_bQ} $$$\, \,$Hence, follow Thales theorem, we see that $XV_b$ $\|$ $NB'$. Similar, $XV_c$ $\|$ $MC'$ $\, \,$But from claim $\textbf{1/}$, we already have that $B'N$ $\|$ $C'M$, so $X$, $V_b$, $V_c$ are colinear $\textbf{Finally, from these claims, we have : }$ $\, \,$ Cuz $\triangle XMI \sim \triangle XIP$ and $AX = XI$, so $\triangle XMA \sim \triangle XAP$ $\, \,$Lead to $$\angle MAX = \angle APM = \angle APB - \angle MPB = 180^\circ - \angle ACB - \angle MTC = \angle LNC = \angle BCY = \angle BAY $$$\, \,$Similar, $\angle XAN = \angle CAY$. Therefore $A$, $X$, $Y$ are colinear, done

Claim: $Y$ lies on $(ABC)$. Proof: \[\measuredangle BYC= 180-\measuredangle CBY-\measuredangle YCB=180-(\measuredangle IMB-\frac{\measuredangle B}{2})-(\measuredangle CNI-\frac{\measuredangle C}{2})=270-\frac{\measuredangle A}{2}-\measuredangle CIB=180-\measuredangle A\]Which gives that $Y\in (ABC)$.$\square$ Claim: $I$ is the $A-$excenter on $\triangle AMN$. Proof: Let $O$ be the circumcenter of $(IMN)$. Since $\measuredangle MIN=\measuredangle MBI+\measuredangle ICN=90-\frac{\measuredangle A}{2},$ we have $\measuredangle MON=180-\measuredangle MAN$ hence $O$ lies on $(AMN)$. If $AI$ intersects $(IMN)$ at $J$ for second time, then \[\measuredangle IMB=\measuredangle MIJ+\frac{\measuredangle A}{2}=\measuredangle MNJ+\measuredangle ONM=\measuredangle ONJ=\measuredangle NJO=\measuredangle NMI\]Similarily we get $\measuredangle CNI=\measuredangle INM$ so $I$ is the $A-$excenter on $\triangle AMN$.$\square$ Claim: $M,N,P,Q$ are concyclic. Proof: Let $MN$ and incircle of $ABC$ be tangent to each other at $S$. Take the inversion centered at $I$ with radius $IS$. Let $D,E,F$ be the tangency points of the incircle with $BC,CA,AB$ respectively. $A^*,B^*,C^*,M^*,N^*$ are the midpoints of $EF,FD,DE,SF,SE$. $P^*,Q^*$ are the intersections of $B^*M^*,C^*N^*$ with $(A^*B^*C^*)$. Note that $M^*B^*\parallel SD\parallel N^*C^*$ and $M^*N^*\parallel EF\parallel B^*C ^*$. \[\measuredangle M^*P^*Q^*=180-\measuredangle Q^*P^*B^*=\measuredangle B^*C^*N^*=\measuredangle N^*M^*P^*\]Thus, $N^*M^*P^*Q^*$ is an isosceles trapezoid which yields $M,N,P,Q$ lie on a circle.$\square$ Claim: $A,X,Y$ are collinear. Proof: Let $PM$ and $QN$ intersect $(ABC)$ at $K,L$ for second time. \[\measuredangle YCA=\measuredangle YCN=\measuredangle CQN=\measuredangle CQL \]\[\measuredangle ABY=\measuredangle MBY=\measuredangle MPB=\measuredangle KPB\]These give that $AY=CL$ and $AY=BK$. Hence $YK\parallel AB$ and $YL\parallel AC$. Also \[\measuredangle KLN=\measuredangle KLQ=\measuredangle KPQ=\measuredangle MPQ=\measuredangle MNL\]Thus, $MN\parallel KL$. By Desargues on $\triangle AMN$ and $\triangle YKL$, since $AM\cap YK=AB_{\infty},MN\cap KL=MN_{\infty},NA\cap LY=AC_{\infty}$ are collinear, they are perspective. $AY,MK,NL$ are concurrent. $MK$ and $NL$ intersect at $X$ so $A,X,Y$ are collinear.$\blacksquare$

Why did I solve a G8 synthetically??? Let's denote with $t$ the common tangent of $w_b$ and $w_c$. Also let $U,V$ be the intersections of $BC$ with $w_b$ and $w_c$ Claim 1: $Y$ lies on $(ABC)$ Proof. $\angle BIU=\angle BCY=\phi$ and $\angle CIV=\angle CBY=\psi, \angle UIt=\angle IBC=\frac{\beta}{2}, \angle VIt=\angle ICB=\frac{\gamma}{2}$ Now: $$90+\frac{\alpha}{2}=\angle BIC=\angle BIU+\angle UIV+\angle CIV= \phi+\psi+90-\frac{\alpha}{2}$$So $\angle BYC=180-\phi-\psi=180- \alpha=\angle BAC \text{. }\square$ Claim 2: $MU$ || $NV$ || $t$ Proof. $\angle MUI=\angle MBI= \angle UBI = \angle UIt \Rightarrow $ $MU$ || $t$ and similarly $NV$ || $t$. $\square$ Claim 3: $MUVN$ is an isosceles trapezoid Proof. As $I$ is the midpoint of arcs $MU$ and $NV$ in $w_b,w_c$, this means that $MI=UI, IN=IV \Rightarrow \triangle MIN$ and $\triangle UIV $ are congruent $\Rightarrow MN=UV$. $\square$ Let $Q'$ and $P'$ be the intersections of $QX$ and $PX$ with $(ABC)$. Claim 4: $BQ'$ || $QC$ || $t$ Proof. Reims theorem on $(ABC), (NQCV)$ and lines $QX,CV$ gives that $BQ'$ || $NV$ and similarly $CP'$ || $UM$. $\square$ Let $K;G$ be the intersections of $AB;BC$ and $NV;MU$ Claim 5: $PMAG$ and $QNAK$ are cyclic Proof. $\angle KNQ=\angle VCQ=180-\angle BQ'Q=180-\angle BAQ=\angle KAQ$ and similarly $\angle PMG=\angle PAG$. $\square$ Claim 6: $(PMAG)$ and $(NAKQ)$ are tangent at $A$ Proof. As $GMNK$ is a trapezoid we have that the circumcenters of $GMA$ and $KAN$ lie on a line with $A$. $\square$ Claim 7: $PMNQ$ is cyclic Proof. $\angle MPQ= \angle BPQ-\angle BPM=(180-\angle BCQ)-\angle MUV=$ $(180-\angle KNQ)-\angle UMN=\angle XNK-\angle MNK=\angle MNX$. $\square$ Claim 8: $A,X,Y$ are collinear Proof. By radical axises on $(PMNQ), (PMAG), (NAKQ)$ we get that $AX$ is the tangent of $(PMAG)$ and $(NAKQ)$. $\angle YAC=\angle YBC= \angle BMU=\angle Q'BM=\angle Q'QA=\angle XAC$. $\square$

Amazing problem! also my first G8 We show $AY$ is tangent to $\odot(AMP), \odot(ANQ)$ $$\angle YAM+\angle MAP = 180^{\circ} - \angle YBP=180^{\circ}-\angle PBI-\angle BPI = \angle BIP = 180^{\circ} - \angle PMA$$Note that $Y\in{\odot(ABC)}$as \begin{align*} & \angle BIC=\angle IMB+\angle INC=\angle IBY+\angle ICY=90^{\circ}+\frac{\angle A}{2}\\ & \angle BYC=360-\angle YBI-\angle BIC-\angle ICY=360^{\circ}-2\left(90^{\circ}+\frac{\angle A}{2}\right)=180^{\circ}-\angle A \end{align*}By another angle chase we show $\odot(I)$ is the incircle of $MNCB$. $$\angle MIN = \angle MBI + \angle NCI = \angle IBC + \angle ICB = 180^{\circ}-\angle BIC=90-\frac{\angle A}{2}$$If we show $\odot(MPNQ)$ we get $X$ as the radical center of $(MNQP)$, $(AMP)$, and $(ANQ)$, which will clearly lie on radax of $(AMP)$, and $(ANQ)$ which is clearly $\overline{AY}$ implying $\overline{A-X-Y}$ is collinear. So finally we go on to another angle chase to prove $\odot(MPNQ)$ which took me a lotta time to figure out \begin{align*} \angle PQN=\angle CQN-\angle CQP&= 180^{\circ}-\angle CIN-\angle CQP\\&=180^{\circ}-(180^{\circ}-\angle INC-\angle ICN)-\angle CQP\\&=180^{\circ}-(180^{\circ}-\angle ICY-\angle ICN)-\angle CQP\\&=\angle ICY+\angle ICN-\angle CQP\\&=\angle ACY-\angle CQP\\&=(180^{\circ}-\angle ABY)-(180^{\circ}-\angle CBP)\\&=\angle CBP-\angle ABY\\&=\angle CBA+\angle ABP-\angle CBA-\angle YBC\\&=\angle ABP-\angle YBC\\ \angle PMN=\angle PMB+\angle BMN&=\angle PIB+2\angle BMI\\&=180^{\circ}-(\angle IPB+\angle PBI)+2\angle BMI\\&=180^{\circ}-(\angle IBY+\angle PBI)+2\angle BMI\\&=180^{\circ}-\angle PBY+2\angle BMI\\&=180^{\circ}-\angle PBY+2\angle CBY+\angle CBA\\&=180^{\circ}-\angle PBA+\angle CBY \end{align*}and we are finally done!

Here's my solution (angle-chase+invert+radical-axis+trig) : Claim-1: $Y \in (ABC)$ Notice that: $$\angle ABY+\angle ACY = 360^\circ - (\angle MIB + \angle NIC) = \angle MIC + \angle BIC = \angle MIX+\angle XIN + 90^\circ + \angle A/2$$$$ = \angle MBI + \angle NCI + 90^ + \angle A/2 = 180^\circ.$$ Claim-2: $MNQP$ is cyclic + $(AMP), (ANQ)$ are tangent. Invert around incircle. Let $X'$ denote the inverse of point $X$ in this inversion. Notice that: -$I$ is the orthocenter of $\triangle A'B'C'$. -$M'A'B'I$ and $N'A'C'I$ are cyclic ($A, M, B$ and $A, N, C$ are collinear). -$M', P', B'$ and $N', Q', C'$ are collinear ($MPIB$ and $NIAC$ are cyclic). -$M'P'B' \parallel N'Q'C'$. -$A',P',B',C',Q'$ lie on same circle. We will prove $M'N'Q'P'$ is cyclic. It suffice to show $M'N'=P'Q'$. Note that: $\angle M'IB'+\angle N'IC' = 360^\circ - (\angle B'M'I + \angle C'N'I) - (\angle M'B'I + \angle N'C'I) = 180^\circ.$ Now, we have: $$M'B' = \frac{B'I \sin \angle M'IB'}{\sin \angle M'B'I}$$Note that: $$\frac{B'I}{\sin \angle M'B'I} = 2R = \frac{C'I}{\sin \angle N'C'I}.$$Thus, we have: $\frac{B'I \sin \angle M'IB'}{\sin \angle M'B'I} = \frac{C'I \sin \angle N'IC'}{\sin \angle N'C'I} = N'C'$ which implies $M'B'C'N'$ is a parallelogram. Note that, from the following sub-claim, we will have: $P'Q' \parallel M'N' \parallel BC$ and $(A'M'P'), (A'N'Q')$ to be tangent at $A'$ which finishes the main-claim. Sub-Claim: $\triangle A'M'P'$ and $\triangle A'N'Q'$ are isosceles. Note that: $\angle A'M'B' = 180^\circ - \angle A'IB' = \angle A'C'B'$ and $\angle A'P'M' = \angle A'C'B'$ which shows that $A'M' = A'P'$. Similarly, $A'N'=A'P'$ and we are done. Hence, Claim-2 is true. Thus, from Claim-2, $AX$ is the radical axis of $(APM), (ANQ)$ along with $AX$ being tangent to both circles $(APM), (ANQ)$. Note that, the following claim finishes the problem: Claim-3: $AY$ is tangent to both circles $(APM), (ANQ)$. Let $R$ be a point of line $AP$ such that $A, R$ are on different sides with-respect to $P$. We will show it with an one-liner angle-chase: $$\angle BAY = 180^\circ - \angle AYB - \angle ABY = 180^\circ - \angle RPB - \angle BPM = \angle APM.$$Thus, $AY$ is tangent to $(APM)$. Similarly, $AY$ is tangent to $(ANQ)$ and we are done.

We prove that $Y$ lies on $(ABC)$ through angle chasing - $\angle ABY+\angle ACY=360-\angle MIB-\angle NIC=180^\circ.$ Next, $YA$ is tangent to both $(ANQ)$ and $(AMP)$ and is thus the radical axis, as $\angle APM=\angle BPA-\angle BPM=(180-\angle BYA)-\angle YBA=\angle YAB=\angle YAM$ and the same works for $(ANQ).$ Finally, we show $X$ lies on the radical axis by showing $MNQP$ are concyclic to finish. Let $\omega_b$ and $\omega_c$ intersect $BC$ at $R,S$ respectively. Observe by Fact 5 $\triangle IRS \cong \triangle IMN$ so $\angle INM=\angle ISR=\angle INC.$ \begin{align*} \angle QPM &= \angle MPB-\angle QPB \\ &= \angle NIC+\angle QCB-180 \\ &= \angle NIC+\angle BCI-\angle QNI \\ &=\angle NIC+\angle NCI-\angle QNI \\ &=180-\angle INC-\angle QNI \\ &=180-\angle INM-\angle QNI \\ &=180-\angle QNM \end{align*}so we are done.

Part 1 - The spiral similarity First, note that \[\angle BIC=\angle BMI+\angle CNI=\angle YBI+\angle YCI.\]Therefore, \[\angle BYC=360^{\circ}-\angle BIC-\angle BIC=180^{\circ}-\angle A.\]Therefore, $Y\in (ABC)$. Now note that \[\angle PBY=180^{\circ}-\angle PMB=\angle PMA\]\[\angle PYB=\angle PAB=\angle PAM\]so $\triangle PBY\sim\triangle PMA$. So there exists a spiral similarity centered at $P$ sending $AM$ to $YB$. Part 2 - Redefining $X$ Redefine $X$ to be the intersection of the perpendicular bisector of $AI$ with $AY$. Then $\angle IXY=2\angle IAX$, so $IX$ is parallel to the isogonal of $AY$ wrt angle $BAC$. In particular, \[\angle(BA,IX)=\angle YAC=\angle YBC=\angle(YB,CB)\Longrightarrow \angle(BI,IX)=\angle(YB,IB)\Longrightarrow \angle BIX=180^{\circ}-\angle YBI=180^{\circ}-\angle BPI.\]Therefore, $XI$ is tangent to $(BPI)$. Part 3 - The finish By our spiral similarity from earlier, \[\angle YAM=\angle YAB=\angle YPB=\angle APM.\]Therefore, $AY$ is tangent to $(APM)$. Then the powers from $X$ to $(APM)$ and $(BPM)$ are $AX^2$ and $IX^2$, respectively, which are equal, so $X$ lies on their radical axis $PM$. Similarly, $X$ lies on $QN$, as desired. $\blacksquare$