[asy][asy] defaultpen(fontsize(11pt)); size(9cm); pair A,B,C,I,O,N,M,T,D,J,O1,Q,K,E,X,S,U,V; A=dir(120); B=dir(210); C=dir(330); M=dir(90); I=incenter(A,B,C); O=circumcenter(A,B,C); N=-A+2*foot(O,A,I); T=-M+2*foot(O,M,I); D=foot(A,B,C); J=foot(I,A,D); O1=circumcenter(M,I,N); Q=-I+2*foot(O1,I,J); K=-N+2*foot(O,N,Q); E=extension(A,K,B,C); X=extension(M,N,A,K); S=-I+2*foot(N,I,N); U=-E+2*foot(E,B,S); V=-E+2*foot(E,C,S); draw(circumcircle(A,B,C),red); draw(A--B--C--cycle,blue); draw(A--N); draw(M--T); draw(A--M); draw(circumcircle(M,I,N),dotted+purple); draw(I--Q); draw(A--K); draw(T--K); draw(M--N); draw(circumcircle(M,I,A),dotted+purple); draw(N--Q); draw(circumcircle(E,U,V)); draw(C--V); draw(B--U); dot("$A$" , A , dir(A)); dot("$B$" , B , dir(B)); dot("$C$" , C , dir(C)); dot("$M$" , M , dir(M)); dot("$I$" , I , dir(210)); dot("$T$" , T , dir(T)); dot("$N$" , N , dir(N)); dot("$Q$" , Q , dir(30)); dot("$K$" , K , dir(K)); dot("$E$" , E , dir(240)); dot("$X$" , X , dir(60)); [/asy][/asy] $\textbf{LEMMA 1:-}$ $ABC$ be a triangle and let $P,Q$ be two points on $\odot(ABC)$ such that $\overline{AP}$ and $\overline{AQ}$ are Isogonal WRT $\Delta ABC$. Then $\overline{PQ}\parallel\overline{BC}$. Proof:- Let $\overline{AQ}\cap\overline{BC}=\{R\}$. Now $\angle BAP=\angle CAQ$ and $\angle APB=\angle ACR$ implies $\Delta ABP\stackrel{+}{\sim}\Delta ARC$ where $(\stackrel{+}{\sim}$) means that both the triangles are directly similar. So, $\angle ARB=\pi-\angle ABP=\angle AQP\implies\overline{PQ}\|\overline{BC}$. Now come back to the problem. Let $\overline{MI}\cap\odot(ABC)=\{T\}$ and $\overline{IQ}\cap\overline{MN}=\{X\}$. Also let $\overline{AX}\cap\odot(ABC)=\{K\}$. Now notice that, $\overline{IA}\perp\overline{AM}$ and $\overline{IX}\perp\overline{MN}$, so $A,I,X,M$ is a cyclic quadrilateral which implies $\measuredangle MIX=\measuredangle MAX=\measuredangle MAK=\measuredangle MTK\implies\overline{IX}\parallel\overline{BC}\parallel\overline{KT}$. So, $\{\overline{AT},\overline{AK}\}$ are Isogonal Lines WRT $\Delta ABC$ from $\textbf{LEMMA 1}$. Now, it's well known that $\{T\}$ is the $A-$ Mixtillinear Incircle touch point with $\odot(ABC)$ and that $\{\overline{AT},\overline{AE}\}$ are Isogonal Lines WRT $\Delta ABC$. So summing up these facts implies that $\{E\}\in\overline{AX}$. Some notation here, Directed angles are denoted by $\measuredangle$ and are assumed to be positive when taken in the counterclockwise direction. So, $\measuredangle MNK=\measuredangle MAX=\measuredangle MIX=\measuredangle MIQ=\measuredangle MNQ$. So, $N,K,Q$ are collinear. Hence, $\overline{AE}\cap\overline{NQ}\in\Omega$. $\blacksquare$

Let the incircle touch $\overline{BC}$ at $D$, and let $D_1$ be the antipode of $D$. It is known that $E$ lies on line $\overline{AD_1}$. Line $\overline{AE}$ meets $\Omega$ again at $X$, and $\overline{AE}$ and $\overline{IQ}$ intersect at $P$. [asy][asy] size(10cm); defaultpen(fontsize(9pt)); pair A = dir(135), B = dir(210), C = dir(-30), I = incenter(A, B, C), M = dir(90), N = dir(-90), D = foot(I, B, C), D1 = 2*I-D, E = B+C-D, P = extension(M, N, A, E), O = circumcenter(A, B, C), X = -A+2*foot(O, A, E), Q = extension(I, P, N, X); draw(unitcircle, blue); draw(circumcircle(M, I, N), orange); draw(arc(circumcenter(A, X, Q), circumradius(A, X, Q), 20, 160, CW), heavycyan); draw(incircle(A, B, C), lightblue); draw(circumcircle(A, I, M), springgreen); draw(A--B--C--cycle, red); draw(A--X, fuchsia); draw(A--N, heavymagenta); draw(M--N, heavyred); draw(D--D1); draw(I--Q, deepcyan); draw(N--Q, dashed); string[] names = {"$A$", "$B$", "$C$", "$D$", "$D_1$", "$E$", "$I$", "$M$", "$N$", "$P$", "$Q$", "$X$"}; pair[] points = {A, B, C, D, D1, E, I, M, N, P, Q, X}; pair[] ll = {A, B, C, D, dir(240), NE, SW, M, dir(250), SW, Q, dir(-80)}; for (int i=0; i<names.length; ++i){dot(names[i], points[i], dir(ll[i]));} [/asy][/asy] Claim: $M,P,N$ are collinear. Proof. Focus on $\triangle D_1DE$. Let $M_1$ be the midpoint of $\overline{BC}$. It is known that $M_1$ is also the midpoint of $\overline{DE}$. Since $I$ is the midpoint of $\overline{DD_1}$ and $\overline{IP} \parallel \overline{DE}$, $\overline{PM_1} \perp \overline{BC}$. But since $\overline{MN}$ passes through $M_1$ and is perpendicular to $\overline{BC}$, it must also pass through $P$, as claimed. $\square$ Now $AMPI$ is cyclic, we $\measuredangle MAI = \measuredangle MPI = 90^\circ$. By power of a point, we have \[ PA \cdot PX = PM \cdot PN = PI \cdot PQ, \]so $AIXQ$ is cyclic. Finally, we show that $Q, X, N$ are collinear by angle chase: \[ \measuredangle AXQ = \measuredangle AIQ = \measuredangle AMP = \measuredangle AXM, \]as desired.

This problem was proposed by Anant Mudgal, India.

[asy][asy] size(13cm); defaultpen(fontsize(11pt)); import olympiad; pair excenter(pair A=(0,0), pair B=(0,0), pair C=(0,0)) { return extension(A,bisectorpoint(B,A,C),B,rotate(90,B)*bisectorpoint(C,B,A)); } pair A = dir(130), B = dir(210), C = dir(-30), M = dir(90), N = dir(270); pair I = incenter(A,B,C); pair Ia = excenter(A,B,C); pair O = (0,0); pair E = foot(Ia, B, C); pair P = 2*foot(O,A,E) - A; pair Q = extension(I, I+10*C -10*B, N, P); pair V = 2*foot(P,M,N) - P; draw(A--B--C--cycle, royalblue); draw(circumcircle(M,I,N), fuchsia); draw(circumcircle(A,B,C), purple); draw(circumcircle(E, foot(Ia,A,3*B-2*A), foot(Ia,A,3*C-2*A)), dashed+springgreen); draw(N--Q, fuchsia); draw(A--P, fuchsia); draw(I--Q, dashed+cyan); draw(N--M, brown); draw(V--M--P, orange); draw(A--Ia, brown); dot("$A$",A, dir(A)); dot("$B$",B, dir(B)); dot("$C$",C, dir(30)); dot("$I$",I,dir(180)); dot("$E$",E,dir(40)); dot("$I_a$",Ia,dir(Ia)); dot("$P$",P,dir(P)); dot("$Q$",Q); dot("$M$",M,dir(130)); dot("$N$",N,dir(230)); dot("$V$",V,dir(180)); [/asy][/asy] Define $P$ as the intersection of $AE$ and $(ABC)$ and redefine $Q$ as the intersection of $NP$ and the line through $I$ parallel to $BC$. We wish to show that $Q$ lies on $MIN$, or $\angle NQI = \angle NMI$ Let $V$ be where the A-mixtillinear circle touches $(ABC)$, and recall that $MIV$ are collinear. Also recall that $AV$ and $AP$ are isogonal. Together, this gives that $\angle IMN = \angle VMN = \angle VAN = \angle PAN = \angle PMN$ Also, note that $\angle PMN = 90 - \angle PNM = \angle(NP,BC) = \angle NQI$ $\blacksquare$

$\newline \underline{\textbf{Solution:}}$ Let the line parallel through $I$ to $BC$ intersect $(BIC)$ at $Y ; AP \cap (ABC) = X ,$ where $P$ is the midpoint of $IY$ $\newline$ Claim: $AE$ and $MY$ meet on $(ABC)$ $\newline$ ${\textbf{\underline{Proof:}}}$ We know that $NP\perp IY$ as $N$ is the center $(BIC)$ by ${\textbf{Incenter-Excenter lemma}}.$ Hence, there exists a Homothety sending $NP$ to $EI_{A}$ (where $I_{A}$ is the $A-$excenter of $\triangle ABC$ ) centered at $A$ as $NP \parallel EI_{A}$. Let $IP\cap (ABC) = K,L$ , where $K$ lies on minor arc $AB$. Let $MX\cap KL = Y'$. As we know that $AX$ and $MN$ intersect at $P,$ we know that $\overline{A-I-N}$ and $\overline{M-Y'-X}$, but by $\textbf{Butterfly Theorem}$ we can also say that $P$ is the midpoint of $IY'.$ But $P$ is the midpoint of $IY$ . Hence, $Y' \equiv Y $ $\newline \newline$ Claim: $Y$ is the orthocenter of $\triangle QMN$ $\newline$ ${\textbf{\underline{Proof:}}}$ Note that $IQ\perp MN$, as $I\in (QMN)$ and also the reflection of $Y$ over feet of $Q-$altitude in $\triangle QMN$, by converse of 'Reflection of Orthocenter over its feet of altitude lies on the circumcircle' we conclude that $Y$ is the orthocenter of $\triangle QMN$. [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 = -16.26014752425227, xmax = 15.677042404757444, ymin = -7.628717846328894, ymax = 11.466886575977151; /* image dimensions */ pen ffcctt = rgb(1,0.8,0.2); pen ffzzqq = rgb(1,0.6,0); pen ffzztt = rgb(1,0.6,0.2); /* draw figures */ draw((-4.771377946595897,6.637510030563354)--(-5.351220989346846,1.9280531223665875), linewidth(0.8) + ffcctt); draw((-5.351220989346846,1.9280531223665875)--(2.16,2.05), linewidth(0.8) + ffcctt); draw((2.16,2.05)--(-4.771377946595897,6.637510030563354), linewidth(0.8) + ffcctt); draw(circle((-1.6259832895844184,3.85981471966658), 4.196319671871327), linewidth(0.8) + ffzzqq); draw(circle((-1.5578637849485213,-0.33595201856165724), 4.417610005230225), linewidth(0.8) + linetype("2 2") + ffzztt); draw(circle((2.411731725234182,3.92536820817784), 5.823790740512733), linewidth(0.8) + linetype("2 2") + ffzztt); draw((-1.6941027942203146,8.05558145789482)--(1.3674241777130995,0.9189627133150626), linewidth(0.8)); draw((-4.771377946595897,6.637510030563354)--(1.3674241777130995,0.9189627133150626), linewidth(0.8)); draw((-5.816385030225462,3.637032223570495)--(8.235211273092387,3.8651640143119916), linewidth(0.8)); draw((-1.6941027942203146,8.05558145789482)--(-1.5578637849485202,-0.3359520185616602), linewidth(0.8)); draw((-4.771377946595897,6.637510030563354)--(0.2909967504490434,-4.3480578096593465), linewidth(0.8)); draw((-1.5578637849485202,-0.3359520185616602)--(8.235211273092387,3.8651640143119916), linewidth(0.8)); draw((-1.6941027942203146,8.05558145789482)--(8.235211273092387,3.8651640143119916), linewidth(0.8)); /* dots and labels */ dot((-4.771377946595897,6.637510030563354),dotstyle); label("$A$", (-5.023760205886583,6.7763723892753385), NE * labelscalefactor); dot((-5.351220989346846,1.9280531223665875),dotstyle); label("$B$", (-5.732549016321524,1.7314637973560558), NE * labelscalefactor); dot((2.16,2.05),dotstyle); label("$C$", (2.4185223036803025,1.9190843648241283), NE * labelscalefactor); dot((-3.406724320346087,3.6761537725360323),linewidth(2pt) + dotstyle); label("$I$", (-3.3143283689552536,3.8369834989422027), NE * labelscalefactor); dot((0.2909967504490434,-4.3480578096593465),linewidth(2pt) + dotstyle); label("$I_A$", (0.3755427912501771,-4.189007442747565), NE * labelscalefactor); dot((0.18764228722721313,2.0179781935600145),linewidth(2pt) + dotstyle); label("$E$", (0.2713091426568034,2.1900918511669), NE * labelscalefactor); dot((1.3674241777130995,0.9189627133150626),linewidth(2pt) + dotstyle); label("$X$", (1.459572736621264,1.0852151760771394), NE * labelscalefactor); dot((-1.5578637849485202,-0.3359520185616602),linewidth(2pt) + dotstyle); label("$N$", (-1.4798161537118757,-0.16558860704334388), NE * labelscalefactor); dot((-1.6941027942203146,8.05558145789482),linewidth(2pt) + dotstyle); label("$M$", (-1.6048965320239241,8.214796739863894), NE * labelscalefactor); dot((-1.623471538445449,3.7051054049127514),linewidth(2pt) + dotstyle); label("$P$", (-1.5006628834305504,3.878676958379552), NE * labelscalefactor); dot((0.15978124345518563,3.7340570372894737),linewidth(2pt) + dotstyle); label("$Y$", (0.2504624129381286,3.8995236880982267), NE * labelscalefactor); dot((-5.816385030225462,3.637032223570495),linewidth(2pt) + dotstyle); label("$K$", (-5.732549016321524,3.795290039504853), NE * labelscalefactor); dot((2.5694419533345614,3.7731785862550105),linewidth(2pt) + dotstyle); label("$L$", (2.647836330585725,3.941217147535576), NE * labelscalefactor); dot((8.235211273092387,3.8651640143119916),linewidth(2pt) + dotstyle); label("$Q$", (8.318146814065257,4.024604066410276), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $MY\cap QN = X'$. Therefore, $\angle MX'Q = \angle MX'N = 90^{\circ}$ which means that $X' \in (ABC)$, But $MY\cap (ABC) = X$, Hence, $X' \equiv X$ which means $\overline{MY}$ and $\overline{QN}$ intersect on $(ABC)$. $\newline \rule{\textwidth}{0.5pt} \newline$ Now, note that pair of lines $( AE, MY )$ and $( MY , QN)$ meet on $(ABC)$; but as $M$ lies on $(ABC)$ they pass through a common point on $(ABC) $ which is $X$. Hence, $AE$ and $QN$ meet on $(ABC)$

Let,$AE$ meet $\Omega$ and $MN$ at $H$ and $F$ respectively.Let incircle of $ABC$ touches $BC$ at $D$ and $D'$ be the reflection of $D$ across $I$.Obviously,$A,D',E$ are collinear. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(16.cm); 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 = 0., xmax = 16., ymin = -5., ymax = 6.; /* image dimensions */ pen xfqqff = rgb(0.4980392156862745,0.,1.); pen ffqqff = rgb(1.,0.,1.); /* draw figures */ draw((2.1,2.9)--(1.38,-0.96), linewidth(0.8) + blue); draw((6.96,-0.96)--(1.38,-0.96), linewidth(0.8) + blue); draw((2.1,2.9)--(6.96,-0.96), linewidth(0.8) + blue); draw(circle((4.17,0.5167357512953366), 3.1567148238562512), linewidth(0.8) + blue); draw(circle((3.0300961532151622,0.41076655831815934), 1.3707665583181594), linewidth(0.8)); draw((3.0300961532151622,1.7815331166363189)--(3.0300961532151627,-0.96), linewidth(0.8)); draw((2.1,2.9)--(6.1358077813556005,-1.953172798816503), linewidth(0.8) + xfqqff); draw((4.17,3.6734505751515876)--(2.2041922186443967,-1.9531727988165017), linewidth(0.8) + xfqqff); draw(circle((7.96603869825883,0.5167357512953362), 4.93707993431668), linewidth(0.8) + ffqqff); draw((2.1,2.9)--(2.2041922186443967,-1.9531727988165017), linewidth(0.8) + xfqqff); draw((2.153441071596689,0.4107665583181613)--(12.901981243302496,0.41076655831813724), linewidth(0.8) + xfqqff); draw((1.2657415013864235,1.7537204126329868)--(6.1358077813556005,-1.953172798816503), linewidth(0.8) + xfqqff); draw((1.2657415013864235,1.7537204126329868)--(4.17,-2.6399790725609145), linewidth(0.8) + xfqqff); draw(circle((2.591768612405925,1.665367343787867), 1.328967329896817), linewidth(0.8) + dotted + red); draw((4.17,3.6734505751515876)--(4.17,-2.6399790725609145), linewidth(0.8) + xfqqff); /* dots and labels */ dot((2.1,2.9),linewidth(2.pt) + dotstyle); label("$A$", (1.98,3.), NE * labelscalefactor); dot((1.38,-0.96),linewidth(2.pt) + dotstyle); label("$B$", (1.04,-1.24), NE * labelscalefactor); dot((6.96,-0.96),linewidth(2.pt) + dotstyle); label("$C$", (7.04,-1.2), NE * labelscalefactor); dot((4.17,3.6734505751515876),linewidth(2.pt) + dotstyle); label("$M$", (4.,3.82), NE * labelscalefactor); dot((4.17,-2.6399790725609145),linewidth(2.pt) + dotstyle); label("$N$", (4.06,-3.04), NE * labelscalefactor); dot((3.0300961532151622,0.41076655831815934),linewidth(2.pt) + dotstyle); label("$I$", (2.76,0.08), NE * labelscalefactor); dot((3.0300961532151627,-0.96),linewidth(2.pt) + dotstyle); label("$D$", (2.9,-1.34), NE * labelscalefactor); dot((3.0300961532151622,1.7815331166363189),linewidth(2.pt) + dotstyle); label("$D'$", (2.7,1.42), NE * labelscalefactor); dot((2.2041922186443967,-1.9531727988165017),linewidth(2.pt) + dotstyle); label("$T$", (1.8,-2.26), NE * labelscalefactor); dot((5.309903846784835,-0.96),linewidth(2.pt) + dotstyle); label("$E$", (5.38,-0.88), NE * labelscalefactor); dot((6.1358077813556005,-1.953172798816503),linewidth(2.pt) + dotstyle); label("$H$", (6.2,-2.24), NE * labelscalefactor); dot((4.17,0.4107665583181568),linewidth(2.pt) + dotstyle); label("$F$", (3.9,0.12), NE * labelscalefactor); dot((12.901981243302496,0.41076655831813724),linewidth(2.pt) + dotstyle); label("$Q$", (13.04,0.32), NE * labelscalefactor); dot((2.153441071596689,0.4107665583181613),linewidth(2.pt) + dotstyle); label("$G$", (2.24,0.5), NE * labelscalefactor); dot((1.2657415013864235,1.7537204126329868),linewidth(2.pt) + dotstyle); label("$J$", (0.96,1.7), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We claim that $IF||BC$.Indeed,since $MN$ passes through midpoint of $DE$,hence midpoint theorem implies that it passes through the midpoint of $D'E$.So $F$ is the midpoint if $D'E$.Hence,$IF||BC$. Let,$T$ be the $A$-mixtilinear intouch point with $\Omega$.Consider the inversion $\Psi$ with center $I$ with radius $IA\times IN$.Therfore,$\Psi$ maps $\Omega$ to itsel.Since $\Psi(M)=T$ and $\Psi(N)=A$ so $\odot(MIN)\to AT$.Therefore,$\Psi(Q)=AT\cap QI$.Call this point $G$.So,$F-I-G$ are collinear. Let $AE$ meet $\Omega$ at $H$.Now,applying converse of pascal theorem on $\{A,T,M,N,HI\cap NG,H\}$ and collinear points $G-I-F$, we get that $J=HI\cap NG\in\Omega$. But then we have, $$\angle IJH=\angle HJN=\angle NMT=\angle NAT=\angle IAG$$. Hence $IGJA$ are concyclic.But $\Psi(G)=Q;\Psi(J)=H;\Psi(A)=N$.Hence,$N-H-Q$ are collinear. $\blacksquare$

[asy][asy] size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -21.05546249792678, xmax = 23.442401419725854, ymin = -27.48358203809951, ymax = 10.315678709153804; /* image dimensions */pen qqwuqq = rgb(0,0.39215686274509803,0); pen fuqqzz = rgb(0.9568627450980393,0,0.6); pen zzttff = rgb(0.6,0.2,1); /* draw figures */draw(circle((-0.4217377473531326,-6.1691286147145705), 9.658051152268548), linewidth(0.8) + blue); draw(circle((6.603683428815336,-6.4481757175063095), 11.946227942240538), linewidth(0.8) + qqwuqq); draw((-8.455708915509009,-0.8087872702593557)--(-9.133543760132753,-10.338347732910771), linewidth(0.8)); draw((-9.133543760132753,-10.338347732910771)--(7.931945269217903,-11.016182577534511), linewidth(0.8) + fuqqzz); draw((7.931945269217903,-11.016182577534511)--(-8.455708915509009,-0.8087872702593557), linewidth(0.8)); draw((-7.99101738605814,-12.167791616661058)--(-0.03842581269693579,3.481313034276731), linewidth(0.8)); draw((-8.455708915509009,-0.8087872702593557)--(6.647917490837926,-12.749244637472346), linewidth(0.8)); draw((-0.8050496820093294,-15.819570263705872)--(18.46359471220613,-7.881656236407207), linewidth(0.8) + linetype("4 4") + zzttff); draw((-5.332560501827347,-6.936481847064757)--(18.46359471220613,-7.881656236407207), linewidth(0.8) + fuqqzz); draw((-7.99101738605814,-12.167791616661058)--(6.647917490837926,-12.749244637472346), linewidth(0.8) + fuqqzz); /* dots and labels */dot((-8.455708915509009,-0.8087872702593557),dotstyle); label("$A$", (-9.612015415161272,-1.0878957356926606), NE * labelscalefactor); dot((-9.133543760132753,-10.338347732910771),dotstyle); label("$B$", (-10.449340811461187,-11.295291042967817), NE * labelscalefactor); dot((7.931945269217903,-11.016182577534511),dotstyle); label("$C$", (8.729398027598775,-10.577583560425033), NE * labelscalefactor); dot((-5.332560501827347,-6.936481847064757),linewidth(4pt) + dotstyle); label("$I$", (-6.2228411920425675,-6.470701854763544), NE * labelscalefactor); dot((-0.8050496820093294,-15.819570263705872),linewidth(4pt) + dotstyle); label("$N$", (-1.4381246417573381,-17.036950903310093), NE * labelscalefactor); dot((-0.03842581269693579,3.481313034276731),linewidth(4pt) + dotstyle); label("$M$", (-1.2387614521621202,4.334783021297266), NE * labelscalefactor); dot((18.46359471220613,-7.881656236407207),linewidth(4pt) + dotstyle); label("$Q$", (18.617812231521583,-7.54726307857772), NE * labelscalefactor); dot((6.647917490837926,-12.749244637472346),linewidth(4pt) + dotstyle); label("$D$", (6.536402942051378,-13.966757783543738), NE * labelscalefactor); dot((4.2718571853908545,-10.87080524710213),linewidth(4pt) + dotstyle); label("$E$", (3.4662098222850224,-11.853507973834427), NE * labelscalefactor); dot((-7.99101738605814,-12.167791616661058),linewidth(4pt) + dotstyle); label("$F$", (-8.176600450075703,-13.368668214758083), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Suppose $MI$ and $AE$ intersect $\Omega$ for the second time at $F$ and $D$ respectively. It is well-known that $F$ is the touching point of the $A$-mixtillinear circle, and that $AF,AD$ are isogonals with respect to $\angle BAC$ (this can be proved easily via an inversion at $A$ with radius $\sqrt{AB\cdot AC}$ with a reflection in $AI$.) Therefore, $\angle BCF=\angle BAF=\angle CAD=\angle CFD$ so $FD\|BC\|IQ$. This implies $$\angle MND=\angle MFD=\angle MIQ=\angle MNQ$$hence $N,D,Q$ are collinear as desired.

Cool problem Let $AE$ meet the perpendicular bisector of $BC$ at $P$. Let the incircle meet $BC$ at $D$ and let the midpoint of $BC$ be $Z$. We have that $DX = EX$ and since $AE || IZ$, we also have that $\angle IZD = \angle PED = \angle PDE = \angle PDZ$ and because $\angle IDZ = \angle PZD = 90^\circ$, we see that $IPZD$ is a rectangle. So, we have that $I,P,Q$ are collinear and also $AIXM$ is cyclic because $\angle IAM = \angle IXM = 90^\circ$. This means $\angle MNK = \angle MAK = \angle MAX =\angle MIX = \angle MIQ = \angle MNQ$ and so $N,K,Q$ are collinear, as desired. $\blacksquare$

Define X as the projection of I onto MN. Since triangles MAN and MXQ are similar there must exist a spiral similarity that sends AN to XQ. If we define R to be the intersection of AX and QN, It is known that the center of said spiral similarity (M) will be the intersection of the circumcircles of triangles ANR and MNR. That is, AX and Qn concur on (ABCM). It is pretty straightforward to see that triangles AXN and AEJ (J is the excenter) are similar (can be done by SAS, verifying that AN/ AJ=XN/ EJ). Then, a homothety h takes AXN to AEJ, and A-X-E are colinear. The conclusion follows

Let $R$ be the midpoint of $BC$. Let $D$ be the foot from $I$ to $BC$. It is well-known that $R$ is the midpoint of $DE$ and that $AE$ passes through antipode of $D$ wrt $(I)$. Let $H$ be the point such that $IHRD$ is a rectangle. Let $F=(ABC)\cap AE$. Thus, by similarity $H$ lies on $AE$. Also $H$ lies on $MN$ and $IQ$ obviously. As $\measuredangle MHI=90^\circ=\measuredangle MAI$, we have $AMHI$ is cyclic. Hence, $$\measuredangle QNM=\measuredangle QIM=\measuredangle HIM=\measuredangle HAM=\measuredangle FAM=\measuredangle FNM,$$we conclude that $N,F,Q$ are collinear. [asy][asy] size(9cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;pen dark=darkmagenta;pen heavy=heavymagenta; pair O,A,B,C,I,D,N,M,E,R,F,H,Q; O=(0,0);A=dir(120);B=dir(200);C=dir(340);I=incenter(A,B,C);path w=circumcircle(A,B,C);D=foot(I,B,C); N=intersectionpoints(w,A--100I-99A)[1];M=-N;E=2midpoint(B--C)-D;R=midpoint(B--C);F=intersectionpoints(w,A--100E-99A)[1]; H=I+R-D;Q=extension(I,H,N,F); draw(A--B--C--cycle,deep);draw(w,heavyblue);draw(circumcircle(M,I,N),heavyblue);draw(A--N,deep);draw(A--F,deep);draw(M--N,deep);draw(I--Q,deep);draw(N--Q,deep+dashed);draw(circumcircle(I,H,A),heavyblue);draw(I--D,deep); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$I$",I,dir(I)); dot("$D$",D,dir(D)); dot("$N$",N,dir(N)); dot("$M$",M,dir(M)); dot("$E$",E,dir(E)); dot("$R$",R,dir(R)); dot("$F$",F,dir(F)); dot("$H$",H,dir(H)); dot("$Q$",Q,dir(Q)); [/asy][/asy]

$F=AN \cap BC, K=MN \cap BC, MI \cap BC=T, MI\cap (ABC)=P, X=PN \cap BC, L=IQ\cap MN, S=(AI)\cap (ABC),U=NQ\cap (ABC),D$ is the touch point between $(I)$ and $BC$. $NI^2=NK\cdot NM=NP\cdot NX \implies \angle XIN=90$. Radical axis theorem with $(AI), (BIC), (ABC)$ gives $S\in AX$. It is well known that $S\in ND$. $\angle XAN=\angle SAN=\angle TDA=\angle NEX\implies AENX$ is cyclic. $\angle ALI=\angle AMP=\angle ANP=\angle XEA \implies A-L-E$ are collinear. $\angle MNU=\angle MIL=\angle MAL=\angle MAE \implies A-E-U$ are collinear as desired.

An easy solution Let MI meet circumcircle at G and AE meet at F. we know that AI is also bisector of ∠GAF so GF || BC || IQ. we need to prove Q,F,N are collinear. ∠IQN = ∠IMN = ∠GMN = ∠GFN and we have GF || IQ ---> Q,F,N are collinear and we're Done.

Suppose $AE$ and $MI$ meet $\Omega$ again at $K, T_a$ respectively. Redefine $Q$ as the point on $NK$ such that $IQ \parallel BC$. Let $IQ$ meet $MN$ at $F$. It's well-known that the $A$-Mixtilinear Incircle contains $T_a$, so $AE$ and $AT_a$ are isogonal in $\angle BAC$, which implies $K, T_a$ are symmetric wrt $MN$. It's easy to see $(AIFM)$ and $(T_aIFN)$ are cyclic by Thales'. Hence, Radical Axes imply $AM, FI, NT_a$ concur at some point $P$. Since $\Omega$ has diameter $MN$, we know $I$ is the orthocenter of $MNP$. Now, the Orthocenter Reflection Lemma finishes, as aforementioned symmetry yields $FP = FQ$. $\blacksquare$ Remark: Inverting about $(BIC)$ also looks promising, as the Diameter of the Incircle Lemma gives $IM^* \parallel AE$.

Neat problem! First we forget about point $E$. Let $QN \cap \Omega \equiv T$ and $QI \cap MN \equiv V$. Clearly $AIVM$ is cyclic Claim $A,V,T$ are collinear. (Proof)Let $AT$ intersects $MN$ at $V'$. $$\angle V'AM=\angle TAM=\angle TNM=\angle QNM=\angle QIM=\angle VIM=\angle VAM$$$$\angle V'AM=\angle VAM$$Now some position arguments give that $V' \equiv V$ $\blacksquare$ Now we let ray $MI$ hit $\Omega$ again at $U$. It is well known that $U$ is the A-mixtilinear intouch point. $$\angle TAN=\angle VAI=\angle VMI=\angle NMU=\angle NAU$$Therefore we have that $AU$ and $AT$ are isogonal lines. We also know $AU$ and $AE$ are isogonal lines. Therefore $E$ lies on $AT$, as required.

Claim: $\overline{IQ},$ $\overline{AE},$ and $\overline{MN}$ concur. Proof. Let $D$ be the foot from $I$ to $\overline{BC},$ let $D'$ be the antipode of $D$ with respect to the incircle, and let $L$ be the midpoint of $\overline{BC}.$ It is well known that $D'$ lies on $\overline{AE}$ and $L$ is the midpoint of $\overline{DE}.$ Hence, $\overline{IQ}$ and $\overline{MN}$ are the perpendicular bisectors of $\overline{DD'}$ and $\overline{DE},$ and therefore concur at the center of $\triangle DD'E,$ which is the midpoint of $\overline{D'E}.$ $\blacksquare$ Let $Y=\overline{AX}\cap\Omega.$ Notice $AIXM$ is cyclic as $\measuredangle MXI=\measuredangle MLB=90=\measuredangle MAI.$ Thus, $$\measuredangle YNM=\measuredangle YAM=\measuredangle XIM=\measuredangle QIM=\measuredangle QNX.$$$\square$

Nice Problem Let $AE \cap \odot(ABC)=F$ and let $MI \cap \odot(ABC)=T$ Claim: $IQ,AE,MN$ concur. Proof: Let $MN \cap IQ = P$, trivially $AMPI$ is cyclic which means that we have to prove $\angle AEB = \angle AHI = \angle AMI = \angle AMT$ which boils down to proving $T$ and $E$ are isogonal which is true as $T$ is the mixtillinear touchpoint (Well known, EGMO Problem 4.39) $\square$ Now it can be easily verified that $\angle AIQ = \angle AFQ=180^{\circ}-\angle NFA$ which finishes the problem $\blacksquare$

$\sqrt{bc}$ inversion ftw. Let $MI \cap \Omega=T_A$ (its known that $T_A$ isthe A-mixtilinear intouch point with $\Omega$) and $AE \cap \Omega=L$, let $(AIM) \cap (T_AIN)=D$, now perform a $\sqrt{bc}$ inversion, $N$ is sent to $AI \cap BC=N'$, $M$ is sent to $AM \cap BC=M'$, $D$ is sent to $M'I_A \cap (N'EI_A)=D'$, since $MN$ is diameter in $\Omega$ we have $M,N,D$ colinear which gives that $AM'N'D'$ are cyclic, now its known that $M'AEI_A$ is cyclic and now by angle chasing: $$\angle M'AT_A=\angle MNT_A=\angle MNL=\angle MAE=\angle EI_AM'=\angle M'N'D'=\angle M'AD' \implies A,T_A,D' \; \text{colinear}$$And inverting that we have that $A,D,E$ are colinear, now redefine $Q$ as $ID \cap NL$, let $AT_A \cap BC=T$, now $Q$ is sent to the miquel point of $D'TN'I_A$, which means that $M'Q'N'I_A$ is cyclic, and inverting back that gives $MINQ$ cyclic, and since $ID \perp MN$ we have that $IQ \parallel BC$ hence the $Q$ we re-made is exactly the $Q$ of the problem statement, thus we are done

amar_04 wrote: In triangle $ABC$ with circumcircle $\Omega$ and incenter $I$, point $M$ bisects arc $BAC$ and line $\overline{AI}$ meets $\Omega$ at $N\ne A$. The excircle opposite to $A$ touches $\overline{BC}$ at point $E$. Point $Q\ne I$ on the circumcircle of $\triangle MIN$ is such that $\overline{QI}\parallel\overline{BC}$. Prove that the lines $\overline{AE}$ and $\overline{QN}$ meet on $\Omega$. Let $X=AE\cap (ABC),Q'=IQ\cap NX$ and $T$ the tangent point of the A-mixtilinear. We are going to prove that $M,I,N,Q'$ are cyclic. $\angle IQ'N=180-\angle NIQ'-\angle INQ'=180-\angle AKB-\angle ANX=180-\angle C-\frac{\angle A}{2}-\angle B-\angle XAC=\frac{\angle A}{2}-\angle XAC=\angle TAN=\angle IMN$

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