[asy][asy] size(8cm); defaultpen(fontsize(10pt)); pair A = dir(125), B = dir(210), C = dir(330), M = dir(90), N = dir(270), D = dir(55), I = incenter(A,B,C), T = intersectionpoint(unitcircle,I--I+dir(M--I)*100), F = extension(A,T,I,I+dir(0)), G = F+dir(N--F)*abs(A-F)*abs(T-F)/abs(N-F), L = extension(A,G,I,F), K = extension(D,I,A,T), S = extension(M,L,N,K), T1 = 2*foot(T,M,N)-T; draw(A--B--C--A--N^^unitcircle, heavyblue); draw(D--A^^L--I^^B--C^^T--T1, heavyblue+linewidth(1.2)); draw(A--T--M^^D--K, orange); draw(N--G^^A--L, purple); draw(M--S^^N--S, red); draw(L--S, red+dashed); draw(circumcircle(A,G,F), cyan); draw(circumcircle(M,F,I), magenta); draw(S--T1, heavygreen+dashed); dot("$A$", A, dir(120)); dot("$B$", B, dir(210)); dot("$C$", C, dir(330)); dot("$D$", D, dir(60)); dot("$M$", M, dir(90)); dot("$N$", N, dir(270)); dot("$I$", I, dir(330)); dot("$F$", F, dir(45)); dot("$K$", K, dir(210)); dot("$G$", G, dir(135)); dot("$S$", S, dir(150)); dot("$L$", L, dir(210)); dot("$T$", T, dir(225)); dot("$T'$", T1, dir(315)); [/asy][/asy] Let $T = \overline{MI} \cap \overline{AF}$ be the $A$-mixtilinear intouch point, and let $S = \overline{NK} \cap (ABC)$. Since $\angle NSM = 90^\circ$, we just want $\overline{AG}$, $\overline{MS}$, $\overline{IF}$ concur. By (degenerate) Reim's we have $AGFI$ cyclic, so by radical axis it suffices to show $MIFS$ cyclic. Let $T' = \overline{SF} \cap (ABC)$. Then $$(AN;TT') \overset{S}= (AK;TF) \overset{I}= (AD;\overline{TM} \cap \overline{AD}, \infty_{AD}) = \frac{TA}{TD},$$which implies $\overline{TT'} \parallel \overline{BC}$, from where it follows by Reim's that $MIFS$ is cyclic, as desired. @2below it's because $(AN;TT')$ only depends on $T'$, and when $\overline{TT'} \parallel \overline{BC}$ the equation is true. Technically the harmonic conjugate of $T'$ would also work, but I was too lazy to address that (should be clear from $G-F-N$ that $T'$ must lie on the opposite side of $\overline{AN}$ as $T$).

MP8148 wrote: [asy][asy] size(8cm); defaultpen(fontsize(10pt)); pair A = dir(125), B = dir(210), C = dir(330), M = dir(90), N = dir(270), D = dir(55), I = incenter(A,B,C), T = intersectionpoint(unitcircle,I--I+dir(M--I)*100), F = extension(A,T,I,I+dir(0)), G = F+dir(N--F)*abs(A-F)*abs(T-F)/abs(N-F), L = extension(A,G,I,F), K = extension(D,I,A,T), S = extension(M,L,N,K), T1 = 2*foot(T,M,N)-T; draw(A--B--C--A--N^^unitcircle, heavyblue); draw(D--A^^L--I^^B--C^^T--T1, heavyblue+linewidth(1.2)); draw(A--T--M^^D--K, orange); draw(N--G^^A--L, purple); draw(M--S^^N--S, red); draw(L--S, red+dashed); draw(circumcircle(A,G,F), cyan); draw(circumcircle(M,F,I), magenta); draw(S--T1, heavygreen+dashed); dot("$A$", A, dir(120)); dot("$B$", B, dir(210)); dot("$C$", C, dir(330)); dot("$D$", D, dir(60)); dot("$M$", M, dir(90)); dot("$N$", N, dir(270)); dot("$I$", I, dir(330)); dot("$F$", F, dir(45)); dot("$K$", K, dir(210)); dot("$G$", G, dir(135)); dot("$S$", S, dir(150)); dot("$L$", L, dir(210)); dot("$T$", T, dir(225)); dot("$T'$", T1, dir(315)); [/asy][/asy] Let $T = \overline{MI} \cap \overline{AF}$ be the $A$-mixtilinear intouch point, and let $S = \overline{NK} \cap (ABC)$. Since $\angle NSM = 90^\circ$, we just want $\overline{AG}$, $\overline{MS}$, $\overline{IF}$ concur. By (degenerate) Reim's we have $AGFI$ cyclic, so by radical axis it suffices to show $MIFS$ cyclic. Let $T' = \overline{SF} \cap (ABC)$. Then $$(AN;TT') \overset{S}= (AK;TF) \overset{I}= (AD;\overline{TM} \cap \overline{AD}, \infty_{AD}) = \frac{TA}{TD},$$which implies $\overline{TT'} \parallel \overline{BC}$, from where it follows by Reim's that $MIFS$ is cyclic, as desired. "which implies $\overline{TT'} \parallel \overline{BC}$". Why does it imply they are parallel?

If $\overline{TT'}||\overline{BC}$, then by angle bisector theorem $(AN;TT')=\frac{TA}{T'A}$. $\frac{TA}{TA'}=\frac{TA}{TD}$ since $TT'DA$ is an isosceles trapezoid.

Let $T = \overline{AF} \cap \odot(O) \ne A$ (it is well known that $T \in \overline{MI}$), $X = \overline{ML} \cap \overline{NK}$, $Y = \overline{MF} \cap \odot(O) \ne M$, $Z = \overline{GI} \cap \odot(O) \ne G$ and $U = \overline{MT} \cap \overline{NG}$. By repeated application of Reims we obtain: points $\{A,G,F,I\},\{T,Y,F,I\}$ are concyclic ; $Y \in \overline{DI}$ and $\overline{TZ} \parallel \overline{BC}$. [asy][asy] size(250); pair A=dir(115),B=dir(-160),C=dir(-20), D = dir(65), N = dir(-90),M=dir(90),I = incenter(A,B,C), T = intersectionpoint(unitcircle,I--I+dir(M--I)*100),F= extension(A,T,I,foot(I,M,N)),G = IP(F--3*F-2*N,unitcircle),L= extension(A,G,F,I),K = extension(D,I,A,F),X=extension(M,L,N,K),Y=IP(F--3*F-2*M,unitcircle),Z=IP(I--100*I-99*G,unitcircle),U=extension(M,T,N,G); draw(unitcircle,cyan+linewidth(0.8)); dot("$A$",A,dir(A)); dot("$B$",B,dir(140)); dot("$C$",C,dir(C)); dot("$N$",N,dir(N)); dot("$M$",M,dir(M)); dot("$F$",F,dir(F)); dot("$T$",T,dir(T)); dot("$I$",I,dir(I)); dot("$G$",G,dir(G)); dot("$D$",D,dir(D)); dot("$K$",K,dir(210)); dot("$L$",L,dir(L)); dot("$X$",X,dir(-90)); dot("$Y$",Y,dir(Y)); dot("$Z$",Z,dir(Z)); dot("$U$",U,dir(20)); draw(A--D^^L--I^^B--C^^T--Z,green); draw(B--A--C^^A--T--M,purple); draw(M--Y--D,fuchsia); draw(Z--G--N,fuchsia); draw(circumcircle(I,F,Y)^^circumcircle(A,G,F),dotted+brown); draw(A--L,fuchsia); draw(L--M^^X--N,purple); draw(T--L,linewidth(0.7)); draw(L--U,dashed+linewidth(0.7)); [/asy][/asy] By radical axes on $\odot(AGFI),\odot(TYFI),\odot(O) ~ \implies ~ \boxed{L \in \overline{TY}}$. Our problem is clearly equivalent to $X \in \odot(O)$, which by Pascal on $MXNGAT$ is further equivalent to points $K,L,U$ being collinear. By Ceva-Menelaus confi in $\triangle ILT$, this is equivalent to $(T,I ; U,M) = -1$. Indeed, $(T,I ; U,M) \stackrel{G}{=} (T,Z ; N,M) = -1$, completing the proof. $\blacksquare$

harmonics :heart_eyes: China TST 2021/4/2 refactored wrote: Scalene $\triangle ABC$ with incenter $I$ is inscribed in circle $\Omega$, with $M$, $N$ as the respective midpoints of arcs $\widehat{BAC}$, $\widehat{BC}$. Let $D$ be the reflection of $A$ in the perpendicular bisector of $\overline{BC}$, $E$ is the extouch point on $\overline{BC}$, and $F$ is defined so that $\overline{FI}\parallel\overline{BC}$ and $\angle BAF=\angle EAC$. Define $G=\overline{NF}\cap\Omega \enskip(\neq N)$, $L=\overline{AG}\cap\overline{IF}$, and $K=\overline{AF}\cap\overline{DI}$. Prove that $\overline{ML}, \overline{NK}$ meet on $\Omega$. [asy][asy] //setup; size(12cm); pen blu,grn,blu1,blu2,lightpurple; blu=RGB(102,153,255); grn=RGB(0,204,0); blu1=RGB(226,235,255); blu2=RGB(196,216,255); lightpurple=RGB(237,186,255); // blu1 lighter //defn pair M,N,I,A,X,P,Z; M=(7, 17.35); N=(7,-2.82); I=(2.59, 4.5); A=foot(M,N,I); X=foot(N,M,I); P=foot(I,M,N); Z=orthocenter(M,N,I); pair reflect(pair P,pair A,pair B){return 2*foot(P,A,B)-P;} pair F,G,L,D,G1,Z1,K; F=extension(A,X,Z,P); G=foot(M,N,F); L=extension(A,G,Z,I); D=reflect(A,M,N); G1=reflect(G,M,N); Z1=2*P-Z; K=extension(D,I,A,X); //draw filldraw(Z--M--N--cycle,blu1,blu); draw(circumcircle(A,M,N),blu); draw(Z--Z1--M--X--A--N,blu); draw(A--Z1^^Z--D--X^^D--extension(D,I,N,X)^^N--G, purple); draw(A--L,magenta); draw(M--L^^N--foot(M,N,K),red); //label void pt(string s,pair P,pair v,pen a){filldraw(circle(P,.18),a,linewidth(.3)); label(s,P,v);} pt("$M$",M,dir(90),blu); pt("$N$",N,dir(-90),blu); pt("$A$",A,dir(130),blu); pt("$I$",I,dir(90),blu); pt("$X$",X,-dir(70),blu); pt("$P$",P,dir(130),blu); pt("$Z$",Z,dir(130),blu); pt("$F$",F,dir(50),blu); pt("$G$",G,dir(130),purple); pt("$L$",L,-dir(50),magenta); pt("$D$",D,dir(50),purple); pt("$G'$",G1,dir(50),purple); pt("$Z'$",Z1,dir(-90),blu); pt("$K$",K,-dir(40),blu); pt("",foot(M,N,K),(0,0),red); pt("",extension(D,I,N,X),(0,0),purple); label("$\Omega$",(13, 16.7),blu); [/asy][/asy] Define: $X=\overline{MI}\cap\overline{AF}\cap\Omega$, and $P$ as the foot from $I$ to $\overline{MN}$; $Z=\overline{AM}\cap\overline{IP}\cap\overline{NX}$ as the orthocenter of $\triangle IMN$, Note that: By Ceva-Menelaus $(ZI;FP)=-1$ meanwhile $(F,\overline{DX}\cap\overline{MN};I,Z) \overset X=(AD;MN)=-1$, so $X,P,D$ collinear; $(M,\overline{ZD}\cap\Omega;A,X)\overset Z=(AD;MN)=-1$ while $(MG;AX)\overset N=(PF;IZ)=-1$, so $Z,G,D$ also collinear; Let $G',Z'$ be the respective reflections of $Z,G$ in $\overline{MN}$, so that $A,G',Z'$ collinear. Now we do yet more harmonic chasing: \[(A,X;\overline{NK}\cap\Omega,D) \overset N=(I,\overline{NX}\cap\overline{DI};K,D) \overset X=(IZ;FP)=-1\text{ and}\]\[(A,X;\overline{ML}\cap\Omega,D)\overset M=(ZI;LZ')\overset A=(MN;GG')=-1,\]so the points coincide as desired!

Obviously we need to prove $ML$ and $NK$ are concurrent on $ABC$. Let $AF$ and $AE$ meet $ABC$ at $S$ and $S'$. It's well known (I have proved them in "Korea winter program 2022 P6") that $S$ is $A$-mixtilinear intouch point so $M,I,S$ are collinear and $S'$ is reflection of $S$ about $MN$. Let $AE$ meet $MN$ at $X$. Claim $: AGFI$ is cyclic. Proof $:$ Note that $\angle GFI = \angle GFX = \angle 180 - \angle GMX = \angle 180 - \angle GMN = \angle 180 - \angle GAI$. Claim $: PFIM$ is cyclic. Proof $:$ Note that $\angle FIS = \angle ISS' = \angle MSS' = \angle MPS' = \angle MPF$. Now By Radical Axis Theorem on $ABC$ and $AGFI$ and $PMIF$ we have that $AG,PM,IL$ are concurrent so $L,P,M$ are collinear. Now we only need to prove $N,K,P$ are collinear. Let $PK$ meet $ABC$ at $N'$. $\frac{FK}{SK} . \frac{SA}{FA} = \frac{S'N'}{SN'} . \frac{\sin{ASP}}{\sin{AFP}} . \frac{\frac{SA}{AP}}{\frac{FA}{AP}} = \frac{S'N'}{SN'} . \frac{\sin{ASP}}{\sin{AFP}} . \frac{\frac{\sin{APS}}{\sin{ASP}}}{\frac{\sin{APF}}{\sin{AFP}}} = \frac{S'N'}{SN'} . \frac{AS}{AS'}$ Also Note that $\frac{SA}{SN} . \frac{S'N}{S'A} = \frac{SA}{S'A} = \frac{SA}{SD} = \frac{SA}{SK} . \frac{SK}{SD} = \frac{SA}{SK} . \frac{KI}{ID} = \frac{SA}{SK} . \frac{FK}{FA}$ so $\frac{SA}{SN} . \frac{S'N}{S'A} = \frac{S'N'}{SN'} . \frac{AS}{AS'}$ so $\frac{S'N}{SN} = \frac{S'N'}{SN'}$ so $N'$ is $N$ which proves that $P,K,N$ are collinear. we're Done.

Lemma :Let $ABCD$ be a cyclic quadrilateral which its diagonals $AC$ and $BD$ intersect at point $R$ and also $P , Q$ are intersection points of $AD , BC$ and $AB , CD$ respectively. So the Miquel point of quadrilateral $ABCD$ , is the foot of altitude from point $R$ to the line $PQ$. Proof :Let $S$ be the Miquel point of quadrilateral $ABCD$ , then since $\angle ASQ + \angle ASP=\angle ABC + \angle ADC =180$ , points $P , S , Q$ are collinear and if $O$ be the center of circumcircle of $ABCD$ , one can see that : $$\angle ASC=\angle ASP - \angle PSB=\angle ABC - \angle ADC=180 - 2\angle ADC=180-\angle AOC$$Thus quadrilaterals $SAOC$ and $SBOD$ are cyclic and by concurrency of radical axes of circles , lines $AC , BD , OS$ are concurrent at point $R$. As the result , since $PQ$ is polar of $R$ wrt the circumcircle of $ABCD$ , we have $OS \perp PQ$ and we're done. Firstly , Since $AF$ passes trough the $A-$mixtilinear point of triangle $\triangle ABC$ , name that as $P$ , we know that points $M , I , P$ are collinear. Now , by angle chasing one can see that : $$\angle AGF=\angle AGN=\angle ABN=\angle B + \frac{\angle A}{2}=180 - \angle AIF$$Thus quadrilateral $AGFI$ is cyclic and by Lemma , point $T$ , the second intersection point of line $LN$ and circle $\omega$ is the Miquel point of this quadrilateral. Now since quadrilateral $TFIN$ is cyclic , we can get : $$\angle FTN = \angle AIF = \angle B + \frac{\angle A}{2} = \angle ACN = \angle DTN$$So points $D , F , T$ are collinear. Now if line $DI$ intersects the circle $\omega$ at point $X$ for second time , then applying Pascal Theorem on $MXDAPM$ , gives us that points $M , K , X$ are collinear and also again by Pascal on $XDTAPM$ , points $P , X , L$ are collinear too. So note that by concurrency of $AG , IF , PX$ and since quadrilateral $AGFI$ is cyclic , thus $IFXP$ is cyclic too and by Lemma , point $S$ , the second intersection point of line $ML$ and circle $\omega$ , is the Miquel point of this quadrilateral and hence , $KS \perp ML$ and $NK$ is perpendicular to the line $ML$ at point $S$ , so we're done.

Let $Z=MN\cap IF$, $P=\odot(IFT)\cap\odot(ABC)$, $Q=\odot(MIF)\cap\odot(ABC)$, $X=AM\cap IF$, $Y=NG\cap DI$, $J=MT\cap KL$, $T$ be the $A$-mixtilinear intouch point and $T'=AE\cap\odot(ABC)$. Firstly it is well known that $\overline{M-I-T}$ are collinear. Let $M'=PF\cap\odot(ABC)$. Now Reim's on $\left\{\odot(PFIT),\odot(ABC)\right\}$ with lines $\left\{TI,PF\right\}$ gives that $IF\parallel MM'$ and as $IF\parallel BC$, we get that $M'\equiv M$ which gives that $\overline{P-F-M}$ are collinear. Let $A'=TF\cap\odot(ABC)$. Now this time Reim's on $\left\{\odot(PFIT),\odot(ABC)\right\}$ with lines $\left\{TF,PI\right\}$ gives that $IF\parallel A'D$ which gives $A'\equiv A$ and so $\overline{A-F-T}$ are also collinear. Now the from the Diameter of Incircle Lemma, we have that $\overline{A-Z-E-T}$ are collinear (can be derived from simple length computation and then showing that $\operatorname{dist}(Z,BC)=\operatorname{dist}(BC)$). Now note that the $A$-nagel cevian and the $A$-symmedian are isogonal conjugates ($\sqrt{bc}$ Inversion proves the fact). This gives that $TT'\parallel BC$ and so $AN$ is the angle-bisector of $\angle TAT'$. Let $T''=QF\cap\odot(ABC)$ Now again ( ) Reim's on $\left\{\odot(MIFQ),\odot(ABC)\right\}$ with lines $\left\{QF,MI\right\}$ gives that $TT''\parallel BC\implies T''\equiv T'$. So $F$ is basically the intersection of the $A$-symmedian with $IF$. Now we begin with the actual solution. Now, $MN\perp BC$ and $IF\parallel BC\implies MN\perp IF\implies\measuredangle MZI=90^\circ=\measuredangle MAN=\measuredangle MAI\implies MAIZ$ is cyclic and similarly $NTIZ$ is also cyclic. Also, $\measuredangle AGF=\measuredangle AGN=\measuredangle ABN=\measuredangle (AN,BC)=\measuredangle AIF\implies AGFI$ is also cyclic. Finally, $\measuredangle T'PI=\measuredangle T'PD=\measuredangle T'TD=\measuredangle AT'T=\measuredangle AZI=\measuredangle T'ZI\implies PIZT'$ is also cyclic. Now Radax on $\left\{\odot(ABC),\odot(MAIZ),\odot(NITZ)\right\}$ gives that $X=AM\cap IF\cap NT$. Also, Radax on $\left\{\odot(ABC),\odot(MZIA),\odot(PIZT')\right\}$ gives $X=AM\cap IF\cap NT\cap PT'$. Then Radax on $\left\{\odot(ABC),\odot(MIFQ),\odot(AGFI)\right\}$ gives $L=AG\cap IF\cap MQ$ and then another Radax on $\left\{\odot(ABC),\odot(MIFQ),\odot(PFIT)\right\}$ finally also gives $L=AG\cap IF\cap MQ\cap PT$. Now $\measuredangle MAI=90^\circ$ simply means that $MA$ is the external-angle-bisector of $\angle BAC$ which is same as the external-angle-bisector of $\angle FAZ$ from the fact that $AN$ is the internal-angle-bisector of $\angle TAT'$. Combining these two pieces of information, we get that \[-1=(X,I;F,Z)\overset{T'}{=}(P,G;Q,A).\]Now finally we also have that $(M,J;I,T)=-1$ (due to the complete quadrilateral of $PFIT$). Now \[-1=(X,I;F,Z)\overset{N}{=}(T,I;NG\cap MT,M),\]which finally gives that $J=LK\cap MT\cap NG$. Now then \[-1=(T,I;J,M)\overset{F}{=}(K,I;Y,P)=(P,Y;K,I)\overset{N}{=}(P,G;NK\cap\odot(ABC),A),\]which on combining with the result we derived at the end of the previous paragraph finishes the proof. Remark: There are many missing cyclic quads in the diagram which I decided not to show because it was just making the diagram way too complex. More Remark: Bro, this application of converse of Pascal is sooo cleverrr!!!!! Arjun bro ORZ. The quote mentioned below follows from the fact that the 6 points lie on a conic and as 5 points determine a conic, which are already on a circle, the conic is itself the circle and thus the 6th point lies on the circle. Thank you soo much for this lesson!! guptaamitu1 wrote: By radical axes on $\odot(AGFI),\odot(TYFI),\odot(O) ~ \implies ~ \boxed{L \in \overline{TY}}$. Our problem is clearly equivalent to $X \in \odot(O)$, which by Pascal on $MXNGAT$ is further equivalent to points $K,L,U$ being collinear.

This problem is great. I like it. Solved after 6 hours of english project work (despair) Observe that the condition is equivalent to showing that $ML$ and $NK$ are concurrent on $\omega$, the circumcircle of $ABC$. Define $T$ as the mixtilinear intouch point (hence collinear with $A$, $F$, and $K$) and $T'$ the reflection of $T$ over the perpendicular bisector of $BC$. A quick angle chase left as an exercise to the reader reveals that $AGIF$ is cyclic, whence $\measuredangle AGI$ = $\measuredangle AFI$ = $\measuredangle ATT'$, revealing that $G$, $I$, and $T'$ are collinear. Let $X = \overline{NK} \cap \omega \neq N$. Then pascal's on cyclic hexagon $MXNAGT$ reveals that $\overline{MX} \cap \overline{AG}$, $\overline{XN} \cap \overline{GT}$, and $\overline{NA} \cap \overline{MT}=I$ are collinear, so if $\overline{XN} \cap \overline{GT}$ lies on line $IF$ we're done. Let $P = \overline{DI} \cap \omega \neq D$. It can easily be proven that $PFIT$ is cyclic, whence $PT$, $IF$, and $AG$ concur. Now pascal's on $GTMPAN$ shows that $\overline{GT} \cap \overline{PA}$, $\overline{TM} \cap \overline{AN} = I$, and $\overline{MP} \cap \overline{GN} = F$ are collinear so we have that $GT$ and $AP$ concur on $IF$. Observe that \[ -1 = (A, D; M, N) \overset{I}{=} (N, P; T, A). \]Furthermore if $XGAP$ is harmonic then the problem is completed (uniqueness of harmonic conjugate, perspectivity thru AP cap GT, which lies on IF). Now observe: \[ (X, A; G, P) \overset{N}{=} (K, A; F, \overline{NP} \cap \overline{AT}) = \overset{P}{=} (D, A; \overline{PF}\cap\omega, N), \]and in particular if $P$, $F$, and $M$ are collinear this is enough. Indeed, notice that \[ \measuredangle MPT = \measuredangle MAT = \measuredangle ATM + \measuredangle TMA = \measuredangle MTD + \measuredangle TMA = \measuredangle MAD + \measuredangle TMA = \measuredangle FIM = \measuredangle FIT = \measuredangle FPT, \]so $M$, $F$, and $P$ are collinear, as desired.

Solved this while gradually losing my sanity. Note that it is equivalent to show that $\overline{ML} \perp \overline{NK}$. Let $T$ be the $A$-mixtilinear touch point. Then it is well-known that $AT$ is isogonal to $AE$, so we can delete $E$. Let $P = \overline{IF} \cap \overline{MN}$. Note that quadrilaterals $AMPI$, $TNPI$ are cyclic due to right angles. Claim: $P$ lies on $TD$. Proof. Angle chase to get \[ \measuredangle TPN = \measuredangle TIN = \measuredangle MIA = \measuredangle MPA = \measuredangle DPM = \measuredangle DPN \]. $\blacksquare$ Claim: $AM$, $IP$, $TN$, $GD$ concur at some point $R$. Proof. Quadrilateral $GIPD$ is cyclic since \[ \measuredangle GIP = \measuredangle GIF = \measuredangle GAF = \measuredangle GAT = \measuredangle GDT = \measuredangle GDP \]then radical axis on $(AMPI)$, $(TNPI)$, $\Omega$ and $(GIPD)$ gives the result. are cyclic due to right angles. $\blacksquare$ Claim: Let $Q = \overline{GD} \cap \overline{AT}$. Then, $L$, $Q$, $M$ are collinear. Proof. Note that $\overline{GF}$, $\overline{RT}$, $\overline{AI}$ concur at $N$. Then Desargues' on $\triangle GRA$ and $\triangle FTI$ gives the result. $\blacksquare$ Note that $AMDN$ is a harmonic quadrilateral. Claim: $S = \overline{MF} \cap \overline{KID}$ lies on $\Gamma$. Proof. Note that \[ (NM;AD) \overset{I}= (A,T;N,\Gamma \cap \overline{KID}) = -1 \]and that \[ (NM;AD) \overset{T}= (RI;FP) \overset{M}= (A,T;\Gamma \cap \overline{MF},N) = -1 \]$\blacksquare$ Claim: $\overline{MQL} \cap \overline{NK}$ lies on $\Gamma$. Proof. Follows as \[ (A,T;D, \Gamma \cap \overline{MQL}) \overset{Q}= (TA;GM) \overset{N}= (RI;FP) = -1 \]and that \[ (A,T;D, \Gamma \cap \overline{NK}) \overset{K}= (AT;SN) = -1 \]$\blacksquare$

I never want to touch a harmonic bundle again [asy][asy] unitsize(3cm); pair A = dir(120); pair B = dir(205); pair C = dir(335); pair I=incenter(A, B, C); pair M=dir(90); pair N=dir(270); pair D=dir(60); pair T=IP(Line(I, M, 10), unitcircle, 0); pair TTT =foot(T, M, N); pair P=foot(I, M, N); pair TT=2*TTT-T; pair F= extension(A,T,I,P); pair G=IP(Line(N, F, 10), unitcircle, 1); pair K=extension(A, F, D, I); pair L=extension(A, G, I, F); pair X=extension(A, M, G, D); pair Q=IP(Line(M, F, 10), unitcircle, 1); pair Y=extension(A, T, G, D); pair Z=extension(L, M, N, K); draw(unitcircle, heavygreen); draw(A--B--C--cycle, heavygreen); draw(A--T, heavygreen); draw(N--G, heavygreen); draw(A--L, heavygreen); draw(A--K--D, heavygreen); draw(M--N, heavygreen); draw(N--Z, heavygreen); draw(M--L, heavygreen); draw(M--X--D, red); draw(P--X--N, red); draw(A--TT, red); draw(D--T, red); draw(M--Q, red); draw(D--Q, red); draw(A--N, heavygreen); draw(M--T, heavygreen); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$I$", I, N); dot("$M$", M, dir(M)); dot("$N$", N, dir(N)); dot("$D$", D, dir(D)); dot("$T$", T, dir(T)); dot("$T'$", TT, dir(TT)); dot("$F$", F, NW); dot("$G$", G, dir(G)); dot("$K$", K, dir(K)); dot("$L$", L, NW); dot("$X$", X, dir(X)); dot("$P$", P, E); dot("$Q$", Q, dir(Q)); dot("$Y$", Y, SE); dot("$Z$", Z, dir(Z)); [/asy][/asy] Let $T$ be the $A$-mixtilinear touch point; it's well-known that $AT$ and $AE$ are isogonal (consider a $\sqrt{bc}$ inversion), so we can ignore $E$ from now on. It's also well-known that $T$, $I$, and $M$ are collinear. Additionally, denote $T'$ as the reflection of $T$ over $MN$. Claim: $AM$, $GD$, $FI$, and $TN$ are concurrent. Proof: Let $X = \overline{MA} \cap \overline{NT}$. By Pascal's on $MMANNT$, we find that $X$ lies on line $FI$. Then by Pascal's on $NNGDAT$ we see that $X$ lies on line $GD$, too. From this point onwards, refer to line $FI$ as $\ell$. Claim: $\ell$, $MN$, and $TD$, and $T'A$ are concurrent Proof: By Pascal's on $AMNGDT$ we find that $\overline{MN} \cap \overline{DT}$ lies on $\ell$. Since $\overline{DT}$ and $\overline{AT'}$ are reflections across $\overline{MN}$, this concurrency point, say $P$, lies on $\overline{AT'}$ too. Claim: $MF$ and $DK$ meet on $\Omega$ Proof:Let $Q = \overline{MF} \cap \Omega$ and $Q' = \overline{DIK} \cap \Omega$. Note that $\angle TAN = \angle NAT'$ and $\overline{XA} \perp \overline{AI}$, so by the right angles and angle bisectors lemma we have $(X, I; F, P) = -1$. Also, $AMDN$ is a harmonic quadrilateral from symmetry. Therefore, we get $$\begin{aligned} -1 &= (X, I; F, P) \overset{M}{=} (A, T; Q, N), \\ -1 &= (N, M; D, A) \overset{I}{=} (A, T; Q', N). \end{aligned}$$Therefore $Q \equiv Q'$, as needed. Claim: $AT$, $GD$, and $LM$ are concurrent Proof: Since $AD$ and $FX$ concur at the point at infinity along line $\ell$, this follows by Descargue's on $\triangle AMD$ and $\triangle FLX$. Denote this concurrency point as $Y$. Claim: (problem statement lol) $LM$ and $NK$ concur on $\Omega$ Proof: Let $Z = \overline{LM} \cap \Omega$ and $Z' = \overline{NK} \cap \Omega$. Then from our earlier harmonic bundles we have $$\begin{aligned} -1 &= (A, T; Q, N) \overset{F}{=} (T, A, M, G) \overset{Y}{=} (A, T, Z, D), \\ -1 &= (T, A; N, Q) \overset{K}{=} (A, T; Z', D), \end{aligned}$$and this implies $Z \equiv Z'$, so we're finally done.

I mean its not thattt bad. Let $T$ be the $A$-mixtilinear touch point. Claim: $\angle TAB=\angle CAE$, meaning $T$ lies on line $A-F-K.$ Proof: Consider a $\sqrt{bc}$ inversion. Notice that it swaps lines $AB$ and $AC,$ and sends $\Omega$ to line $BC$. Thus, it sends the $A$-mixtillinear incircle to the $A$-excircle and sends $T$ to $E,$ proving the claim. Claim: Quadrilateral $GAMT$ is harmonic. Proof: Call line $I-F-L$ line $\ell.$ Pascal on $NNTMMA$ gives $X=AM \cap NT$ lies on $\ell.$ Pascal on $ATMGNA$ lives $Y=AA\cap GM$ is on $\ell.$ Pascal on $AANTTM$ gives $AA\cap TT$ is on $\ell.$ Thus, $AA,TT,MG$ intersect on $\ell$ impplying the claim. Now, projecting this bundle onto $\ell$ from $N$ gives $(X,I;F,J=MN\cap \ell).$ But as $\angle XTI=\angle MTN=90,$ ray $TI$ bisects $\angle ATJ$ hence $T,J,D$ are collinear. Then, pascal on $DGNMAT$ gives $X$ is on line $DG.$ Also, Pascal on $MMHDAT$ where $H$ is the intersection of $MF$ with $\Omega$ gives $H,I,D$ collinear. Now, let $P=NK\cap \Omega \neq N.$ Then, \[-1=(AT;GM)\overset{F}{=}(TA;HN)\overset{K}{=}(AT;PD)\]Now, let $X'$ be the reflection of $X$ about $J$ which by symmetry lies on line $MD.$ Then, lines $AX'$ and $XD=CD$ are symmetric about $MN.$ Thus, if $AX'$ hits $\Omega$ at $Y\neq A,$ we have $\angle BAG=\angle CAX'$ so \[\angle LAI=\angle GAB+\angle BAI=\angle CAI+\angle CAX'=\angle IAX'.\]Then, as $MN$ is a diameter, $\angle XAI=90$ so $-1=(XI;LX').$ Now, from the above harmonic bundle we have $-1=(AT;PD)\overset{M}{=}(X,I;MP\cap \ell,X')$ implying $M,P,L$ collinear. Thus lines $ML,NK$ intersect at $P,$ which is on $\Omega$ as desired.

A really nice projective practice problem, got it in 24 mins, i give it a 9/10 since at the end it gets a little messy. Let $AF \cap (ABC)=T_A$ and $AE \cap (ABC)=T_A'$, let $LM \cap (ABC)=Q$, now before going for the claims notice some trivial propeties, first $T_A$ is the A-mixtilinear intouch point of $\triangle ABC$ by $\sqrt{bc}$ inversion and by the same it holds that $M,I,T_A$ are colinear, also $T_AT_A' \parallel BC$ so $DT_A=AT_A'$. Claim 1: $AGFI$ is cyclic. Proof: This follows trivially by Reim's on $(ABC)$ and $FI$ as $N$ is the midpoint of arc $BC$ in $(ABC)$ so its tangent to $(ABC)$ is parallel to $BC$. Claim 2: $Q,F,T_A'$ are colinear. Proof: By PoP we get $LF \cdot LI=LA \cdot LG=LQ \cdot LM$ so $QFIM$ is cyclic hence by Reim's on $(QFIM), (ABC)$ we get $Q,F,T_A'$ colinear as desired. A projective Finishing: We do some projections to show that $Q,K,N$ colinear: $$(A, K; T_A, F) \overset{I}{=} (A, D; IT_A \cap AD, \infty_{BC}) = \frac{T_AA}{DA}=\frac{T_AA}{AT_A'} = (A, N; T_A, T_A') \implies NK, T_A'F \; \text{meet at} \; Q$$Thus we are done .

Projective has scarred me. 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clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $T = AF \cap \Omega$. Note that $T$ is the $A$-mixintillinear touch point. Thus the points $M$, $I$ and $T$ are collinear. Furthermore by Reims on antiparallel $AG$ and $NN$ we find that $AGFI$ is cyclic. Now let $T'$ be the reflection of $T$ about $MN$. We will now let $X = NK \cap \Omega$. Then $X$, $F$ and $T'$ are collinear. To see this let $F' = XT' \cap AT$. Now we will show that $\frac{KF'}{KA} = \frac{KI}{KD}$. Consider projecting, \begin{align*} (F'T,KA) \overset{X}{=} (T'T, NA) \implies \frac{F'K}{F'A} \cdot \frac{TA}{TK} = \frac{T'N}{T'A} \cdot \frac{TA}{TN} \end{align*}However this reduces to, \begin{align*} \frac{F'K}{F'A} = \frac{TK}{T'A} = \frac{TK}{TD} = \frac{KI}{KD} \end{align*}as desired. Now by Reims on antiparallel $TT'$ and $AG$ we find that the points $M$, $I$, $F$ and $X$ are concyclic. Therefore by radical axis on $(AGFI)$, $(MXFI)$ and $(ABC)$ we indeed have $AG$, $MX$, and $IF$ are concurrent. However this means we are done.

Let $T$ be the mixtilinear incircle touch point and $U$ be the reflection of $T$ over $MN$. It is well known that $T$ lies on $AK$, $U$ lies on $AE$, and $TI$ passes through $M$. Denote $X = LM \cap (ABC)$. Note $AIFG$ is cyclic, as \[\measuredangle GFI = \measuredangle GNN = \measuredangle GAI.\] Power of a point also gives us another cyclic quadrilateral $XMIF$, and \[\measuredangle MXF = \measuredangle MIF = \measuredangle TNM = \measuredangle MNU = \measuredangle MXU,\] so $X$, $F$, and $U$ are collinear. As $TI$ bisects $\angle ATD$, if we let $P = TI \cap AD$, we use angle bisector theorem to get \[(AN;UT) = \frac{UA}{TA} = \frac{TD}{TA} = \frac{PD}{PA} = (AD; \infty P) = (AK;FT).\] Thus Prism lemma tells us $FU$ and $NK$ meet on the circumcircle, or $X$, as desired. $\blacksquare$

Let T be the A-mixtillinear touch point. It is known that AT and AE are isogonal. Also $T \in IM$. Let T' be the reflection of T across MN. We will now prove that AM, GD, FI and TN are concurrent. Let $MA \cap NT = X$. By Pascal on MMANNT we get that $X \in FI$. By Pascal on NNGDAT we get that $X \in GD$ $\Rightarrow$ AM, GD, FI and TN are concurrent at X. Now we will show that FI, MN, TD and T'A are concurrent. By Pascal on AMNGDT we get that $MN \cap DT \in FI$. Since $S_{AD} \equiv S_{TT'} \equiv MN$ and AD = TT', T'A, TD and MN are concurrent $\Rightarrow$ FI, MN, TD and T'A are concurrent in H for example. Now we will show that $MF \cap DK \in \Omega$ by phantom points. Let $MF \cap \Omega = Y$ and $DK \cap \Omega = Y'$. Now $\angle TAN = \angle NAT'$ and $AX \perp AI$ $\Rightarrow$ by right angle and bisectors we have that $(X,I;F,H) = -1$. Also AMDN is harmonic, since it is a kite. Now projecting we get $(X,I;F,H)\stackrel{M}{=}(A,T;Y,N) = -1$. Also $(N,M;A,D)\stackrel{I}{=}(A,T;Y',N) = -1$ $\Rightarrow$ $(A,T;Y,N) = (A,T;Y',N) = -1$ $\Rightarrow$ $Y \equiv Y'$ $\Rightarrow$ $MF \cap DK \in \Omega$ (at Y). Now we will show that AT, GD and LM are concurrent. This directly follows from Desargues theorem for $\triangle AMD$ and $\triangle FLX$ and let $AT \cap GD \cap LM = J$. Lastly we want to show that $LM \cap NK \in \Omega$ by phantom points. Let $LM \cap \Omega = P$ and $NK \cap \Omega = P'$. Now projecting we get $(A,T;Y,N)\stackrel{F}{=}(T.A;M,G)\stackrel{J}{=}(A,T;P,D) = -1$ and $(T,A;N,Y)\stackrel{K}{=}(A,T;P',D) = -1$ $\Rightarrow$ $(A,T;P,D) = (A,T;P',D) = -1$ $\Rightarrow$ $P \equiv P'$, which is what we wanted to initially prove $\Rightarrow$ we are ready.

Let $P$ be the intersection of $MN$ and $IF$. Let $T$ be the $A$-mixtilinear touch point. It is well known that $F$, $I$, and $P$ lie on $TA$, $TM$, and $TD$ respectively.

Let $X$ be the intersection of $(FIM)$ and $(ABC)$. We will show $X$ is on $ML$ and $NK$. By inversion about $N$, $AGFI$ is cyclic. By the radical center theorem, $AG$, $FI$, and $MX$ concur. Therefore, $X$ is on $ML$. Let $I'$, $F'$, and $T'$ be the reflections of $I$, $F$, and $T$ about $MN$. Claim: $NK$ and $T'F$ intersect on $(ABC)$ Let $R$ be the exsimilicenter of $FI$ and $II'$. By Monge's theorem on $AD$, $FI$, and $II'$, $R$ is on $NK$. Because $(RP;II')=(FI';PF')$ and $A=NI\cap T'P$, $M=NP\cap T'I'$, and $D=NI'\cap T'F'$ are on $(ABC)$, $NR$ and $T'F$ intersect on $(ABC)$ by the prism lemma. Let $X'=NK\cap T'F$. Then, $\angle MX'F=\angle MX'T'=90-\angle T'X'N=90-\angle PMI = \angle MIP=\angle MIF$. So, $MIFX'$ is cyclic, so $X'=X$.

here is a collection of results obtained using solely Pascal's theorem which i couldnt be bothered to sort out (in some order they give the solution ) surely a huge fakesolve idk in total Pascal's is used $\boxed{6}$ times

Let $T$ be $A-$mixtilinear touch point with $(ABC)$. Since $AT$ and $AE$ are isogonals, $A,F,T$ are collinear. Also note that $M,I,T$ are collinear. $ND\cap IF=J,NM\cap IJ=P,AE\cap (ABC)=U,TJ\cap (ABC)=W,GT\cap IF=V$. We have $TU\parallel BC$. Observe that $I,J$ are symetric to $MN$. $A,P,U$ are collinear. Claim: $VFI\sim VIJ\iff VI^2=VF.VJ$. Proof: Let's show that $(\overline{TV},\overline{TU}),(\overline{TI},\overline{TI}),(\overline{TA},\overline{TJ})$ is an involution. Project onto $(ABC)$. Proving $(G,U),(M,M),(A,W)$ is equavilent to showing the concurrency of $GU,AW$ and the tangent to $(ABC)$ at $M$. Since $\measuredangle MAN=90=\measuredangle NPI$, $(M,I,N,IP\cap TN)$ is an orthogonal system thus, $MA,IP,TN$ are concurrent. \[(A,T;G,M)\overset{N}{=} (I,NT\cap IJ;F,P)\overset{A}{=} (N,M;T,U)=-1\]Hence $MM,GG,AT$ are concurrent. Apply pascal on $GGUTAW$ to see that $GG\cap TA,GU\cap AW,BC_{\infty}$ are collinear. Since the line passing through both $GG\cap AT$ and $BC_{\infty}$ is $MM$, we get that $GU,AW,MM$ are concurrent.$\square$ We have \[\frac{NJ}{ND}.\frac{KD}{KI}.\frac{VI}{VJ}=\frac{NJ}{ND}.\frac{AD}{IF}.\frac{VF}{VI}=\frac{NJ}{IF}.\frac{IJ}{NJ}.\frac{VF}{IV}=\frac{IJ}{IV}.\frac{VF}{IF}=1\]Thus, $N,K,V$ are collinear. Let $\overline{NKV}$ intersect $(ABC)$ at $S$. Applying pascal on $MSNAGT$ to get $MS\cap AG,V,I$ are collinear. Hence $L,S,M$ are collinear which is equavilent to $ML\perp NK$ as desired.$\blacksquare$

Solved while talking to MathLuis, kind of a mess of a solution Let $AE \cap (ABC)=J$, $O= AF \cap (ABC)$, Let, $H= ML \cap (ABC)$. Claim: $(AGFI)$ Cyclic Proof: $$\measuredangle AGF = \measuredangle AGN = \measuredangle ABC + \measuredangle BAN = \measuredangle (AB, IF) + \measuredangle BAI = \measuredangle AIF. \square$$ Also, since $LM \cdot LH = LG \cdot LA = LF \cdot LI$ implies $(HFIM)$ cyclic. Pascal on $(AGNMHJ)$ gives us: $L - P - HJ \cap NG$. So $J-F-H$. Pascal on $AJHDOA$ gives us gives us $I-P-F-S$ where $P$ is $MN \cap IF \cap AE$ (well known incenter config). S is where the tangent line from $A$ on $(ABC)$ intersects $HD$. Pascal $(OONAAM)$ gives the intersections to the tangents of $(ABC)$ at $O$, and $A$ , $ON \cap AM$, and $I$ are collinear. Observe that by right angles $(AMPI)$ cyclic, and $(IPON)$ cyclic. So by radical axis $I-P- AA \cap OO$. So $S = AA \cap OO$. Now we have $-1=(NM; AD) \stackrel{O} = (KD \cap ON,I; K, D) $, and $-1 = (OH; AD)$ So by prism lemma $N-H-K$ collinear. Finishing.

Probably the last solution to this problem for this year! Used a projective transformation just to show that I can, I have a funny feeling like you dont need to. Didn't take me long, and it wasn't as hard as medium China TST geometry problems are made out to be. Let $\Omega$ denote the circumcircle of $\triangle ABC$. Notice how the problem is equivalent to showing that $\overline{ML}$ and $\overline{KN}$ concur on $\Omega$. Let $T$ denote the $A-$mixtilinear intouch point and let $\ell$ denote the line through $I$ parallel to $\overline{BC}$. The angle condition simply tells that $F$ is the intersection of $\overline{AT}$ and $\ell$. Let $D'$ be the second intersection of line $\overline{DI}$ with $\Omega$ and $T'$ the reflection of $T$ across the perpendicular bisector of segment $BC$. Note that since arcs $TN$ and $T'N$ are equal by symmetry, it follows that points $A$ , $E$ and $T'$ are collinear. We start off by making the following observation. Claim : Points $G$ , $I$ and $T'$ are collinear. Proof : Note that by Pascal's Theorem on hexagons $AAMTTN$ and $ANNTMM$, it follows that points $AN \cap MT = I$ , $AA \cap TT$ , $AM \cap NT$ and $NN \cap MM = \infty$ are collinear. Thus, the tangents to $\Omega$ at $A$ and $T$ intersect on $\ell$. Further note that by Pascal's Theorem on $AATMGN$, points $AT \cap GN = F$ , $MT \cap AN = I$ and $MG \cap AA$ are collinear. Thus, $\overline{MG}$ passes through the intersection of the tangents to $\Omega$ through $A$ and $T$ implying that $AGTM$ is harmonic. Thus, \[-1=(AT;GM)\overset{I}{=}(N,M;,T,GI \cap \Omega)\]which implies that points $G$ , $I$ and $T$ are indeed collinear as claimed. Take a homography sending $ATNM$ to a rectangle while preserving its circumcircle. [asy][asy] import geometry; size(10cm); defaultpen(fontsize(9pt)); pair foot(pair P, pair A, pair B) { return foot(triangle(A,B,P).VC); } pen pri; pri=RGB(24, 105, 174); pen sec; sec=RGB(217, 165, 179); pen tri; tri=RGB(126, 123, 235); pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair A=dir(110); pair T = dir(250); pair N = dir(290); pair M= dir(70); pair I = intersectionpoint(line(A,N),line(T,M)); pair F = midpoint(A--T); pair G = intersectionpoints(circle(A,T,N),line(N,F))[1]; pair L = intersectionpoint(line(A,G),line(F,I)); pair D = intersectionpoints(line(foot(G,M,N),G),circle(A,M,N))[1]; pair D_1= intersectionpoints(line(D,I),circle(A,M,N))[0]; pair K = intersectionpoint(line(A,T),line(D,D_1)); pair X = intersectionpoint(line(L,M),line(K,N)); filldraw((path)(A--T--N--M--cycle), white+0.1*pri, pri); filldraw(circumcircle(A,N,M), tfil, tri); draw(A--L,sec); draw(L--I,sec); draw(G--N,pri); draw(D_1--D,pri); draw(D_1--M,pri+dashed); draw(L--M,dashed+tri); draw(X--N,dashed+tri); dot("$A$",A,dir(A)); dot("$T$",T,dir(T)); dot("$N$",N,dir(N)); dot("$M$",M,dir(M)); dot("$I$",I,dir(90)); dot("$F$",F,dir(170)); dot("$G$",G,dir(G)); dot("$L$",L,dir(L)); dot("$D$",D,dir(D)); dot("$D'$",D_1,dir(D_1)); dot("$K$",K,dir(340)); [/asy][/asy] By Pascal's Theorem on hexagon $AMMTNN$, points $AM \cap NT$ , $MM \cap NN = \infty$ and $AN \cap MT = I$ are collinear. Thus, lines $\overline{AM}$ , $\overline{NT}$ and $\ell$ concur. Thus, $\ell$ maps to the line through $I$ parallel to sides $AM$ and $NT$. We start off by making the following observation, which is a lot easier to see after the projective transformation. Claim : Lines $\overline{DG}$ , $\overline{AT}$ and $\overline{LM}$ concur. Proof : Since $\ell$ is parallel to the sides of the rectangle with center $I$, $F$ is the midpoint of $AT$. Let $F'$ denote the reflection of $F$ across $I$. Thus, $F'$ is also the midpoint of $MN$. Thus, $FN \parallel AF'$. Let $S = LM \cap AT$. Then, \[\frac{GL}{AG}=\frac{FL}{F'F}= \frac{FL}{AM} = \frac{FS}{AS}\]which implies that $GS \parallel \ell$. Further since the diagonals of quadrilateral $GD'T'D$ intersect at the center $I$, it must be a rectangle and thus, $GD \perp DT' \parallel MN$ which implies that $GD \parallel \ell$ as well. This implies that points $G$ , $S$ and $D$ are collinear, which proves the claim. Claim : Points $M$ , $F$ and $D'$ are collinear. Proof : Since in the original diagram, $D$ is the reflection of $A$ across $\overline{MN}$ , quadrilateral $AMDN$ is harmonic. Thus, \[-1=(MN;AD)\overset{I}{=}(TA;ND')\overset{M}{=}(T,A;\infty , AT \cap MD')\]which implies that $\overline{MD'}$ bisects segment $AT$, which implies the claim. Now, let $X = \overline{KN} \cap \Omega$. By Pascal's Theorem on hexagon $XNGDD'M$, it follows that points $XN \cap DD' = K$ , $NG \cap D'M = F$ and $GD \cap MX$ are collinear. Thus, points $X$ , $S$ and $M$ are collinear. We already proved that points $L$ , $S$ and $M$ are collinear, which implies that $X$ also lies on line $\overline{LM}$, which implies that indeed lines $\overline{NK}$ and $\overline{LM}$ intersect on $\Omega$, as desired.

Average Chinese TST problem: write a good geometry problem, then add a bunch of lines and points to hide the original problem. We will use a ridiculous number of phantom points. Let $Q$ be the tangency point of $\Omega$ with the $A$-mixtilinear incircle, $P = \overline{DI} \cap \Omega$, $R = \overline{NQ} \cap \overline{AP}$, and $F' = \overline{AK} \cap \overline{RI}$. Then, in triangle $ANP$, the cevians $\overline{NR}$, $\overline{PI}$, $\overline{AF'}$ concur at $K$. Claim: $\overline{AA}$, $\overline{NP}$, $\overline{RI}$ meet at a point $U$. Proof: Let $U_1 = \overline{NP} \cap \overline{RI}$; then $(U_1F'; RI) = -1$ by definition. On the other hand, setting $U_2 = \overline{RI} \cap \overline{AA}$ and using the fact that $\overline{QI}$ passes through $M$ (see #3 here), \[-1 = (AD; NM) \stackrel I= (NP; AQ) \stackrel A= (IR; F'U_2).\]So $U_1=U_2$, as needed. $\blacksquare$ Claim: $\overline{UI} \parallel \overline{BC}$. Proof: We use another phantom point. Let $\ell$ be the line through $I$ parallel to $\overline{BC}$, and let $U_3 = \ell \cap \overline{AA}$. Now, $\measuredangle U_3IP = \measuredangle ADP = \measuredangle ANP$ (or Reim's theorem ) so $\overline{U_3I}$ is tangent to $(IPN)$. Next, let $\omega$ be the circle tangent to $\overline{U_3A}$ at $A$ and tangent to $\overline{U_3I}$ at $I$. As $\overline{U_3I}\parallel \overline{NN}$, a homothety at $A$ takes $\omega$ to $\Omega$. So $\overline{AA}$ is the radical axis of $\omega$ and $\Omega$, i.e. $U_3$ is the radical center of $(IPN)$, $\omega$, and $\Omega$. It follows that $U_3$ lies on $\overline{PN}$, so $U_3 = U$. $\blacksquare$ The previous claim and definition of $Q$ together imply $F' = F$. Now let $T = \overline{NR} \cap \Omega$. Claim: $F$ lies on $\overline{PM}$. Proof: Let $M' = \overline{PF} \cap \Omega$. Then $-1 = (UF; RI) \stackrel P= (NM';AD)$ so $AM'=M'D$, i.e. $M' = M$. $\blacksquare$ Now by Pascal on $TNGAPM$, $R = \overline{TN} \cap \overline{AP}$, $F = \overline{NG} \cap \overline{PM}$, $L' = \overline{GA} \cap \overline{TM}$ are collinear along the line $\ell$. But $L$ also lies on $\overline{AG}$ and $\ell$, hence $L = L'$. Then $\overline{LM}$ passes through $T$, where it intersects $\overline{NK}$ on $\Omega$, as needed.