$\omega$ be the incircle of $\triangle ABC$. Let $T' = AT\cap \omega\ (\neq T)$, it's well-known that $T'$ is the point diametrically opposite $D$ w.r.t. $\omega$. Therefore, $\angle ATD$ is $90^{\circ}$. Now consider the radical axes of $\odot ADT$, $\odot AEFI$ and $\omega$.

Let $\ell$ be the line through $A$ parallel to $BC$ it is well known that $\ell$ is the polar of midpoint of $DE$ and also this midpoint lies on $A-$ median then the rest is trivial

Sharygin 2013 along with the fact that the antipode of the incircle touch point lies on the Nagel cevian.

Vrangr wrote: $\omega$ be the incircle of $\triangle ABC$. Let $T' = AT\cap \omega\ (\neq T)$, it's well-known that $T'$ is the point diametrically opposite $D$ w.r.t. $\omega$. Therefore, $\angle ATD$ is $90^{\circ}$. Now consider the radical axes of $\odot ADT$, $\odot AEFI$ and $\omega$. Expanding upon my previous sketch. $\measuredangle$ refers to directed angles. Let $I$ be the incentre and $\omega$ be the incircle of $\triangle ABC$.

Let $F'$ be the point diametrically opposite $F$ in the incircle and let $X=FI \cap DE$. $K'$ be the intersection of $DE$ and the line through $A$ parallel to $BC$. Let $M$ be the midpoint of $BC$. As above $A,F,T,N$ colinear. It's well-known $X$ lies on the $A$-median so: $$-1=(\infty_{BC},M;C,B) \stackrel{A}{=} (K',X;E,F)$$Also: $$-1=(T,F';E,F) \stackrel{F}{=}(K,X;E,F)$$So $K=K'$ as desired.

Let AN intersect the incircle in F' and T and let FF' intersect DE at P .. it is wellknown that FF' is a diameter in the incircle and AP is median in ABC . we have $(D,E;F',T)=-1$ so $(D,E;P,K)=-1$ and project from A we get thd desired result.

I solved this in my sleep My Solution: We will prove that the line parallel to $BC$ through $A$, $FT$ and $DE$ are concurrent. Let $I$ be the incenter of $ABC$. It is a well known result that $AN, FI$ meet on the incircle at the antipode of $F$. Let the antipode of $F$ on the incircle be $F'$. Then we can say $A,T,F'$ are collinear. Now, let $FF'$ meet the line parallel to $BC$ at $X$. As $IF \perp BC$, we get $\angle FXA=\angle IXA=90^\circ (\spadesuit)$. Now see that as $FF'$ is a diameter, $\angle FTA =90^\circ$. So, $FTXA$ is cyclic from $(\spadesuit)$. Note that $\angle IDA = \angle IEA = \angle IXA = 90^\circ$ from $(\spadesuit)$. So, $DEAX$ is cyclic and $I$ also lies on this circle. Now, Applying Radical Axis theorem on circles $DETF, AXTF, AXDE$, we get $AX, DE, FT$ are concurrent. Hence $K$ lies on $AX$, so $AK\parallel BC$. $\blacksquare$.

Too easy for Sharygin ! Just note that , if $G$ is diametrically op. to $F$ then $DEGT$ is a harmonic quad. Since $A,G,T,N$ collinear So projecting by $F$ we have $(D,E;X,K)=-1$ .Where $X$ is the intersection pt. of $DE$ & $A-median$ . Again projecting by $A$ we have $AK,AM,AB,AC$ are harmonic pencils ! where $M$ is the intersection pt of $A-median$ & $BC$ . Since $M$ is the midpoint of $BC$ so $AK$ is parallel to $BC$

Here's the solution that I submitted during the actual exam.

Here is my solution for this problem Solution Let $G$, $P$ be second intersections of $AN$, $AF$ with the incircle of $\triangle$ $ABC$, respectively; $Q$ $\equiv$ $FG$ $\cap$ $DE$ It's easy to see that: $AN$ $\perp$ $FK$ at $T$ Since: $DFEP$, $DTEG$ are harmonic quadrilaterals, we have: $\dfrac{FD}{FE}$ = $\dfrac{PD}{PE}$, $\dfrac{TD}{TE}$ = $\dfrac{GD}{GE}$ So: $\dfrac{KD}{KE}$ = $\dfrac{FD}{FE}$ . $\dfrac{TD}{TE}$ = $\dfrac{FD}{FE}$ . $\dfrac{TD}{TE}$ = $\dfrac{PD}{PE}$ . $\dfrac{GD}{GE}$ or $K$, $G$, $P$ are collinear But: $GP$ $\perp$ $AF$ then: $G$ is orthocenter of $\triangle$ $AFK$ or $FG$ $\perp$ $AK$ Combine with: $FG$ $\perp$ $BC$, we have: $AK$ $\parallel$ $BC$

The hardest part of the problem is dealing with the strange point labels. Let $\overline{AT}$ meet the incircle again at a point $L$, and suppose that $\overline{DE}$ and $\overline{AM}$ meet at a point $X$. Let $Y$ be the intersection of lines $AK$ and $BC$, possibly at infinity. It is well-known that $\overline{FIXL}$ is collinear, so $-1 = (DE;LT) \stackrel{F}{=} (DE;FK) \stackrel{A}{=} (BC;MY)$, so $\overline{AK} \parallel \overline{BC}$, as desired.

Let $L = \overline{ET} \cap \overline{DF}$. Pascal on $FFTEED$ $\implies$ $\overline{LAK}$ collinear. Since $TD$ is a diameter, $T$ is the orthocenter of $\triangle DKL$. So, $\overline{TD} \perp \overline{AK}$, as desired.

Relabel the points such that the incircle touches $BC, CA, AB$ at $D, E, F$ respectively. Let $l$ be the line through $A$ parallel to $BC$, $D'$ be the antipode of $D$ wrt the incircle, and $N = l \cap DD'$. The desired conclusion is equivalent to showing $l$, $EF$, and $DG$ are concurrent. The angle condition implies $AG$ passes through $D'$. It's also easy to see $l \perp DD'$ by parallel lines. Claim: $ANEIF$ and $ANGD$ are cyclic. Proof. Because $N, D', I, D$ are collinear, $$\angle AND = \angle ANI = 90^{\circ} = \angle AFI = \angle AEI$$proving the first claim. Observe $\angle AGD = \angle AND = 90^{\circ}$ which proves our second claim. $\square$ Now, it follows that $AN = l$, $EF$, and $DG$ are concurrent the Radical Center of $(ANEIF)$, $(ANGD)$, and $(DGEF)$. $\blacksquare$ Note: If the centers of the $3$ circles are collinear, then the $3$ lines concur at infinity. Also, I should've used projective... lol.

Let $K = \overline{AP_\infty} \cap \overline{DE}$ be a point such that $\overline{AK} \parallel \overline{BC}$. EGMO 9.49 from Sharygin 2013 establishes that $\overline{AM}$ is the polar of $K$ with respect to the incircle, where $M$ is the midpoint of $BC$. But now, $$-1 = (FN; MP_\infty) \stackrel A= (F, T; \overline{AM} \cap \overline{FT}, \overline{AP_\infty} \cap \overline{FT}),$$so $\overline{AM} \cap \overline{FT}$ lies on the polar of $\overline{FT} \cap \overline{AP_\infty}$. But $K$ also lies on the polar by La Hire's, and since the polar is a line, such $K$ is unique. From here $\overline{AP_\infty} \cap \overline{FT}=K$ so we are done.

Let $M$ be the midpoint of $BC$.It is well known that $FT$ is the polar of $M$.By La Hire ,$AM$ is the polar of $K$.Now,it is well known that $FI,AM,DE$ concur ,at $Q$.By La Hire,$AK$ is the polar of $Q$.Now $IQ \perp AK, IF \perp BC$,so done.

Let $F'$ be the antipode of $F,$ let $M$ be the midpoint of $\overline{BC},$ and let $Y=\overline{AM}\cap\overline{DE}.$ It is well-known that $Y\in\overline{FF'}.$ Notice $$-1=(DE;F'T)\stackrel{F}=(DE;YK)\stackrel{A}=(BC;M,\overline{AK}\cap\overline{BC}),$$so $\overline{AK}\cap\overline{BC}$ is the point at infinity along $\overline{BC}.$ $\square$

$M$ is the pole of $DT$ wrt. the incircle by symmetry. $A$ is the pole of $EF$, and thus $K$ is the pole of $AM$. Now $P_{\infty}$, the point at infinity of line $BC$ is the pole of $DI$, and since $EF$, $AM$ and $DI$ concur, by the polar transformation, it's poles are colineal, which implies the problem.

Let $FI \cap DE=G$ then its known that if u let $M$ the midpoint of $BC$ then $A,G,M$ are colinear, now taking polars w.r.t. $\omega$ we have that this is $\mathcal P_A, \mathcal P_G, \mathcal P_M$ concurrent and that means $DE,FT$ and the line through $A$ parallel to $BC$ concurrent thus we are done

Alright, finally back from hospital after 6 straight days T__T Let's get back to work... Also woahhh!! This is so similar to the Sharygin 2023 P22. :no_mouth: Firstly, let $M$ be the midpoint of $BC$, $P=AM\cap DE$ and $Q=AM\cap FT$. Also $\ell$ denote the line through $A$ parallel to $BC$. Now it is well known that $M$ is also the midpoint of $FN$. Also from the Diameter of the Incircle Lemma, we have that $\angle FTN=90^\circ$ which along with the fact that $M$ is the midpoint of $FN$ we have that $MF=MT$ and as $MF$ is tangent to the incircle, we have that $MT$ is tangent to the incircle. Now $-1=(F,N;M,\infty_{BC})\overset{A}{=}(F,T;Q,FT\cap\ell)$ and $-1=(B,C;M,\infty_{BC})\overset{A}{=}(D,E;P,DE\cap\ell)$. Now firstly note that the polar of $A$ w.r.t. the incircle is the line $DE$ so $DE\cap\ell$ lies on the polar of $A$ w.r.t. the incircle and by La Hire's Theorem, we thus have that $A$ lies on the polar of $DE\cap\ell$ and also as $(D,E;P,DE\cap\ell)=-1$, we also have that $P$ also lies on the polar of $DE\cap\ell$ and thus the line $AP$ becomes that polar of $DE\cap\ell$. Now using a similar logic, we see that the polar of $FT\cap\ell$ is the line $MQ\equiv AP$. Now two points have the same polar means that both the points are identical and we thus have that $K=\ell\cap DE\cap FT$ and we are done.

Relabel points so that $D$ is the $\overline{BC}$-intouch point, etc. Here are two solutions. Solution 1: Let $P$ be the $D$-antipode with respect to the incircle and $Q$ the point on $\overline{DP}$ such that $\overline{AQ} \parallel \overline{BC}$. It is well-known that $A,P,T,N$ are collinear, so $AEFIQ$ and $AQTD$ can easily be seen to be cyclic, and by radical center on the incircle, $(AEFIQ)$, and $(AQTD)$, we find that $\overline{EF}$, $\overline{DT}$, and $\overline{AQ}$ concur, hence $K$ lies on $\overline{AQ}$. $\blacksquare$ Solution 2: Define $P$ as before and let $X=\overline{EF} \cap \overline{DP}$. Since $\overline{TP}$ passes through $A$, $$-1=(F,E;T,P)\stackrel{D}{=}(F,E;K,X)\stackrel{A}{=}(B,C;\overline{AX} \cap \overline{BC},\overline{AK} \cap \overline{BC}).$$It is well-known that $\overline{AX} \cap \overline{BC}$ is just the midpoint of $\overline{BC}$, so $\overline{AK} \cap \overline{BC}=P_{\infty \overline{BC}}$ which implies the desired result. $\blacksquare$

Let $I$ be the incenter of $\triangle ABC$, $X = \overline{AM} \cap \overline{DE},$ and $Y$ be the second intersection of $\overline{AN}$ and the incircle. Then from EGMO chapter 4, we know that $A$, $I$, $X$, and $Y$ are collinear. Claim: $AM$ is the polar of $K$ wrt the incircle Proof: Since $DE$ is the polar of $A$, by La Hire's we find that $A$ lies on the polar of $K$. However, by taking a homothety at $A$, we see that $Y$ is the $F$-antipode wrt the incircle, so $\overline{YN} \perp \overline{FT}$. By taking a homothety at $F$ we see that $\overline{IM} \perp \overline{FT}$. But since $\overline{MF}$ is tangent to the incircle, this is enough to imply that line $FT$ is the polar of $M$; therefore, by La Hire's we see that $M$ lies on the polar of $K$, too, which proves the claim. Now, since $X$ lies on $\overline{AM}$, we have $$-1 = (D, E; X, K) \overset{A}{=} (B, C; M, \overline{AK} \cap \overline{BC}).$$But since $(B, C; M, P_{\infty}) = -1$, where $P_{\infty}$ is the point at infinity along line $BC$, this implies that $\overline{AK} \parallel \overline{BC}$, as needed.

Relabel the points so $D=BC \cap \omega$, $E=AC \cap \omega$, and $F=AB \cap \omega$. Let the midpoint of $BC$ be $M$ and $AM \cap \omega=\{P,Q\}$. Projecting the reciprocal pairs $\{B,C\}, \{D,N\}, \{M,M\}$ onto $\omega$ yields $DT, EF, PP,$ and $QQ$ are concurrent. Let $PQ \cap EF = X$ so that $K$ and $X$ are conjugated. Then, $\mathcal{H} (F,E;X,K) = \mathcal{H}(B,C;M,AK \cap BC)$. The harmonic conjugate of $M$ is the point at infinity, thus $AK \parallel BC$.

Let $M$ be the midpoint of $BC$. It is well known that $\overline{AM}$, $\overline{DE}$, and $\overline{IF}$ concur at some point $P$. Let $S$ be the other intersection point of $AN$ and the incircle. It is well known that $SF$ is a diameter of the incircle. As such, it follows that \[ (DE;ST) \overset{F}= (DE;PK) \overset{A}= (B,C;M,\overline{AK} \cap \overline{BC}) = -1 \]which implies that $\overline{AK} \cap \overline{BC}$ intersect at $\infty$, or that $\overline{AK} \parallel \overline{BC}$.

Let $M$ be the midpoint of $BC$. Let $\Gamma$ denote the incircle. Define $P = AN \cap \Gamma$. It is well known that $PX$ is a diameter of $\Gamma$. Let $Q = PX \cap YZ$. Now we claim that $(ZY, QK) = -1$. Indeed we find, \begin{align*} -1 = (ZY, PT) \overset{X}{=} (ZY, QK) \end{align*}Now projecting through $A$ we have, \begin{align*} -1 = (ZY,QK) \overset{A}{=} (BC,M\infty) \end{align*}so we indeed have $AK \parallel BC$.

We take note of the following: $XK \perp AN$ through Diameter of the Incircle lemma. $K = XT \cap YZ$ is the pole of $AM$, so $IK \perp AM$. $XI$, $AM$, and $YZ$ concur by an incircle concurrence lemma. Combining this with $AT \perp YZ$, we get that the concurrence is the orthocenter of $\triangle AIK$. Thus $XI \perp AK$, which finishes. $\blacksquare$

Let $M$ be the midpoint of $BC$ (and therefore the midpoint of $XN$, by a property I am too lazy to find the name of), let $P$ be the other intersection of $AT$ with $\omega$, let $Q=AM\cap YZ$, and let $\omega$ be the incircle. Claim. $AN\perp XT$. Proof. Note that $P$ is the "top point" of $\omega$ and $X$ is the "bottom point" of $\omega$ (assuming that $BC$ is the "bottom tangent" of $\omega$), $PX$ is a diameter of $\omega$, so $\angle PTX=90^{\circ}$. Claim. Line $XT$ is the polar of $M$ with respect to $\omega$. Proof. Since $MX$ is tangent to $\omega$, we just need to prove that $MX=MT$, which is true because triangle $XTN$ is a right triangle. Claim. Line $AM$ is the polar of $K$ with respect to $\omega$. Proof. $YZ$ is the polar of $A$ wrt $\omega$, and $XT$ is the polar of $M$ wrt $\omega$, so we can finish by La Hire's. Claim. $P,Q,X$ are collinear. Proof. We employ barycentric coordinates with reference triangle $ABC$. Then $I=(a:b:c)$, $X=(0:a+b-c:a-b+c)$, $Y=(a+b-c:0:-a+b+c)$, and $Z=(a-b+c:-a+b+c:0)$. Thus line $YZ$ is \[(-a+b+c)x+(-a+b-c)y+(-a-b+c)z=0,\]line $XI$ is \[(a-b-c)(c-b)x+a(a-b+c)y+a(-a-b+c)z=0,\]and line $AM$ is \[(0)x+(1)y+(-1)z=0.\]Now we plug everything into the concurrence lemma: \begin{align*} &\begin{vmatrix} 0 & 1 & -1 \\ -a+b+c & -a+b-c & -a-b+c \\ (a-b-c)(c-b) & a(a-b+c) & a(-a-b+c) \end{vmatrix} \\ =& \begin{vmatrix} 0 & 1 & 0 \\ -a+b+c & -a+b-c & -2a \\ (a-b-c)(c-b) & a(a-b+c) & 2a(c-b) \end{vmatrix} \\ =& -\begin{vmatrix} -a+b+c & -2a \\ (a-b-c)(c-b) & 2a(c-b) \end{vmatrix} \\ =& -2a(c-b)\begin{vmatrix} -a+b+c & -1 \\ a-b-c & 1 \end{vmatrix} \\ =& \; 0, \end{align*}as desired. Therefore, \[ -1=(Z,Y;P,T)\stackrel{X}{=}(Z,Y;Q,K). \]However, since \[ -1=(B,C;M,\infty_{BC})\stackrel{A}{=}(Z,Y;Q,(A\infty_{BC}\cap ZY)), \]we indeed have $K=A\infty_{BC}\cap ZY$, as desired. $\blacksquare$

Let M be the midpoint of BC. Let l be the line which is parallel to BC trough A. Let $AM \cap DE = X$ and $AM \cap FT = Y$. Now it follows that $(B,C;M,P\infty_{BC}) = -1$ and also $(B,C;M,P\infty_{BC})\stackrel{A}{=}(D,E;X,DE \cap l) = -1$. Also its well known that BF = NC and since BM = MC, it follows that FM = MN $\Rightarrow$ $(F,N;M,P\infty_{BC}) = -1$. Now projecting trough A we get $(F,N;M,P\infty_{BC})\stackrel{A}{=}(F,T;Y,FT \cap l) = -1$. Now from the diameter of the incircle lemma, we have that $\angle NTF = 90^{\circ}$ and since FM = MN, then MT = MF and as MF is tangent to the incircle, we get that MT is also tangent to the incircle. Now its obvious that the polar of A is DE $\Rightarrow$ $DE \cap l \in DE$ $\Rightarrow$ $DE \cap l$ lies on the polar of A. Now by La Hire, A should lie on the polar of $DE \cap l$. Now from $(D,E;X,DE \cap l) = -1$ and since D and E lie on a circle, we have that X lies on the polar of $DE \cap l$. Since A should lie on the polar of $DE \cap l$ and X lies on the polar of $DE \cap l$, then AX is the polar of $DE \cap l$. Similarly the polar of $FT \cap l$ is the line $MY \equiv AX$ $\Rightarrow$ the polar of $DE \cap l$ and the polar of $FT \cap l$ is AX $\Rightarrow$ since the two points have the same polar, the points are identical $\Rightarrow$ we have that $K = DE \cap FT \cap l$ $\Rightarrow$ $K\in l$ $\Rightarrow$ $AK \parallel BC$ and we are ready.

Realise that $T$ is the same as $G$ in GOTEEM 2020 P1 and then the same solution follows.

Let $M$ be the midpoint of $BC$ and $XN,$ so $MT$ is tangent to incircle. Let $R=(AYIZ) \cap AM$, so $$90=\angle IRA=\angle IRM=\angle IXM,$$so $IRMX$ is cyclic. Thus $IRTMX$ is also cyclic, so $R=inverse(K)$ (follows by taking inverses of lines $XT$ and $YZ$ wrt incircle). All of this means $AM=polar(K).$ \textit{Lemma:} $XI, YZ, AM$ concur. \textit{Proof:} Follows from Simson lines and homothethy. Let $S=XI \cap YZ \cap AM.$ Since (where $P_{\infty}$ is point at infinity wrt $BC$) $XI=polar(P_{\infty})$ and $YZ=polar(A)$ and $AM=polar(K),$ we know $polar(S) = \overline{AKP_{\infty}}$ so $A, K, P_{\infty}$ are collinear, and we are done.

Basically the same as the above ones. Nonetheless a fun projective geometry exercise.

Suppose that $X'$ is the antipode of $X$, which lies on $\overline{AN}$, and let $P = \overline{IX} \cap \overline{YZ}$. It is well-known that $P$ lies on the $A$-median, which we will call $\overline{AM}$. Finally, let $Q = \overline{AK} \cap \overline{BC}$. Since $T'YTZ$ is harmonic, we have \[-1 = (Z,Y;T',T) \overset{X}{=} (Z,Y;P,K) \overset{A}{=} (B,C;M,Q).\] However, we also know that $(B,C,M,\infty) = -1$, so $Q$ is the point at infinity. This implies that $\overline{AK} \parallel \overline{BC}$. $\square$

Let $M$ be the midpoint of $BC$, I will prove that if $R=XY\cap XI$, $A-R-M$, clearly $R$ lies on the polar of $A$, as well as this $R$ lies on the polar of the intersection of $BC$ and the line tangent to the incircle and passing through the antipode. Thus the polar of $R$ is the line through $A$ tangent to $BC$. Thus if we let the second intersection of $XI$ with the incircle be $P$ and the intersection of $XI$ with the polar of $R$ be $Q$, we get that $-1=(XP;RQ)\overset{\mathrm{A}}{=}(XN;M'\infty)$, Thus $M'$ is the midpoint of $XN$ so $M'=M$ and $A-R-M$ now let $AK\cap BC$ be $J$. Note that $-1=(PT;YZ)\overset{\mathrm{X}}{=}(YZ;RK)\overset{\mathrm{A}}{=}(BC;MJ)$ Thus as $M$ is the midpoint of $BC$ $J$ a point at infinity so $AK\parallel BC$.

did someone say projective nuke? Rename the intouch triangle to $XYZ$ as per the OTIS problem. Let $D$ be the antipode of $X$ on the incircle, and recall that $\overline{ADTN}$ are collinear (proof by a homothety sending the incircle to the $A$-excircle). Now let $PQR$ be the $D$-cevian triangle in $XYZ$. Brocard on $XYDZ$ implies $QR \perp \overline{XPD} \perp BC$. Pascal on $XYYDZZ$ implies $\overline{QAR}$ collinear. So, the goal is now to show that $K \in \overline{QAR}$. Pascal on $XTDZYY$ says $K$, $A$, $Q$ are collinear, done. $\blacksquare$