Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

I don't like this solution, but I couldn't find a better one this late at night (or this early in the morning; it's 4:15 AM here ). Let $S=KA\cap \Omega$, and let $T$ be the antipode of $K$ on $\Omega$. Let $X,Y$ be the touch points between $\Omega$ and $CA,AB$ respectively. The line $AD$ is parallel to $KT$ and is cut into two equal parts by $KS,KN,KD$, so $(KT,KN;KS,KD)=-1$. This means that the quadrilateral $KTSN$ is harmonic, so the tangents to $\Omega$ through $K,S$ meet on $NT$. On the other hand, the tangents to $\Omega$ through the points $X,Y$ meet on $KS$, so $KXSY$ is also harmonic, meaning that the tangents to $\Omega$ through $K,S$ meet on $XY$. From these it follows that $BC,XY,TN$ are concurrent. If $P=XY\cap BC$, it's well-known that $(B,C;K,P)=-1$, and since $\angle KNP=\angle KNT=\frac{\pi}2$, it means that $N$ lies on an Apollonius circle, so $NK$ is the bisector of $\angle BNC$. From here the conclusion follows, because if $B'=NB\cap \Omega,\ C'=NC\cap \Omega$, we get $B'C'\|BC$, so there's a homothety of center $N$ which maps $\Omega$ to the circumcircle of $BNC$.

Grobber, I like your solution! Just to clarify a few points which took me some time to understand: grobber wrote: and since $\angle KNP=\angle KNT=\frac{\pi}2$, This is because the segment KT is a diameter of $\Omega$. grobber wrote: From here the conclusion follows, because if $B'=NB\cap \Omega,\ C'=NC\cap \Omega$, we get $B'C'\|BC$, Is this trivial? The only explanation I have is to use the intersecting secant and tangent theorem, which yields $BB^{\prime} \cdot BN = BK^2$ and $CC^{\prime} \cdot CN = CK^2$, from what we conclude $\frac{BB^{\prime} \cdot BN}{CC^{\prime} \cdot CN} = \frac{BK^2}{CK^2}$, but since the line NK bisects the angle BNC, we have $\frac{BK}{CK} = \frac{BN}{CN}$, so that we get $\frac{BB^{\prime} \cdot BN}{CC^{\prime} \cdot CN} = \frac{BN^2}{CN^2}$, and thus $\frac{BB^{\prime}}{CC^{\prime}} = \frac{BN}{CN}$, what immediately implies B'C' || BC. As for another solution of the problem, see http://www.mathlinks.ro/Forum/viewtopic.php?t=14741 . Darij

I like very mutch Grobber's solution too. Another way to see that B'C'//BC, once we know that <BNK = <CNK is the following: <B'C'K = <B'NK = <C'NK = <C'KC.

Indeed, I was stupid... Darij

An alternate solution using some computation and inversion : Let us prove that for a circle through $B<C$ tangent to the incircle of triangle $ABC$ at point $N'$, $K, M, N'$ are collinear. Now, it is enough if we prove that angles $BKM$ and $BKN'$ are equal. After a few computations, we get $\tan {BKM} = \frac { cos(\frac{B}{2}) cos(\frac{C}{2})}{sin(\frac{B-C}{2})}$ Now apply an inversion with center $K$ and radius $BK$. Thereagain, after a few computations, we get $tan(BKN') = tan(BKM) = \frac { cos(\frac{B}{2}) cos(\frac{C}{2})}{sin(\frac{B-C}{2})}$. So we are done.

Dear Mathlinkers, this problem was already posted, but where? In order to have a complete synthetic proof, we can observe that NK goes through the A-excenter... Sincerely Jean-Louis

Dear jayme, I too on trying for a synthetic solution, tried by the same collinearity. That is the midpoint of the altitude , tangency point of the incircle with the corresponding side and the corresponding excenter are collinear. This is true by considering the diametrically opposite point of the tangency point of the excircle with $BC$ and drawing a parallel through it to $BC$ and using homothety and semiprojection result. But I was not able to finish the problem using this synthetic idea. So, could you please tell your complete solution?

I observed now this old and nice problem. I''ll search its synthetical proof and return soon. Yes Jayme, your remark is very interesting. Indeed, if denote the point $L$ where the $A$-exincircle touches the side $[BC]$ , then $M\in I_aK\cap IL$ because $\{\begin{array}{c} \frac {KD}{KL}=\frac {s-a}{a}=\frac {h_a}{2r_a}=\frac {MD}{LI_a}\\\\ \frac {LD}{LK}=\frac sa=\frac {h_a}{2r}=\frac {MD}{IK}\end{array}$.

Let $N'$ be the tangency point of the circle $\omega$ passing through $B,C$ with $\Omega.$ Let $U$ denote the antipode of $K$ WRT $ \Omega$ and $V$ the tangency point of the A-excircle $(I_a)$ with $BC.$ According to this topic, $N'U,N'K$ bisects $\angle AN'V$ internally and externally. Let $N''$ be the image of $N'$ under the homothety with center $A$ that takes $\Omega$ and $(I_a)$ into each other. Then $UN' \parallel VN''$ $\Longrightarrow$ $N'K \perp N''V$ $\Longrightarrow$ $\triangle N'VN''$ is isosceles with apex $N',$ which implies that $I_a \in N'K,$ due to $I_aV=I_aN''=r_a.$ But since $M \in KI_a,$ we deduce that $N \equiv N'.$

Here is a somewhat longer solution...

Let $I$ be the incenter and $I_A$ be the $A$-excenter of triangle $\triangle{ABC}$. Let $\omega$ be the incircle, $\Gamma$ be the $A$-excircle and let $\Omega$ be the circle with diameter $AD$. The homothety with center $A$ sending $\omega$ to $\Gamma$ takes $K$ to $K'$ where the tangent at $K'$ to $\Gamma$ is parallel to $BC$, perpencular to $AD$ and hence parallel to the tangent to $\Omega$ at $A$. Hence the negative homothety taking $\Omega$ to $\Gamma$ takes $A$ to $K'$ and therefore has center on $AK'$. The center of this negative homothety also lies on the common tangent $BC$ to $\Omega$ and $\Gamma$ and therefore the center of the negative homothety is $K$. This implies that $M$, $I_A$ and $K$ are collinear. Now let the circle $\gamma$ with diameter $II_A$ intersect $\omega$ at $X$ and $Y$. Note that this implies that $XI_A$ and $YI_A$ are tangent to $\omega$ and hence that $XY$ is the polar of $I_A$ with respect to $\omega$. Now let $XY$ intersect $BC$ at $T$. Let $N'$ denote the point on $\omega$ such that $TN'$ is tangent to $\omega$ and $N' \neq K$. Now note that since $T$ lies on the polar of $I_A$ with respect to $\omega$, $I_A$ lies on the polar of $T$ with respect to $\omega$, which is $N'K$. Hence $I_A$, $N'$, $K$ and $M$ are collinear and thus $N=N'$. Since $BI_A$ and $CI_A$ bisect the exterior angles of the triangle at $B$ and $C$, respectively, $\gamma$ passes through $B$ and $C$. Hence $BCYX$ is cyclic and, by power of a point with respect to $\gamma$ and $\omega$, $TN^2 = TX \cdot TY = TB \cdot TC$. This implies that the circumcircle of $\triangle{BCN}$ is tangent to $\omega$ at $N$, as desired.

WLOG, let $AB<AC$ (the case $AB=AC$ is trivial.) Let $a=BC$, $b=CA$, $c=AB$. Let $I$ be the the center of $\Omega$, let $r$ be its radius, let $\Omega$ be tangent to $AB$ and $AC$ at $X$ and $Y$, respectively, and let $Z$ be the point where $XY$ meets $BC$. Let $\omega_A$ be the $A$-excircle of $ABC$, let $I_A$ be its center, $r_A$ be its radius, and let $J$ be its tangency point with $BC$, We first claim that $M$, $K$, and $J$ are collinear. Let $J'$ be its antipode in $\omega_A$. A homothety centered at $A$ taking $\Omega$ to $\omega_A$ sends $K$ to $J'$. Thus, a homothety centered at $K$ mapping $\Omega$ to $\omega_A$ must send $A$ to $J'$ as well. Line $AD$ is sent to a line parallel to it passing through $J'$, i.e., line $JJ'$. Since $J,D$ both line on $BC$, the homothety must map $D$ to $J$. Thus, the homothety must map the midpoint of $AD$ to the midpoint of $JJ'$, so $M$, $K$, and $I_A$ must be collinear. Now let $P$ be the midpoint of $ZK$. Because $(Z,B,K,C)$ is harmonic, we must have $PK^2 = PB \cdot PC$ (this can also be verified by computing the lengths directly, given $ZB$, which can be found through Menelaus.) Consequently, if the tangent to $\Omega$ from $P$ distinct from $PB$ is tangent to $\Omega$ at $N'$, we have $PN'^2 = PK^2 = PB \cdot PC$, whence $PN'$ is tangent to the circumcircle of $\triangle N'BC$, i.e., $N'$ is the tangency point of the circle through $B,C$ tangent to $\Omega$. We wish to show that $N'$, $K$, and $I_A$ are collinear. We claim that $\triangle PIK \sim \triangle I_a KJ$. The result would then follow, for we would have $\angle I_a KJ = \angle ZIK = 90^{\circ} - \angle N'KI = \angle N'KD$. Since both triangles are right, it suffices to prove $PK/KI = I_aJ / KJ \iff PK \cdot KJ = r r_A$. Now, $rr_A = \frac{K^2}{s(s-a)} = (s-b)(s-c)$, where $K$ is the area and $s$ is the semiperimeter of $\triangle ABC$. Also, $KJ = a - BK - CJ = a - 2(s-b) = b-c$, and \[ \frac{ZB}{ZB + a} = \frac{KB}{KC} = \frac{s-b}{s-c} \implies ZK = ZB + s-b = \frac{(s-b)(a+b-c)}{b-c}. \] Thus, \[ PK \cdot KJ = \frac{ZK}{2} \cdot KJ = \frac{(s-b)(s-c)}{b-c} \cdot (b-c) = (s-b)(s-c) = r r_A, \] so we are done.

Let $I_A$ be the $A$-excenter with tangency points $X_A$, $X_B$, and $X_C$ to $BC$, $CA$ and $AB$, respectively. Define $P$ to be the midpoint of $KI_A$. Let $r$ be the radius of the incircle and $R$ the radius of the $A$-excircle. It is well-known that $M$, $K$ and $I_A$ are collinear. We claim that $NBPC$ is cyclic; it suffices to prove that $2BK \cdot KC = 2KP \cdot KN = KN \cdot KI_A$. On the other hand, by Power of a Point we have that \[ I_AK \left( I_AK + KN \right) = II_A^2 -r^2 \implies KN \cdot KI_A = II_A^2 - r^2 - I_AK^2 \] Now we need only simplify the right-hand side using the Pythagorean Theorem; it is $\left( (r+R)^2 + KX_A^2 \right) - r^2 - \left( R^2 + KX_A^2 \right) = 2Rr$. So it suffices to prove $Rr = (s-b)(s-c)$, which is not hard. Now, since $P$ is the midpoint of minor arc $\widehat{BC}$ of $(NBC)$ (via $BK=CX_A$), while the incircle is tangent to segment $BC$ at $K$, the conclusion follows readily. [asy][asy]/* DRAGON 0.0.9.6 Homemade Script by v_Enhance. */ import olympiad; import cse5; size(11cm); real lsf=0.8000; real lisf=2011.0; defaultpen(fontsize(10pt)); real xmin=-4.18; real xmax=3.94; real ymin=-2.38; real ymax=2.34; /* Initialize Objects */ pair A = (-1.5, 2.0); pair B = (-2.0, 0.0); pair C = (0.5, 0.0); pair I = incenter(A,B,C); pair D = foot(A,B,C); pair M = midpoint(A--D); pair K = foot(I,B,C); pair N = (2)*(foot(I,M,K))-K; pair I_A = (2)*(IntersectionPoint(circumcircle(A,B,C),Line(A,I,lisf),1))-I; pair X_B = foot(I_A,A,B); pair X_C = foot(I_A,A,C); path CircleNBC = circumcircle(N,B,C); pair X_A = foot(I_A,B,C); path NI_A = N--I_A; pair P = IntersectionPoint(CircleNBC,NI_A,1); /* Draw objects */ draw(A--B, rgb(0.0,0.6,0.6)); draw(B--C, rgb(0.0,0.6,0.6)); draw(C--A, rgb(0.0,0.6,0.6)); draw(incircle(A,B,C), rgb(0.0,0.8,0.0) + linewidth(1.2)); draw(A--D, dotted); draw(CircleNBC, rgb(0.2,0.8,0.0)); draw(CirclebyPoint(I_A,X_A), rgb(0.0,0.8,0.0) + linewidth(1.2) + dashed); draw(B--X_B, rgb(0.6,0.6,0.6) + linetype("4 4")); draw(C--X_C, rgb(0.6,0.6,0.6) + linetype("4 4")); draw(NI_A, rgb(0.0,0.0,0.8)); /* Place dots on each point */ dot(A); dot(B); dot(C); dot(I); dot(D); dot(M); dot(K); dot(N); dot(I_A); dot(X_B); dot(X_C); dot(X_A); dot(P); /* Label points */ label("$A$", A, lsf * dir(135)); label("$B$", B, lsf * dir(135)); label("$C$", C, lsf * dir(45)); label("$I$", I, lsf * dir(45)); label("$D$", D, lsf * dir(135)); label("$M$", M, lsf * dir(45)); label("$K$", K, lsf * dir(45)); label("$N$", N, lsf * dir(115)); label("$I_A$", I_A, lsf * dir(45)); label("$X_B$", X_B, lsf * dir(170)); label("$X_C$", X_C, lsf * dir(45)); label("$X_A$", X_A, lsf * dir(45)); label("$P$", P, lsf * dir(45)); /* Clip the image */ clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy]

Interestingly, this problem can be solved using almost exclusively harmonic ratios. Again, let $X,Y$ be the tangent points of $\Omega$ with $AB$ and $AC$ respectively and $S,T$ be the intersections of $\Omega$ with $AK$ and $KI$ (extended). Let $V$ be the point at infinity on $AD$. Then $(V,M;A,D)=-1$, so by projection through $K$ onto $\Omega$, we have, as previously noted, that $KNST$ is a harmonic quadrilateral. Let $N',S',T'$ be the intersections of the tangents at $N,S,T$ with $BC$. Then $(T',N';S',K)=-1$, or $N'$ is the midpoint of $S'K$. $A$ lies on the pole of $S'$, so $S'$ lies on the pole of $A$, which is $XY$. Thus $(S',K;B,C)=-1$ as well. This means that $B$ and $C$ are inverses with respect to the circle with diameter $S'K$, or that $N'B*N'C=N'K^2=N'N^2$, which makes $N'N$ a tangent of circle $BNC$.

ThinkFlow wrote: Let the tangent to $\Omega$ at $N$ intersect line $BC$ at $P$. It suffices to show that aopsnowrap][math=inline]$PN^2 = PB\cdot BC$[/math].[/aopsnowrap Here's a nice way to finish, inspired by the official solution to RMM 2013 Problem 3. Let $Q=PI\cap NMK$ be the inverse of $P$ w.r.t. $(I)$, so $PQ\cdot PI = PK^2 = PN^2$, whence it's enough to show $BQIC$ is cyclic. On the other hand, it's well-known that $II_a$ is a diameter of $(BIC)$ and $M,K,I_a$ are collinear, so $\angle{IQI_a} = 90^\circ$ implies that $Q$ lies on the circle with diameter $II_a$, as desired.

ThinkFlow wrote: Let the tangent to $\Omega$ at $N$ intersect line $BC$ at $P$. It suffices to show that $PN^2 = PB\cdot BC$. The way I did this was to let $R$ be the inverse of $M$ w.r.t. $(I)$, $S = PR \cap EF, Q = EF \cap BC$. Let $L$ be the point diametrically opposite to $K$, and let the tangent at $L$ intersect $EF$ at $T$. Let $U$ be $AL \cap BC$. By the properties of harmonics, it suffices to prove $P$ is the midpoint of $QK$. Now, $KT$ is the polar of $U$, hence $KT \perp IU$. On the other hand, $P$ is the pole of $KN$, so the polar of $M$ passes through $P$. As it also passes through $R$, so $PR$ is the polar of $M$. Hence, the pole of $AM$ is $EF \cap PR$, which is $S$. Hence, $IS \perp AM \implies IS || BC$. The other thing to notice is that $PR$ is the polar of $M$ implies $PR \perp IM \implies PS \perp IU$ because $I,M, U$ are collinear. Together with $KT \perp IU$ we get $PS \perp KT$. Now, we need to prove $P$ is the midpoint of $QK$, so it suffices to prove $S$ is the midpoint $QT$. But this is clear, as the perpendicular distance from $S$ to $BC$ is $r$ (as $IS || BC$), and the perp distance from $T$ is $2r$ (as $LT || BC$).

Let the tangent at $N$ to the incircle intersect $BC$ at $P$. It remains to prove that $PN$ is tangent to $(BCN)$. We proceed with the following well known result: Lemma: Points $N$, $M$, $K$, and $I_a$ are collinear.

$\Box$ Define $Q = PI \cap NK$. From kite $NPKI$, we see $\angle PQN = 90$, so \[PN^2 = PQ \cdot PI.\] Because $\angle PQN = IQI_A = 90$, we know $Q$ lies on the circle with diameter $II_A$, or $(IBI_AC)$. Using the power of $P$, we find \[PB \cdot PC = PQ \cdot PI = PN^2,\] and thus $PN$ is tangent to $(BNC)$ at $N$. $\blacksquare$ [asy][asy] size(270); defaultpen(linewidth(0.4)+fontsize(10)); pair A, B, C, D, I, K, M, IA; A = dir(130); B = dir(210); C = dir(330); D = foot(A, B, C); I = incenter(A, B, C); K = foot(I, B, C); M = .5A + .5D; IA = extension(B, rotate(90, B)*I, C, rotate(90, C)*I); pair N = IP(2M-K--M, incircle(A, B, C)); pair P = extension(B, C, N, rotate(90, N)*I); pair Q = extension(P, I, N, K); draw(A--B--C--cycle); draw(incircle(A, B, C)); draw(circumcircle(B, N, C), blue); draw(circumcircle(I, B, C), heavygreen); draw(I--P--B--P--N--IA--cycle^^N--I--K--D--A); draw(C--P--I--P--N, magenta); label("$A$", A, dir(90)); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, S); label("$I$", I, dir(90)); label("$K$", K, SW); label("$N$", N, 2dir(135)); label("$I_A$", IA, S); label("$P$", P, W); label("$Q$", Q, dir(90)); dot(A); dot(B); dot(C); dot(D); dot(I); dot(K); dot(N); dot(IA); dot(P); dot(Q); [/asy][/asy]

Let the incenter be $I$. Let $(BIC)$ Intersect $(KM)$ at $Q$ and $I_A$, such that $I_A$ is outside $ABC$. Let $(BCN)$ Intersect $KM$ again at $P$. It suffices to show that $P$ is on the perpendicular bisector of $BC$ due to a homothety. It is well known (from the "Midpoints of Altitudes Lemma") that $I_A$ must be the $A$-excenter. Then from Incenter/Excenter Lemma, $II_A$ is a diameter of $(BIC)$. It follows that \[ 90 = \angle IQI_A = \angle IQK \]but since $KN$ is a chord of the incircle, this implies $Q$ is the midpoint of $KN$. Now taking the power of a point from $K$, we see that \[ KN \cdot KP = KB \cdot KC = KQ \cdot KI_A \]But this implies $PI_A = PK$. However, since the $A$-excircle is tangent to $BC$ at the reflection of $K$ across the midpoint of $BC$, then dropping altitudes shows that $P$ lies on the perpendicular bisector of $BC$, as desired. $\blacksquare$

As always, we can remove $M$ from our diagram by noting that line $MK$ is the same as line $KI_A$. Let $Q$ be the midpoint of $KI_A$. We claim $BNCQ$ is cyclic. Let $S$ be the midpoint of $NK$. Since $\angle ISI_A = \angle IBI_A = 90$ (well known), we know $S$ lies on the circle containing $B$, $I$, $C$, and $I_A$ (this circle being from a common configuration). By PoP, $KS \cdot KI_A = KB \cdot KC$. However, we know $KS \cdot KI_A = KN \cdot KQ$. Thus, $BNCQ$ is cyclic. Let $P$ be the circumcenter of $BCN$. Notice that since $BK = XC$, we have $QB = QC$ and thus $QP$ is the perpendicular bisector of $BC$. In other words, $Q$ is the arc midpoint of arc $BC$ on the circumcircle of $BCN$. Consider a homothety at $N$ that takes $K$ to $Q$. This homothety must also take $I$ to $P$, finishing the proof.

By a well known lemma $KM$ passes through the $A$-excenter $J.$ If $P$ is the midpoint of $KN$ then $\angle IPJ=90^\circ$ where $I$ is the incenter so $B,I,C,J,P$ are concyclic by fact 5. Thus $BK\cdot CK=PK\cdot JK=NK\cdot XK$ where $X$ is the midpoint of $JK$ so $B,N,C,X$ are concyclic. But then $X$ is the bottommost point on this circle since $K$ and the foot from $J$ to $BC$ are equidistant from the midpoint of $BC.$ Thus the homothety at $N$ sending $K$ to $X$ must send the incircle to circle $(BNCX),$ done.

It is well known that $\overline{KM}$ passes through $I_A$ the $A$-excenter. Now, consider the circle $\omega$ through $B$ and $C$ orthogonal to $\Omega$. This contains $B'$ and $C'$ the inverses of $B$ and $C$ with respect to $\Omega$. Then, the center of $\omega$ is the intersection of $\ell_1$ the perpendicular bisector of $\overline{BC}$ and $\ell_2$ the perpendicular bisector of $\overline{B'C'}$. Homothety at $K$ with factor $2$ sends $\ell_1$ to the perpendicular from $I_A$ to $\overline{BC}$ and $\ell_2$ to $\overline{AI}$, which intersect at $I_A$, so the center of $\omega$ is the midpoint of $\overline{KI_A}$. Therefore, inversion around $\omega$ swaps $(BKC)$ and $(BNC)$ fixing $\Omega$. Since $(BKC)$ is tangent to $\Omega$, we see that $(BNC)$ is tangent to $\Omega$.

Let $S$ be the antipode of $K$ in $\Omega$ and $T=EF\cap BC$ where $E,F$ are the $B-,C-$ touchpoints of $\Omega$. Claim. $S,T,N$ are collinear. Proof. Projecting $(A,D;M,\infty_{AD})=-1$ from $K$ onto $\Omega$ gives $(AK\cap\Omega,K;N,S)=-1$. Let $L=AK\cap\Omega$. So, $LNKS$ is harmonic quadrilateral. So, tangents at $L,K$ meet on line $SN$. But since tangents to $E,F$ meet on line $LK$, it means $LEKF$ is harmonic quadrilateral. Hence, $T=EF\cap BC$ is the meeting point of tangents at $L,K$. So, $T\in SN$, as claimed. $\blacksquare$ Since $KN\perp NS$, it means $N$ is the foot from $K$ to $ST$. Extend $NK$ to meet $(BCN)$ again at $P$. Note that $(T,K;B,C)=-1$ which means $(BCN)$ is the Apollonius circle for $KT$ as $\angle TNK=90^\circ$. Hence, $\frac{BN}{NC}=\frac{BK}{KC}$, which means $NK$ is the angle-bisector of $\angle BNC$. So, $NK$ extended to meet the circle $(BNC)$ is the midpoint of arc $BC$ not containing $N$. Since $\Omega$ is tangent to $BC$, $\Omega$ is tangent to $(BNC)$ by Shooting Lemma.

We can remove $AD$ by replacing $KM$ with $KI_A$, where $I_A$ is the $A$-excenter. By shooting lemma, it would suffice to show that $TK \cdot TN = TB^2 = TC^2$ where $T$ is the midpoint of arc $\widehat{BC}$ of $(BCN)$. We will show this using phantom points. Let $T'$ be the intersection of the perpendicular bisector of $BC$ and line $KI_A$. Let $A'$ be the midpoint of $BC$ and $X$ be the $A$-extouch point. Since $T'A'$ and $IK$ are both perpendicular to $BC$, they are parallel, and hence $\angle{NKI} = \angle{KT'A'} = \angle{T'I_AX} = x$ (say). By the converse of midpoint theorem in $\Delta{KI_AX}$, $T'$ is the midpoint of $KI_A$. This yields: $KT' \cdot KN = \frac{1}{2} KI_A \cdot KN = \frac{1}{2} \cdot KI_A \cdot \frac{\sin{(180^{\circ} - 2x)}}{\sin{x}} \cdot r = KI_A \cdot \cos{x} \cdot r = r_a \cdot r = (s - b) \cdot (s - c) = KB \cdot KC$ implying that $N, B, C, T'$ are concyclic as desired.

Xoink phone write ups Let $MK$ meet the perpendicular bisector of $BC$ at $E$ note that by the shooting lemma $E \in (BCN)$. Let $O$ be the circumcenter of $(BCN)$. Then note that since $ON=OE$ and $IK=IN$ and $\angle NKI = \angle NEO$. Since $\triangle KNI \sim \triangle ENO$ and $N$, $K$, and $E$ are collinear. This implies $O$, $I$, $N$ collinear, and we are finished.

#attachments[url][/url]

MELSSATIMOV40 wrote: #attachments[url][/url] 2 nd page

By the midpoints of altitudes lemma, $KN$ passes through the $A$ excenter $I_A$ so now we remove the extraneous altitude. Now we take the tangent to the incircle at $N$ and show it is tangent to $(BCN)$ as well. Let the tangent intersect $BC$ at $T$ and let $IT \cap KN=X$. Since $IT$ is the perpendicular bisector of $KN$, X is the midpoint of $KN$ and it lies on the circle with diameter $II_A$ which is $(BIC)$. Since $\Delta IKT$ is right angled at $T$ and $X$ is foot of altitude from $K$ to the hypotenuse, it follows by similarity that $TK^2=TX.TI$. Now by Power of Point, $$TN^2=TK^2=TX.TI=TB.TC$$So $TN$ is indeed tangent to the circle $(BCN)$.

Someone PLEASE stop me from becoming too reliant on projective stuff Relabel several of the points to match with the diagram.

Let $A’=(DEF)\cap\overline{AD}\neq D$, $T’$ the pole of $\overline{A’D}$ and $T$ the midpoint of $AD$. I claim that $\overline{KT}$ tangent to $(DEF)$, and $KT^2=TB\cdot TC$ which will finish. First notice that $\overline{KD’}$ passes through $T$, since $(A’D;KD’)\stackrel{D}{=}(AX;M\infty_{AX})=-1$, however since $A$ is the pole of $\overline{EF}$, the polar of $A$ which is just $\overline{EF}$ also passes through $T’$, the pole of $\overline{A’D}$. Hence by the midpoints of harmonic bundles lemma $TD^2=TD\cdot TT’=TB\cdot TC$. So it suffices to show $KT=TD$. Now $\measuredangle D’KD=90^{\circ}$, so $\triangle T’KD$ has circumcentre $T$. Hence we indeed have $KT=DT$. Since $K\in(DEF)$, we must also have that $KT$ tangent to $(DEF)$ and $(BKC)$ as desired.

Let $I$ be the incentre, $\Delta KPQ$ be the intouch triangle, and let $R=PQ \cap BC$. Let $K'$ be the antipode of $K$ in $\Omega$. Let $\Omega_A$ be the $A$-excircle with centre $I_A$, $L$ be the $A$-extouch point, $L'$ be the antipode of $L$ in $\Omega_A$. Recall that $A-K-L'$ and $A-K'-L$. Since $I_A$ is the midpoint of $LL'$, there exists negative homothety centred at $K$ takes $D$ to $L$, $A$ to $L'$, and so $M$ to $I_A$. Thus $M-K-I_A$. Note that $AK$ is the polar of $R$ wrt $\Omega$, so $AK \perp IR$. It follows that $\Delta IKR \sim \Delta KLL'$. Let $T$ be the midpoint of $KR$, then we have $\Delta IKT \sim \Delta KLI_A$, so $IT \perp KI_A$, and this implies that $KI_A$ is the polar of $T$ wrt $\Omega$. Now $N$ lies on $KI_A$, so $TN$ is tangent to $\Omega$ at $N$. Then since $(B,C;K,R)=-1$, we have $TB \cdot TC = TK^2 = TN^2$. Thus $TN$ is also tangent to $(BCN)$, so we are done. $\square$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(25cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(15); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -27.52499633361368, xmax = 35.607853428992016, ymin = -28.02653819614295, ymax = 1.5669851300784052; /* image dimensions */ pen ffqqtt = rgb(1,0,0.2); pen ffcctt = rgb(1,0.8,0.2); pen qqccqq = rgb(0,0.8,0); pen qqttcc = rgb(0,0.2,0.8); /* draw figures */ draw((-6.329572706245745,-9.094920005090973)--(10.432798083220217,-8.920793369494598), linewidth(1) + ffqqtt); draw((xmin, 2.8494321797757816*xmin + 8.940768148316133)--(xmax, 2.8494321797757816*xmax + 8.940768148316133), linewidth(1) + ffqqtt); /* line */ draw((xmin, -0.6801847296747661*xmin-1.8245634255080363)--(xmax, -0.6801847296747661*xmax-1.8245634255080363), linewidth(1) + ffqqtt); /* line */ draw((-2.95328979009002,-9.059847353829664)--(-3.05,0.25), linewidth(1) + ffcctt); draw(circle((-1.187149883734938,-5.409450816082839), 3.631854017871503), linewidth(1) + qqccqq); draw(circle((5.424266233632817,-25.495331056367387), 16.52161792320652), linewidth(1) + qqccqq); draw((xmin, -2.5030417550537365*xmin-11.918166183056378)--(xmax, -2.5030417550537365*xmax-11.918166183056378), linewidth(1) + ffcctt); /* line */ draw(circle((2.0503484157693896,-8.886150964311547), 8.382521251006963), linewidth(1) + linetype("4 4") + gray); draw(circle((2.118558174948934,-15.452390936225115), 10.573001088906159), linewidth(1) + qqttcc); draw((5.424266233632817,-25.495331056367387)--(5.252649785554102,-8.974604480481936), linewidth(1) + ffcctt); /* dots and labels */ dot((-3.05,0.25),dotstyle); label("$A$", (-2.9294902802652074,0.5805343525376935), NE * labelscalefactor); dot((-6.329572706245745,-9.094920005090973),dotstyle); label("$B$", (-6.184777846149563,-8.75786634151438), NE * labelscalefactor); dot((10.432798083220217,-8.920793369494598),dotstyle); label("$C$", (10.552003679457885,-8.593457878590927), NE * labelscalefactor); dot((-1.187149883734938,-5.409450816082839),linewidth(4pt) + dotstyle); label("$I$", (-1.055233802937851,-5.140880157198436), NE * labelscalefactor); dot((5.424266233632817,-25.495331056367387),linewidth(4pt) + dotstyle); label("$I_{A}$", (5.553986406584933,-25.23159432644427), NE * labelscalefactor); dot((-1.1494244085796241,-9.041108894103633),linewidth(4pt) + dotstyle); label("$K$", (-1.0223521103531603,-8.790748034099071), NE * labelscalefactor); dot((5.252649785554102,-8.974604480481936),linewidth(4pt) + dotstyle); label("$K'$", (5.389577943661481,-8.72498464892969), NE * labelscalefactor); dot((-2.95328979009002,-9.059847353829664),linewidth(4pt) + dotstyle); label("$D$", (-2.830845202511136,-8.790748034099071), NE * labelscalefactor); dot((-3.00164489504501,-4.404923676914827),linewidth(4pt) + dotstyle); label("$M$", (-2.8637268950958266,-4.154429379657723), NE * labelscalefactor); dot((-3.662196151950729,-2.751536299526585),linewidth(4pt) + dotstyle); label("$N$", (-3.521360746789636,-2.4774630578385133), NE * labelscalefactor); dot((2.1374209125265966,-17.26821997523551),linewidth(4pt) + dotstyle); label("$P$", (2.2658171481158864,-17.01117118027167), NE * labelscalefactor); dot((-2.405810280265177,-5.896322596815108),linewidth(4pt) + dotstyle); label("$Q$", (-2.271856428571398,-5.634105545968792), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Lemma 1: If $P'$ is the midpoint of the arc $\overarc{BC}$ in $(BNC)$, then $(BNC)$ is tangent to $\Omega\iff N-K-P'$. Proof: Inverting by $P'$ fixing $B$ and $C$ sends $BC\longleftrightarrow (BNC)$ and $\Omega$ is tangent to $(BNC)\iff\Omega\longleftrightarrow\Omega\iff K\longleftrightarrow N\iff N-K-P' \ $ $_{\blacksquare}$ Lemma 2: $M-K-I_A$. Proof: If $K''$ is the antipode of $K'$ on the excircle, then the homothety from $A$ sending the incircle to th excircle yields $A-K-K''$ thus the homothety from $K$ mapping $AD\mapsto K'K''$ maps $M\mapsto I_A\Rightarrow M-K-I_A \ $ $_{\blacksquare}$ Lemma 3: If $P=(BNC)\cap MK$, then $KP=I_AP$ . Proof: Let $Q=(BIC)\cap MK$. $\angle IQK=\angle IQI_A=\angle IBI_A=90^{\circ}\implies NQ=KQ$. But $NK\cdot KP=BK\cdot CK=KQ\cdot KI_A=\frac{NK}{2} \cdot KI_A\implies KP=\frac{KI_A}{2}=I_AP \ $ $_{\blacksquare}$ Now notice that $P$ lies on the perp. bissector of $KK'$ because $K'I_A\perp BC$, thus it lies on the perp. bissector of $BC\implies \boxed{P=P'}$, but that means we are done because clearly $N-K-P' \ $ $_{\blacksquare}$.

If $AB = AC$ then the conclusion is obvious since $N$ lies on the perpendicular bisector of $\overline{BC}$; thus without loss of generality assume that $AB < AC$. Let $I_A$ be the $A$-excentre, which is known to lie on line $MK$, let $I$ be the incentre, and let the tangent to $\Omega$ at $N$ intersect line $BC$ at $T$. Further let $M_A$ be the midpoint of arc $BC$ of $(ABC)$ not containing $A$, let $\ell$ be the tangent to $(ABC)$ at $M_A$, let $K'$ be the foot from $I_A$ to $BC$ and let $L$ be the midpoint of $\overline{AC}$. As $T$ is the pole of line $NK$ in $\Omega$, we have $TI \perp KI_A$. Orient $BC$ parallel to the $x$-axis; now $\operatorname{slope}(TI) = \frac{IK}{TK} = \frac{r}{TK}$ and \[\operatorname{slope}(KI_A) = \frac{I_A K}{-KK'} = \frac{\operatorname{dist}(I_A, \ell) + M_A L}{-KK'} = -\frac{\operatorname{dist}(I, \ell) + M_A L}{2KL} = -\frac{IK + 2M_A L}{2(BL - BK)} = -\frac{r + 2BL \tan \angle M_ABC}{2(\frac a2 - (s - b))} = -\frac{r + a \tan(\frac{\alpha}{2})}{b - c}.\]Since $\operatorname{slope}(TI) \cdot \operatorname{slope}(KI_A) = -1$, we have \[-\frac{r}{TK} \cdot \frac{r + a \tan(\frac{\alpha}{2})}{b - c} = -1 \implies TK = \frac{r(r + a \tan(\frac{\alpha}{2}))}{b - c}.\]Let $E$ be the foot from $I$ to $AC$; now $\tan\left(\frac{\alpha}{2}\right) = \tan(\angle IAE) = \frac{IE}{AE} = \frac{r}{s - a}$. As such, \begin{align*} TK &= \frac{r(r + a \tan(\frac{\alpha}{2}))}{b - c} \\ &= \frac{r^2(1 + \frac{a}{s - a})}{b - c} \\ &= \frac{(\frac{A}{s})^2(\frac{s}{s - a})}{b - c} \\ &= \frac{\frac{s(s - a)(s - b)(s - c)}{s^2} \cdot \frac{s}{s - a}}{b - c} \\ &= \frac{(s - b)(s - c)}{b - c}. \end{align*}Now we can calculate \begin{align*} TK &= \frac{(s - b)(s - c)}{b - c} \\ TK((s - c) - (s - b)) &= (s - b)(s - c) \\ TK^2 &= TK^2 + TK((s - c) - (s - b)) - (s - b)(s - c) \\ TK^2 &= (TK - (s - b))(TK + (s - c)) \\ TK^2 &= (TK - BK)(TK + CK) \\ TK^2 &= TB \cdot TC. \end{align*}But $TN = TK$, so $TN^2 = TB \cdot TC$, and therefore $TN$ is tangent to $(NBC)$. Since $TN$ is by definition also tangent to $\Omega$, we're done.