Suppose that $\omega_b$ is tangent to $\overline{AB}$ at $D$ and $\omega_c$ is tangent to $\overline{AC}$ at $E$. The homothety at $S_b$ sending $\omega_b$ to $\Omega$ sends $D$ to $S_c$, so $D$ lies on $\overline{S_bS_c}$. Similarly, $E$ lies on $\overline{S_bS_c}$. Since $S_bA=S_bI$ and $S_cA=S_cI$ by the incenter-excenter lemma, $\overline{S_bS_c}$ is perpendicular to $\overline{AI}$, so $AD=AE$. Thus, $A$ lies on the radical axis of $\omega_b$ and $\omega_c$. Let $\overline{IN_a}$ intersect $\Omega$ again at $X$, and let $S_a$ be the antipode of $N_a$ in $\Omega$. Let the tangents to $\Omega$ at $S_b$ and $S_c$ intersect at $T$. We have $TS_b=TS_c$, so $T$ lies on the radical axis of $\omega_b$ and $\omega_c$. Also, notice that \[-1=(B,C;S_a,N_a) \overset{I}{=} (S_b,S_c;A,X),\]so $T$ lies on $\overline{AX}$. Since $A$ and $T$ both lie on the radical axis of $\omega_b$ and $\omega_c$, we know that $X$ lies on the radical axis, as desired.

Let $N_aI$ meet $\Omega$ again at $T$, the tangents to $\Omega$ at $S_b$ and $S_c$ meet at $P$, $\omega_b$ touch $AB$ at $X$, and $\omega_c$ touch $AC$ at $Y$. Clearly, $P$ lies on the Radical Axis of $\omega_b$ and $\omega_c$. Fact 5 implies $S_bS_c$ is the perpendicular bisector of $AI$. Thus, $AI \perp S_bS_c$. But the Shooting Lemma yields $X \in S_bS_c$ and $Y \in S_bS_c$ respectively, so $AX = AY$ follows immediately. Hence, $A$ clearly belongs the the Radical Axis of $\omega_b$ and $\omega_c$ as well. Now, if ray $AI$ meets $\Omega$ again at $M$, symmetry gives $$-1 = (B, C; M, N_a) \overset{I}{=} (S_b, S_c; A, T).$$Thus, $AS_bTS_c$ is harmonic, whence $T \in AP$, i.e. $T$ belongs to the Radical Axis of $\omega_b$ and $\omega_c$. $\blacksquare$

Nice problem. Rewrite $S_b$ and $S_c$ as $E$ and $F$. Say $EF$ intersects $AB, AC$ at $B', C'$. By homothety, we find that in fact $\omega_b$ is tangent to $AB$ at $B'$, similarly for $\omega_c$ and $C'$. Now, let $T$ be the $A$-mixtilinear touchpoint. It is well known that $T$ is the second intersection of $IN_a$ and $\Omega$. It suffices to prove that $T$ is on the radical axis of $\omega_b$ and $\omega_c$. But since $AB' = AC'$ we have that $A$ is on this radical axis, and by radical axis theorem applied to $\omega_b, \omega_c, \Omega$ we find that the intersection $K$ of the tangents to $\Omega$ from $E$ and $F$ also lies on this radical axis. Moreover, it is well-known that $(T, A; E, F) = -1$ (see i.e. APMO 2012 4), so thus we find that $T$, $A$, and $K$ are collinear, as desired.

Really nice one. Sketch of solution with a friend As carteacoes não param.. sério, eu só fui carteando um monte de ponto em alguns deram certo kkkk First, it is a well known lemma (can be proved by angle chase) that the mixtilinear contact with the circumcircle is on the line $N_aI$. Let the mixtilinear touch the circumcircle in $T$. Thus we just need $T$ to be on the radical axis of both circles. Let the tangents from $S_c,S_b$ to $(ABC)$ touch themselves in $R$. Notice $R$ is on the radical axis of the two circles. Now, by Death Star lemma, the contacts of the circles $w_b,w_c$ with $AB,AC$ (call those $P,Q$, respectively) are on line $S_bS_c$, as this line is the perp bisector of $AI$. After noticing this, angle chase proves $AP=AQ$ which implies by the tangency that $A$ is on the radical axis, as the power of $A$ to $w_b,w_c$ are $AQ^2=AP^2$, respectively. Now we just need $AS_bS_cT$ to be harmonic, because this would imply $R,A,T$ collinear. To prove this, just project through $I$ from the circle to the circle, which is possible by involution, and you'll end up with needing $BN_aCM$ to be harmonic, where $M=AI\cap (ABC)$, which is true as $BM/MC=BN_a/CN_a$. $\blacksquare$

my and jrsbr solution. Let $D = \omega_B \cap AB$, $E = \omega_C \cap AC$ and $l$ be the radical axis of $\omega_B$ and $\omega_C$. First, by Death Star Lemma (Brazilian name), we have that $S_C$, $D$, $E$, $S_B$ are collinear, so, $DE \perp AI \implies AD = AE\implies A \in l$. So, let ${A, T} = l \cap \Omega$. Now, we make a $\sqrt{bc}$ inversion. By the angle condition, $S_b' \in BC$ and $\angle BS_b'A = \dfrac{\angle B}{2} \implies AB = S_b'B$. Similarly, $AC = S_c'C$. $T' \in BC$ and $T' \in$ radical axis of $\omega_B'$ and $\omega_C' \therefore T'S_b' = T'S_c' \therefore BT' + c = CT' + b \therefore$ $T'$ is the contact point of $A$-exincircle $\therefore T$ is the $A-$mixtilinear tangency point. It is known fact that $T$ is on line $IN_a$, then the problem follows. $_\blacksquare$

Let $IN_a$ intersect $\Omega$ again at $T$. $X=TS_C\cap AB$, $K=TS_C\cap \omega_C\neq S_C$, $Y=S_BT\cap AC$, $L=TS_B\cap\omega_C$, $P=S_BS_C\cap AB$, $Q=S_BS_C\cap AC$. It is well-known that $I$ is the midpoint of $XY$ and by homothety at $S_B$ and $S_C$, respectively, we have $P\in\omega_B$ and $Q\in\omega_C$. Now PQXY is a isosceles trapezoid and thus cyclic. Also, $\angle XKQ=0.5\angle S_CO_CQ=0.5\angle S_COS_B=\angle S_CTS_B=\angle S_CPX$ so $K\in (PQXY)$. Similarly, $L\in (PQXY)$. Since $XY\perp AI\perp S_BS_C$, we have $KL$ and $S_BS_C$ antiparallel and thus $KLS_BS_C$ is cyclic. This implies the desired result.

Very nice problem Just show that $A ,P ,Q ,K$ collinear [if you are feeling difficulty in this then read chapter 4 of evan chen book(EGMO)] $K$ is intersection of tangent at $S_b ,S_c$ in $\Omega$ also notice that $T$ is tangent point of $A-$ mixitilinear incircle with $\Omega$ [to prove this use $\sqrt{bc}$ inversion] now we left with to proof that if $TS_bAS_c$ is harmonic quadrilateral then we are done which is true by $$-1 = (B, C; M, N_a) \overset{I}{=} (S_b, S_c; A, T).$$hence this follow $T ,A,P,Q$ are collinear and we are done. Great thanks to Malay for telling that $T$ is tangent point of $A-$ mixtilinear incircle with $\Omega$ and that $I ,N_a ,T$ are collinear .

Nice one! Let $T$ be the A-mixtilinear touch point, which is infact the desired point. It is well known that $T$ lies on $IN_a$. We will prove that $T$ lies on the radical axis of $\omega_b, \omega_c$. Let $\omega_b$ touch $AB$ at $X$ and let $\omega_c$ touch $AC$ at $Y$. By shooting lemma, $X,Y$ lie on $S_bS_c$. By Fact 5, $S_bS_c$ is perpendicular bisector of $AI$. Hence, $XY \perp AI$ and as $AI$ bisects $\angle XAY$, we have $AX=AY$. Hence, $\text{Pow}_{\omega_b}A = AX^2 = AY^2 = \text{Pow}_{\omega_c}A$. Hence $A$ lies on the radical axis of $\omega_b, \omega_c$ Radax on $\omega_b, \omega_c, (ABC) \implies S_cS_c \cap S_bS_b=Z$ lies on radical axis of $\omega_b, \omega_c$. (Here $S_bS_b$ means $S_b$-tangent to $(ABC)$).

Hence $Z-A-T$. As $A,Z$ lie on radical axis $\omega_b, \omega_c$, hence so does $T$! Hence we are done!

Denote by $T$ the second intersection of $IN_a$ with $\odot (ABC)$. Let $\omega_b$ hit $AB$ at $E$, and define $F$ similarly. Note that due to homothety reasons, $E,F \in S_bS_c$. Note that $AE=AF$, as $AI$ is the perpendicular bisector of $EF$ and also the angle bisector of $\angle EAF$. This means $A$ lies on the radical axis of $\omega_b, \omega_c$. Note that we have : $$-1=(BC;N_aS_a) \overset {I}{=} (S_bS_c; TA)$$ Hence $AT$ passes through $S_bS_b \cap S_cS_c$. However by radical axis on $\omega_b, \omega_c, \odot (ABC)$, the radical axis of the first two circles also passes through that point. Done. $\square$

This is quite nice ! Say $T$ be the $\text{A-mixtillinear touch point}$,it is well known that $I,N_a,T$ are collinear(hence the importance of $I,N_a$ is gone).Now note that if $\omega_b,\omega_c$ meet $AB,AC$ at $E,F$ respectively,then $S_b,S_c,E,F$ are collinear.Since $S_bS_c \perp AI$ we have $AE=AF$, thus $A$ lies on the radax of $\omega_b,\omega_c$. So it requires to prove that $AT$ is the radax of $\omega_b,\omega_c$. Now if the tangent to $(ABC)$ at $S_b,S_c$ meet at $X$ then $X$ also lies on the radax of $\omega_b,\omega_c$, as the circles are tangent to $(ABC)$ at $S_b,S_c$ respectively.So $XA$ is the radax of $\omega_b$ and $\omega_c$. Now ,its well known that $$(S_b,S_c;T,A)=-1$$This gives that $T \in XA$ which is the radical axis of $\omega_b$ and $\omega_c$ , which is exactly what we wanted $\blacksquare$

Nice problem . Firstly let $T=N_aI\cap\odot(ABC)$. It is clear that $T$ is the $A-$mixtilinear touchpoint. It just suffices to prove that $T$ lies on the Radical Axis of $\left\{\odot(\omega_B),\odot(\omega_C)\right\}$. Firstly, let $X=S_bS_c\cap AB$ and $Y=S_bS_c\cap AC$. Note that due to homothety, we have that $\odot(\omega_B)$ and $\odot(\omega_C)$ are tangent to $AB$ and $AC$ at $X$ and $Y$ respectively. Now it is well known that $S_bS_c$ is the perpendicular bisector of $AI$. This means that $AX=AY$ that is $A$ lies on the Radical Axis. Now Radax on $\left\{\odot(ABC),\odot(\omega_B),\odot(\omega_C)\right\}$ gives that that tangents to $\odot(ABC)$ at $S_b$ and $S_c$ meet at some point say $Z$ which lies on the Radical Axis too. Thus we are actually done if we show that $T$ lies on $AZ$, which is equivalent of showing that $(A,T;S_b,S_c)=-1$. But this clearly follows by taking a homothety centered at $T$ that maps $\odot(ABC)\mapsto A-\text{mixtilinear incircle}$ and so we are done.

Lemma 1 Let $ABC$ be a triangle .$N$ be midpoint of arc $BAC$ and $I$ be incenter of $ABC$.$T$ be the tangency point of $A$ mixtilinear incircle with $(ABC)$. Then $N-I-T$ are collinear. Lemma 2 Let $AB$ be a chord in circle $\gamma$. A circle $\omega$ is internally tangent to $\gamma$ at $J$ and tangent to $AB$ at $K$ then $JK$ pass through midpoint of arc $AB$. Lemma 3 Let $ABC$ be a triangle with $I$ incenter. $BI$ meet $(ABC)$ again at $M$. We have $MA=MI=MC$. Sol:- We introduce the $A$ mixtilinear incircle. Let it be tangent to $(ABC),AB,AC$ at $T,K,J$ respectively.By lemma $1$ , $T-I-N_A$ are collinear.By lemma $2$ , $T-K-S_c$ are collinear and $T-J-S_b$ are also collinear. $D,E$ be the tangency points of $\omega_c$ with $AC$ and $\omega_b$ with $AB$ respectively. By lemma $2$ we find that $D,E$ lie on $S_bS_c$. By lemma $3$ , $S_cA=S_cI$ and $S_bA=S_bI \implies DE$ is perpendicular bisector of $AI$. Since $AI$ bisects $\angle EAD$ we get $ AE=AD \implies A$ lies on the radax of $\omega_b, \omega_c$.Let the tangents at $S_b,S_c$ meet at $F$. $F$ lies on radax of $\omega_b, \omega_c$ so $AF$ is the radax of the $2$ circles. Note that $S_cF \parallel KA$ and $S_bF \parallel JA$ and by homothety at $T$ sending mixtilinear incircle to $(ABC)$ we also have $KJ \parallel S_cS_b$.Hence $T$ is the homothetic center of $AKJ$ and $FS_cS_b \implies T-A-F$ are collinear so we are done.

First, we get by homothety from point $S_C$ that the contact point of $\omega_C$ with $AC$ lies on $S_BS_C$. Similarly, a homothety from $S_B$ shows that the contact point of $\omega_B$ with $AB$ lies on $S_BS_C$. A quick angle chase then shows that $A$ is equidistant from these two contact points, so $A$ lies on the radical axis of $\omega_B$ and $\omega_C$. Let $IN_A$ cut $(ABC)$ at a point $T\ne N_A$, so that $T$ is actually the mixtilinear intouch point and it suffices to show that $T$ also lies on the radical axis. To this end, we also let $X$ be the intersection of the common tangents from $S_C$ and $S_B$ to $\omega_C,(ABC)$ and $\omega_B,(ABC)$ respectively, which clearly lies on the radical axis. We thus only need to prove that points $X,A$ and $T$ are collinear, or in other words that points $S_C,S_B,A$ and $T$ are harmonic. Let us invert from point $A$ with radius $\sqrt{bc}$, then reflect across the angle bisector of $\angle BAC$. One can check that $\omega_B$ and $\omega_C$ go to the $C$ and $B$-excircles respectively, and the mixtilinear incircle goes to the $A$-excircle. Point $S_C$ thus goes to point $D$ : the $B$-excircle contact point with $BC$; $S_B$ goes to point $E$ : the $C$-excircle contact point with $BC$, and finally $T$ goes to point $F$ : the $A$-excircle contact point with $BC$. Since inversion preserves cross-ratios, we must show that $E,D,F$ and the point at infinity (image of $A$) are harmonic, or in other words that $F$ is the midpoint of $DE$, but this follows easily from all the tangencies. $\square$

Let $X$, $Y$ be the touchpoints of $\omega_b$ and $\omega_c$ with $\overline{AB}$ and $\overline{AC}$ and let $D$, $E$ and $O$ be the center of $\omega_b$ and $\omega_c$ and $(ABC)$. Suppose $K=\overrightarrow{S_bX} \cap (ABC)$ then by homothety $XD \parallel KO$ but $DX \perp AB \implies OK \perp AB \implies K = S_c$ and similarly we have $\overline{S_c - X - Y - S_b}$. $\measuredangle AXY = \measuredangle AS_cS_b + \measuredangle BAS_c = \measuredangle ABI + \measuredangle BCI = \measuredangle ICA + \measuredangle IBC = \measuredangle S_cS_bA + \measuredangle S_bAC = \measuredangle XYA \implies AX = AY$ so by POP $A$ lies on the radical axis of $\omega_b$ and $\omega_c$. By Radax on $(ABC) , \omega_b, \omega_c$ we have that, if the radical axis of $\omega_b$ and $\omega_c$ intersect $(ABC)$ at $T$ then $(AT; S_bS_c) = -1$ thus all we gotta prove is that the A mixtilinear touchpoint $V$ is a point such that $(AV; S_bS_c) = -1 \iff N_aV$ passes through the midpoint of $\overline{S_bS_c}$ but this is easy from the mixtilinear config. Suppose the A mixtilinear incircle touches $\overline{AB}$ and $\overline{AC}$ at $P$ and $Q$ then it is well known that $I = (P+Q)/2$ and $S_bS_c \parallel PQ$ so by homothety at $V$, $VN_a$ passes through the midpoint of $S_bS_c$.

The problem was proposed by Eytan Tirosh, Israel.

Note that $A$ lies on the radax.Let $S_bS_c$ meet $AB$ at $P$ and $AC$ at $Q$.Now let the tangent to $\omega _c,\Omega$ at $S_b$ and that to $\omega_b,\Omega $ at $S_c$ meet at $X$.Then $XS_c \parallel AB \implies \angle XS_cS_b =\angle APQ= \angle AQP= \angle XS_bS_c$ , so $XS_b=XS_c$ and thus $X$ lies on the radax as well. Now , let the line perpendicular to $AI$ at $I$ meet $AB$ at $K$ and $AC$ at $L$.Then , since corresponding sides are parallel to each other, $XS_bS_c$ and $ALK$ are homothetic , Thus $AX$ , or the radax of $\omega_c$ And $\omega_b$ passes through $KS_c \cap LS_b =IN_a \cap \Omega =T$ , the mixtilinear touchpoint.

Solved with Archit Manas, Kanav Talwar, Sidharth Suresh. By homothety, it follows that the points at which $\omega_b$ is tangent to $AB$ and $\omega_c$ tangent to $AC$ are actually $F = S_bS_c \cap AB$ and $E = S_bS_c \cap AC$. But also, by fact 5, $S_bI = S_bA$ and $S_cI = S_cA$, $S_bS_c$ is the perpendicular bisector of $AI$. So, $E,F$ also lie on the perpendicular bisector of $AI$, implying that $AEIF$ is a rhombus, so $AE^2 = AF^2$ so $A$ lies on the radical axis of $\omega_b$ and $\omega_c$. Note that $T = IN_a \cap (ABC)$ is the mixtilinear touchpoint so its inverse is the extouch point, say $E$. Say the inverses of $S_b$ and $S_c$ are $X$ and $Y$. It suffices to show that $E$ lies on the radical axis of the inverted circles, or that $EX^2 = EY^2$. But this follows since $EX = BX + BE = c + (s-c) = s = b+(s-b) = CY + CE = EY$, so we're done. $\blacksquare$

Nice problem! I will skip some easy to check details. By homothecy tangency points are on the segment $S_bS_c$. Tangents from $A$ to both circles are equal, so $A$ is in the radical axis. Consider the inversion centered at $A$ with radius $\sqrt{AX\cdot AY}$. It is easy to see that $S'_bS'_c$ is the common tangent. By power of point $AX$ is the $A$-median in $\triangle AS'_bS'_c\; (*)$. Now, it is easy to check that $S'_bIS'_cN_a$ is a parallelogram so $N_aI$ is the $N_a$-median in $\triangle N_aS_bS_c\; (**)$. Finally $\triangle N_aS_bS_c = \triangle AS_cS_b \sim AS'_bS'_c$, this combined with $(*),(**)$ show that $\angle S'_cAX = \angle S_cN_aI$, meaning that $AX$ and $N_aI$ meet at $\Omega$.

let the tangents of the two circles on $AB$ and $AC$ be $X$ and $Y$. Note that using Shooting Lemma shows that $S_B, X, Y, S_C$ are collinear. In particular, since $S_BS_C \perp AI$, then $AX = AY$, so $A$ has equal power to both circles and lies on the radical axis. Let $X$ be the intersection of the tangents to the circle $(ABC)$ at $S_B$ and $S_C$. Then note that $X$ must also have equal power to both circles, so $AX$ is the radical axis. Suppose $T$ is the intersection of the $A$-mixtilinear incircle and $(ABC)$. It is well-known that $TI$ passes through $N_A$. Thus, it would suffice to show that $T$ lies on $AX$. But this is equivalent to showing that $TS_BS_CA$ is harmonic. To see this is true, consider a homothety centered at $T$ bringing the circumcircle to the $A$-mixtilinear incircle. Suppose $S_B$ and $S_C$ map to $K$ and $L$ on $AC$ and $AB$, respectively. It is well known that $K$ and $L$ are the tangents of the mixtilinear incircle to $AC$ and $AB$. But then $TA$ is the $T$-symmedian wrt $TKL$. So $TA$ must be the $T$-symmedian wrt $TS_CS_B$, as desired. $\blacksquare$

it is commonly known that $\overline{N_AI}$ intersects the circumcircle again at a point $T$, which is the tangency-point of the mixtilinear circle with respect to the circumcircle and $\triangle ABC$ Now we define a bunch of points: the tangents through $S_C, S_B$ meet at $K$. let the two smaller circles be tangent to lines $\overline{AB}, \overline{AC}$ at points $E,F$ respectively. Finally, let $\overline{AB}$ meet the tangent through $S_B$ again at point $P$, define $Q$ similarely. Let $G,H$ be the intersection points of the two smaller circles. We can quickly note that $S_C, S_B, E, F$ are collinear by well-known theory It is clear that $\overline{GH}$ is the radical axis of the two smaller circles. It can also easily be seen that $K$ lies on that line too. we now claim that $A$ also lies on that line. That, however, by Power of a Point is equivalent to saying $AE=AF$ It is quite easy to see that $APKQ$ is a parallelogram, meaning $AP=KQ$ and $AQ=KP$ also, since they are tangents, $KS_C = KS_B$ Finally, because of parallel lines, $\triangle S_BEP$ is isosceles, equally for $S_CFQ$, but by those equalities, $AE=AF$ follows immediately Now it remains to show that $P$ also lies on this line. But that is quite easy to see using obscure theory about mixtilinear circles If we consider the mixtilinear circle tangent to $\Omega$ at $T$, it is tangent to $\overline{AB}, \overline{AC}$ at two points $X,Y$. But we know that $T,X,S_C$ lie on a line, same for $T,Y,S_B$. Therefore, a homothety centered at $T$ sends the mixtilinear circle to $\Omega$ and the two tangency points to the midpoints of the arcs. But by the fact that $\overline{AX}, \overline{AY}$ are both tangent to the mixtilinear circles, ergo $A$ is the intersection of the tangents, $A$ gets mapped to $K$. We therefore conclude that $T$ also lies on the desired line

Let $D,E$ be the tangency points of $\omega_b,\omega_c$ with $\overline{AB}$ and $\overline{AC}$, respectively. Let $T$ be the intersection point of the radical axis of $\omega_b,\omega_c$ with $\Omega$. We show that $I,N_a,T$ are collinear. Claim: $S_b,D,E,S_c$ are collinear. Proof. Let $S',S_b$ be the intersection of $S_bD$ with $\Omega$. We claim that $S'\equiv S_c$. Note that $S'A^2=S'D\cdot S'S_b$ and $S'B^2=S'D\cdot S'S_b$. So, $S'A=S'B$, which means $S'\equiv S_c$. Hence, $D\in\overline{S_bS_c}$. Similarly, we get that $E\in\overline{S_bS_c}$. So, $S_b,D,E,S_c$ are collinear. $\blacksquare$ Claim: $A$ lies on the radical axis of $\omega_b,\omega_c$. Proof. The following angle chase, \[\angle S_cAB=\angle ABS_c=\angle DBS_c, \ \ \angle BS_cD=\angle BS_cS_b=\angle BAS_b\]\[\Longrightarrow \angle ADS_c=\angle DBS_c+\angle BS_cD=\angle S_cAB+\angle BAS_b=\angle S_cAS_b\]Similarly, we get $\angle S_bEA=\angle S_cAS_b$. It follows that $\angle AED=\angle EDA$. Hence, $AD=AE$. It follows that \[Pow_{\omega_b}(A)=AD^2=AE^2=Pow_{\omega_c}(A),\]which means that $A$ lies on the radical axis of $\omega_b,\omega_c$. $\blacksquare$ Claim: $(AS_cTS_b)$ is a harmonic quadrilateral. Proof. Note that the tangent to $\Omega$ at $S_b$ is the radical axis $\Omega,\omega_b$ and tangent to $\Omega$ at $S_c$ is the radical axis of $\Omega,\omega_c$. Let $K=S_bS_b\cap S_cS_c$. So, $K$ is the radical center of $\Omega,\omega_b,\omega_c$. So, $K$ lies on the radical axis of $\omega_b,\omega_c$, which means that $K,A,T$ are collinear. So, the tangent to $\Omega$ at $S_b,S_c$ and $AT$ concur at $K$. The claim follows. $\blacksquare$ Note that $\angle CN_aS_b=\angle CBS_b=\frac{\angle B}2$, while $\angle N_aCA=90^\circ-\frac{\angle A}2-\angle C=\frac{\angle B-\angle C}2$, while $\angle ACS_c=\frac{\angle C}2$. Adding, we get $\angle N_aCA+\angle ACS_c=\angle N_aS_bS_c=\frac{\angle B}2=\angle CS_cS_b$. Hence, we have that $\overline{N_aS_b}\parallel\overline{S_cC}$. Finally, we had that $(AS_cTS_b)$ is a harmonic quadrilateral. Note that projecting this onto line $S_cC$ which is also the $C-$ angle bisector, from perspective at $N_a$, takes $A$ to $I_c$, as $AN_a$ is the $A-$external angle bisector, while $S_b$ goes to $\infty_{S_cC}$ as $S_cC\parallel\overline{S_bN_a}$. Let $I'=N_aT\cap S_cC$. Then, $T$ goes to $I'$ under this projection. Hence, we have that $\left(I',I_C;S_c,\infty_{S_cC}\right)=-1$. It follows that $I'$ is the reflection of $I_C$ about $S_c$. Hence, $I'\equiv I$ from Fact 5. We conclude that $N_a,I,T$ are collinear. $\blacksquare$

Let $T$ be the $A$-mixtilinear touch point, so that $T$ lies on $IN_a$. We need to show that $T$ lies on $l$, the radical axis of $\omega_b$ and $\omega_c$. Let $E=S_bS_c \cap AC$, $F=S_bS_c \cap AB$. It is well-known that $\omega_b$ is tangent to $AB$ at $F$, $\omega_c$ is tangent to $AC$ at $E$. Also, since $AI \perp S_bS_c$, we have that $AE=AF$. So $A$ lies on $l$. Let $P$ be the radical centre of $\Omega$, $\omega_b$, $\omega_c$, so $P$ lies on $l$. Note that $PS_b$ and $PS_c$ are tangent to $\Omega$. It suffices to show that $P,A,T$ are collinear, or that $(A,T;S_b,S_c)=-1$. Let $S_a$ be the midpoint of arc $BC$ not containing $A$. Then $(A,T;S_b,S_c) \stackrel{I}{=} (S_a,N_a;B,C) =-1$, which finishes the proof. $\square$

Let $S_BS_C$ intersect lines $AB$ and $AC$ at $F,E$ respectively. As $S_BS_C \perp AI$, it follows that $AE=AF$. But by shooting lemma $\omega_B$ is tangent to $AB$ at $F$ and $\omega_C$ is tangent to $AC$ at $E$, so $A$ lies on the radical axis. Now, if the tangents to $\omega_B,\omega_C$ at $S_B,S_C$ intersect at $D$, then $DS_C=DS_B$ since they are tangent to $\Omega$, so $D$ lies on the radical axis. It suffices to show $DA$ passes through $T$ but this is show after projecting $-1=(N_A,M_A; B,C)$ through $I$.

oops overly complicated length chase, but i think the idea is quite straightforward and motivated basically you break apart the problem and analyze where $IN_a$ and the radical axis hit the circumcircle separately, and redefining the radical axis of the two circles as a simpler object allows us to calculate where it hits the circumcircle more easily

edit: the projective sol is really nice, i think the main takeaway here is to always be on the lookout for simpler definitions of objects. the main idea of that solution is that the radical axis is the $A$-symmedian of $\triangle AS_bS_c$, so then it is suffices to show that $(AX_2;S_bS_C)$ is harmonic which is easy by projecting through $I$. Although in the above solution I did find a simpler definition of the radical axis of $\omega_b$ and $\omega_c$, it wasn't as simple as it could have been, which led to a less elegant solution. also, ratios involving $\sin^2$ should scream symmedians anyways.

We define some points: Let the $A$-Mixtilinear touch point be $T$. Let $\omega_b$ cut $AB$ at $V$ and $\omega_c$ cut $AC$ at $U$. Let the midpoint of arc $BC$ not containing $A$ be $M_a$. Let $\omega_b$ and $\omega_c$ cut at points $F$ and $G$ and finally let the tangents at $S_b$ and $S_c$ to $\Omega$ meet at $X$. $\underline{Claim-1}:$ $\overline{S_b-U-V-S_c}$ are collinear $\underline{Proof}:$ By shooting lemma $\overline {S_b-V-S_c}$ and $\overline{S_b-U-S_c}$ are collinear. $\underline{Claim -2}:$ $A$ and $X$ lie on $FG$. $\underline{Proof}:$ It is well known that $AI$ is perpendicular to $S_bS_c$ i.e. $UV$. Since $AI$ bisects $\angle UAV$ $\implies AU=AV$. $\implies$ $AU^2=AV^2=Pow(A,\omega_b)=Pow(A,\omega_c)$ so $A$ lies on the radical axis which is $FG$. Also by radical center theorem on $\omega_b, \omega_c, \Omega$ ; $X$ also lies on the radical axis which is $FG$. Now it remains to show that $T$ lies on $AX$ or that $(S_bS_c; AT)=-1$ It is well known that $IN_a$ passes through $T$, so by projecting through $I$ onto $\Omega$, we get:- $$(S_bS_c;AT)= (BC;M_aN_a)$$And since $(BC;M_aN_a)=-1$, the problem is solved.

EGMO 2023 p6 Let $T$ be the touchpoint of the mixtilinear. Let $R_1$ and $R_2$ be the intersection of the $\omega_c$ and $\omega_b$. And Let ray $R_1R_2$ intersect $\Omega$ at $T’$. Let $X$ and $Y$ be the intersections of $S_bS_c$ with $AB$ and $AC$. By shooting lemma we have that $S_b-X-Y-S_c$ are colinear. Now let $Z$ be the intersection of the tangents at $S_b$ And $S_c$. By radical axis we get that $Z-R_1-R_2$ are colinear. Easily one can show that $\triangle AXY \sim \triangle ZS_bS_c$ so $AX=AY$ so by POP $A$ lies on line $R_1R_2$. Finally we know that $AS_bTS_c$ is harmonic. But we have that $AS_bT’S_c$ is harmonic as well. Hence $T=T’$ as desired

Take $\sqrt{bc}$ inversion and reflect over the angle bisector of $\measuredangle CAB$. $S_B^*,S_C^*$ lie on $BC$ such that $\measuredangle BAS_B^*=\frac{\measuredangle B}{2}$ and $\measuredangle S_C^*AC=\frac{\measuredangle C}{2}$. $T^*$ is the tangency point of $A-$excircle with $BC$ since $T$ is the $A-$mixtilinear touchpoint. Let $I_A$ be the $A-$excenter. Note that $I_A,A,C,S_B^*$ and $I_A,B,A,S_C^*$ are concyclic. \[\measuredangle S_C^*S_B^*I_A=\measuredangle CS_B^*I_A=\measuredangle CAI_A=\measuredangle I_AAB=\measuredangle I_AS_C^*B=\measuredangle I_AS_C^*S_B^*\]Hence we conclude that $I_AS_B^*=I_AS_C^*$. Since $T^*$ is the altitude from $I_A$ to $S_B^*S_C^*,$ we get that $T^*$ is the midpoint of $S_B^*S_C^*$. \[Pow_{\omega_B}(T^*)=T^*{S_B^*}^2=T^*{S_C^*}^2=Pow_{\omega_C}(T^*)\]Thus, $T^*$ lie on the radical axis of $\omega_B$ and $\omega_C$ which gives the desired result.$\blacksquare$

Let $\omega$ be the A-Mixtilinear circle. Let $\omega$ touch $\Omega$ at $T$, $AB$ at $Q$ and $AC$ at $P$. Also let the tangents at $S_b$ and $S_c$ to $\Omega$ meet at $S$, $\omega_b$ meet $AB$ at $E$ and $\omega_c$ meet $AC$ at $D$. It is well known that $\overline{N_a-I-T}$ so we need to prove that $T$ lies on the radical axis of $\omega_b$ and $\omega_c$. We claim that the radical axis is $AS$. Notice that the homothety centered at $T$ taking $\omega$ to $\Omega$ maps $P\mapsto S_b$ and $Q\mapsto S_c$, hence $PQ\parallel S_bS_c$. Using the same principle we get $\overline{S_c-E-D-S_b}$. Since $PQ\parallel S_bS_c$ and $\triangle PAQ$ is isosceles we have that $\triangle DAE$ is also isosceles so $AD=AE$, hence $A$ lies on the radical axis of these circles. We also have that $S$ lies on the radical axis since $SS_b=SS_c$ and they are tangent to $\omega_b$ and $\omega_c$ respectively so it remains to prove $\overline{S-A-T}$. Since $AP$ and $AQ$ are tangent to $(TPQ)$, it follows that $TA$ is the T-symmedian in $\triangle TPQ$ so from the homothety centered at $T$ sending $P\mapsto S_b$ and $Q\mapsto S_c$ we get that $TA$ is the T-symmedian in $\triangle TS_bS_c$. Also notice that $SS_b$ and $SS_c$ are tangent to $(TS_bS_c)$ so $S$ lies on this symmeadian as well. Therefore $\overline{S-A-T}$. $\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 6.429191956037488, xmax = 32.90877674654499, ymin = -12.419702613298554, ymax = 4.402461592605151; /* image dimensions */ pen qqzzff = rgb(0.,0.6,1.); draw((15.832053323280904,-2.0275292737901105)--(14.,-9.)--(22.,-9.)--cycle, linewidth(0.8) + qqzzff); /* draw figures */ draw((15.832053323280904,-2.0275292737901105)--(14.,-9.), linewidth(0.8) + qqzzff); draw((14.,-9.)--(22.,-9.), linewidth(0.8) + qqzzff); draw((22.,-9.)--(15.832053323280904,-2.0275292737901105), linewidth(0.8) + qqzzff); draw(circle((18.,-6.324094837667738), 4.812532434986434), linewidth(0.8)); draw((13.345461509593461,-5.101090389956919)--(22.,-9.), linewidth(0.8)); draw((21.6045723535254,-3.1354389161238636)--(14.,-9.), linewidth(0.8)); draw((13.345461509593461,-5.101090389956919)--(16.137097925760145,-10.761443629411442), linewidth(0.8)); draw((16.137097925760145,-10.761443629411442)--(21.6045723535254,-3.1354389161238636), linewidth(0.8)); draw((18.,-1.5115624026813064)--(16.137097925760145,-10.761443629411442), linewidth(0.8)); draw(circle((17.269019990821878,-8.065256278930772), 2.9241535168772543), linewidth(0.8)); draw((13.345461509593461,-5.101090389956919)--(21.6045723535254,-3.1354389161238636), linewidth(0.8)); draw(circle((15.768880896443848,-5.73785656879408), 2.5056800855402903), linewidth(0.8)); draw(circle((18.781655760607542,-5.632631045234103), 3.768929107884769), linewidth(0.8)); draw((15.635059631574881,3.6127148674749225)--(13.345461509593461,-5.101090389956919), linewidth(0.8)); draw((15.635059631574881,3.6127148674749225)--(21.6045723535254,-3.1354389161238636), linewidth(0.8)); draw((15.635059631574881,3.6127148674749225)--(16.137097925760145,-10.761443629411442), linewidth(0.8)); /* dots and labels */ dot((15.832053323280904,-2.0275292737901105),dotstyle); label("$A$", (15.209871044927434,-1.8496625940424607), NE * labelscalefactor); dot((14.,-9.),dotstyle); label("$B$", (13.557389804439255,-9.386829927178836), NE * labelscalefactor); dot((22.,-9.),dotstyle); label("$C$", (22.161661314595182,-9.232345184811535), NE * labelscalefactor); dot((21.6045723535254,-3.1354389161238636),linewidth(4.pt) + dotstyle); label("$S_b$", (21.657677447697665,-3.021084554939399), NE * labelscalefactor); dot((13.345461509593461,-5.101090389956919),linewidth(4.pt) + dotstyle); label("$S_c$", (12.835466968072537,-4.941671723386705), NE * labelscalefactor); dot((18.,-1.5115624026813064),linewidth(4.pt) + dotstyle); label("$N_a$", (18.061684451455907,-1.318436355961291), NE * labelscalefactor); dot((16.950033863118854,-6.724966903399473),linewidth(4.pt) + dotstyle); label("$I$", (16.99923197529357,-6.517077551181163), NE * labelscalefactor); dot((16.137097925760145,-10.761443629411442),linewidth(4.pt) + dotstyle); label("$T$", (15.832127798274004,-11.230341682932418), NE * labelscalefactor); dot((14.440865525365025,-7.322143831328071),linewidth(4.pt) + dotstyle); label("$Q$", (13.997919444234876,-7.588833429057954), NE * labelscalefactor); dot((19.459202200872692,-6.127789975470877),linewidth(4.pt) + dotstyle); label("$P$", (19.805196207209487,-6.167578426650942), NE * labelscalefactor); dot((15.136459424322965,-4.674836552559095),linewidth(4.pt) + dotstyle); label("$E$", (14.815190579744367,-4.519414970040134), NE * labelscalefactor); dot((17.645627762076796,-4.0776596246304955),linewidth(4.pt) + dotstyle); label("$D$", (17.693912440476634,-3.974567546367139), NE * labelscalefactor); dot((15.635059631574881,3.6127148674749225),linewidth(4.pt) + dotstyle); label("$S$", (15.691598158478383,3.721402313013908), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Let $S_bS_c \cap AB = P$ and $S_bS_c \cap AC = Q$. Notice that $P \in \omega_b$ and $Q \in \omega_c$ due to homothety, $AP = AQ$ from $AI \perp PQ$, and thus $A$ lies on the radical axis of $\omega_b$, $\omega_c$. If we let this radical axis intersect $PQ$ at $X$, then $X$ is the center of homothety sending $PQ \mapsto S_cS_b$. Hence it suffices to show $AX$ and the excentral-cevian are isogonals, which we can bash. $\blacksquare$

Note that the line passing through the intersections of $\omega_b$ and $\omega_c$ is the radical axis of the two. We make the following claim. *** Claim 1. This radical axis is the line $AT$. Proof. First, let $T$ be the intersections of the tangents to $(ABC)$ at $S_b$ and $S_c$. Additionally, since $\omega_b$ and $\omega_c$ are internally tangent to $(ABC)$ at $S_b$ and $S_c$, we have that the tangents to $(ABC)$ at $S_b$ and $S_c$ are also tangent to $\omega_b$ and $\omega_c$, respectively. However, by the tangencies from $(ABC)$, we have that $TS_b=TS_c$, which gives us that \[pow_{\omega_b}(T)=TS_b^2=TS_c^2=pow_{\omega_c}(T),\]so $T$ lies on the radical axis of $\omega_b$ and $\omega_c$. Now, let $\omega_b$ and $\omega_c$ intersect $AB$ and $AC$ at $T_b$ and $T_c$, respectively. By the Death Star Lemma, we have that $S_cT_c$ passes through $S_b$ and $S_bT_b$ passes through $S_c$, so $S_c$, $T_b$, $T_c$, and $S_b$ are collinear, in that order. However, since $S_b$ and $S_c$ are arc midpoints, we have that \[\angle AT_cT_b=\angle AT_cS_c=\frac{1}{2}(\stackrel{\frown}{AS_c}+\stackrel{\frown}{CS_b})=\frac{1}{2}(\stackrel{\frown}{BS_c}+\stackrel{\frown}{CS_b})=\angle AT_bS_b=\angle AT_bT_c,\]so $\triangle AT_bT_c$ is isosceles with $AT_b=AT_c$. This then gives us that \[pow_{\omega_b}(A)=AT_b^2=AT_c^2=pow_{\omega_c}(A),\]meaning that $A$ also lies on the radical axis of $\omega_b$ and $\omega_c$. This means that $AT$ is indeed the radical axis of $\omega_b$ and $\omega_c$, as desired. This proves our claim. *** Now, let $AT$ intersect $(ABC)$ at $X\neq A$. We now make the following claim. *** Claim 2. $AT$ and $IN_a$ intersect at point $X$. Proof. Since we know that $X\in AT$ by definition, it suffices to show that $X$ lies on $IN_a$. First, note that since $TS_b$ and $TS_c$ are both tangent to $(ABC)$, $ABXC$ is a harmonic quadrilateral, or $(AX;BC)=-1$. Now, let $N_aI\cap (ABC)=X'$ so that $X'\neq N_a$, and let the arc midpoint of $\stackrel{\frown}{BC}$ not containing $A$ be $S_a$. If we can show $X=X'$, then we are done. First, notice that \[BS_a=CS_a, BN_a=CN_a \implies BS_a*CN_a=BN_a*CS_a \implies (N_aS_a;BC)=-1.\]However, projecting this with respect to $I$ back onto the circumcircle $(ABC)$, we have that \[-1=(N_aS_a;BC)\overset{I}{=}(X'A;BC),\]and since $(X'A;BC)=-1$, we also have that $(AX';BC)=(X'A;BC)=-1$. However, by the uniqueness of the harmonic conjugate, since we had that $(AX;BC)=-1$, this means that we must have $X=X'$, as desired. This means that $AT$ and $IN_a$ do indeed intersect at the point $X$, proving our claim. *** This means that $IN_a$ and the line containing the intersections of $\omega_b$ and $\omega_c$, or the radical axis of $\omega_b$ and $\omega_c$, $AT$, intersect on $(ABC)$ at point $X$, as desired. This completes our proof.

We uploaded our solution https://calimath.org/pdf/EGMO2023-6.pdf on youtube https://youtu.be/SfpkS9Ifv_M.

Easy for a p6 tbh, but nice