ISL 2023 G6 ahh like diagram fr. Let $N,P$ be midpoints of $AC,AB$ respectively and let $AB \cap (BMX)=U$ and $AC \cap (CMY)=V$, also let $UV$ hit $MN, MP$ at $E,F$ respectively and let $XA \cap (BMX)=X'$ and $YA \cap (CMY)=Y'$. Claim 1: $S,A,M$ colinear. Proof: Notice from double reim's and antiparallels that $PNXUVY$ is cyclic, therefore from Reim's again $BVUC$ is cyclic and by PoP and radax we can easly conclude. Claim 2: $E$ lies on $(CMY)$ and $F$ lies on $(BMX)$. Proof: Proving one is enough by symetry, so notice that from Reim's we get $MN \parallel CY'$ which means that $AX=XY'$ and also in $(CMY)$ notice that by the parallels we get arcs $CY,MV$ are equal but from parallels again we get $\angle VEM=\angle YY'C$ which is enough to conclude. Finish: Consider inversion centered at $A$ with radius $-\sqrt{AS \cdot AM}$ then all we need now is to prove that $(MX'Y'), (MUV)$ are tangent at $M$, but now in angles this can be written as $\angle EVM=\angle Y'X'M+\angle UMY'$ but notice that: \[\angle Y'X'M+\angle UMY'=\angle UX'M+\angle ACY=\angle Y'AC+\angle ACY=\angle CSY'+2\angle ACY=\angle MSE=\angle EVM \]Therefore we are done .

We have $\angle CAX=\angle CBA=\angle CMX$ and $\angle YAB=\angle ACB=\angle YMB$, so $AMCX$ and $AMBY$ are cyclic. Hence $\angle MYX=\angle MYA=\angle MBA=\angle CMX$. Consider the inversion with centre $M$ and radius $MB=MC$. Then $\angle Y'X'M=\angle MYX=\angle CMX=\angle CMX'$, so $X'Y'\parallel BC$. We also have $S'=BX'\cap CY'$, so $(S'X'Y')$ and $(S'BC)$ are tangent. Since inversion preserves tangencies, this gives that $(SXY)$ and $(SBC)$ are tangent, as desired.

Inversion helps well.

xoinks invert at $M$ swapping $B,C$ (this includes reflection) now by easy angle chase $AMCX, AMBY$ cyclic, so essentially, our problem is: let $A'BC$ be a triangle and let the circle through $A',M$ tangent to $(A'BC)$ intersect $A'B$ and $A'C$ at $X',Y'$. if $BY'\cap CX'=S'$, prove that $(X'S'Y')$ is tangent to $(BS'C)$. first of all, $X'Y'$ parallel to $BC$ by homothety at $A'$, so $S'$ lies on $A'M$ by ceva well known. but now tangency is obvious by homothety at $S'$ sending $X'Y'\to BC$.

Here's a simple direct approach. Let lines $SX$ and $SY$ hit $(BSC)$ again at $P$ and $Q$. We will show that $\overline{PQ}\parallel \overline{XY}$, which implies the result by homothety at $S$. [asy][asy] defaultpen(fontsize(10pt)); size(250); pair A, B, C, M, N, L, O, X, Y, S, P, Q; A = dir(120); B = dir(210); C = dir(330); O = (0,0); M = (B+C)/2; N = (A+C)/2; L = (A+B)/2; X = extension(A, rotate(90,A)*O, M, N); Y = extension(A, rotate(90,A)*O, M, L); S = IP(circumcircle(B, M, X), circumcircle(C, M, Y), 0); P = extension(S, X, C, C+N-M); Q = extension(S, Y, B, B+L-M); draw(A--B--C--cycle, orange); draw(circumcircle(A, B, C), red); draw(X--M--Y--cycle, lightblue); draw(circumcircle(B, M, X), heavygreen+dotted); draw(circumcircle(C, M, Y), heavygreen+dotted); draw(circumcircle(S, B, C), heavycyan+dashed); draw(P--S--Q, heavycyan); draw(B--Q--P--C, lightblue); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$M$", M, dir(270)); dot("$X$", X, dir(70)); dot("$Y$", Y, dir(180)); dot("$S$", S, dir(120)); dot("$P$", P, dir(0)); dot("$Q$", Q, dir(230)); [/asy][/asy] By Reim's, we have $\overline{PC}\parallel \overline{MX}\parallel \overline{AB}$ and $\overline{QB}\parallel \overline{MY}\parallel \overline{AC}$. In addition, we can angle chase to obtain \[ \angle CPQ = \angle CSQ = \angle CSY = \angle YMB = \angle C \]and $\angle MXY = \angle BAY = \angle C$. Together these imply $\overline{PQ}\parallel \overline{XY}$, as desired.