contest-mathgeometry
Let the circumcircle of triangle $ABC$ intersect the circle with diameter $AI$ at $S$ where $I$ is the incentre of triangle $ABC$.
Let $M$ be the midpoint of major arc $BAC$. Let the circle $BIC$ intersect circumcircle $SIM$ at $X$. Let $MX$ and $MI$ meet $BC$ at $G$ and $F$ respectively. Prove that circle $SFG$ is tangent to circle $ABC$.
The way I had in mind was to use power of a point wirh a point circle at $S$, sadly I couldn't find the solution. Please help..
Let the midpoints of $CB$ and $CA$ be $A'$ and $B'$. Now we place a point on $D$ on $AB$ such that $AD=AB'$. And as $2c=a+b$, we also have $BD=BA'$.
Thus, by SAS congruence, $\triangle BIA'\cong \triangle BID$ and $\triangle AIB'\cong \triangle AID$. Thus, $\angle BDI= \angle IA'B$ and $\angle ADI =\angle IB'A$. Hence, $\angle IB'A=180^{\circ}-\angle IA'B=\angle IA'C$.
Thus, $C$, $B'$, $A'$ and $I$ lie on a circle.
Let $DP$ intersect circle $I$ at $K$.
Notice that $RP\times PI =FP\times PE = KP\times PD$, therefore $R,K,I,D$ are co-cyclic. Also notive $IK=ID$, therefore all the red angles are equal immediately.
Now drop a perpendicular line from $A$ to BC, we have the two pink angles being equal as $ID$ is also perpendicular to $BC$.
Since the two cyan angles are equal and the ninty degree angles are equal, we have the two green angle at vertex $A$ being equal. Therefore the top red angle is equal to the top pink angle.
Look at the red and pink angles sharing edge $ID$, we know $PD$ is parallel to $AI$. Therefore $PD$ is perpendicular to $EF$.
Best Answer
Here's a solution using some advanced tools of projective geometry. I apologise if this is not at all helpful -- perhaps tomorrow I'll see if there's an easier solution...
Let $N$ be the midpoint of minor arc $BC$. Let the incircle touch sides $BC$, $CA$, $AB$ at $P$, $Q$, $R$.
First, an inversion centre $N$ radius $NB=NI=NC$ shows that $N$, $P$ and $S$ are collinear. By the radical axis theorem, $MS$, $IX$ and $BC$ concur at $K$. Then $$-1=S(B,C;M,N)=(B,C;K,P),$$ which implies that $QR$ goes through $K$ too.
Consider cyclic quadrilateral $BICX$. By Brokard's theorem, the polar of $K$ in circle $(BIC)$ passes through $Y=BX\cap CI$ and $Z=BI\cap CX$. But the polar of $M$ in circle $(BIC)$ is the line $BC$, so by La Hire's theorem the polar of $K$ goes through $M$ too. Now $(B,C;P,K)=-1$ implies that the polar of $K$ passes through $P$ as well. So $M,P,Y,Z$ are collinear.
Now by the dual of Desargues' involution theorem on $BICX$ and point $M$, we have that $(MB,MC)$, $(MI,MX)$, $(MY,MZ)$ are pairs of an involution. This induces the involution $(B,C)$, $(G,F)$, $(P,P)$ on line $BC$. But the unique involution that swaps $B$, $C$ and fixes $P$ is harmonic conjugation in the points $K$, $P$. Hence $(K,P;F,G)=-1$.
Together with $\angle KSP=90^{\circ}$, this implies that $SP$ is the angle bisector of $\angle FSG$. Since $SP$ goes through $N$, a homothety at $S$ implies the desired result.