60th Belarusian Mathematical Olympiad

Let $O_1,O_2,O_3$ be respectively the centers of the circles ${\Gamma }_1,{\Gamma }_2,{\Gamma }_3$, and $R_1,R_2,R_3$ be their radii. Let $\Gamma$ be the circumscribed circle of the triangle $O_1{O}_2{O}_3$ (see Fig. 1), and let $\Gamma$ touch the lines $O_3{O}_1,O_3{O}_2,O_1{O}_2$ at points $K_1,K_2,K_3$ respectively. Then $$O_1{K}_1=O_1{K}_3,O_2{K}_2=O_2{K}_3⟹O_1{K}_1+O_2{K}_2=O_1{O}_2=R_1+R_2.(1)$$

Moreover $$(R_3-R_1)+O_1{K}_1=O_3{O}_1+O_1{K}_1=O_3{K}_1=O_3{K}_2=O_3{O}_2+O_2{K}_2=(R_3-R_2)+O_2{K}_2.$$

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Therefore, if $O_1{K}_1>R_1$ ($O_1{K}_1<R_1$), then $O_2{K}_2=O_3{K}_2-(R_3-R_2)=(R_3-R_1)+O_1{K}_1-(R_3-R_2)=R_2+(O_1{K}_1-R_1)>R_2$, ($O_2{K}_2<R_2$), which gives $O_1{K}_1+O_2{K}_2>R_1+R_2$ ($O_1{K}_1+O_2{K}_2<R_1+R_2$), contrary to (1). Therefore, $O_{1}K=R_1$, $O_2{K}_2=R_2$. So $K_1,K_2,K_3$ coincide with $M_1,M_2,M_3$ respectively. It means that $\Gamma$ is circumcircle of the triangle $M_1{M}_2{M}_3$ (see Fig. 2). So $S{M}_1\perp O_3{O}_1$ (as the radius of the inscribed circle), thus $S{M}_1$ is the tangent to ${\Gamma }_3$.