Belarusian Mathematical Olympiad

Let $I$ be the center of the circle $\Gamma$ passing through $C,M,N$; let $r$ be the radius of $\Gamma$. Let $P$ and $Q$ be the intersection points of segments $AI,BI$ and $\Gamma$; let $H$ be the foot of the perpendicular from $I$ on $AB$. Set $a=BC,b=AC,c=AB,x=AP,y=BQ,l=IH$.



Fig. 3 It is evident that $l=r$ if and only if $\Gamma$ touches the line $AB$ ($H$ is a point of tangency. If $H$ lies on the side $AB$, then $c=AH+HB$ (see Fig. 1-3). By the power of a point theorem, $$b\cdot \frac{b}{2}=x(x+2r),a\cdot \frac{a}{2}=y(y+2r),(1)$$

If $H$ lies on the side $AB$, then $c=AH+HB$. So $$\begin{array}{rl}c=& \sqrt{(x+r{)}^2-l^2}+\sqrt{(y+r{)}^2-l^2}=\sqrt{x^2+2xr+r^2-l^2}+\\ & \sqrt{y^2+2yr+r^2-l^2}=\sqrt{x(x+2r)+r^2-l^2}+\sqrt{y(y+2r)+r^2-l^2}\end{array}(1)$$ $$=\sqrt{\frac{a^2}{2}+r^2-l^2}+\sqrt{\frac{b^2}{2}+r^2-l^2}.(2)$$ It is evident that for $l>r$ from (2) it follows $c<\frac{a}{\sqrt{2}}+\frac{b}{\sqrt{2}}$, for $l<r$ from (2) it follows $c>\frac{a}{\sqrt{2}}+\frac{b}{\sqrt{2}}$. Therefore, $c=\frac{a}{\sqrt{2}}+\frac{b}{\sqrt{2}}$ if and only if $l=r$, i.e. the circle passing through $C,M,N$, touches the side $AB$.