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,
$$b \cdot \frac{b}{2} = x(x + 2r), \quad a \cdot \frac{a}{2} = y(y + 2r), \qquad (1)$$

If $H$ lies on the side $AB$, then $c = AH + HB$. So
$$\begin{aligned} c = & \sqrt{(x+r)^2 - l^2} + \sqrt{(y+r)^2 - l^2} = \sqrt{x^2 + 2xr + r^2 - l^2} + \\ & \sqrt{y^2 + 2yr + r^2 - l^2} = \sqrt{x(x+2r) + r^2 - l^2} + \sqrt{y(y+2r) + r^2 - l^2} \end{aligned} \quad (1)$$
$$= \sqrt{\frac{a^2}{2} + r^2 - l^2} + \sqrt{\frac{b^2}{2} + r^2 - l^2}. \quad (2)$$
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 $AB$.