Belarusian Mathematical Olympiad

Consider $360^{\circ }$ rotation of $\Gamma$ (together with all segments) about its center. Under this rotation each of the segments covers some ring with the center at the center of $\Gamma$. If we show that the sum of the areas of all these rings is not less than $\pi$, then the statement will be proved.

We estimate the area $S_{AB}$ of the ring covered by the segment $AB$ under the rotation. Let the altitude from the center $O$ of $\Gamma$ onto the line containing $AB$ meet the line at point $C$. Consider two cases depending on the position of $C$.

Let $C$ belong to $AB$. Then the area $S_{AB}$ is equal to $$S_{AB}=\pi \max \{O{A}^2-O{C}^2,O{B}^2-O{C}^2\}=\pi \max \{A{C}^2,B{C}^2\}\ge \pi A{B}^2/4.$$

Now let $C$ do not belong to $AB$. Then $$\begin{array}{rl}S=\pi |O{B}^2-O{A}^2|& =\pi |C{B}^2-C{A}^2|=\pi (A{B}^2+2AB\cdot AC)\ge \\ & \ge \pi A{B}^2>\pi A{B}^2/4.\end{array}$$

Let $d_1,\dots ,d_N$ be the lengths of the segments. Then the sum of the areas of all rings is not less than $\frac{\pi }{4}(d_1^2+\cdots +d_N^2)\ge \frac{\pi }{4N}(d_1+\cdots +d_N{)}^2\ge \pi$, as required.