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Belarusian Mathematical Olympiad

Geometry Difficulty 4.6 AIME Prove it Belarus

Several small circles are arranged inside a unit circle $\Gamma$ . The sum of the perimeters of all these small circles is not less than $\pi$ and none of them includes the center of $\Gamma$ .
Prove that there exists a concentric with $\Gamma$ circumference intersecting at least two of these small circles.

Solution

Let $r_1, r_2, \dots, r_k$ be radii of small circles. By condition,
$2\pi(r_1 + r_2 + \dots + r_k) \ge \pi,$
i. e.
$r_1 + r_2 + \dots + r_k \ge 1/2. \quad (*)$
Consider $360^\circ$ rotation of $\Gamma$ (together with all small circles) about its center. Under this rotation each of small circles covers some ring with the center at the center of $\Gamma$ . The width $d_i$ of the ring covered by the small circle with the radius $r_i$ is equal to $2r_i$ . If all covered rings have no common points, then
$d_1 + d_2 + \dots + d_k = 2(r_1 + r_2 + \dots + r_k) < 1,$
contrary to $(*)$ .

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