Hellenic Mathematical Olympiad

Let we can put $N$ non overlapping circles $C_i(K_i, r)$ into the given circular ring tangent to its border. The circle $C(O, R-r)$ has length greater than the perimeter $\ell_{K_1...K_N K_1}$ of the polygon with vertices the centers of the circles $C_i(K_i, r)$, and therefore:
$$N \cdot 2r < \ell_{K_1...K_N K_1} < 2\pi(R-r) \Rightarrow N < \pi \left( \frac{R}{r} - 1 \right). \quad (1)$$

Figure 2

Figure 3

Let $OA$ is tangent from $O$ to one of the circles $C_i(K_i, r)$. Then
$$OA^2 = R(R-2r) \Leftrightarrow OA = \sqrt{R(R-2r)}.$$
The circle $C(O, OA)$ is tangent to the sides of the polygon line $K_1K_2,...,K_{N-1}K_N$ and we have
$$2\pi\sqrt{R(R-2r)} < N \cdot 2r + 2r \Rightarrow \pi\sqrt{\frac{R}{r}\left(\frac{R}{r}-2\right)} - 1 \le N \quad (2)$$
From (1) and (2) it follows that:
$$\pi\sqrt{\frac{R}{r}\left(\frac{R}{r}-2\right)} - 1 \le N < \pi\left(\frac{R}{r}-1\right),$$
from which, because of the hypothesis $R = 11r$, we find
$$\pi\sqrt{11(11-2)} - 1 \le N < \pi(11-1) \\ \Leftrightarrow \pi\sqrt{99} - 1 \le N < 10\pi \approx 31.4 \Leftrightarrow N = 31.$$