24th 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: $$N\cdot 2r<{\ell }_{K_1...K_N{K}_1}<2\pi (R-r)\Rightarrow N<\pi \left(\frac{R}{r}-1\right).(1)$$

Figure 2 Figure 3

Let $OA$ is tangent from $O$ to one of the circles $C_i(K_i,r)$. Then $$O{A}^2=R(R-2r)\iff OA=\sqrt{R(R-2r)}.$$ The circle $C(O,OA)$ is tangent to the sides of the polygon line $K_1{K}_2,...,K_{N-1}{K}_N$ and we have $$2\pi \sqrt{R(R-2r)}<N\cdot 2r+2r\Rightarrow \pi \sqrt{\frac{R}{r}\left(\frac{R}{r}-2\right)}-1\le N(2)$$ 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 $$\begin{gathered} \pi \sqrt{11(11-2)}-1\le N<\pi (11-1) \\ \iff \pi \sqrt{99}-1\le N<10\pi \approx 31.4\iff N=31. \end{gathered}$$