, 2010

Consider the regular $n$-gon with vertices at the centers of the balls (Fig. 16). Its edges are of length $2r$ and its perimeter is $n \cdot 2r$. The radius of the circumcircle of the $n$-gon is $R-r$ and the length of the circumcircle is $2\pi(R-r)$. Since a chord of a circle is always shorter than the corresponding arc of the circle, we have $n \cdot 2r < 2\pi(R-r)$ or $nr + \pi r < \pi R$, which implies $r/R < \frac{\pi}{n+\frac{\pi}{n}}$.

Fig. 16

Consider the isosceles triangle with vertices at the centers of two adjacent balls and at the closest point to them on the common axis of the cylinders (Fig. 17). The two equal sides of the triangle are of length $R-r$, the base is of length $2r$ and the vertex angle is $\frac{2\pi}{n}$. The altitude drawn onto the base divides the triangle into two equal right triangles with the hypotenuse $R-r$, one of the legs $r$ and the opposite angle $\frac{\pi}{n}$. Hence $\frac{r}{R-r} = \sin \frac{\pi}{n} < \frac{\pi}{n}$, whence $nr < \pi R - \pi r$, which implies $\frac{r}{R} < \frac{\pi}{n+\frac{\pi}{n}}$.

Fig. 17