## Task 15/78

1. Given is an annulus in which $n$ circles are inscribed such that each circle touches the inner and outer boundary of the annulus as well as two adjacent circles. The task is to find the ratio of the sum of the areas of the circles to the area of the annulus.
2. Correspondingly, the task is to solve for the ratio of a) the surface areas, b) the volumes when $n$ spheres are inscribed in a torus in the same manner.
3. In each case, find the limit of the ratio as $n \rightarrow \infty$.
1) If $A_{k}$ is the sum of the areas of the circles and $r$ is the radius of the inscribed circles, then $A_{k}=n r^{2} \pi$. If $R$ is the mean radius of the annulus, then due to the tangency conditions, for the area $A_{R}$ of the annulus

$$
A_{R}=\left[(R+r)^{2}-(R-r)^{2}\right] \pi=4 R r \pi \quad \text { and } \quad \frac{r}{R}=\sin \frac{2 \pi}{2 n}=\sin \frac{\pi}{n}
$$

Thus, we get

$$
\frac{A_{k}}{A_{R}}=\frac{n r^{2} \pi}{4 R r \pi}=\frac{1}{4} n \sin \frac{\pi}{n}
$$

2a) Now let $A_{k}$ be the sum of the surface areas of the spheres, and $A_{T}$ be the surface area of the torus. Then

$$
A_{k}=n \cdot 4 r^{2} \pi \quad ; \quad A_{T}=2 R \pi \cdot 2 r \pi=4 R r \pi^{2}
$$

(according to the 1st Pappus-Guldin theorem). Thus, we have

$$
\frac{A_{k}}{A_{T}}=\frac{4 r^{2} \pi n}{4 R r \pi^{2}}=\frac{n}{\pi} \sin \frac{\pi}{n}
$$

(since here also $\frac{r}{R}=\sin \frac{\pi}{n}$)

2b) If $V_{k}$ is the sum of the volumes of the spheres and $V_{T}$ is the volume of the torus, then

$$
V_{k}=n \cdot \frac{4}{3} r^{3} \pi \quad ; \quad V_{T}=2 R \pi \cdot r^{2} \pi=2 R r^{2} \pi^{2}
$$

(according to the 2nd Pappus-Guldin theorem). Thus, we get

$$
\frac{V_{k}}{V_{T}}=\frac{\frac{4}{3} r^{3} \pi n}{2 R r^{2} \pi^{2}}=\frac{2 n}{3 \pi} \sin \frac{\pi}{n}
$$

3) For the three limits, we get

$$
\begin{gathered}
\lim _{n \rightarrow \infty} \frac{1}{4} n \cdot \sin \frac{\pi}{n}=\lim _{n \rightarrow \infty} \frac{\pi}{4} \frac{\sin x}{x}=\frac{\pi}{4} \\
\lim _{n \rightarrow \infty} \frac{n}{\pi} \cdot \sin \frac{\pi}{n}=\lim _{n \rightarrow 0} \frac{\sin x}{x}=1 \quad ; \quad \lim _{n \rightarrow \infty} \frac{2 n}{3 \pi} \cdot \sin \frac{\pi}{n}=\lim _{n \rightarrow 0} \frac{2}{3} \frac{\sin x}{x}=\frac{2}{3}
\end{gathered}
$$

where $\pi=n x$ was set.

Surprisingly, in the first case, a transcendental value is obtained, while in the second and third cases, algebraic (even rational) values are obtained.