## Task A-3.3.

In the sphere $S$ of radius $R$, a ring is inscribed composed of eight smaller spheres of equal radius, each touching two adjacent ones, and all touching the sphere $S$ along the same circle of radius $R$. The sphere $S_{1}$ touches all eight smaller spheres and the sphere $S$. Determine the radius of the sphere $S_{1}$ in terms of $R$. Write the final solution without using trigonometric functions.
## Solution.

Let $r$ be the radius of each of the eight small spheres, and $\rho$ be the desired radius of the sphere $S_{1}$. The large sphere $S$ has a radius $R$.

Consider the cross-section of the sphere $S$ and the given eight spheres by a plane that contains their points of tangency with the sphere $S$. We want to determine the radius of each of the eight small spheres.

By examining the triangle $A O B$ (see the figure), we conclude: $\frac{r}{R-r}=\sin \frac{\pi}{8}$

so we get

$$
r=R \cdot \frac{\sin \frac{\pi}{8}}{1+\sin \frac{\pi}{8}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_30_a9c12f6adaba3761a080g-14.jpg?height=592&width=1338&top_left_y=284&top_left_x=380)

Now consider the cross-section by a plane passing through the center $O$ of the sphere $S$, the center $O_{1}$ of the sphere $S_{1}$, and the center of one of the small spheres.

From the triangle $A O O_{1}$ (see the figure), we have

$$
(r+\rho)^{2}=(R-r)^{2}+(R-\rho)^{2}
$$

From this, we get $\rho=R \cdot \frac{R-r}{R+r}$.

Using the derived expression for $r$, we get

$$
\rho=R \cdot \frac{1-\frac{\sin \frac{\pi}{8}}{1+\sin \frac{\pi}{8}}}{1+\frac{\sin \frac{\pi}{8}}{1+\sin \frac{\pi}{8}}}=\frac{R}{1+2 \sin \frac{\pi}{8}}
$$

Finally, let's calculate $\sin \frac{\pi}{8}$:

$$
\sin \frac{\pi}{8}=\sqrt{\frac{1-\cos \frac{\pi}{4}}{2}}=\frac{\sqrt{2-\sqrt{2}}}{2}
$$

Thus, $\rho=\frac{R}{1+\sqrt{2-\sqrt{2}}}$.