4. Four balls with the same radius $r$ are placed on a plane so that each one touches two others. Another ball with radius $R$ is placed so that it touches all four balls. What is the distance $d$ from the center of this ball to the plane? Discuss all possibilities.
Solution. Let's connect the center $O$ of the fifth sphere with the centers $O_{i}$, $i=1,2,3,4$, of the given four spheres (see diagram). Then we will have

$$
\begin{aligned}
d=H+r & =r+\sqrt{(R+r)^{2}-\bar{O}_{1}^{2}} \\
& =r+\sqrt{(R+r)^{2}-2 r^{2}} \\
& =r+\sqrt{R^{2}+2 r R-r^{2}}
\end{aligned}
$$

For $R=r(\sqrt{2}-1)$, we have $d=r$, i.e., the fifth sphere is among the four spheres.

For $R>r(\sqrt{2}-1)$, the fifth sphere is

![](https://cdn.mathpix.com/cropped/2024_06_05_9925ad5ba00e33a79b8cg-2.jpg?height=553&width=603&top_left_y=562&top_left_x=862)
placed above the four spheres.

## II year