3. Four identical spheres with radius $R$ are given, each touching the other three. Determine the radius of the sphere that touches the given spheres.
Solution. The centers $A_{1}, A_{2}, A_{3}$, and $A_{4}$ of the given spheres are vertices of a regular tetrahedron with edge $2 R$. Let $x$ be the radius of the circumscribed sphere around the tetrahedron $A_{1} A_{2} A_{3} A_{4}$, and $O$ its center. Let $O_{1}$ be the projection of vertex $A_{4}$ onto the base $A_{1} A_{2} A_{3}$. Then $\overline{A_{1} O_{1}}=R \frac{2 \sqrt{3}}{3}$, and

$$
h^{2}={\overline{A_{4} O_{1}}}^{2}=(2 R)^{2}-\left(R \frac{2 \sqrt{3}}{3}\right)^{2}=\left(R \frac{2 \sqrt{6}}{3}\right)^{2}
$$

From the right triangle $\mathrm{A}_{1} \mathrm{O}_{1} \mathrm{O}$ we get

$$
\begin{aligned}
& {\overline{A_{1} O_{1}}}^{2}+{\overline{O O_{1}}}^{2}={\overline{A_{1} O}}^{2} \\
& \frac{4 R^{2}}{3}+\left(\frac{2 R \sqrt{6}}{3}-x\right)^{2}=x^{2} \\
& x=R \frac{\sqrt{6}}{2}
\end{aligned}
$$

The radii of the desired spheres are

$$
\begin{aligned}
& R_{1}=x+R=R\left(\frac{\sqrt{6}}{2}+1\right) \\
& R_{2}=x-R=R\left(\frac{\sqrt{6}}{2}-1\right)
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_17f3bf65aefe5b309f24g-6.jpg?height=581&width=526&top_left_y=255&top_left_x=958)