Inside a larger sphere, four unit-radius spheres are arranged such that each touches this larger sphere and the other three unit-radius spheres as well. What fraction of the larger sphere's volume is occupied by the four unit-radius spheres together?
Solution. Let the center of the large sphere be $O$, and the centers of the small spheres be $O_{1}, O_{2}, O_{3}, O_{4}$. The four centers are the four vertices of a regular tetrahedron (Figure 1). A sphere can be circumscribed around the tetrahedron, and clearly the center of this sphere coincides with the center $O$ of the large sphere.

![](https://cdn.mathpix.com/cropped/2024_05_02_da9e3e5341a0397b319dg-1.jpg?height=418&width=420&top_left_y=279&top_left_x=842)

Figure 1

![](https://cdn.mathpix.com/cropped/2024_05_02_da9e3e5341a0397b319dg-1.jpg?height=518&width=526&top_left_y=749&top_left_x=794)

Let the radius of the sphere circumscribed around the tetrahedron be $r$, and the radius of the large sphere be $R$. Consider the triangle $O_{1} O_{2} O_{3}$ as the base of the tetrahedron (Figure 2). The length of each side of the triangle is 2 units. The orthogonal projection of the vertex $O_{4}$ onto the base is $P$. The center $O$ of the sphere circumscribed around the tetrahedron lies on the segment $O_{4} P$.

In the right triangle $O_{2} \mathrm{PO}_{4}$, $O_{4} O_{2}=2$,

$$
P O_{2}=\frac{2}{3} \cdot \frac{2 \sqrt{3}}{2}=\frac{2 \sqrt{3}}{3}
$$

since $O_{2} P$ is two-thirds of the median (and also the altitude) in an equilateral triangle with side length 2. In the right triangle $O_{2} P O_{4}$, we can apply the Pythagorean theorem to find the height $O_{4} P$ of the tetrahedron:

$$
O_{4} P^{2}=O_{4} O_{2}^{2}-P O_{2}^{2}, \quad \text { that is } \quad O_{4} P=\sqrt{2^{2}-\left(\frac{2 \sqrt{3}}{3}\right)^{2}}=\sqrt{\frac{8}{3}}=\frac{2 \sqrt{6}}{3}
$$

We also apply the Pythagorean theorem to the triangle $\mathrm{OPO}_{2}$, where $O O_{2}=r$, $O P=\frac{2 \sqrt{6}}{3}-r$, and $P O_{2}=\frac{2 \sqrt{3}}{3}$:

$$
r^{2}=\left(\frac{2 \sqrt{6}}{3}-r\right)^{2}+\left(\frac{2 \sqrt{3}}{3}\right)^{2}
$$

from which $r=\frac{\sqrt{6}}{2}$. The radius of the large sphere is then

$$
R=r+1=\frac{\sqrt{6}}{2}+1=\frac{\sqrt{6}+2}{2}
$$

The volume of the large sphere is:

$$
V_{1}=\frac{4}{3} \pi\left(\frac{\sqrt{6}+2}{2}\right)^{3}
$$

The combined volume of the four unit-radius spheres is:

$$
V_{2}=4 \cdot \frac{4 \pi}{3} \cdot 1^{3}=\frac{16 \pi}{3}
$$

The ratio of the volumes is:

$$
\frac{V_{2}}{V_{1}}=\frac{\frac{16 \pi}{3}}{\frac{4 \pi}{3}\left(\frac{\sqrt{6}+2}{2}\right)^{3}}=\frac{4}{\frac{(\sqrt{6}+2)^{3}}{8}}=\frac{32}{(\sqrt{6}+2)^{3}} \approx 0.36
$$