## Problem II - 3

Two spheres of radius $r$ are externally tangent. Three spheres of radius $R$ are externally tangent to each other, each tangent to the other two. Each of these spheres is also externally tangent to the first two.

Find the relationship between $R$ and $r$.
Solution.

The centers $O_{1}, O_{2}, O_{3}$ of the three spheres of radius $R$ are the vertices of an equilateral triangle.

The centers of the two spheres of radius $r$ determine a line perpendicular to the plane determined by the centers of the spheres of radius $R$; the point of tangency $T$ of the spheres of radius $r$ must be the center of the previous equilateral triangle. The five centers are the vertices of a double tetrahedron. Let $C_{1}$ be one of the centers of the spheres of radius $r$ and consider the tetrahedron $\mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3} \mathrm{C}_{1}$.

$T$ is the foot of the perpendicular from $C_{1}$ and the following relationships hold:

$$
O_{1} O_{2}=2 R, O_{1} C_{1}=r+R, C_{1} T=r
$$

The triangle $C_{1} O_{1} T$ is a right triangle at $T$, and we have

$$
O_{1} T=\frac{2}{3} R \sqrt{3}
$$

An application of the Pythagorean theorem in this last triangle allows us to write

$$
(r+R)^{2}=\frac{4 R^{2}}{3}+r^{2}
$$

which simplifies to give

$$
2 R r=\frac{1}{3} R^{2} \Leftrightarrow R=6 r
$$