\section*{Problem 6 - 191046}

Four spheres with a given radius \(r\) are arranged in space such that each one touches the other three externally.

The four tangent planes, each touching three of these spheres and not intersecting the fourth, form a regular tetrahedron.

Calculate the volume of this tetrahedron in terms of \(r\)!
}

The four centers \(M_{1}, M_{2}, M_{3}, M_{4}\) of the spheres are the vertices of a regular tetrahedron with side length \(2 r\).

Let \(Z\) be the center of this regular tetrahedron, i.e., the intersection of the perpendiculars dropped from each vertex to the opposite face. Each such perpendicular has, according to the formula for the height of a regular tetrahedron, the length

\[
h=\frac{2}{3} r \sqrt{6}
\]

\(Z\) divides each perpendicular in the ratio \(3:1\). (2)

This statement (2) can be proven, for example, through volume considerations (by decomposing the tetrahedron into four congruent tetrahedra) or by applying the intercept theorem. (One may also assume (2) as known.)

The four faces of the tetrahedron \(A B C D\) formed by the four tangent planes are each parallel to the corresponding faces of the tetrahedron \(M_{1} M_{2} M_{3} M_{4}\) at a distance \(r\). Therefore, \(A B C D\) is a regular tetrahedron with the center \(Z\). It is now necessary to determine the height \(h^{\prime}\) of the tetrahedron \(A B C D\) in terms of \(h\), because with this, the volume \(V^{\prime}\) of the tetrahedron \(A B C D\) can then be calculated in terms of \(h\) and, according to (1), in terms of \(r\).

We provide two methods for this:

1. \(A B C D\) is obtained from \(M_{1} M_{2} M_{3} M_{4}\) through central dilation with the dilation center \(Z\). The dilation ratio \(K\) is, due to (2) (with (D)

\[
K=\left(\frac{1}{6} r \sqrt{6}+r\right): \frac{1}{6} r \sqrt{6}=1+\sqrt{6}
\]

2. We extend the edges \(M_{4} M_{1}, M_{4} M_{2}, M_{4} M_{3}\) until they intersect the plane through \(A, B, C\) at points \(A^{\prime}, B^{\prime}, C^{\prime}\).

The tetrahedron \(M_{4} A^{\prime} B^{\prime} C^{\prime}\) is again regular due to the parallelism of the planes through \(A, B, C\) and \(M_{1}, M_{2}, M_{3}\) and has the height \(h+r\). If we repeat this process three more times (with respect to the other faces), we finally obtain the tetrahedron \(A B C D\) with

\[
h^{\prime}=h+4 r
\]

Using (3) or (4), the volume is calculated as

\[
V^{\prime}=\frac{2}{3} r^{3} \sqrt{2}(1+\sqrt{6})^{3}
\]

Solutions from the 4th round of 1979 taken from [5]

\subsection*{7.22 20th Olympiad 1980}

\subsection*{7.22.1 1st Round 1980, Class 10}