}

The centers of the three large spheres all lie in a plane parallel to the tabletop at a distance of \(r\). They form an equilateral triangle \(\triangle M_{1} M_{2} M_{3}\) with side length \(2 r\), and let \(S\) be the centroid of this triangle.

Then \(\left|M_{1} S\right|=\frac{2 \sqrt{3}}{3} r\), since the centroid divides the medians in the ratio \(2: 1\) and each median in an equilateral triangle is also an altitude, whose length can be determined by applying the Pythagorean theorem as \(\sqrt{(2 r)^{2}-r^{2}}\).

Let \(M\) be the center of the fourth sphere. Since it touches the three large spheres, \(M\) must, due to symmetry, lie on the perpendicular from \(S\) to the tabletop. This forms a right triangle \(\triangle M_{1} S M\) with a right angle at \(S\).

If \(R\) is the desired radius of the fourth sphere, then on the one hand \(|S M|=r-R\), since the fourth sphere also touches the tabletop from above and the radii of all spheres perpendicular to the tabletop, including that of the fourth sphere, lie on the line \(M S\). On the other hand, the fourth and the first sphere also touch each other, so their centers \(M_{1}\) and \(M\) have the distance \(r+R\).

Substituting this into the equation given by the Pythagorean theorem for the right triangle \(\triangle M_{1} S M\), we get

\[
(r+R)^{2}=\left|M_{1} M\right|^{2}=\left|M_{1} S\right|^{2}+|S M|^{2}=\frac{4}{3} r^{2}+(r-R)^{2}
\]

or \(4 r R=(r+R)^{2}-(r-R)^{2}=\frac{4}{3} r^{2}\), and thus, due to \(r>0\), finally \(R=\frac{1}{3} r\).
\section*{Problem 1 - 071231}

Three equally sized wooden spheres with a radius of length \(r\), which touch each other in pairs, lie on a flat tabletop.

How large is the radius of a fourth sphere that touches all three spheres and the tabletop simultaneously?