Given three unit-radius spheres, each pair of which touch each other and the plane $S$. Determine the radius of the sphere whose center is on the plane $S$ and touches the spheres.

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Solution. The centers of the three given unit spheres $O_{1}, O_{2}, O_{3}$ are the three vertices of a regular triangle with a side length of 2 units. The plane $S_{1}$ of the triangle is parallel to the plane $S$, and is 1 unit away from it. Let the radius of the sphere that externally touches the three spheres be denoted by $r$ ($r<1$), and its center by $O_{4}$. Since $O_{4}$ is in the plane $S$, its distance from $S_{1}$ is 1 unit. The four centers determine a regular pyramid with a triangular base. Let the projection of $O_{4}$ onto the plane $S_{1}$ be denoted by $P$. Since the triangle $O_{1} O_{2} O_{3}$ is regular, $P$ is the center and also the centroid of the triangle.

![](https://cdn.mathpix.com/cropped/2024_05_02_f1f65d7c839b1b595576g-1.jpg?height=342&width=420&top_left_y=360&top_left_x=847)
Using the Pythagorean theorem in the right triangle $O_{1} P O_{4}$, where $O_{4} P=1, O_{1} O_{4}=1+r$, and $O_{1} P=\frac{2}{3} \sqrt{3}$, we have:

$$
(1+r)^{2}=\left(\frac{2 \sqrt{3}}{3}\right)^{2}+1
$$

From this, $r=\sqrt{\frac{7}{3}}-1 \approx 0.5275$ units is the radius of the sphere that externally touches the given spheres.

However, there is another possibility, where the sought sphere is internally tangent to the given spheres, and let its radius be $R$. In this case, similarly to the previous one:

$$
R-1=\sqrt{\frac{7}{3}}, \quad \text { i.e., } \quad R=\sqrt{\frac{7}{3}}+1 \approx 2.5275 \text { units. }
$$

Judit Nagy (Szombathely, Keresk. és Vendéglátói Szakképző Isk., 12th grade)

Remark. In the solution, we accepted from "intuition" that the sought sphere either externally touches all three given unit spheres or internally touches each of them. However, if we do not completely trust our intuition (which is especially justified in a spatial geometry problem!), we can very easily verify this. Suppose the sought sphere externally touches the sphere centered at $O_{1}$ and internally touches the sphere centered at $O_{2}$. Then (denoting the sought radius by $r^{\prime}$) $O_{4} O_{1}=r^{\prime}+1, O_{4} O_{2}=r^{\prime}-1$, so

$$
O_{1} O_{2}=2=\left(r^{\prime}+1\right)-\left(r^{\prime}-1\right)=O_{4} O_{1}-O_{4} O_{2}
$$

Thus, the triangle $O_{1} O_{2} O_{4}$ is degenerate, meaning $O_{4}$ lies on the line $O_{1} O_{2}$. However, this line, being in the plane parallel to $S$, does not intersect $S$, of which $O_{4}$ is a point; this is a contradiction.