A circle of radius $r$ is tangent to four other circles, each also of radius $r$, and any two of these four circles have no common interior points. What is the radius of the smallest circle that can contain all these circles?
Let the center of the tangent circle be $O$, and the centers of the tangent circles be $O_{1}, O_{2}, O_{3}, O_{4}$. We will use the additional notations from the diagram. It is obvious that a circle with radius $3r$ centered at $O$ contains all five circles, but a circle with a radius smaller than $3r$ centered at $O$ does not suffice. We will show that a circle with a radius smaller than $3r$ and a different center also does not work. Consider, for example, the sector $O_{1} O O_{2}$. This angle is at least $60^{\circ}$, and since there are three more such sectors, it is not greater than $180^{\circ}$. Suppose there exists a point $O^{\prime}$ such that a circle with a radius smaller than $3r$ can be drawn around it, containing all five circles. Since $O \neq O^{\prime}$, there is a sector $O_{i} O O_{i+1}$ (where $i=1, \ldots, 4, O_{5}=O_{1}$) that contains a point different from $O$. Let this sector be $\mathrm{O}_{2} \mathrm{OO}_{3} \varangle$. Suppose for now that $O^{\prime}$ is an interior point of this sector. This sector is not adjacent to the sector $F_{1} O F_{4} \varangle$, which, according to the previous, is not greater than $180^{\circ}$. Therefore, $F_{4} O O^{\prime} \varangle + F_{1} O O^{\prime} \varangle \geq 180^{\circ}$, so one of these two angles is at least a right angle. Suppose $F_{1} O O^{\prime} \geq 90^{\circ}$. Then in the triangle $F_{1} O O^{\prime}$, $F_{1} O^{\prime}$ is the largest side, and since $F_{1} O = 3r$, $F_{1} O^{\prime} > 3r$. This is a contradiction.

Now consider the case where $O^{\prime}$ lies on one of the sides of the sector it contains. Without loss of generality, we can assume that $O^{\prime} \in O F_{2}$. If $F_{4} O F_{2} \varangle = 180^{\circ}$, then $O^{\prime} F_{4} > O F_{4} = 3r$. If $F_{4} O F_{2} \varangle \neq 180^{\circ}$, then in the triangle $F_{4} O O^{\prime}$, the angle $F_{4} O O^{\prime} \varangle$ is at least $120^{\circ}$, so the side $F_{4} O^{\prime}$ is the largest, and again $F_{4} O^{\prime} > F_{4} O = 3r$. This also leads to a contradiction, showing that there is no point $O^{\prime}$ such that a circle with a radius smaller than $3r$ can be drawn around it, containing all five circles. Therefore, the radius of the smallest circle is $3r$.

Siket István (Szeged, Ságvári E. Gimn., IV. o.t.)

Remarks. 1. Szobonya László (Fazekas M. Főv. Gyak. Gimn., IV. o.t.) proved the following: if a circle with radius $r$ and center $O$ is circumscribed by $\frac{\pi}{2 \cdot \operatorname{arc} \sin (R /(r+R))} + 1$ (the upper integer part) tangent circles with radius $R$ and no common interior point, then the center of the smallest circle containing all circles is $O$, and its radius is $r + 2R$. This also implies the statement of our problem.

2. The problem can be extended to spheres: it can be proven that if a sphere with radius $r$ is tangent to at least nine spheres with radius $r$ and no common interior point, then the radius of the smallest sphere containing all spheres is $3r$.

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