Let $\mathcal{K}$ be a right circular cone and $\mathcal{G}$ its inscribed sphere. The centers of the spheres $g_{1}, g_{2}, \ldots, g_{n}$ with radius $r$ form a regular $n$-sided polygon with side length $2r$. Furthermore, each sphere $g_{i}$ touches the lateral surface and the base of $\mathcal{K}$, as well as $\mathcal{G}$. What values can $n$ take?
Solution. Let the vertex of $\mathcal{K}$ be $C$, and the centers of the spheres $\mathcal{G}$ and $g_{i}$ be $O$ and $O_{i}(i=1,2, \ldots, n)$, respectively. Consider the intersection of $\mathcal{K}$ with the plane $C O O_{1}$. This is an isosceles triangle $A B C$ whose inscribed circle has center $O$ and radius $R$, and the circle with center $O_{1}$ and radius $r$ touches the base $A B$, the side $A C$, and the inscribed circle (Figure 1). Let the intersection of the line through $O_{1}$ parallel to $A B$ and the line $C O$ be $Q$, and the length of the segment $O_{1} Q$ be denoted by $d$.

![](https://cdn.mathpix.com/cropped/2024_05_02_d60c34cda092c487551ag-1.jpg?height=667&width=596&top_left_y=325&top_left_x=754)

If $\angle C A B = 2 \alpha$, then $\angle O A B = \alpha$, because the center of the inscribed circle lies on the angle bisectors of the triangle. Since $O_{1} Q$ is parallel to $A B$, $Q$ and $O_{1}$ are at the same distance from the line $A B$, from which we get that $O Q = R - r$. We know that for externally tangent circles, the distance between the centers of the circles is equal to the sum of their radii, so $O O_{1} = R + r$. Therefore, from the right triangle $O O_{1} Q$, we get

$$
\sin \alpha = \frac{R - r}{R + r}
$$

from which (since $\alpha \frac{1}{2}\left(-1 + \frac{2}{1 + (\sqrt{2} - 1)}\right) = \frac{\sqrt{2} - 1}{2}
$$

follows. We know that the function $x \mapsto \sin x$ is strictly increasing on the interval $\left[0, \frac{\pi}{2}\right]$, so from inequalities (2) and (3) for $\varphi = \frac{180^{\circ}}{n}$, we get the following bounds:

$$
\arcsin \frac{\sqrt{2} - 1}{2} < \frac{180^{\circ}}{n} < \arcsin \frac{1}{2}, \quad \text { that is } \quad 11.95^{\circ} < \frac{180^{\circ}}{n} < 30^{\circ},
$$

which means $6 < n < 16$.

![](https://cdn.mathpix.com/cropped/2024_05_02_d60c34cda092c487551ag-2.jpg?height=466&width=512&top_left_y=1586&top_left_x=796)

We will show that for $n = 7, 8, \ldots, 15$, there exist corresponding cones. From equation (1), for each value of $n$, we can determine the corresponding angle $\alpha_{n}$, which, due to the inequalities for $n$, will be greater than $0^{\circ}$ but less than $45^{\circ}$. Then, knowing $r$, we can also determine $R$, and thus the cone.

Remark. It is easy to see that for $n = 6$, our calculation yields $R = r$, meaning that in this case, the sphere of radius $R$ just fits between the 6 spheres of radius $r$, without protruding, so we do not get a suitable cone. If, however, $n = 16$, then the sphere of radius $R$ protrudes beyond the narrowest cylinder containing the small spheres and whose generators are perpendicular to the plane of their centers, so we do not get a suitable cone.