6. On the table, there are three cones standing on their bases, touching each other. The radii of their bases are 32, 48, and 48, and the vertex angles are $\frac{\pi}{3}$, $\frac{2 \pi}{3}$, and $\frac{2 \pi}{3}$ respectively (the vertex angle of a cone is the angle between its generators in the axial section). A sphere is suspended above the table, touching all the cones. It turns out that the center of the sphere is equidistant from the centers of the bases of all the cones. Find the radius of the sphere.
Answer: $13(\sqrt{3}+1)$.
![](https://cdn.mathpix.com/cropped/2024_05_06_bd69057b1c245517edf6g-10.jpg?height=506&width=1302&top_left_y=1686&top_left_x=434)

Solution. Let $O_{1}, O_{2}, O_{3}$ be the centers of the bases of the cones, $O$ be the center of the sphere, $C$ be its projection on the table, and $R$ be the radius of the sphere. Point $C$ is equidistant from $O_{1}, O_{2}, O_{3}$, meaning it is the center of the circumcircle of triangle $O_{1} O_{2} O_{3}$ (see the right figure). Let $r$ be the radius of this circle. Note that

$$
A O_{1}=\sqrt{O_{1} O_{2}^{2}-A O_{2}^{2}}=\sqrt{80^{2}-48^{2}}=64
$$

from which

$$
r^{2}-A O_{2}^{2}=A C^{2}=\left(A O_{1}-r\right)^{2} \Longleftrightarrow r^{2}-48^{2}=r^{2}-128 r+64^{2} \Longleftrightarrow r=50
$$

Then $C B=18$ and $C D=2$, where $B$ and $D$ are the points of intersection of segments $C O_{1}$ and $C O_{2}$ with the bases of the cones.

The tangency of the sphere with the first cone means that the perpendicular from point $O$ to the generatrix of the cone, lying in the plane $C O O_{1}$, is equal to $R$ (see the left figure). Indeed, in this case, the sphere and the cone touch the plane passing through this generatrix perpendicular to $C O O_{1}$ and lie on opposite sides of it. Therefore,

$$
R=C B \cdot \cos \frac{\pi}{6}+O C \cdot \sin \frac{\pi}{6} \Longleftrightarrow O C=2 R-18 \sqrt{3}
$$

Applying these arguments to the other cone, we get

$$
R=C D \cdot \cos \frac{\pi}{3}+O C \cdot \sin \frac{\pi}{3}=1+\sqrt{3}(R-9 \sqrt{3}) \Longleftrightarrow R=13(\sqrt{3}+1)
$$

It is also necessary to check that the sphere touches the generatrices of the cones, not their extensions beyond the vertex. This is equivalent to the conditions

$$
R \cos \frac{\pi}{6} \leqslant C O_{1}=r=50 \quad \text { and } \quad R \cos \frac{\pi}{3} \leqslant 50
$$

which are obviously satisfied.

## Variant 4