6. On the table, there are three cones standing on their bases, touching each other. The radii of their bases are 72, 28, and 28, and the angles at their vertices are $-\frac{\pi}{3}$, $\frac{2 \pi}{3}$, and $\frac{2 \pi}{3}$ respectively (the angle at the vertex 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 all points of contact with the cones. Find the radius of the sphere.
Answer: $\frac{\sqrt{3}+1}{2}$.

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. The point $C$ is equidistant from the points of tangency of the bases of the cones, so it lies at the intersection of the angle bisectors of triangle $O_{1} O_{2} O_{3}$ (see the right figure). Thus, $C$ is the center of the inscribed circle of $\triangle O_{1} O_{2} O_{3}$, and the segment $A C$ is the radius of this circle. Note that the perimeter of triangle $O_{1} O_{2} O_{3}$ is 256 and

$$
A O_{1}=\sqrt{O_{1} O_{2}^{2}-A O_{2}^{2}}=\sqrt{100^{2}-28^{2}}=96
$$

from which

$$
A C=\frac{A O_{2} \cdot A O_{1}}{128}=\frac{28 \cdot 96}{128}=21
$$

Let $B$ and $D$ be the points of intersection of segments $C O_{1}$ and $C O_{2}$ with the bases of the cones. Then

$$
B C=A O_{1}-A C-B O_{1}=96-21-72=3 \quad \text { and } \quad C D=\sqrt{A O_{2}^{2}+A C^{2}}-D O_{2}=\sqrt{28^{2}+21^{2}}-28=7
$$

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,

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

Applying these arguments to the other cone, we get

$$
R=C D \cdot \cos \frac{\pi}{3}+O C \cdot \sin \frac{\pi}{3}=\frac{7}{2}+R \sqrt{3}-\frac{9}{2}=R \sqrt{3}-1 \Longleftrightarrow R=\frac{1}{2}(\sqrt{3}+1)
$$

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

$$
R \cos \frac{\pi}{6} \leqslant C O_{1}=75 \quad \text { and } \quad R \cos \frac{\pi}{3} \leqslant C O_{2}=35
$$

which are obviously satisfied.
![](https://cdn.mathpix.com/cropped/2024_05_06_bd69057b1c245517edf6g-15.jpg?height=508&width=1304&top_left_y=1525&top_left_x=434)

## Variant 5