6. On the table, there are three cones standing on their bases, touching each other. The radii of their bases are 1, 12, and 12, and the vertex angles are $-4 \operatorname{arctg} \frac{1}{3}, 4 \operatorname{arctg} \frac{2}{3}$, and $4 \operatorname{arctg} \frac{2}{3}$ respectively (the vertex angle of a cone is the angle between its generators in the axial section). A sphere is placed on the table, touching all the cones. Find the radius of the sphere.
Answer: $\frac{40}{21}$.
![](https://cdn.mathpix.com/cropped/2024_05_06_bd69057b1c245517edf6g-07.jpg?height=522&width=1304&top_left_y=1007&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, $R$ be the radius of the sphere, $C$ be the point of tangency of the sphere with the table, and $2 \alpha$ and $2 \beta$ be the angles at the vertices of the first and second cones. On the left diagram, the section of the first cone by the plane $C O O_{1}$ is shown. The tangency of the sphere with the first cone means that the perpendicular $O E$, dropped from point $O$ to the generatrix $P B$, is equal to $R$. Indeed, in this case, the sphere and the cone touch the plane passing through the generatrix $P B$ perpendicular to the section and lie on opposite sides of it. Therefore, the sphere touches the lines $B C$ and $B E$ at points $C$ and $E$, respectively, from which

$$
O C=O E=R, \quad B C=B E \quad \text { and } \quad \angle B O C=\frac{1}{2} \angle E O C=\frac{\pi}{4}-\frac{\alpha}{2} .
$$

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

$$
B C=R \cdot \operatorname{tg}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)=R \cdot \frac{1-\operatorname{tg} \frac{\alpha}{2}}{1+\operatorname{tg} \frac{\alpha}{2}}=R \cdot \frac{1-\frac{1}{3}}{1+\frac{1}{3}}=\frac{R}{2}, \quad C D=R \cdot \operatorname{tg}\left(\frac{\pi}{4}-\frac{\beta}{2}\right)=R \cdot \frac{1-\operatorname{tg} \frac{\beta}{2}}{1+\operatorname{tg} \frac{\beta}{2}}=\frac{R}{5}
$$

On the right diagram, the section of the cones by the plane of the table is shown. Note that $A O_{1}=5$ and

$$
\left(12+\frac{1}{5} R\right)^{2}=C O_{2}^{2}=A O_{2}^{2}+\left(A O_{1}-B O_{1}-B C\right)^{2}=144+\left(4-\frac{1}{2} R\right)^{2} \Longleftrightarrow 21 R^{2}-880 R+1600=0
$$

from which $R=\frac{40}{21}$ or $R=40$. It is necessary that the sphere touches the generatrices of the cones, not their extensions beyond the vertex. This means that $R \cos \alpha \leqslant C O_{1}$ and $R \cos \beta \leqslant C O_{2}$. The first condition is equivalent to

$$
R \cos \alpha \leqslant B O_{1}+B C \Longleftrightarrow \frac{4}{5} R \leqslant 1+\frac{1}{2} R \Longleftrightarrow R \leqslant \frac{10}{3}
$$

and it is satisfied only by $R=\frac{40}{21}$. The second condition reduces to the inequality $\frac{5}{13} R \leqslant 12+\frac{1}{5} R$, which is true for $R=\frac{40}{21}$.

## Variant 3