6. On the table, there are three cones standing on their bases, touching each other. The heights of the cones are the same, and the radii of their bases are 1, 2, and 3. A sphere is placed on 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: 1.
![](https://cdn.mathpix.com/cropped/2024_05_06_bd69057b1c245517edf6g-33.jpg?height=486&width=1298&top_left_y=1112&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 contact of the sphere with the table, $2 \alpha$ and $2 \beta$ be the angles at the vertices of the first and second cones, $A$ and $B$ be the points of intersection of the segments $C O_{1}$ and $C O_{2}$ with the bases of the cones. The left diagram shows the section of the first cone by the plane $C O O_{1}$. The contact of the sphere with the first cone means that the perpendicular $O E$, dropped from point $O$ to the generatrix $P A$, is equal to $R$. Indeed, in this case, the sphere and the cone touch the plane passing through the generatrix $P A$ perpendicular to the section, and lie on opposite sides of it. Then the sphere touches the lines $A C$ and $A E$ at points $C$ and $E$, from which

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

Therefore, $A C=R \cdot \operatorname{tg}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)$ and, similarly, $B C=R \cdot \operatorname{tg}\left(\frac{\pi}{4}-\frac{\beta}{2}\right)$. Let

$$
\lambda=\operatorname{tg}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right), \quad \mu=\operatorname{tg}\left(\frac{\pi}{4}-\frac{\beta}{2}\right)
$$

Since $O_{1} O_{2}=3, O_{1} O_{3}=4, O_{2} O_{3}=5$, the triangle $O_{1} O_{2} O_{3}$ is a right triangle. By the condition, point $C$ is equidistant from the points of contact of the bases of the cones. Then $C$ lies at the intersection of the angle bisectors of triangle $O_{1} O_{2} O_{3}$ and, therefore, is the center of its inscribed circle, the radius of which is 1. By calculating $C O_{1}^{2}$ and $C O_{2}^{2}$ in two ways, we get

$$
\left\{\begin{array} { l } 
{ ( 1 + \lambda R ) ^ { 2 } = 2 , } \\
{ ( 2 + \mu R ) ^ { 2 } = 5 }
\end{array} \Longleftrightarrow \left\{\begin{array} { l } 
{ \lambda ^ { 2 } R ^ { 2 } + 2 \lambda R = 1 , } \\
{ \mu ^ { 2 } R ^ { 2 } + 4 \mu R = 1 }
\end{array} \Longleftrightarrow \left\{\begin{array}{l}
\lambda^{2} R^{2}+2 \lambda R-1=0, \\
\left(\lambda^{2}-\mu^{2}\right) R^{2}=2 R(2 \mu-\lambda)
\end{array}\right.\right.\right.
$$

(we subtracted the second equation from the first). Note that

$$
\frac{1}{\lambda}-\lambda=\frac{1+\operatorname{tg} \frac{\alpha}{2}}{1-\operatorname{tg} \frac{\alpha}{2}}-\frac{1-\operatorname{tg} \frac{\alpha}{2}}{1+\operatorname{tg} \frac{\alpha}{2}}=\frac{4 \operatorname{tg} \frac{\alpha}{2}}{1-\operatorname{tg}^{2} \frac{\alpha}{2}}=2 \operatorname{tg} \alpha \quad \text { and, similarly, } \quad \frac{1}{\mu}-\mu=2 \operatorname{tg} \beta
$$

By the equality of the heights of the cones, the relation $\operatorname{tg} \beta=2 \operatorname{tg} \alpha$ holds. Therefore,

$$
1-\mu^{2}=\frac{2 \mu}{\lambda}\left(1-\lambda^{2}\right) \quad \text { and } \quad \lambda^{2}-\mu^{2}=\lambda^{2}-1+1-\mu^{2}=\frac{\left(1-\lambda^{2}\right)(2 \mu-\lambda)}{\lambda}
$$

The second equation of $(*)$ gives

$$
R=\frac{2(2 \mu-\lambda)}{\left(\lambda^{2}-\mu^{2}\right)}=\frac{2 \lambda}{\left(1-\lambda^{2}\right)} \Longleftrightarrow R-\lambda^{2} R=2 \lambda
$$

From the first equation of the system $(*)$, it follows that $\lambda R=\sqrt{2}-1$. Excluding $\lambda$, we get

$$
R-\frac{(\sqrt{2}-1)^{2}}{R}=\frac{2(\sqrt{2}-1)}{R} \Longleftrightarrow R^{2}=(\sqrt{2}-1)(\sqrt{2}+1)=1 \Longleftrightarrow R=1
$$