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 $2r$, $3r$, and $10r$. A sphere with a radius of 2 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 $r$.
Answer: 1.
![](https://cdn.mathpix.com/cropped/2024_05_06_bd69057b1c245517edf6g-37.jpg?height=488&width=1298&top_left_y=1569&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

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

Therefore, $A C=2 \operatorname{tg}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)$ and, similarly, $B C=2 \operatorname{tg}\left(\frac{\pi}{4}-\frac{\beta}{2}\right)$. 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 $O_{1} O_{2}=5 r, O_{1} O_{3}=12 r, O_{2} O_{3}=13 r$, 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 $2 r$. Calculating $C O_{1}^{2}$ and $C O_{2}^{2}$ in two ways, we get

$$
\left\{\begin{array} { l } 
{ ( 2 r + 2 \lambda ) ^ { 2 } = 8 r ^ { 2 } , } \\
{ ( 3 r + 2 \mu ) ^ { 2 } = 1 3 r ^ { 2 } }
\end{array} \Longleftrightarrow \left\{\begin{array} { l } 
{ 4 \lambda ^ { 2 } + 8 \lambda r = 4 r ^ { 2 } , } \\
{ 4 \mu ^ { 2 } + 1 2 \mu r = 4 r ^ { 2 } }
\end{array} \Longleftrightarrow \left\{\begin{array}{l}
\lambda^{2}+2 \lambda r-r^{2}=0 \\
\lambda^{2}-\mu^{2}=r(3 \mu-2 \lambda)
\end{array}\right.\right.\right.
$$

(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 $\operatorname{tg} \beta=\frac{3}{2} \operatorname{tg} \alpha$ holds. Therefore,

$$
1-\mu^{2}=\frac{3 \mu}{2 \lambda}\left(1-\lambda^{2}\right) \quad \text { and } \quad \lambda^{2}-\mu^{2}=\lambda^{2}-1+1-\mu^{2}=\frac{\left(1-\lambda^{2}\right)(3 \mu-2 \lambda)}{2 \lambda}
$$

The second equation of $(*)$ gives

$$
r=\frac{\lambda^{2}-\mu^{2}}{3 \mu-2 \lambda}=\frac{1-\lambda^{2}}{2 \lambda}=\frac{1}{2}\left(\frac{1}{\lambda}-\lambda\right)
$$

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

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