6. On the table, there are three cones standing on their bases, touching each other. The radii of their bases are $2 r, 3 r$, and $10 r$. A frustum of a cone is placed on the table with its smaller base down, and it shares a common generatrix with each of the other cones. Find $r$ if the radius of the smaller base of the frustum is 15.
Answer: 29.
![](https://cdn.mathpix.com/cropped/2024_05_06_bd69057b1c245517edf6g-30.jpg?height=520&width=1302&top_left_y=776&top_left_x=434)

Solution. Let $C$ be the center of the smaller base of the truncated cone, and $O_{1}, O_{2}, O_{3}$ be the centers of the bases of the other cones, with $R=15$. Denote by $\mathcal{K}_{0}$ the cone that completes the truncated cone to a regular cone, and by $\mathcal{K}_{1}$ the cone with the center of its base at $O_{1}$. On the left diagram, the section of $\mathcal{K}_{0}$ and $\mathcal{K}_{1}$ by the plane $\Pi$ passing through points $O_{1}$ and $C$ and perpendicular to the table is shown. By the problem's condition, $\mathcal{K}_{0}$ and $\mathcal{K}_{1}$ have a common generatrix, which lies in $\Pi$ since it passes through the vertices of the cones. Let $B$ be the point of intersection of this generatrix with the table. Then $B$ lies on the boundaries of the bases of $\mathcal{K}_{0}$ and $\mathcal{K}_{1}$, as well as on the segment $C O_{1}$ connecting the centers of the bases. Hence, the bases of $\mathcal{K}_{0}$ and $\mathcal{K}_{1}$ touch each other at point $B$, i.e., $B C=R$. Similarly, it can be verified that the distance from $C$ to the bases of the other two cones is also $R$. Note that

$$
O_{1} O_{2}=5 r, \quad O_{1} O_{3}=12 r, \quad O_{2} O_{3}=13 r
$$

which means that triangle $O_{1} O_{2} O_{3}$ is a right triangle. Direct the coordinate axes along the rays $O_{1} O_{3}$ and $O_{1} O_{2}$ (see the right diagram). Let the coordinates of point $C$ be $(x, y)$. Since

$$
C O_{1}=B O_{1}+B C=2 r+R, \quad C O_{2}=3 r+R, \quad C O_{3}=10 r+R
$$

the following equalities hold:

$$
(2 r+R)^{2}-x^{2}=y^{2}=(10 r+R)^{2}-(12 r-x)^{2} \Longleftrightarrow 4 r^{2}+4 r R=20 r R+24 r x-44 r^{2} \Longleftrightarrow 3 x=6 r-2 R
$$

and also

$$
(2 r+R)^{2}-y^{2}=x^{2}=(3 r+R)^{2}-(5 r-y)^{2} \Longleftrightarrow 4 r^{2}+4 r R=6 r R+10 r y-16 r^{2} \Longleftrightarrow 5 y=10 r-R
$$

Since $x^{2}+y^{2}=C O_{1}^{2}=(2 r+R)^{2}$, we get

$$
225(2 r+R)^{2}=(30 r-10 R)^{2}+(30 r-3 R)^{2} \Longleftrightarrow 225 r^{2}-420 R r-29 R^{2}=0
$$

This equation in terms of $r$ has a unique positive solution $r=\frac{29}{15} R=29$.

## Variant 9