6. On the table, there are three cones standing on their bases, touching each other. The radii of their bases are 10, 15, and 15. A truncated cone is placed on the table with its smaller base down, and it shares a generatrix with each of the other cones. Find the area of the smaller base of the truncated cone.
Answer: $4 \pi$.
![](https://cdn.mathpix.com/cropped/2024_05_06_bd69057b1c245517edf6g-21.jpg?height=516&width=1306&top_left_y=450&top_left_x=434)

Solution. Let $C$ be the center of the smaller base of the truncated cone, and $r$ be the radius of this base. Let $O_{1}, O_{2}, O_{3}$ be the centers of the bases of the other cones. 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 base center 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 $r$. Therefore, the following equalities hold:

$$
r = C O_{1} - 10 = C O_{2} - 15 = C O_{3} - 15
$$

Then $\mathrm{CO}_{2} = \mathrm{CO}_{3}$, which means that point $C$ lies on the common tangent $A O_{1}$ to the bases of the larger cones (see the right diagram). Note that

$$
A O_{1} = \sqrt{O_{1} O_{2}^{2} - A O_{2}^{2}} = \sqrt{25^{2} - 15^{2}} = 20
$$

Therefore,

$$
(15 + r)^{2} = C O_{2}^{2} = 15^{2} + A C^{2} = 15^{2} + (A O_{1} - r - 10)^{2} = 15^{2} + (10 - r)^{2}
$$

from which $r = 2$ and $\pi r^{2} = 4 \pi$.

## Variant 7