6. On the table, there are three cones standing on their bases, touching each other. The radii of their bases are 23, 46, and 69. 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 radius of the smaller base of the truncated cone.

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# Answer: 6.

![](https://cdn.mathpix.com/cropped/2024_05_06_bd69057b1c245517edf6g-25.jpg?height=514&width=1304&top_left_y=1456&top_left_x=434)

Solution. Let $C$ be the center of the smaller base of the truncated cone, $R$ be the radius of this base, and $O_{1}, O_{2}, O_{3}$ 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 $R$. Let $r=23$ and note that

$$
O_{1} O_{2}=3 r, \quad O_{1} O_{3}=4 r, \quad O_{2} O_{3}=5 r
$$

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

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

the following equalities hold:

$$
(r+R)^{2}-x^{2}=y^{2}=(3 r+R)^{2}-(4 r-x)^{2} \Longleftrightarrow r^{2}+2 r R=6 r R+8 r x-7 r^{2} \Longleftrightarrow 2 x=2 r-R
$$

and also

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

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

$$
36(r+R)^{2}=(6 r-3 R)^{2}+(6 r-2 R)^{2} \Longleftrightarrow 23 R^{2}+132 r R-36 r^{2}=0
$$

This equation in terms of $R$ has a unique positive solution $R=\frac{6}{23} r=6$.

## Variant 8