Problem 6. (30 points) Ten balls of the same radius are arranged in the form of a triangular pyramid such that each ball touches at least three others. Find the radius of the ball inscribed in the center of the pyramid, touching six identical balls, if the radius of the sphere in which the pyramid of balls is inscribed is $5 \sqrt{2}+5$.

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# Solution.

When ten identical spheres are arranged in this way, the centers $A, B, C, D$ of four of them are located at the vertices of a regular tetrahedron, and the points of contact are located on the edges of this tetrahedron. Therefore, the edge of the tetrahedron is equal to four radii of these spheres, the radius of the outer sphere is greater than the radius of the sphere circumscribed around the tetrahedron by a quarter of the length of the tetrahedron's edge, and the radius of the inner sphere is smaller than the distance from the center of the tetrahedron to its face by the same amount. Consider the section of the tetrahedron by the plane $A B M$ (Fig. 3).

![](https://cdn.mathpix.com/cropped/2024_05_06_424f80418dd801e5b28ag-23.jpg?height=651&width=662&top_left_y=1907&top_left_x=366)

Fig. 2

![](https://cdn.mathpix.com/cropped/2024_05_06_424f80418dd801e5b28ag-23.jpg?height=691&width=759&top_left_y=1867&top_left_x=1114)

Fig. 3

Let the length of the edge of the tetrahedron be $a$, the radius of the sphere circumscribed around the pyramid of spheres be $R$, and the radius of the sphere inscribed in the center of the pyramid of spheres be $r$.

In triangle $\triangle A B M: A M = B M = \frac{a \sqrt{3}}{2} ; M E = M H = \frac{1}{3} A M = \frac{a \sqrt{3}}{6}$;

$A H = B E = \frac{2}{3} A M = \frac{a \sqrt{3}}{3}$

$A E = B H = \sqrt{A M^{2} - M E^{2}} = \frac{2 a}{\sqrt{6}} = \frac{a \sqrt{6}}{3}$.

From the similarity of triangles $\triangle A E M$ and $\triangle A H O$ we have $\frac{A O}{A M} = \frac{A H}{A E} = \frac{a \sqrt{3}}{a \sqrt{6}} = \frac{\sqrt{2}}{2}$;

$A O = B O = \frac{\sqrt{2}}{2} A M = \frac{a \sqrt{6}}{4}$.

In triangle $\triangle A B O: S_{A B O} = \frac{A H \cdot B O}{2} = \frac{A B \cdot F O}{2} \Rightarrow F O = \frac{A H \cdot B O}{A B} = \frac{a^{2} \sqrt{18}}{12 a} = \frac{a \sqrt{2}}{4}$.

Then $R = A O + A L = \frac{a \sqrt{6}}{4} + \frac{a}{4} = \frac{a(\sqrt{6} + 1)}{4} ;$ and $r = F O - F K = \frac{a \sqrt{2}}{4} - \frac{a}{4} = \frac{a(\sqrt{2} - 1)}{4}$.

$$
\frac{r}{R} = \frac{(\sqrt{2} - 1)}{(\sqrt{6} + 1)} = \frac{(\sqrt{6} - 1)(\sqrt{2} - 1)^{4}}{5} ; r = \frac{(\sqrt{6} - 1)(\sqrt{2} - 1)}{5} R = (\sqrt{6} - 1)
$$

Answer. $(\sqrt{6} - 1)$.

## Variant-31