335. What is the greatest number of balls of radius 7 that can simultaneously touch, without intersecting, a ball of radius 3?

Transition to space.
335. We will prove that there cannot be more than six such spheres. Suppose there are seven. We connect the centers of all seven spheres with the center of the given sphere and denote by $O_{1}, O_{2}, \ldots, O_{7}$ the points of intersection of these segments with the surface of the given sphere. For each point $O_{i}$, we consider the set of points on the sphere for which the distance (along the surface of the sphere) to the point $O_{i}$ is no greater than the distance to any other point $O_{k}, k \neq i$. The sphere will be divided into seven spherical polygons. Each polygon is the intersection of six hemispheres containing the point $O_{i}$, the boundaries of which are great circles, along which the plane passing through the midpoint of the segment $O_{i} O_{k}$ and perpendicular to it intersects the sphere.

Each of the resulting polygons contains a circle inside it, the spherical radius of which is seen from the center of the original sphere at an angle $\alpha, \sin \alpha=0.7$.

Let $K$ and $N$ be the number of sides and vertices of the resulting partition, respectively. (Each side is a common side of two adjacent polygons and is counted once. The same applies to vertices.) It is easy to see that for such a partition, Euler's formula holds (see problem 324). In our case, this gives $K=N+5$. On the other hand, $K \geqslant \frac{3}{2} N$, since at least three sides emanate from each vertex, and each side is counted twice.

It is not difficult to obtain that $K \leqslant 15, N \leqslant 10$. In problem 325, we proved that among all spherical $n$-gons containing a given circle, the smallest area is that of a regular $n$-gon. Moreover, it can be shown that the sum of the areas of regular $n$- and $(n+2)$-gons is greater than twice the area of a regular $n$-gon. (We consider polygons circumscribed around the same circle.) It is also obvious that the area of a regular circumscribed $n$-gon decreases with increasing $n$. From this, it follows that the sum of the areas of the resulting seven polygons cannot be less than the sum of the areas of five regular quadrilaterals and two regular pentagons circumscribed around circles with a spherical radius corresponding to the central angle $\alpha=\arcsin 0.7$. The area of the corresponding regular pentagon will be

$$
s_{5}=9\left[10 \arccos \left(\cos \alpha \sin \frac{\pi}{5}\right)-3 \pi\right]
$$

and the area of the regular quadrilateral

$$
s_{1}=9\left[8 \arccos \left(\frac{\sqrt{2}}{2} \cos \alpha\right)-2 \pi\right]
$$

It is not difficult to prove that $2 s_{5}+5 s_{4}>36 \pi$. Thus, seven spheres of radius 7 cannot simultaneously, without intersecting, touch a sphere of radius 3. At the same time, it is easy to show that six spheres can be arranged in such a way.