3. In a cubic box of size $1 \times 1 \times 1$, there are 8 wooden balls. Can the sum of their radii be greater than 2?

(V. N. Dubrovsky, K. A. Krop)
## Answer: can.

Place two equal spheres of radius $R$ at the opposite corners of the bottom of the box, each touching the other and two walls of the box. (These spheres will be inscribed in the triangular prisms into which the box is divided by a vertical diagonal plane.) Consider the spheres symmetric to these two with respect to the axis passing through the centers of two opposite side walls. Clearly, they will be inscribed in the upper corners of the box, and all 4 spheres will touch each other; the centers of all 4 spheres form a tetrahedron.

Now place equal spheres of smaller radius $r$ in the two free lower corners of the box so that each touches two walls and two larger spheres: we can first roll a sphere of very small radius into the corner, and then start "inflating" it until it touches one (and therefore the other) large sphere. Place similar spheres in the last two free corners - under the lid.

Estimate the sum of the radii of the spheres. Consider the 4 lower spheres (Fig. 6). Divide them into two pairs of touching large and small spheres. Let $O_{1}$ and $O_{2}$ be their centers, $T_{1}$ and $T_{2}$ be the points of their contact with the parallel side walls. The length of the broken line $T_{1} O_{1} O_{2} T_{2}$ is $R+(R+r)+r=2(R+r)$. At the same time, it is greater than the edge of the cube, since its length is obviously equal to the minimum distance between points of parallel faces. Thus, $2(R+r)>1$, and the sum of all 8 radii $4(R+r)>2$.