Kosukhin O.N.

Can a cube with an edge of 1 be cut into pieces by four planes so that for each of the pieces the distance between any two of its points is:

a) less than $4 / 5$;

b) less than $4 / 7$?

It is assumed that all planes are drawn simultaneously, and the cube and its parts do not move.
a) Let's choose three edges of the cube that have a common vertex. We will draw the first two planes perpendicular to the first edge so that it is divided into three equal parts by these planes. The third plane will be drawn perpendicular to the second edge through its midpoint. The fourth plane will be drawn perpendicular to the third edge through its midpoint. Then the cube will be divided into 12 identical rectangular parallelepipeds with sides $1 / 2, 1 / 2$, and $1 / 2$, as shown in the figure. The distance between any two points of the rectangular parallelepiped does not exceed its diagonal. The lengths of the diagonals of all 12 parts into which the cube is cut are $\sqrt{\frac{1}{3^{2}}+\frac{1}{2^{2}}+\frac{1}{2^{2}}}=\sqrt{\frac{11}{18}}4 / 7 \quad(13 \cdot 49=637>576=16 \cdot 36$, so, ${ }^{13} / 36>16 / 49$ ).

Suppose the cube is cut into parts by some four planes. It is easy to see that three planes can cut the cube into no more than 8 parts. In this case, the fourth plane can cut each of these parts into no more than two new parts. Therefore, four planes can cut the cube into no more than 16 parts. (In fact, there can be no more than 15 such parts.) This means that some two black vertices will fall into one of the parts. Therefore, with any cutting of the unit cube by four planes, there will be such a part and such two of its points that the distance between these points is greater than $4 / 7$.

## Answer

a) Yes; b) No.