Problem 972

I. solution. a) First, let's examine how many parts a pyramid's apex through 4 planes divides the space. In the solution to problem 650 (K. M. L. X. volume, p. 5), we showed that three planes intersecting at a single point divide the space into eight parts, while a fourth plane, which is in a general position relative to the previous ones - that is, it does not pass through the common point of the three planes and is not parallel to any of the previous planes or the intersection lines of any two previous planes - cuts through 7 of the 8 parts, resulting in a total of 15 parts, of which one is finite and the rest are infinite. If the fourth plane passes through the common point of the three planes, the finite part degenerates into a point, resulting in 14 parts. (Imagine, for example, on page 151, the plane of $ABC$ being replaced by a plane parallel to $ABC$ passing through point $D$.)

Finally, consider the plane of the base. This is intersected by the four side planes in two pairs of parallel lines, which divide it into nine parts. Each plane part borders a newly formed part of space, so a total of $14+9=23$ parts of space are formed.

b) The octahedron is bounded by four pairs of parallel planes $(1,2 ; 3,4 ; 5,6 ; 7,8)$ (Figure 1).

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Figure 1

The first pair of planes divides the space into three parts; the first and second pairs together into nine parts; the first, second, and third pairs together into 27 parts. (These three pairs of planes form a finite part: a parallelepiped. Up to this point, the problem essentially matches the first part of problem 650.) Consider one plane of the fourth pair, the hatched triangular plane marked with 7 (Figure 2).

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Figure 2

This plane is intersected by the previous six planes in three parallel lines, as shown in the figure, for example, the line marked 1 is the intersection line of the plane marked 1 with the hatched plane marked 7, and so on. The resulting intersection lines - as can be seen from the figure - divide the plane into 16 parts, so this plane cuts through 16 parts of space, as does the parallel plane marked 8. Therefore, a total of $27+2 \cdot 16=59$ parts of space are formed.

Imre Csiszár (Petőfi g. III. o. t.)

II. solution. We follow the same procedure as in the second solution of the cited problem 650.

a) From the square pyramid, stepping out on the faces, edges, and apex, we reach $5+8+5$ new parts of space. However, we must notice that these parts of space, together with the original pyramid, do not yet completely fill the space; 4 parts of space are missing. We can directly reach these by stepping out from the quadrilateral solid angle formed at the apex of the pyramid through the faces of the solid angle. (When visualizing the parts of space, do not overlook the intersection lines of two opposite side planes, which pass through the apex and are parallel to one of the base edges.) Therefore, a total of $1+(5+8+5)+4=23$ parts of space are formed.

b) By extending the faces of the octahedron, a tetrahedron is built on each face. Therefore, when counting the parts of space, we must also take into account the new parts of space formed by stepping out from the tetrahedra. Stepping out from the octahedron on the faces, edges, and apex, we reach $8+12+6$ new parts of space. Stepping out from the faces of the tetrahedra, we do not reach new parts of space. (We either return to the octahedron or enter a part of space that can be reached by stepping out from an edge of the octahedron.) Stepping out from the base edges of the tetrahedra (which are common edges of the tetrahedra and the octahedron) also does not lead to new parts of space, but to a tetrahedron built on a face. Stepping out from the side edges of the tetrahedra, we reach parts of space that can also be reached by stepping out from the apex of another tetrahedron. Therefore, for each tetrahedron, we must take into account the four parts of space that can be reached by stepping out from the apex. The total number of parts of space is: $1+(8+12+6)+8 \cdot 4=59$, which fill the entire space.