Let's determine how many regions the space is divided into by a) the six planes of the cube's faces?

b) the four planes of the tetrahedron's faces?
# I. solution:

a) First, we consider the planes of two parallel faces of the cube. These two planes divide the space into 3 parts. A plane perpendicular to these and belonging to one of the other faces divides each of these space parts in half, so the three planes delimit 6 space parts. The plane of the face parallel to the previous one divides three of these 6 space parts in half, so the four planes delimit 9 space parts. The two planes of the third pair of faces divide each of these space parts into 3 parts, so in the end, 27 space parts are formed, of which only the one that the cube fills is finite.

![](https://cdn.mathpix.com/cropped/2024_05_02_03006199ff90eba78debg-1.jpg?height=780&width=665&top_left_y=350&top_left_x=730)

b) The plane of one face of the tetrahedron $ABCD$ (e.g., the unmarked plane $ACD$) divides the space into 2 parts. Another face plane (e.g., the densely dashed $ABD$) divides each of the previous 2 space parts in half, so the two intersecting planes delimit 4 space parts. The third plane (the sparsely dashed $BCD$) divides each of the previously formed space parts in half, as it is not parallel to either of the first two planes and does not pass through their intersection line. Thus, the three planes delimit 8 space parts, as shown in the figure. The fourth plane $(ABC)$, which we have not separately visualized, divides each of the previously formed space parts into two, except for the one labeled 8, which is bounded by the tetrahedron's vertex trihedron at vertex $D$. Thus, a total of $8+7=15$ space parts are formed, of which only the one that the tetrahedron fills is finite.

Anna Mestyán (Debrecen, Svetits lg. IV. o. t.)

II. solution: If we connect a point inside the cube to one of its vertices, then this point, intersecting the 3 planes meeting at the vertex (i.e., passing through the mentioned vertex), ends up in a space part delimited by these 3 planes, and remains in this space part because it does not intersect any new plane; thus, a space part is connected to each vertex of the cube. If we draw a line from a point inside the cube perpendicular to one of its edges, then this line, after intersecting the edge, or rather the intersection of the two planes meeting at the edge, ends up in a space part whose two bounding planes are the planes of the two cube faces meeting at the edge. This space part is also infinite because the line does not intersect any new plane after intersecting the edge. Finally, if we draw a perpendicular from a point inside the cube to one of its faces, then this line, after intersecting a single face, ends up in an infinite space part. Considering that we end up in the space part connected to a vertex after intersecting 3 planes, while we end up in the space part connected to an edge after intersecting only 2 planes, these two space parts cannot be the same. They can have two bounding planes in common (if the vertex is exactly on the edge in question), but even then, they are separated by the plane that we did not intersect when we ended up in the space part connected to the edge. Similarly, it can be shown that the space part connected to a face is different from the previous two. Since the number of space parts connected to a vertex is equal to the number of vertices, the number of space parts connected to an edge is equal to the number of edges, and the number of space parts connected to a face is equal to the number of faces, the total number of space parts is: $8+12+6$ infinite +1 finite space part, or a total of 27 space parts. Our visualization shows that the counted space parts fill the entire space, so we have indeed counted all the space parts in the manner mentioned.

b) In the case of the tetrahedron, using a completely similar line of thought, we arrive at the number of space parts being $4+6+4+1=15$.

Károly Legéndy (Bp. VIII., Apáczai Csere g. IV. o. t.)

Note: After the counting method described in the second solution, it is not superfluous to check whether the mentioned space parts fill the entire space.

(See problem 690 in this issue).