A cube is divided by planes that are perpendicular to each of its body diagonals and pass through their midpoints. How many parts does these 4 planes divide the surface of the cube into? - How many parts are created if perpendicular planes are placed at both one-third points of each body diagonal? (8 planes).
I. Let the two parallel faces of the cube be $ABCD$ and $EFGH$ such that $AE$ and $BF$ are parallel edges. First, consider the planes perpendicular to the space diagonal $AG$ at its midpoint and trisection points. We claim that the latter two planes are the planes $S_1$ and $S_2$ passing through the points $B, D, E$ and $C, F, H$, respectively, i.e., the planes through the endpoints of the edges starting from $A$ and $G$.

![](https://cdn.mathpix.com/cropped/2024_05_02_29cad6fbb9905701a3e6g-1.jpg?height=716&width=631&top_left_y=344&top_left_x=736)

This follows from the fact that, on the one hand, the altitudes from the vertices $B, D$, and $E$ in the congruent triangles $ABG$, $ADG$, and $AEG$ intersect $AG$ at the same point, so $S_1$ is the plane perpendicular to $AG$ at this point, and similarly, $S_2$ is also perpendicular to $AG$. On the other hand, by erecting planes $S_a$ and $S_g$ perpendicular to $AG$ at $A$ and $G$, respectively, the adjacent pairs of faces intersect in parallel (and equal) cube edges: $AB$ between $S_a$ and $S_1$, $DC$ between $S_1$ and $S_2$, and $HG$ between $S_2$ and $S_g$. These pairs of planes are therefore equidistant from each other, and thus divide $AG$ into three equal parts. $S_1$ and $S_2$ are indeed the planes perpendicular to $AG$ at its trisection points, uniquely determined.

![](https://cdn.mathpix.com/cropped/2024_05_02_29cad6fbb9905701a3e6g-1.jpg?height=664&width=616&top_left_y=1397&top_left_x=749)

The plane $S_0$ perpendicular to $AG$ at its midpoint is located between $S_1$ and $S_2$, parallel to them and at an equal distance from each, thus bisecting the edges

$$
BF, \quad FE, \quad EH, \quad HD, \quad DC, \quad CB
$$

of the cube, i.e., all the edges that do not have $A$ or $G$ as endpoints.

II. First, consider the second part of the problem. $S_1$ and $S_2$ intersect the cube's surface in the diagonals of the three faces meeting at $A$ and $G$, excluding $A$ and $G$. Considering the pairs of planes perpendicular to the space diagonals $BH$, $CE$, and $DF$ and trisecting them, every vertex of the cube appears at $A$ and $G$, and every face diagonal also appears as a dividing line, and no other dividing line appears. The face diagonals divide the cube's 6 faces into 24 isosceles right triangles. Two such triangles share a hypotenuse, which is one edge of the cube that is not a dividing line, forming a connected part, so the 4 pairs of planes divide the cube's surface into 12 parts.

In the first part of the problem, $S_0$ is involved, which cuts an isosceles right triangle from each face of the cube, with the right angle at the vertex opposite to $A$ or $G$, and the hypotenuse connecting the midpoints of the sides leading to it, forming the intersection line of $S_0$. Considering the planes perpendicular to the other 3 space diagonals and bisecting them, every vertex of the cube appears at $A$ and $G$, and the right angle of the small triangle also appears, and indeed 3 times, and every midpoint of an edge also appears as the endpoint of a hypotenuse, 2 times. Thus, each face of the cube is divided into 4 small triangles and 1 square. The perimeter of the latter is a dividing line, so they are separate parts of the surface. The small triangles, however, meet in 3's at each vertex of the cube, connecting to each other in pairs with their legs, which are not dividing lines, so 3 small triangles form a connected part of the cube's surface, whose boundary is given by the 3 hypotenuses. According to this, the planes perpendicular to the space diagonals and bisecting them divide the cube's surface into $6+8=14$ parts.

Csaba Sükösd (Budapest, József A. g. III. o. t.) Zoltán Csörnyei (Veszprém, Lovassy L. g. IV. o. t.)