## Task 1 - 160811

A plane cut is to be made through a cube $A B C D E F G H$ (see figure) such that the resulting section is an equilateral triangle, all of whose vertices are also vertices of the cube.

Give all possibilities for such a cut, and depict a cube with one such cut in cavalier perspective!

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0682.jpg?height=297&width=348&top_left_y=454&top_left_x=1391)
Assume that two vertices of a section figure are the endpoints of the same body diagonal. Then the third vertex would have to be at a distance $AG$ from each of these two endpoints, which is not possible.

Assume that two vertices of a section figure are the endpoints of the same edge of the cube. Then the third vertex would have to be at a distance $AB$ from each of these two endpoints, which is not true for any vertex of the cube.

Therefore, a section triangle that meets the requirements of the problem can only exist if it has three face diagonals of the cube as its three sides. Each such triangle is equilateral, since all face diagonals of the cube are of equal length, so each such section figure meets the conditions.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0682.jpg?height=377&width=439&top_left_y=1262&top_left_x=817)

The vertices of each such triangle are the endpoints of the three edges of the cube that emanate from the same vertex of the cube. By this property, each vertex is assigned exactly one triangle, and conversely, each such triangle is assigned exactly one vertex.

Thus, there are exactly 8 equilateral triangles of the required type, namely the triangles $B D E, A C F$, $B D G, A C H, A F H, B E G, C F H, D E G$.
![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0682.jpg?height=510&width=1402&top_left_y=1926&top_left_x=336)