Consider the planes that pass through the midpoints of the edges originating from each vertex of a cube with a side length of 2 units. What is the volume of the convex body bounded by these planes?
Solution. Let $K$ be a vertex of the given cube, and $A, B, C$ be the midpoints of the edges emanating from it (figure). Every plane passing through the midpoints of the three edges emanating from a vertex of a cube with an edge length of 2 units cuts off a regular triangular pyramid from the cube, whose lateral faces are pairwise perpendicular to each other. If these pyramids are reflected over one of their edges (e.g., the pyramid $ABCK$ over the edge $AB$), then the fourth vertex of the pyramid lands at the center of the face of the cube containing the edge ($K$'s image is $K'$), and due to the perpendicularity, the image of the side edge opposite the mirror line is precisely a 1 unit long perpendicular segment standing on the face at the center of the face ($KC$'s image is $K'C'$).

![](https://cdn.mathpix.com/cropped/2024_05_02_196934c277b6107d8707g-1.jpg?height=518&width=509&top_left_y=473&top_left_x=797)

Consider the perpendicular bisector planes of the cube's edges. These three planes (parallel edges belong to the same plane) divide the 2-unit edge cube into eight 1-unit edge cubes. Since the intersection lines of these planes in pairs are precisely the perpendicular lines standing on the faces at the centers of the faces, according to the previous, these planes also divide the convex body mentioned in the problem into 8 congruent parts.

One such part can be obtained by cutting off a pyramid (the $ABCK$) from a 1-unit edge cube and then attaching three pyramids congruent to the cut-off one (the reflections of the cut-off pyramid over the edges $AB$, $BC$, and $CA$). Since the three edges of the pyramid meeting at one vertex are 1 unit long and these edges are pairwise perpendicular to each other, the volume of the pyramid is $\frac{1}{6}$. Thus, the volume of one eighth part is $1 + (3-1) \cdot \frac{1}{6} = \frac{4}{3}$, so the volume of the resulting convex body (which is otherwise a regular octahedron) is $8 \cdot \frac{4}{3} = \frac{32}{3}$.