Consider the perpendicular bisecting planes of the edges of a regular tetrahedron with unit edge length. How many parts do these planes divide the tetrahedron into, and what is the volume of each part?
We denote the vertices of the tetrahedron as \(A, B, C, D\), and its center as \(O\). The perpendicular bisector planes of the edges contain those points that are equidistant from the two endpoints of the chosen edge. Since \(O A = O B = O C = O D\), each perpendicular bisector plane passes through the point \(O\). Since all edges of our tetrahedron are equal, each perpendicular bisector plane also passes through the two endpoints of the opposite edge - for example, the perpendicular bisector plane of the edge \(A B\) passes through \(C\) and \(D\) - that is, it contains the opposite edge. Therefore, the perpendicular bisector planes of the edges coincide with those planes that pass through the center of the tetrahedron and contain one of its edges (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_0a95e70462b668bc5cb5g-1.jpg?height=602&width=718&top_left_y=439&top_left_x=690)

Figure 1

Due to the symmetry of the regular tetrahedron, it is sufficient to examine how these planes divide the pyramid \(A B C O\), because the same parts will be formed in the pyramids \(B C D O\), \(A B D O\), and \(A C D O\). Out of the six perpendicular bisector planes, three - \(A B O\), \(B C O\), and \(C A O\) - do not intersect the pyramid \(A B C O\), but coincide with one of its faces. The other three planes, however, do intersect the pyramid. These are precisely the perpendicular bisector planes of the edges \(A B\), \(B C\), and \(C A\). Each of these planes passes through \(O\), one vertex of the triangle \(A B C\), and the midpoint of the opposite edge. Therefore, these planes divide the triangle \(A B C\) into 6 congruent small triangles, as shown in Figure 2. Consequently, the planes divide the pyramid \(A B C O\) into 6 congruent pyramids. Each of these pyramids has its vertex at \(O\), and the opposite face is one of the aforementioned small triangles.

![](https://cdn.mathpix.com/cropped/2024_05_02_0a95e70462b668bc5cb5g-1.jpg?height=501&width=501&top_left_y=1506&top_left_x=796)

In this way, the tetrahedron \(A B C D\) is divided by the perpendicular bisector planes into \(4 \cdot 6 = 24\) congruent pyramids. Since the volume of a regular tetrahedron with unit edge length is \(\frac{\sqrt{2}}{12}\), the volume of each of the parts is: \(\frac{\sqrt{2}}{12} \cdot \frac{1}{24} = \frac{\sqrt{2}}{288}\).