A tetrahedron has planes passing through each of its edges and the midpoint of the opposite edge. How many parts do these planes divide the tetrahedron into, and what is the volume of these parts?
Let the vertices of the tetrahedron be denoted by \(A, B, C, D\), and the centroid of the face opposite to \(A\) by \(S_A\), the centroid of the face opposite to \(B\) by \(S_B\), and so on. The segments \(A S_A, B S_B, C S_C, D S_D\) are the medians of the tetrahedron. It is known that these segments intersect at a point \(S\), which is the centroid of the tetrahedron, and \(A S: S S_A = 3: 1\) (a proof can be found in the collection of geometry problems, problem 1941). Each of the planes mentioned in the problem passes through the point \(S\), as each plane contains two medians of the tetrahedron. Let's denote, for example, the plane passing through the edge \(A C\) by \(A C S\). Consider the tetrahedron \(B C D S\). The faces of this tetrahedron that meet at the vertex \(S\) are in the planes \(B C S, C D S, B D S\), so only the planes \(A B S\), \(A C S\), and \(A D S\) intersect its interior. These planes contain one median of the triangle \(B C D\) each, and they divide the tetrahedron \(B C D S\) and its base \(B C D\) into six parts. It is easy to see that the medians of a triangle divide it into six equal-area parts, so the mentioned planes divide the tetrahedron \(B C D S\) into six equal-volume parts. The volume of the tetrahedron \(B C D S\) is one-fourth of the volume of \(A B C D\), since \(S\) is four times closer to the face \(B C D\) than \(A\).

The same can be said about the tetrahedrons \(A B C S\), \(A B D S\), and \(A C D S\). Therefore, the planes mentioned in the problem divide the tetrahedron into 24 equal-volume parts.

Jáger Márta (Bp., Veres Pálné Gimn., IV. o.t.) and Terék Zsolt (Fazekas M. Főv. Gyak. Gimn., III. o.t.)

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