\section*{Problem 6 - 131046}

A regular tetrahedron with vertices \(A, B, C\), and \(D\) and edge length \(a\) is intersected by six pairwise distinct planes, each of which contains exactly one edge of the tetrahedron and the midpoint of the opposite edge.

a) How many sub-bodies are created in total if all cuts are imagined to be performed simultaneously?

b) Calculate the volumes of the individual sub-bodies using the edge length \(a\).
}

Since \(A C=A D\), \(A\) lies on the perpendicular bisector plane \(\epsilon\) of \(C D\). Similarly, \(B\) lies on \(\epsilon\).

Therefore, \(\epsilon\) is the plane among the intersection planes that contains \(A B\). Correspondingly, the other intersection planes coincide with the perpendicular bisector planes of the tetrahedron edges.

As a regular tetrahedron, \(A B C D\) has a circumcenter \(S\) with \(S A=S B=S C=S D\), which therefore lies on each of the mentioned perpendicular bisector planes, i.e., on all intersection planes. Consequently, the plane through \(A, B, S\) is the intersection plane that contains \(A B\). Similarly, the other intersection planes coincide with the planes connecting one tetrahedron edge and \(S\).

Each of the faces adjacent to \(S\) of the tetrahedron \(S B C S\) thus lies in one of the intersection planes. The same applies to the tetrahedra \(A B D S, A C D S, B C D S\). The desired decomposition of \(A B C D\) can therefore be obtained by further decomposing the four tetrahedra \(A B C S, A B D S, A C D S, B C D S\) (into which \(A B C D\) can initially be decomposed).

For the further decomposition of \(A B C S\), only those cutting edges are relevant that (besides through \(S\)) pass through interior points of \(A B C S\). These are exactly those that pass through interior points of the triangle \(A B C\), i.e., those that pass through \(D\), one of the vertices \(A, B, C\), and a midpoint of an edge.

They divide the area of the triangle \(A B C\) by its medians into 6 equal-area (even congruent) sub-areas.

Thus, the tetrahedron \(A B C S\) is divided into exactly 6 equal-volume (even congruent) sub-bodies.

The same applies to the tetrahedra \(A B D S, A C D S, B C D S\) which are congruent to \(A B C S\). Therefore, a total of 24 equal-volume tetrahedra are created.

Each of them thus has the volume \(V_{K}=\frac{1}{24} V_{T}\), where \(V_{T}\) is the volume of the tetrahedron \(A B C D\). Since \(V_{T}=\frac{a^{3}}{12} \sqrt{2}\), it follows that

\[
V_{K}=\frac{a^{3}}{288} \sqrt{2}
\]

Adapted from [5]

\subsection*{7.16 XIV. Olympiad 1974}

\subsection*{7.16.1 First Round 1974, Grade 10}