\section*{Problem 5 - 041045}

The length of the edges of a regular tetrahedron is \(a\). A plane \(\epsilon\) is placed through the midpoint of one edge such that it does not contain this edge and is parallel to two non-intersecting edges.

Calculate the area of the cross-section formed by this plane with the tetrahedron!
}

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1217.jpg?height=434&width=417&top_left_y=865&top_left_x=317)

Let \(A, B, C, D\) be the vertices of the tetrahedron, and \(E\) be the midpoint of \(AB\) (see figure). Then \(\epsilon\) can, without loss of generality, be assumed to contain \(E\) and be parallel to \(AD\) and \(BC\). Therefore, \(\epsilon\) does not intersect \(AD\) or \(BC\). Consequently, \(A\) and \(D\) as well as \(B\) and \(C\) lie on the same side of \(\epsilon\), while \(A\) and \(B\) lie on different sides of \(\epsilon\). Thus, \(\epsilon\) intersects the edges \(AC\), \(CD\), and \(DB\). Let the intersection points be denoted in this order by \(F\), \(G\), and \(H\).

Then it follows that

\[
E F \| B C \text { and } \quad G H \| B C \quad \text { as well as } F G \| A D \text { and } \quad H E \| A D
\]

From this, it follows by the second ray theorem that

\[
|E F|=|F G|=|G H|=|H E|=\frac{a}{2}
\]

Thus, the intersection figure is a rhombus. Since, due to the regularity of the tetrahedron, the side faces are congruent equilateral triangles, their heights are also congruent; hence,

\[
|D E|=|C E|=|D F|=|B F|
\]

Since \(|C D|=|B D|\) also holds, it follows by the congruence theorem (sss) that \(\triangle C D E \cong \triangle B D F\) and thus \(\angle D C E \cong \angle D B F\).

Because \(|C G|=|B H|=\frac{1}{2}|C D|\), it follows by the congruence theorem (sws) that \(\triangle C G E \cong \triangle B H F\) and thus \(|G E| = |H F|\). The midpoints of \(G E\) and \(H F\) coincide at the intersection of the diagonals in the rhombus, so \(G E\) and \(H F\) are diameters of the same circle.

Thus, \(EFGH\) is a rhombus and a cyclic quadrilateral, making it a square with the area

\[
I=|E F|^{2}=\frac{a^{2}}{4}
\]

Adapted from \([5]\)
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