\section*{Problem 6 - 021046}

A regular tetrahedron with edge length \(a\) is to be cut by a plane such that a square cross-section is formed.

a) Specify the position of the cutting plane!

b) Why is the cross-section a square?

c) Calculate the area of the square!
}

a) Let \(A, B, C, D\) be the vertices of the regular tetrahedron, and \(E\) be the midpoint of the edge \(A B = a\). One of the sought cutting planes, which cuts the tetrahedron such that a square cross-section is formed, is the plane \(\epsilon\), which passes through the point \(E\) and is parallel to the edges \(A D\) and \(B C\). b) The plane \(\epsilon\) described in a) intersects the edges \(A C, C D\), and \(B D\) in this order at the points \(F, G\), and \(H\).

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1195.jpg?height=428&width=394&top_left_y=837&top_left_x=840)

Then \(E F \| B C\) and \(G H \| B C\) and thus \(E F \| G H\), as well as \(F G \| A D\) and \(H E \| A D\) and thus \(F G \| H E\).

From this, it follows by the intercept theorem that

\[
E F = F G = G H = H E = B H = H D = C G = G D = C F = A F = \frac{a}{2}
\]

The cross-section \(E F G H\) is therefore a rhombus, and the points \(F, G, H\) are, in this order, the midpoints of the edges \(A C, C D, B D\).

Since the four faces of the tetrahedron \(A B C D\) are equilateral triangles and \(E\) is the midpoint of \(A B\), it follows by the Pythagorean theorem that:

\[
C E = D E = \frac{a}{2} \sqrt{3}
\]

Thus, the triangle \(\triangle C D E\) is isosceles. Since \(G\) is the midpoint of \(C D\), it is also the foot of the perpendicular from \(E\) to \(D C\). Therefore, \(E G\) is the height to the base length of the isosceles triangle \(\triangle C D E\). By the Pythagorean theorem, we have:

\[
E G^{2} = D E^{2} - D G^{2} = \frac{3 a^{2}}{4} - \frac{a^{2}}{4} = \frac{a^{2}}{2}
\]

Since the segments \(E H, H G\), and \(E G\) satisfy the relationship

\[
E G^{2} = E H^{2} + H G^{2} \quad \left(\frac{a^{2}}{2} = \frac{a^{2}}{4} + \frac{a^{2}}{4}\right)
\]

it follows by the converse of the Pythagorean theorem that the triangle \(\triangle E G H\) is a right triangle with the right angle at \(H\). Since \(\angle E H G = 90^{\circ}\), the rhombus \(E F G H\) is a square.

c) For the area \(A\) of the square \(E F G H\), we have:

\[
A = |E F| \cdot |E H| = \frac{a}{2} \cdot \frac{a}{2} = \frac{a^{2}}{4}
\]