\section*{Problem 2 - 071242}

Which of all the planes that contain the same body diagonal of a cube with edge length \(a\) cut out a figure of the smallest area from the cube? Calculate the area of such a figure!
}

Without loss of generality, we consider all planes that contain the body diagonal (see figure).

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1848.jpg?height=328&width=328&top_left_y=2023&top_left_x=864)

Let \(\epsilon\) be one of these planes. \(\epsilon\) contains, in addition to \(D\) and \(F\), at least one point \(X_{1}\) on a cube edge \(k_{1}\) different from \(D\) and \(F\).

If \(k_{1}\) emanates from \(F\), then \(\epsilon\) runs through this cube edge as well as through the edge \(k_{2}\) parallel to \(k_{1}\) that contains the point \(D\). Corresponding to the three edges emanating from \(F\), there are three such positions of the plane \(\epsilon\), and the resulting cross-sections \(F_{\epsilon}\) are congruent rectangles with the area

\[
\left|F_{\epsilon}\right|=a \cdot \sqrt{2} a=\sqrt{2} a^{2}
\]

If \(k_{1}\) does not emanate from \(D\) or \(F\), i.e., if \(k_{1}\) is one of the six edges \(A B, B C, C G, G H, H E\), or \(E A\), then there is a parallel edge \(k_{2}\) among these six edges. Since \(\epsilon\) intersects \(k_{1}\), the plane \(\epsilon\) also intersects \(k_{2}\) at a point \(X_{2}\), and \(X_{1} F X_{2} D\) forms a parallelogram with the diagonal \(D F\).

\(X_{1} F X_{2} D\) is the cross-section of \(\epsilon\) with the cube. By appropriately choosing \(X_{1}\) (and thus also \(X_{2}\)) on one of the other pairs of parallel edges, a cross-section congruent to \(X_{1} F X_{2} D\) is formed.

Since the diagonal \(D F\) divides the cross-section into two congruent triangles, the area of the cross-section is minimized exactly when the height \(l\) to the base \(D F\) of the corresponding triangles has the smallest length.

Let \(X_{1}\) be located on \(A B\). If the cube is projected parallel to \(F D\) onto a plane perpendicular to \(F D\), the segments \(l, A C, C H, H A, B G, G E\), and \(E B\) do not distort, and the images of the original points are marked with a prime, then \(A^{\prime}, B^{\prime}, C^{\prime}, G^{\prime}, H^{\prime}, E^{\prime}\) form the vertices of a regular hexagon with the center \(D^{\prime}\).

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1849.jpg?height=494&width=466&top_left_y=869&top_left_x=795)

Since \(l=X_{1}^{\prime} D^{\prime}\), \(l\) clearly has the smallest value when the point \(X_{1}^{\prime}\) bisects the segment \(A^{\prime} B^{\prime}\), i.e., when \(X_{1}^{\prime} D^{\prime}\) is the height in the equilateral triangle \(\triangle A^{\prime} B^{\prime} D^{\prime}\). Then \(|l|=a \frac{\sqrt{2}}{2}\) and for the area of the corresponding cross-section

\[
\left|F_{e}\right|=a \sqrt{3} \cdot a \frac{\sqrt{2}}{2}=\frac{\sqrt{6}}{2} a^{2}
\]

Since \(2 a^{2} > \frac{\sqrt{6}}{2} a^{2}\), this cross-section is indeed minimal.

Corresponding to the positions of \(X_{1}, X_{2}\) on one of the three pairs of parallel edges considered above, there are three positions of \(\epsilon\) in which this minimal cross-section is cut out.

\section*{Adapted from \([5]\)}