What are the limits between which the area and perimeter of the rectangular cross-sections of a regular tetrahedron with a unit edge vary?
Solution. First, we show that if a plane intersects a regular tetrahedron in a quadrilateral, then the intersection is a rectangle if and only if the plane is parallel to two opposite edges of the tetrahedron. Let the intersecting plane be \( S \), and the vertices of the tetrahedron be \( A, B, C, D \). We choose the labeling such that \( S \) does not intersect the edge \( AB\) (this is possible because out of the six edges, \( S \) intersects only four). Consider the planes \( S, ABC \), and \( ABD \). Among the pairwise intersection lines of these three planes, one is the line \( AB \), and the other two are the opposite edges of the quadrilateral cut by \( S \) from the tetrahedron. If the plane section is a rectangle, then these two latter lines are parallel. However, we know (see, for example, Problem 1703 in the Collection of Geometric Problems I) that the pairwise intersection lines of three planes are either parallel or intersect at a point. In our case, therefore, the three intersection lines are parallel. If we consider the planes \( S, ACD \), and \( BCD \), then by the same reasoning, we get that the other two edges of the cut rectangle are parallel to the edge \( CD \). If the vertices of the cut quadrilateral are denoted as \( E, F, G \), and \( H \) as shown in the figure, then \( EH \parallel FG \parallel AB \) and \( EF \parallel GH \parallel CD \). Conversely, if the plane \( S \) is parallel to \( AB \) and \( CD \), then the opposite sides of the quadrilateral \( EFGH \) are parallel to \( AB \) and \( CD \), respectively, and since these two latter lines are perpendicular to each other, the quadrilateral \( EFGH \) is a rectangle in this case.

![](https://cdn.mathpix.com/cropped/2024_05_02_11082969ad7e59f2372eg-1.jpg?height=385&width=382&top_left_y=680&top_left_x=861)

Let's examine the perimeter and area of the rectangular sections. Let \( AG = x \). Then \( AH = GH = BE = BF = EF = x \), because the triangles \( AHG \) and \( BEF \) are equilateral, as \( GH \parallel CD \parallel EF \) makes them similar to the triangles \( ACD \) and \( BCD \), respectively. Similarly, we find that the triangles \( DGF \) and \( CHE \) are also equilateral, which implies \( 1 - x = DG = DF = FG = HE = EC = CH \).

Thus, the perimeter of the rectangle \( EFGH \) is:

\[
K = EF + FG + GH + HE = x + (1 - x) + x + (1 - x) = 2
\]

That is, the perimeter of the rectangular sections is constant, always 2 units. The area of the rectangle is:

\[
T = EF \cdot FG = x(1 - x) = x - x^2
\]

The graph of the function \( f(x) = -x^2 + x \) is a downward-opening parabola, which intersects the \( x \)-axis at the points \( (0, 0) \) and \( (1, 0) \), and its vertex is at the point \( \left(\frac{1}{2}, \frac{1}{4}\right) \). Therefore, the area of the rectangle can be any positive number not greater than \( \frac{1}{4} \). The area of the section is maximal when the vertices of the rectangle coincide with the midpoints of four edges of the tetrahedron (in this case, the section is a square). If the intersecting plane approaches one of the edges of the tetrahedron, then the area of the section approaches 0 (one edge of the rectangle approaches 1, while the other approaches 0).