Let's cut a regular tetrahedron with a plane that is parallel to two opposite edges. Determine the maximum area of the cross-section.

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Let $ABCD$ be the tetrahedron and let the intersecting plane be parallel to the edges $AC$ and $BD$.

Since the intersecting plane is parallel to the edge $BD$, its intersection with the plane $BCD$ is $EF$ and its intersection with the plane $ABD$ is $GH$, both parallel to $BD$, and thus

$$
EF \| GH
$$

Similarly, it can be shown that

$$
GE \| HF
$$

But

$$
GE \| AC \text{ and } AD \perp BD
$$

so

$$
GE \perp BD
$$

From (1), (2), and (3), it follows that the intersection figure is a rectangle.

Let one edge of the tetrahedron be $a$ and $EC = x$, then $EC = EF = x$, because the triangle $BCD$ is equilateral and thus the similar triangle $EFC$ is also equilateral.

Since the triangle $BEG$ is also equilateral,

$$
BE = EG = a - x
$$

Therefore, the area of the rectangle is:

$$
t = x(a - x)
$$

or

$$
t = -x^2 + ax
$$

$t$ is maximized when $x = \frac{a}{2}$. In this case, $t = \frac{a^2}{4}$. The largest area intersection is thus a square, and the intersecting plane passes through the midpoints of the edges.