8. Given a unit cube (see figure), consider the plane containing the edges $A B$ and $C D$ and the plane containing the edges $A E$ and $F D$. These two planes cut the cube into four parts. What is the maximum among the volumes of these parts?
(A) $\frac{1}{4}$
(B) $\frac{1}{3}$
(C) $\frac{\sqrt{2}}{4}$
(D) $\frac{3}{8}$
(E) $\frac{1}{2}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_cf63dde6cb9f14f4f823g-03.jpg?height=399&width=417&top_left_y=106&top_left_x=1453)
8. The answer is (B). Let $G$ and $H$ be the other two vertices of the cube, with $A G$ being an edge of the cube. Fix a Cartesian reference frame, with the origin at $A$ and axes $x, y$, and $z$ along $A B, A E, A G$. The four regions are characterized by the conditions $\{x \leqslant z ; y \leqslant z\},\{x \leqslant z ; y \geqslant z\},\{x \geqslant z ; y \leqslant z\},\{x \geqslant z ; y \geqslant z\}$. Cutting the cube with a third plane passing through $A G$ and $H D$, the resulting regions are six and are characterized by the conditions $\{x \leqslant y \leqslant z\}$ and similar ones, obtained by permuting the variables.

This implies, among other things, that of the four initial regions, two are cut by the third plane (and these will be the ones with the maximum volume), while the other two are not.

The six resulting regions are all congruent, by symmetry, and therefore each has a volume of $\frac{1}{6}$. From this, it follows that of the four regions described in the question, two have a volume of $\frac{1}{3}$ and two have a volume of $\frac{1}{6}$.

## Second Solution

It is easy to verify that the pyramid with base $C D F G$ and vertex at $A$ is not cut by the planes in question. Furthermore, of the five faces of the pyramid, three are also faces of the cube (or parts of them), while the remaining two are coplanar with the cutting planes.

This means that the pyramid is one of the four regions into which the cube is divided. Similarly, the symmetric pyramid $D B A E H$ is another of these regions. Since each of these pyramids has a volume of $\frac{1}{3}$, the remaining two parts cannot have a larger volume, so the maximum of the volumes is $\frac{1}{3}$.