, 2005

Solution:

The answer is (B). Let $G$ and $H$ be the other two vertices of the cube, with $AG$ an edge of the cube. Let us fix a Cartesian reference frame, with origin at $A$ and axes $x, y$, and $z$ along $AB, AE, AG$. The four regions are characterized by the conditions $\{x \leqslant z ; y \leqslant z\},\{x \leqslant z ; y \geqslant z\},\{x \geqslant z ; y \leqslant z\},\{x \geqslant z ; y \geqslant z\}$. Let us cut the cube with a third plane, passing through $AG$ and $HD$. The resulting regions are six and are characterized by the conditions $\{x \leqslant y \leqslant z\}$ and similar ones, obtained by permuting the unknowns.

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

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

Solution:

It is easy to verify that the pyramid with base $CDFG$ and vertex at $A$ is not cut by the planes in question. Moreover, 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 cut. Similarly, the symmetric pyramid $DBAEH$ is another of these regions. Since each of them has volume $\frac{1}{3}$, the remaining two parts cannot have a greater volume, so the maximum of the volumes is $\frac{1}{3}$.