Let $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ be a cube. The midpoint of edge $B^{\prime} C^{\prime}$ is $E$, and the midpoint of edge $C^{\prime} D^{\prime}$ is $F$. The plane $A E F$ divides the cube into two parts. Determine the ratio of the volumes of the two parts.
Solution. Let the edge of the cube be of unit length. The intersection point of the plane $A E F$ and the line $A^{\prime} B^{\prime}$ is point $G$, and the intersection point of the line $A^{\prime} D^{\prime}$ and the plane is point $H$. The intersection points exist because the plane is not parallel to any of the lines. Furthermore, let the intersection points of the plane $A E F$ with the lines $B B^{\prime}$ and $D D^{\prime}$ be points $I$ and $J$, respectively. The volume of the part of the cube cut off by the plane $A E F$ and containing point $A^{\prime}$ can be obtained by subtracting the volumes of the tetrahedra $B^{\prime} G I E$ and $D^{\prime} H J F$ from the volume of the tetrahedron $A A^{\prime} G H$.

![](https://cdn.mathpix.com/cropped/2024_05_02_ae596584e54f792e7c79g-1.jpg?height=458&width=566&top_left_y=427&top_left_x=774)

$E C^{\prime} F$ is an isosceles right triangle, so $\angle H F D^{\prime} = \angle E F C^{\prime} = 45^{\circ}$. Therefore, the right triangle $H D^{\prime} F$ is also isosceles. Thus, $H D^{\prime} = D^{\prime} F = \frac{1}{2}$, and $H A^{\prime} = \frac{3}{2}$. Considering the angle $\angle A H A^{\prime}$ and using the theorem of parallel intercepts:

$$
J D^{\prime} = A A^{\prime} \cdot \frac{H D^{\prime}}{H A^{\prime}} = 1 \cdot \frac{1}{2} : \frac{3}{2} = \frac{1}{3}
$$

By similar reasoning, $\angle B^{\prime} G I E = \angle F E C^{\prime} = 45^{\circ}$, and $B^{\prime} G = B^{\prime} E = \frac{1}{2}$, and $B^{\prime} I = \frac{1}{3}$.

The edges of the tetrahedron $A A^{\prime} G H$ meeting at point $A^{\prime}$ are pairwise perpendicular, so the volume of the tetrahedron is

$$
V_{A A^{\prime} G H} = \frac{3}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \cdot \frac{1}{3} = \frac{3}{8} \text{ volume units.}
$$

The edges of the tetrahedron $B^{\prime} G I E$ meeting at point $B^{\prime}$ are also pairwise perpendicular. Given their lengths, the volume is:

$$
V_{B^{\prime} G I E} = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{3} = \frac{1}{72}
$$

The volume of the congruent tetrahedron $D^{\prime} H J F$ is also $\frac{1}{72}$ volume units.

Based on the above, the volume of the part of the cube cut off by the plane $A E F$ and containing point $A^{\prime}$ is:

$$
V_{A A^{\prime} G H} - V_{B^{\prime} G I E} - V_{D^{\prime} H J F} = \frac{3}{8} - \frac{1}{72} - \frac{1}{72} = \frac{25}{72}
$$

The other part is accordingly $\frac{47}{72}$, and the ratio of the volumes of the two parts is $25: 47$.