Let $A B C D A_{1} B_{1} C_{1} D_{1}$ be a unit cube. The midpoint of edge $B C$ is $F$, and the center of the square $D C C_{1} D_{1}$ is $O$. The plane of triangle $A F O$ divides the cube into two solids. Determine the ratio of the volumes of these two solids.

![](https://cdn.mathpix.com/cropped/2024_05_02_666a4c6d5a7717152565g-1.jpg?height=292&width=325&top_left_y=266&top_left_x=889)
Solution. Let's find the points where the plane of triangle $A O F$ intersects the edges $C C_{1}$ and $D D_{1}$. Reflect the cube across the midpoint $O$ of the face. Due to the central symmetry of the cube, the point $C$ reflects to $D_{1}$, and the point $D$ reflects to $C_{1}$. The reflection of point $F$, denoted as $F^{\prime}$, will be the midpoint of the edge $D_{1} B^{\prime}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_6a933d831f9860146c8bg-1.jpg?height=547&width=596&top_left_y=301&top_left_x=754)

Let $P$ be the point where the plane $\mathcal{S}$ of triangle $A O F$ intersects the edge $C_{1} C$ of the cube, and $Q$ be the point where the plane $\mathcal{S}$ intersects the edge $D_{1} D$. Due to the reflection, $A^{\prime}$ and $F^{\prime}$ lie in the plane $\mathcal{S}$, and therefore the segments $F A^{\prime}$ and $A F^{\prime}$ also lie in the plane $\mathcal{S}$, along with the points $P$ and $Q$.

The right triangles $F C P$ and $F A_{1}^{\prime} A^{\prime}$ are similar. We have $F C = \frac{1}{2}$, $F A_{1}^{\prime} = \frac{3}{2}$, and $A_{1}^{\prime} A^{\prime} = 1$. From the ratio of corresponding sides:

$$
C P: 1 = \frac{1}{2}: \frac{3}{2}, \quad \text{and thus} \quad C P = \frac{1}{3}
$$

From the similar triangles $F^{\prime} D_{1} Q$ and $F^{\prime} A_{1} A$:

$$
Q D_{1}: A A_{1} = D_{1} F^{\prime}: A_{1} F^{\prime}, \quad \text{where} \quad A A_{1} = 1, \quad D_{1} F^{\prime} = \frac{1}{2}, \quad A_{1} F^{\prime} = \frac{3}{2}
$$

Substituting the corresponding values into the ratio:

$$
Q D_{1}: 1 = \frac{1}{2}: \frac{3}{2}, \quad \text{and thus} \quad Q D_{1} = \frac{1}{3}
$$

$P$ and $Q$ divide the parallel segments $C C_{1}$ and $D D_{1}$ in the same ratio, which means that $P$ and $Q$ are reflections of each other across the point $O$, and thus $P, O$, and $Q$ lie on the same line.

We claim that the plane $\mathcal{S}$ divides the cube into two solids, one of which is a frustum of a pyramid defined by the points $A, F, C, P, Q, D$. The planes of the right triangles $A Q D$ and $F P C$ are parallel (since they are from opposite faces of the cube). Both planes are perpendicular to the unit-length segment $D C$. These two triangles will be the base and top faces of the frustum.

From the similarity of the triangles, it follows that the lines $A F$, $D C$, and $Q P$ intersect at a single point. The volume of the frustum is:

$$
V = \frac{m}{3}(T + \sqrt{T t} + t)
$$

where $m$ is the height, and $T$ and $t$ are the areas of the base and top faces. Here,

$$
\begin{gathered}
m = 1, \quad T = \frac{1 \cdot \frac{2}{3}}{2} = \frac{1}{3}, \quad t = \frac{\frac{1}{2} \cdot \frac{1}{3}}{2} = \frac{1}{12} \quad \text{and} \\
V_{1} = \frac{1}{3}\left(\frac{1}{3} + \sqrt{\frac{1}{3} \cdot \frac{1}{12}} + \frac{1}{12}\right) = \frac{1}{3}\left(\frac{1}{3} + \frac{1}{6} + \frac{1}{12}\right) = \frac{7}{36}.
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_6a933d831f9860146c8bg-2.jpg?height=532&width=328&top_left_y=127&top_left_x=888)

The volume of the other solid is the difference between the volume of the cube and the volume of the frustum we just found:

$$
V_{2} = 1^{3} - \frac{7}{36} = \frac{29}{36}
$$

The ratio of the two volumes is: $\frac{7}{36}: \frac{29}{36} = 7: 29$.