In the parallelogram $A B C D$, every vertex is located on the same side of the plane $S$, such that the distances from $A, B$, and $C$ to $S$ are 4, 6, and $8 \mathrm{~cm}$, respectively. The area of the orthogonal projection of the parallelogram onto the plane $S$, which is the quadrilateral $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, is $10 \mathrm{~cm}^{2}$. What is the volume of the solid $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$?
# Reflect the

$$
A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}
$$

solid over the center of the $A B C D$ parallelogram. During the reflection, points $A$ and $C$, as well as points $B$ and $D$, swap places, and the image of the $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ quadrilateral is the $A^{\prime \prime} B^{\prime \prime} C^{\prime \prime} D^{\prime \prime}$ quadrilateral located in the plane $S^{\prime \prime}$ parallel to $S$.

![](https://cdn.mathpix.com/cropped/2024_05_02_a0b28079fd1b140f845eg-1.jpg?height=677&width=697&top_left_y=369&top_left_x=714)

We will show that the

$$
A^{\prime} B^{\prime} C^{\prime} D^{\prime} A^{\prime \prime} B^{\prime \prime} C^{\prime \prime} D^{\prime \prime}
$$

solid is a right prism with a quadrilateral base. The image of the segment $A A^{\prime}$ under the reflection is the segment $C A^{\prime \prime}$, so $A^{\prime \prime} C$ is parallel to $A A^{\prime}$, and thus perpendicular to $S$. But $C C^{\prime}$ is also perpendicular to $S$, so the points $A^{\prime \prime}, C$, and $C^{\prime}$ lie on the same line. Similarly, we can show that the point triples $A^{\prime}, A, C^{\prime \prime}; B^{\prime}, B, D^{\prime \prime}$, and $D^{\prime}, D, B^{\prime \prime}$ also lie on the same line. Therefore, using the parallelism of the planes $S$ and $S^{\prime \prime}$, $A^{\prime} B^{\prime} C^{\prime} D^{\prime} A^{\prime \prime} B^{\prime \prime} C^{\prime \prime} D^{\prime \prime}$ is a right prism. The length of the side edges is $C^{\prime} A^{\prime \prime} = C^{\prime} C + C A^{\prime \prime} = C^{\prime} C + A^{\prime} A = 8 + 4 = 12 \text{ cm}$, so its volume is $V = t_{A^{\prime} B^{\prime} C^{\prime} D^{\prime}} \cdot C^{\prime} A^{\prime \prime} = 10 \cdot 12 = 120 \text{ cm}^3$. The volume of the original $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ solid is half of the volume of the solid, $60 \text{ cm}^3$.

Ádám Reiff (Szolnok, Verseghy F. Gymnasium, 2nd grade)

Note. The volume of the solid does not depend on the distance of points $B$ and $D$ from $S$. Many determined the distance of $D$ from $S$ (6 cm), but this is not necessary for solving the problem.