Given a circle with radius $r$ and a segment $AB$ parallel to the plane of the circle, at a distance $r$ from the plane. The orthogonal projection of the segment $AB$ onto the plane of the circle is a diameter of the circle. From every interior point of the segment $AB$, we draw two half-lines perpendicular to $AB$, each of which intersects the circle. Determine the volume of the solid bounded by the circle and the half-lines.

The problem involves a circle with radius $r$ and a segment $AB$ that is parallel to the plane of the circle and is at a distance $r$ from the plane. The orthogonal projection of the segment $AB$ onto the plane of the circle is a diameter of the circle. From every interior point of the segment $AB$, we draw two half-lines perpendicular to $AB$, each of which intersects the circle. We need to determine the volume of the solid bounded by the circle and the half-lines.
I. solution. Let the given circle be $k$, and the projection of the segment $AB$ onto the plane of $k$ be $A'B'$. Denote the sought solid by $U$, and $V$ by the half-cylinder of height $r$ whose base is obtained by cutting the circle $k$ with the segment $A'B'$, and then translating this half-circle in the plane of $k$ perpendicularly to $A'B'$ to a distance greater than $r$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_b9853616a2374cfe3ec9g-1.jpg?height=585&width=1119&top_left_y=312&top_left_x=492)

Figure 1

Cut the solids $U$ and $V$ with a plane $S$ perpendicular to $AB$. Let $S$ intersect $AB$ at $C$, $A'B'$ at $C'$, the circumference of the circle at $D$ and $E$. $V$ is intersected by the rectangle $C^*C^{**}D^*F$, whose first three vertices are obtained by translating the points $C$, $C'$, and $D$ perpendicularly to $A'B'$ in the plane of $k$. Thus, the area of the triangle $DEC$ of the solid $U$ is equal to the area of the rectangle, regardless of the position of the point $C$ which determines the position of the plane $S$. From this, by Cavalieri's principle, it follows that the volumes of the two solids are equal. The volume of the half-cylinder is $r^3 \pi / 2$, which is also the volume of the solid $U$.

II. solution. Keep the previous notations, and cut the solid with a plane $S_1$ parallel to the plane of $k$, and denote the distance from $AB$ by $z$ (where $0 \leq z \leq r$), and the intersection points with $CD$ and $CE$ by $F$ and $G$, respectively. From the similar triangles $CFG$ and $CDE$,

$$
FG : DE = z : r = \lambda,
$$

independent of the choice of $C$, meaning that the plane section of the solid $U$ in $S_1$ is such that the ratio of each chord cut by any $S$ from this and from $k$ is constant over the entire plane section. This means that the plane section in $S_1$ is the affine image of the base circle with ratio $\lambda$ (on the axis $A'B'$). The area of the plane section is given by the product of the area of the circle and the ratio of the affinity, i.e., $r^2 \pi \lambda = r z \pi$.

The volume of the solid $U$ can be obtained by integrating the area of the plane sections from 0 to $r$:

$$
\int_{0}^{r} r z \pi \, \mathrm{d} z = r \pi \left[ \frac{z^2}{2} \right]_{0}^{r} = \frac{r^3 \pi}{2}
$$

László Horváth (Hódmezővásárhely, Bethlen G. Gymnasium)

Remarks. 1. The Cavalieri principle used can be stated intuitively as follows: if two solids can be placed in space such that when cut by planes parallel to a suitably positioned plane, the areas of the sections in any plane are equal, then the volumes of the two solids are equal. (The mentioned placement merely facilitates the correspondence of the sections.) A beautiful example of such a pair of solids is a hemisphere of radius $r$ and a cylinder of radius $r$ and height $r$ from which a cone has been cut out, whose base circle is the top face of the cylinder, and whose vertex is the center of the bottom face.
![](https://cdn.mathpix.com/cropped/2024_05_02_b9853616a2374cfe3ec9g-1.jpg?height=464&width=1178&top_left_y=2179&top_left_x=454)

Figure 2
3. The above solutions calculated the volume of the solid under the assumption that the sought solid has a volume. Determining or proving that the solid indeed has a volume was not the task of the contestants. The wording of the problem also suggested that the volume exists, and only the value of the volume needed to be determined. However, we would like to emphasize that if we use one of the above methods to determine the volume of a solid, we do not claim that the volume of the solid is this and this, but only that if the solid has a volume, then it can only be this much.

Most of our readers were probably surprised to read these lines, as they have not heard that a solid might not have a volume. This, of course, depends on what we mean by a solid and what we mean by volume. School education does not address this. In this regard, we provide an example of a solid that does not have a volume.

Consider a unit cube, and among the plane sections parallel to the base $ABCD$, translate those whose distance from the base is a rational number by the vector $AB$. The resulting so-called "riddled cube" has every plane section parallel to $ABCD$ as a unit square, so each has an area of one unit, and the solid does not have a volume.