Reflect the vertices of a unit cube onto one of its body diagonals as an axis. What volume is filled by the original and the image points of the cube together?
Consider a cube, and let's denote the vertices of one of its horizontal faces as $A$, $B$, $C$, and $D$, and the vertices above them as $E$, $F$, $G$, and $H$, respectively, with the center of the cube being $O$.

![](https://cdn.mathpix.com/cropped/2024_05_02_121de161e8c54c5e6d25g-1.jpg?height=781&width=854&top_left_y=257&top_left_x=625)

Let's paint red all the edges starting from $A$ and $G$, and paint the rest of the edges blue. Any blue edge's midpoint, together with points $A$ and $G$, forms an isosceles triangle, since the faces of the cube are congruent. Therefore, these midpoints lie in the plane $S$ that is perpendicular to $AG$ and passes through $O$, and they are equidistant from $O$. We call two of these midpoints "adjacent" if they lie on the same face of the cube, so each midpoint has two adjacent midpoints, and they are equidistant from each other. The midpoints of the blue edges thus form a regular hexagon in the plane perpendicular to $AG$ and passing through $O$. Let's paint this hexagon blue as well.

Consider the two pyramids that share the blue hexagon as their base, with their apexes at $A$ and $G$. The line $AG$ is the axis of symmetry for both pyramids, and reflecting them over this line maps them onto themselves. Let's paint these pyramids blue, and the rest of the cube red. This way, we have painted six congruent, triangular-based pyramids red. Consider, for example, the one connected to the edge $AB$. Its other vertices are the midpoints $P$ and $Q$ of the edges $BC$ and $BF$. If we reflect over the line $AG$, $A$ remains in place, $P$ and $Q$ reflect within $S$ to $O$, and to the opposite vertices of the blue hexagon. These are the midpoints $R$ and $T$ of the edges $EH$ and $DH$. Since reflection in space over a line can be replaced by a $180^\circ$ rotation about that line, and a $180^\circ$ rotation about $AG$ maps the blue pyramids onto themselves, the red tetrahedron attached to the $APQ$ face from the outside is mapped to the tetrahedron attached to the $ATR$ face from the outside. The $ATR$ face, however, lies in the plane of the cube's $ADHE$ face, meaning that the reflection of the cube's red edges over $AG$ does not intersect with either the red or blue parts. This implies that the combined volume of the original and reflected cubes is obtained by adding the volume of the red parts to the volume of the original cube.

![](https://cdn.mathpix.com/cropped/2024_05_02_121de161e8c54c5e6d25g-2.jpg?height=975&width=933&top_left_y=125&top_left_x=585)

The area of the $BPQ$ face of the tetrahedron $ABPQ$ is $1/8$, and the height from $A$ to $BPQ$ is 1, so the volume of the tetrahedron is $1/24$. Therefore, the combined volume of the six small tetrahedrons is $1/4$, and the combined volume of the two cubes is $5/4$.