Let's rotate a unit cube around one of its body diagonals by $60^{\circ}$. Calculate the volume of the common part of the original and the rotated cube! (H)
Let's use the notation from the figure. Let $AG$ be the body diagonal around which we rotate, and let its midpoint be $O$. The points $P$, $Q$, and $R$ are the midpoints of the edges $DC$, $CB$, and $BF$, respectively. The point $O$ is the center of symmetry of the cube, and in this symmetry, the image of point $P$ is $P'$, the midpoint of edge $EF$. Similarly, we can derive the points $Q'$ and $R'$. We will show that during a $+60^{\circ}$ rotation around $AG$, the image of $P$ is $Q$, the image of $Q$ is $R$, the image of $R$ is $P'$, and so on. First, we will show that $PO \perp AG$. For this, we calculate the lengths of segments $PO$, $OA$, and $AP$: $PO = \frac{1}{2} PP' = \frac{1}{2} ED = \frac{\sqrt{2}}{2}$, $OA = \frac{\sqrt{3}}{2}$, $AP = \sqrt{1 + \frac{1}{4}} = \sqrt{\frac{5}{4}}$. Since $PO^2 + OA^2 = AP^2$, by the converse of the Pythagorean theorem, $\angle POA = 90^{\circ}$.

Similarly, it can be shown that $QO$ and $RO$ are also perpendicular to $AG$. We have already seen that $PO = \frac{\sqrt{2}}{2}$, and similarly, $QO = PQ = \frac{\sqrt{2}}{2}$, so the triangle $PQO$ is equilateral. Now it follows that the image of $P$ during a $+60^{\circ}$ rotation around $AG$ is $Q$. Similarly, we find that the image of $Q$ is $R$, the image of $R$ is $P'$, and so on.

Next, let's consider the image of the face $ABCD$ after rotation. This face has three non-collinear points $A$, $P$, and $Q$, so the image of the plane of this face is determined by the images of these three points, which are $A$, $Q$, and $R$. The plane determined by points $A$, $Q$, and $R$ cuts a tetrahedron $AQRB$ from the cube. Similarly, it can be shown that the images of the other two faces meeting at vertex $A$, and the three faces meeting at vertex $G$, cut tetrahedra congruent to $AQRB$ from the cube. The volume of such a tetrahedron is $\frac{1}{3} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot 1 = \frac{1}{24}$, so the volume of the common part is $1 - 6 \cdot \frac{1}{24} = \frac{3}{4}$.

Balogh Attila (Szombathely, Nagy Lajos Gymnasium, 11th grade)

Note. The volume of the common part can also be calculated by determining the volume of the tetrahedron with base $PQR P' Q' R'$ and apex $A$ or $G$, and then taking this volume twice.

![](https://cdn.mathpix.com/cropped/2024_05_02_7c7816c1c6921d9e8428g-1.jpg?height=609&width=553&top_left_y=1208&top_left_x=775)