Two unit-edged cubes have been placed such that the endpoints of one of their body diagonals coincide, and the two cubes are rotated $60^{\circ}$ relative to each other along this axis.

What volume do they occupy together, and what is the surface area of the resulting solid?
a) We show that the two edges of the two cubes, which do not run into the endpoints of the common axis, bisect each other in pairs. Based on this, we can easily decompose the jointly filled space and the surface of the shape into parts with calculable volumes and surfaces, respectively.

Let the base of one cube be $ABCD$, one of its side faces be $BCGF$, the midpoint of the $BF$ edge be $P$, and the midpoints of the $BC$ and $CD$ edges be $Q$ and $R$, respectively.
![](https://cdn.mathpix.com/cropped/2024_05_02_e02db587015624d077c5g-1.jpg?height=568&width=1154&top_left_y=350&top_left_x=476)

Each of these is equidistant from $A$ and $G$, because the distance is the hypotenuse of a right triangle whose legs are 1 and $1/2$. The same applies to the other 3 midpoints of the edges that do not run into $A$ or $G$; the length of the hypotenuses is $r = \sqrt{5}/2$. Thus, the 6 midpoints lie on the surface of the spheres with radius $r$ centered at $A$ and $G$, and on the intersection of these spheres, which is also in the plane perpendicular to and bisecting the body diagonal $AG = \sqrt{3}$. The center of this circle, $O$, bisects $AG$, and its radius, for example, from the right triangle $AQO$, is $\sqrt{2}/2$. The distance between any two consecutive midpoints, for example, $QR$ in the face $BCGF$, is the same, $\sqrt{2}/2$. Therefore, the 6 midpoints form the vertices of a regular hexagon inscribed in a circle, with $\angle POQ = \angle QOR = 60^\circ$.

If we now derive the second cube from the original by rotating it $60^\circ$ around $AG$, then $P$ moves to $Q$, and $Q$ moves to $R$, so the new position of $BF$ bisects $BC$, and the new position of $BC$ bisects $CD$, and so on, as we stated.

Cut off the right tetrahedra with edges $1/2, 1/2, 1$ around the vertices $B$ and $C$ from the original cube with the planes $APQ$, $GQR$, etc., and similarly cut off the "neighborhoods" of the other 4 vertices of the cube, excluding $A$ and $G$. The remaining solid is a regular hexagonal double pyramid (bipyramid), which is also inside the cube after the rotation, forming the common part of the interiors of the two cubes. The 6 tetrahedra that are cut off but still attached, however, move outside the original cube, for example, the tetrahedron $APQB$ moves into $AQRB'$, and $B'$ separates the cube from the triangle $AQR$ in the face $ABCD$.

b) According to this, the sought volume is the sum of the volume of the double pyramid and 12 small tetrahedra, or alternatively, the sum of the volume of the original cube and 6 small tetrahedra:

$$
V = 1^3 + 6 \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot 1 = \frac{5}{4} \text{ volume units. }
$$

The surface area is 12 times the sum of the areas of the 3 right faces of the small tetrahedra:

$$
F = 12 \left( \frac{1}{4} + \frac{1}{4} + \frac{1}{8} \right) = \frac{15}{2} \text{ area units. }
$$

Remark. It is interesting that both the volume and the surface area of our shape are $1/4$ larger than those of the original cube.

Gizella Mátrai (Pécs, Széchenyi I. Gymnasium, 3rd grade)