18. Two congruent cubes have a common diagonal, and one is obtained from the other by rotating around this diagonal by an angle of $60^{\circ}$. Find the volume of the intersection of these cubes if the edge length of the cube is $a$.
18. Let a cube $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ be given. It is easy to see that the midpoints of the edges $C D, D D^{\prime}, D^{\prime} A^{\prime}, A^{\prime} B^{\prime}, B^{\prime} B$ and $B C$ are at equal distances from the points $A$ and $C^{\prime}$--the endpoints of the cube's diagonal, so they all lie in the plane of symmetry of points $A$ and $C^{\prime}$, forming the vertices of a regular hexagon (prove this!). When the cube is rotated by $60^{\circ}$ about the diagonal $A C^{\prime}$, this hexagon will coincide with itself and will be contained in the intersection of the cubes (the original and the rotated one). The intersection of the cubes will be the union of two regular hexagonal pyramids with a common base--the considered hexagon. By calculating its area and the length of the cube's diagonal, we can find the desired volume: $V=\frac{3 a^{3}}{4}$.