Volume intersection of two cubes

Given how bad my description below is, I recommend getting hold of a cube first here!

Consider a unit cube oriented with its space diagonal vertical (for example, if two opposite vertices are $A$ and $H$ , orient the cube so that $AH$ is vertical). The cube has rotational symmetry of order $3$ about this vertical axis (if you spin it through $120^\circ$ , it still looks the same).

The second cube in this question has been rotated through half this angle. This transformation is equivalent to a reflection in the horizontal plane that bisects $AH$ .

The section this horizontal plane makes through the cube is a regular hexagon (with sides $\frac{\sqrt2}{2}$ ) (it intersects every edge of the cube except those connected to $A$ or $H$ ).

The solid formed by the intersection of the two cubes is therefore a hexagonal bipyramid (ie two hexagonal pyramids stuck together at their bases); I've attempted an illustration using GeoGebra below:

The hexagonal section of a cube is reasonably familiar, so to make computation easier we can realign the cube with coordinate axes; if $A$ has coordinates $(0,0,0)$ and $H$ coordinates $(1,1,1)$ , then the remaining vertices of the solid are the six permutations of the triple $\left(0,\frac12,1 \right)$ . The intersecting plane is $x+y+z=\frac32$ .

We've carved off six tetrahedra, each with volume $\frac{1}{24}$ , to leave a volume of $\frac34$ , giving the answer $\boxed7$ .

Let the two cubes have their common main diagonal between $(0, 0, 0)$ and $(1, 1, 1)$ . We'll keep cube 1 fixed, and rotate cube 2 by $60^{\circ}$ clockwise about the axis $\mathbf{a} = (1, 1, 1)$ , the rotation matrix corresponding to this rotation is

$R =\large \begin{bmatrix} \frac{2}{3} && \frac{2}{3} && -\frac{1}{3} \\ -\frac{1}{3} && \frac{2}{3} && \frac{2}{3} \\ \frac{2}{3} && -\frac{1}{3} && \frac{2}{3} \end{bmatrix}$

The corner of the second cube will become symmetrically on top of the top plane of the first cube. There are three planes to consider:

$z = 1$ (this is plane $CDHG$ )

$x = 1$ (plane $ABCD$ )

and

$y = 0$ (plane $ADHE$ )

We want to determine the equation of the image of the plane $z = 1$ under the given rotation. Let $\mathbf{p } =(x, y, z)$ be a point on this plane, then its image is $\mathbf{p'} = R \mathbf{p} = (x', y', z')$ (because the rotation axis passes through the origin). Hence, $\mathbf{p} = R^T \mathbf{p'}$ . Now the first equation says that $z = 1$ , therefore, the equation governing $p'$ is

$-\frac{1}{3} x' + \frac{2}{3} y' + \frac{2}{3} z' = 1$

And this is plane $CD' H' G'$

Similarly for $x = 1$ , the equation of its image is

$\frac{2}{3} x' - \frac{1}{3} y' + \frac{2}{3} z' = 1$ (this is plane $A' B' C D'$ )

And finally for the plane $y = 0$ , its image is the plane,

$\frac{2}{3} x' + \frac{2}{3} y' - \frac{1}{3} z' = 0$

(this is plane $A' D' H' E$ )

We can now intersect these three planes with the plane $z' = 1$ , and this gives the base of tetrahedron $IJCD'$ . It can be shown (details omitted) that the coordinates of the vertices of this tetrahedron are: $I = (0.5 , 0, 1) , J = (0, 0.5, 1), C = (1, 1, 1), D' = (\frac{1}{3}, \frac{1}{3} , \frac{4}{3} )$ . Using these coordinates one can easily determine the volume of the tetrahedron as being equal to $\frac{1}{24}$ . Now, there are six of these tetrahedrons on the six faces of first cube, thus the total volume that has to be deducted from the second cube volume is $6 \cdot \frac{1}{24} = \frac{1}{4}$ . Therefore, the common volume is $1 - \frac{1}{4} = \frac{3}{4}$ , and the answer is $3 + 4 = \boxed{7 }$ .