A cube inscribed in a cube

Let $O(0, 0, 0)$ be the center of both cubes, so that one of the fixed points on the rotating cube is $A(\frac{1}{2}, \frac{1}{2}, \frac{1}{2})$ .

Let $N(n, n , n)$ be the center of the circle that $B(-\frac{1}{2}, \frac{1}{2}, \frac{1}{2})$ rotates around. Then $\overrightarrow{ON} = (n, n , n)$ and $\overrightarrow{BN} = (n + \frac{1}{2}, n - \frac{1}{2}, n - \frac{1}{2})$ , and since $\overrightarrow{ON} \perp \overrightarrow{BN}$ , $n(n + \frac{1}{2}) + n(n - \frac{1}{2}) + n(n - \frac{1}{2}) = 0$ , which simplifies to $n = \frac{1}{6}$ .

Let $C(a, b, c)$ be the locus of points of $B$ rotating around $N$ . Then $\overrightarrow{ON} = (\frac{1}{6}, \frac{1}{6} , \frac{1}{6})$ and $\overrightarrow{CN} = (\frac{1}{6} - a, \frac{1}{6} - b, \frac{1}{6} - c)$ , and since $\overrightarrow{ON} \perp \overrightarrow{CN}$ , $\frac{1}{6}(\frac{1}{6} - a) + \frac{1}{6}(\frac{1}{6} - b) + \frac{1}{6}(\frac{1}{6} - c) = 0$ , which simplifies to $c - \frac{1}{2} = a + b$ .

Also, $AC = 1$ , so by the distance equation $(a - \frac{1}{2})^2 + (b - \frac{1}{2})^2 + (c - \frac{1}{2})^2 = 1$ . Substituting $c - \frac{1}{2} = a + b$ gives $(a - \frac{1}{2})^2 + (b - \frac{1}{2})^2 + (a + b)^2 = 1$ , an equation of an ellipse, which rearranges to $b = \frac{1}{4}(-2a + 1 \pm \sqrt{-12a^2 + 4a + 5})$ and has maximum and minimum $a$ -values when the discriminant $-12a^2 + 4a + 5 = (2a + 1)(-6a + 5) = 0$ , so that the minimum is $a = -\frac{1}{2}$ and the maximum is $a = \frac{5}{6}$ .

$a = \frac{5}{6}$ also represents the maximum of half the side length of the larger cube, so the maximum full side length of the larger cube is $s = 2 \cdot \frac{5}{6} = \frac{5}{3}$ , which means $p = 5$ , $q = 3$ , and $p + q = \boxed{8}$ .

Let the side length of the smaller rotating cube be $2 a$ . Then for our unit cube, $a = \frac{1}{2}$ . The fixed vertices of the rotating cube are at $(a, a, a)$ and $(-a, -a, -a)$ . Now let's take the vertex $(a, a, -a)$ and see its position after a rotation by $\theta$ .

The rotation matrix about the unit axis $\mathbf{u} = \frac{1}{\sqrt{3}} (1, 1, 1)$ passing through the origin is

$\mathbf{R}(\theta) = \mathbf{u} \mathbf{u}^T + (\mathbf{I} - \mathbf{u} \mathbf{u}^T ) \cos \theta + \mathbf{S_u} \sin \theta$

where the matrix $\mathbf{S_u}$ is given by

$\mathbf{S_u} = \begin{bmatrix} 0 && - u_z && u_y \\ u_z && 0 && -u_x \\ -u_y && u_x && 0 \end{bmatrix}$

Hence,

$\mathbf{R}(\theta) = \frac{1}{3} \begin{bmatrix} 1 && 1 && 1 \\ 1 && 1 && 1 \\ 1 && 1 && 1 \end{bmatrix} + \frac{1}{3} \cos \theta \begin{bmatrix} 2 && -1 && -1 \\ -1 && 2 && -1 \\ -1 && -1 && 2 \end{bmatrix} + \frac{1}{\sqrt{3}} \sin \theta \begin{bmatrix} 0 && -1 && 1 \\ 1 && 0 && -1 \\ -1 && 1 && 0 \end{bmatrix}$

Hence the image of the vector $\mathbf{p} = (a, a, -a )$ is

$\mathbf{p'} = \mathbf{R}(\theta) \mathbf{p} = \displaystyle a ( \frac{1}{3} (1, 1, 1) + \frac{1}{3} \cos \theta (2,2,-4) + \frac{1}{\sqrt{3}} \sin \theta (-2, 2, 0 ) )$

We can take any of the three coordinates of point $\mathbf{p'}$ and maximize it. Let's take $\mathbf{p'}_y$ :

$\mathbf{p'}_y = \displaystyle \dfrac{ a }{3} ( 1 + 2 \cos \theta + 2\sqrt{3} \sin \theta )$

The maximum of this combination of $\cos \theta$ and $\sin \theta$ is

$\text{Max} ( \mathbf{p'}_y ) = \displaystyle \dfrac{ a}{3} \left( 1 + \sqrt{ 2^2 + \left( 2 \sqrt{3} \right)^2 } \right) = a \left( \dfrac{5}{3} \right)$

Thus the maximum side length is $2 a \left( \dfrac{5}{3} \right) = \dfrac{5}{3}$ , which makes the answer $5 + 3 = \boxed{8}$ .