Given a cube of side length $a$, what is the the radius of the largest circle that can be inscribed in the cube?
My Attempt:
I just assumed that the largest circle is the incircle of the hexagon generated by connecting the midpoints of the edges of the cube in the "correct" sequence. Since the side length of this hexagon is $ s = \dfrac{a}{\sqrt{2}}$, then the radius of the inscribed circle will be $ \frac{1}{2} s \cot 30^\circ $, and this is equal to $ \dfrac{1}{2} a \sqrt{ \dfrac{ 3 }{2} } = a \sqrt{ \dfrac{3}{8} } $
Set up Cartesian coordinates so that the cube is described by $0 \leq x \leq a, 0 \leq y \leq a, 0 \leq z \leq a$. Say the circle has radius $r$, and is in a plane $P$ perpendicular to a unit vector $\hat n = (n_x,n_y,n_z)$.
Obviously if $P$ is parallel to any face of the cube, then $r \leq \frac{a}{2}$, and the circle is not maximal. Assume going forward that's not the case. This implies $n_x \neq \pm 1, n_y \neq \pm 1, n_z \neq \pm 1$.
Since the angle between $\hat \imath$ and $\hat n$ is $\arccos(n_x)$, the angle between plane $P$ and plane $x=0$ (which are not parallel) is also $\arccos(n_x)$. The angle between plane $P$ and plane $x=a$ is the same. Since $x=0$ and $x=a$ are parallel planes, they intersect with $P$ in parallel lines, and the distance between these lines is
$$ \frac{a}{\sin(\arccos(n_x))} = \frac{a}{\sqrt{1-n_x^2}}$$
The circle must be between these parallel lines in $P$, so the diameter $2r$ cannot be more than $\frac{a}{\sqrt{1-n_x^2}}$. The other four limiting planes give corresponding limits in terms of $n_y$ and $n_z$:
$$ \begin{align*} r &\leq \frac{a}{2 \sqrt{1-n_x^2}} \\ r &\leq \frac{a}{2 \sqrt{1-n_y^2}} \\ r &\leq \frac{a}{2 \sqrt{1-n_z^2}} \end{align*} $$
Since $\hat n$ is a unit vector, $n_x^2+n_y^2+n_z^2=1$. If $n_x^2 < \frac{1}{3}$ then $r < a \sqrt{\frac 38}$, which you have already shown is attainable. The same goes for $n_y^2 < \frac{1}{3}$ or $n_z^2 < \frac{1}{3}$. So $r = a \sqrt{\frac 38}$ is in fact the largest possible radius, and it's only possible when the unit vector normal to the circle's plane is
$$ \frac{\pm \hat \imath \pm \hat \jmath \pm \hat k}{\sqrt{3}} $$
which is in the direction of one of the cube's four diagonals.
Now suppose the circle does have this maximum radius, and recall again our parallel lines where $P$ intersects the planes $x=0$ and $x=a$. Since the circle is between the parallel lines and its diameter $a \sqrt{\frac{3}{2}}$ is exactly the distance between the two lines, the circle must be tangent to both lines, and the midpoint of those tangent points must be the circle's center. Since the center is the midpoint of a point on plane $x=0$ and a point on plane $x=a$, the $x$-coordinate of the center is $\frac{a}{2}$. Similarly, the $y$- and $z$-coordinates are also $\frac{a}{2}$; the circle's center coincides with the cube's center.
Therefore there are only four possible inscribed circles with the maximum radius - the different orientations of the solution you described.