Fit cube in a circular hole (only 50% of the time) QUESTION:
You have a hole in the shape of a circle with radius R and a cube with side 1. If the probability that the cube will enter the hole is 50%, what is the  radius R of the circle?
Examples of three different orientations of the cube exactly (if one would connect the center of the cube and the center of the hole, they would form a vertical line) above the hole:


What I've tried:
I know the longest straight line distance in the cube will be √3. And then I thought, starting with that distance representing the diameter of a circle and now rotate it either clock-wise or counter clock-wise (don't matter obv). But here I'm stuck. It seems that I should rotate it x many degrees so that it will fit exactly only 50% of the time. But can't come up with the answer. Maybe it's not even the right approach and you have a different approach and can come up with an answer. Thanks.
 A: The minimum radius of the circle allowing the cube to pass is equal to the maximum distance between a vertex of the cube and the vertical line passing through the center of the cube. That line will intersect two opposite faces of the cube: let's focus on one of those faces and let $P$ be the intersection point.
We can take for instance the origin as center of the cube and $V=(\pm1/2,\pm1/2,1/2)$ as vertices of the face. We can set then $P=(x,y,1/2)$. The distance between line $OP$ and a generic vertex $V$ is given by ${\sqrt3\over2}\sin\theta$, where $\theta=\angle POV$. We can compute $\cos\theta$ from $\vec{OP}\cdot\vec{OV}$ and thus obtain:
$$
\sin^2\theta=1-{(\pm x\pm y +1/2)^2\over3(x^2+y^2+1/4)}.
$$
It is not hard to see that the sign choice giving the maximum value for $\sin\theta$ depends on the quadrant where $P$ lies. Let's focus on the first quadrant, where $x$ and $y$ are both positive and we must take the minus sign for both in the above formula. The locus of $P$ where $\sin\theta$ (and thus $\sin^2\theta$) is constant is then a conic, given by the above equation.
I plotted below the locus (a hyperbola) for $\sin^2\theta=0.983$, corresponding to a minimum radius $r={{\sqrt3\over2}\sin\theta}\approx0.859$. The internal white region corresponds to $\sin^2\theta>0.983$, whereas the brown region corresponds to $\sin^2\theta<0.983$. You can find the picture for the whole face by reflecting this figure about the cartesian axes.
I chose that value for $\sin^2\theta$ because the area of the brown region is very close to the area of the white region (according to GeoGebra). If the orientation of the cube is fixed by choosing a random point uniformly on a face, than a circle with the radius given above would let pass only a half of the cubes. If the orientation is chosen otherwise (for instance, uniformly on a sphere) then the computation is more complicated, but the result shouldn't be too different.

