What is the smallest ball containing four non-adjacent vertices of a cube? 
Let $ABCDA'B'C'D'$ be a cube. What is the smallest ball containing the regular tetrahedron $AB'CD'$?

I know that the ball with center being the center of the cube is a ball containing $A$, $B'$, $C$, and $D'$. But, how to prove that it is the smallest one?
Thanks you for any ideas!
 A: If I understand correctly, your tetrahedron (now with the correct points identified) is the regular tetahedron with side-length sqrt(2). If that is the case, then symmetry confirms that the centre of the circle coincides with the centre of the tetrahedron and the centre of the cube.
The distance from the centre of the cube to any of its vertices is sqrt(3)/2, so that must be the radius of the 'smallest' sphere or ball containing the tetrahedron.
To propose a simmetry arguement: If you rotate by 180 degrees about the centre normal of any face of the cube, then the vertices of the cube rotate onto themselves (that is a well understood symmetry property of any cube). Thus the equivalent vertices of the tetrahedron also rotate onto themselves. Thus the enclosing sphere and its centre point does not move. So the centre points of all three bodies lie at (0,0,0) (assuming the cube is the set of eight points (+/-0.5,+/-0.5,+/-0.5)). The rest follows...
A: Any ball containing all 4 points certainly must have diameter at least as large as the the maximum distance between any two points.
If I'm imagining how you are labeling the vertices correctly, then $A$ and $C'$ are opposite so the sphere you mention attains this bound.
