I am writing about (convex) polyhedra (in 3-dimensional Euclidean space) and am trying to determine some symmetry groups. It would be very good to be able to say that every polyhedron has a unique sphere with minimal radius containing it. If this were true, then the center of this sphere would be preserved by the symmetries of the polyhedron (because they preserve distances) and these symmetries would be easier to determine (since they would have a fixed point).

Now, I have two related questions and I would appreciate also partial answers:

(1) Does such a unique minimal sphere always exist? Is there an appropriate restriction on the type of polyhedra considered that guarantees its existence?

(2) If such a sphere exists, it cannot, in general, touch all vertices of the polyhedron. So can the sphere be characterized in other ways? What kind of point is the center of this sphere?


Let's note first that any compact set $K$ has a so called "center of mass" $C_K$. If an affine invertible map take $K$ to $K'$, it will also map $C_K$ to $C_{K'}$. In particular, isometries of $K$ will fix $C_K$.

Now, about the minimal sphere: for every compact $K$ there exists a sphere of smallest radius containing $K$ and such a sphere is unique. The existence is not that hard ( use a compactness argument). The uniqueness follows is proved like this: the intersection of two distinct spheres of radius $R$ is contained in a sphere of radius $< R$. Again, as you mentioned, every isometry of $K$ preserves the center of this sphere. If $K$ is convex then the center lies in $K$. For a general compact $K$ , the center of the minimal sphere lies in the convex hull of the intersection of the minimal sphere with $K$ ( thanks @Christian Blatter: for correcting an previous erronous statement).

Now, it is not true in general that $C_K$, the center of mass, is the center of the smallest sphere, as one can see already in the case of triangles, but that is the case if the group of isometries of $K$ is large enough ( if there exists only one point fixed by all the symmetries of $K$).

  • $\begingroup$ @Christian Blatter: You are right, it only contains $2$ vertices! Let me erase that statement. Thanks! $\endgroup$ – orangeskid Aug 19 '17 at 13:07
  • $\begingroup$ @Christian Blatter: I think the correct statement is: the center of the minimal sphere is in the convex hull of the intersection of the sphere with $K$. $\endgroup$ – orangeskid Aug 19 '17 at 13:08
  • $\begingroup$ That's almost the answer I wanted, except that I still don't understand why the minimal sphere exists, i.e. I don't get how the compactness argument works. Let $\mathbb{S}$ be the set of all spheres containing a given compact set $K$. The radii of the spheres in $\mathbb{S}$ are bounded below by half the diameter of $K$. But how do I know that the minimum is attained? $\endgroup$ – 57Jimmy Aug 20 '17 at 6:40
  • $\begingroup$ I guess this answer does the job: math.stackexchange.com/q/2100328/356190 $\endgroup$ – 57Jimmy Aug 20 '17 at 6:44

You can reason on the vertices alone, by convexity the edges and faces are wholly included in the minimal sphere.

For vertices in general position, the sphere touches exactly four of them. And the center is... the center of the circumscribed sphere.


Assume there are two distinct minimal spheres. By minimality, they have the same radius. If you move one of the sphere in the direction of the other, one of the touching vertices will certainly exit (the most extreme in the direction of motion).

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  • $\begingroup$ I don't, in general, assume that the polyhedra are convex (I will specify this in the question). But I would be interested to know if convexity is indeed a sufficient condition. You didn't explain why such a minimal sphere should exist or how it is constructed. $\endgroup$ – 57Jimmy Aug 19 '17 at 10:53
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    $\begingroup$ @57Jimmy: I didn't say that the polyhedron had to be convex, it needn't. By the way, the convex hull of the polyhedron is also included in the minimal sphere, so the distiction is irrelevant. $\endgroup$ – Yves Daoust Aug 19 '17 at 10:56
  • $\begingroup$ Ok, got that. So we might assume wlog that the polyhedron is convex! That's good. But does it imply the existence of such a sphere? $\endgroup$ – 57Jimmy Aug 19 '17 at 10:58
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    $\begingroup$ @PeterSheldrick: unfortunately, this is insufficient as you need to justify that among these $n(n-1)(n-2)(n-3)/24$ spheres, there is one which contains all points. $\endgroup$ – Yves Daoust Aug 19 '17 at 11:14
  • $\begingroup$ @Peter not just groups of four vertices. You could have just three on a great circle as in a very flat tetrahedron, or two on a diameter. The "miniball" problem is hard to solve efficiently, see some other comments or clues on how to do it. $\endgroup$ – Oscar Lanzi Aug 19 '17 at 11:32

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