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Is there an equation which would give me the radius of the smallest sphere containing a certain tetrahedron (no need to touch all vertices); given that I know the insphere, circumsphere radii and the longest edge of the tetrahedron.

For 2D example of a triangle:

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Is this a regular tetrahedron? – Alex Becker Jun 12 '12 at 22:19
No it's not, all the edges would be equal if it were. – zamazalotta Jun 12 '12 at 22:34
The answer to the last sentence is no. Consider a "spire-shaped" tetrahedron with an equilateral base and a large altitude; the insphere is centered near the base, but the smallest enclosing sphere is centered about halfway up the altitude. – Rahul Jun 12 '12 at 22:49
Then it should be the barycenter of the vertices? – zamazalotta Jun 12 '12 at 23:00
If you want me to be notified of a comment, you need to put the syntax "@Rahul" somewhere in it. (You always get notified of comments on your own questions.) Anyway, think about it for a minute: does the barycenter of the vertices work for the example I just mentioned? – Rahul Jun 12 '12 at 23:08
up vote 1 down vote accepted

Let $r_\min$ be the radius of the minimal enclosing sphere.

The circumradius is an upper bound on $r_\min$, because the circumsphere encloses the tetrahedron.

$\sqrt{3/8}$ times the length of the longest edge is an upper bound on $r_\min$, because for a given length $\ell$, the tetrahedron with the largest minimal enclosing sphere and no edge longer than $\ell$ is the regular tetrahedron of edge length $\ell$, and its circumradius is $\sqrt{3/8}\ell$.

Putting that together, the value $\min(r_{\text{circ}},\sqrt{3/8}\ell_\max)$ is an upper bound on $r_\min$.

I don't know how the inradius can be used to improve the bound. But I'm pretty sure all you can get are bounds; specifying the circumradius, length of longest edge, and inradius is not sufficient to uniquely specify a tetrahedron.

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Thank you very much for your answer. Do you think I can infer something about the center of this minimal sphere based on the centers of the insphere and circumsphere centers which I already know? – zamazalotta Jun 13 '12 at 0:02
@zamazalotta: I doubt it. Consider a flat "cap-shaped" tetrahedron with one large equilateral face, and the fourth vertex somewhere just slightly above that face. The circumcenter is way off near infinity, and the incenter is just underneath the fourth vertex, but the center of the minimal enclosing sphere is precisely the center of the large equilateral face and doesn't change as you move the fourth vertex. – Rahul Jun 13 '12 at 0:11
@zamazalotta: In general, it's a good idea to check your intuition against the various kinds of "bad" tetrahedra that are possible. These are illustrated, for example, in Figure 54 of Edelsbrunner's "Triangulations and meshes in computational geometry". – Rahul Jun 13 '12 at 0:14
That is a really good article, thanks a lot. One thing that I observe in 2D is that if the diameter of CC is bigger than the largest edge, the center of CC is always outside the triangle. So if this hold for 3D may be I can use it to further constrain such tetrahedra – zamazalotta Jun 13 '12 at 0:35

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