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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: http://demonstrations.wolfram.com/CircumcircleAndIncircleOfATriangle/

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    Your edit conveys no in$f$ormation; any 4 points lie on a sphere.2012-06-12

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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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    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 tetrahedra2012-06-13