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I draw a sphere such that all vertices of a tetrahedron are on the sphere. Is there some general rule to determine whether the circumcenter lies inside or outside of the tetrahedron?

I mean in 2D case the circumcenter is inside of a triangle if the triangle is acute and outside if not. Is there a similar rule? Does the placement depend on the angles?

Blue
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psi
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  • Oops. I mean is the circumcentre inside or outside of the tetrahedron. – psi Aug 10 '20 at 11:48
  • I suspect that there is no criterion that can be checked faster than computing the circumcenter and checking whether it is inside. Or what exactly do you already know about the tetrahedron? Its vertex-coordinates? Its edge-lengths? Its solid angles? Maybe the circumradius? Computing any of the latter two would already be quite some work and is only one step short of solving the whole problem. – M. Winter Aug 10 '20 at 12:57
  • An obtuse tetrahedron will have its circumcenter outside of the tetrahedron. – Math Lover Aug 10 '20 at 13:00
  • Are you sure @MathLover ? It looks like figure 7 is a counterexample in the article "Triangulation of Simple 3D Shapes with Well-Centered Tetrahedra" by Evan VanderZee, Anil N. Hirani, and Damrong Guoy. https://imr.sandia.gov/papers/imr17/VanderZee.pdf – psi Aug 10 '20 at 13:12
  • @M.Winter I know the vertex-coordinates. You are propably right that easiest way is calculate... – psi Aug 10 '20 at 13:13
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    The circumcentre in inside the tetrahedron iff all its barycentric coordinates are positive. There's a formula for the barycentric coordinates here https://math.stackexchange.com/questions/2863613/tetrahedron-centers but it's not pretty. – Angina Seng Aug 10 '20 at 13:29
  • @ellipsi I see your point in figure 3. May be there is no easy way to tell other than finding the coordinates as Angine Seng suggested. – Math Lover Aug 10 '20 at 14:40

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As addition, I'll simply leave the following passage from Eppstein, David, John M. Sullivan, and Alper Üngör. “Tiling Space and Slabs With Acute Tetrahedra.” Computational Geometry 27, no. 3 (March 1, 2004): p. 246. https://doi.org/10.1016/j.comgeo.2003.11.003.

In two dimensions, a triangle is acute if and only if its circumcenter lies inside the triangle. Given an acute (Delaunay) triangulation, the dual Voronoi tessellation thus provides an orthogonal dual mesh whose nodes are inside the corresponding triangles. [...] In three dimensions, an acute tetrahedron can fail to contain its own circumcenter (as with the example in the lemma above), and conversely a tetrahedron that contains its circumcenter can fail to be acute (as with a symmetric sliver).

Max
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A criterion was described in an article by Blumenthal

Blumenthal, Leonard M. Global subsets of the sphere. Arch. Math. (Basel) 7 (1956), 367–373

and involves a certain linear combination of roots of determinants involving the points.

Mikhail Katz
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  • Is there any geometric intuition underlying this criterion? A coordinatized version, while welcome, might not necessarily be super enlightening in its rawest form. – Rivers McForge Sep 26 '25 at 21:01
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    @RiversMcForge, yes, it is very simple. If you have 4 vectors in 3-space, they are necessarily linearly dependent. If the origin lies in the span of the 4 vectors, the coefficients will all be positive. – Mikhail Katz Sep 28 '25 at 07:39
  • Oh, that makes perfect sense! I see exactly where the determinants will come in then. Thanks for explaining! – Rivers McForge Oct 17 '25 at 04:41