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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?

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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
  • 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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