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What are the conditions under which the center of circumscribing sphere of a tetrahedron is located inside(outside, face, edge) of the tetrahedron? In other words, how can we define acute(obtuse) tetrahedron?

Chung. J
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    Have you heard about barycentric coordinates? http://en.wikipedia.org/wiki/Barycentric_coordinate_system – hjhjhj57 Dec 15 '14 at 07:08

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Without loss of generality assume the vertices of the tetrahedron are $0,v_1,v_2$ and $v_3\in\mathbb{R}^3$. Since the these vectors (without $0$) are linearly independent, they form a basis. So every point $x\in\mathbb{R}^3$ can be written as: $$ x = a_1v_1+a_2v_2+a_3v_3, $$ for some real numbers $a_i$. It's not hard to see that if:

1) If $0<a_1+a_2+a_3<1$, $x$ lies in the interior of the tetrahedron.

2) If $a_1+a_2+a_3 = 1$, or $a_i=0$ and $0\leq a_j+a_k\leq 1$, then $x$ lies in the tetrahedron (check the cases where one or two of the scalars are zero).

3) In any other case, $x$ is in the exterior of the tetrahedron.

Now, the formula of the circumsphere ultimately tells us that the center of the sphere is $\frac{1}{2a}\left(D_x,D_y,D_z\right)$. So to know where its center lie w.r.t. the tetrahedron you just need to express this vector in terms of the basis $\{v_1,v_2,v_3\}$ and see what happens to the sum of the coefficients.

hjhjhj57
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  • It is not an answer what I want. Don't move it to coordinate. – Chung. J Dec 15 '14 at 08:08
  • Everybody knows the answer when it moves to cartesian coordinate. – Chung. J Dec 15 '14 at 08:10
  • For triangle A, the biggest angle is smaller than 90 iff Ais acute triangle. – Chung. J Dec 15 '14 at 08:12
  • Well, we could say the the tetrahedron is acute if the center of the circumsphere is inside the tetrahedron, right if it's in a face of the tetrahedron and obtuse if it's outside. Is this what you mean? – hjhjhj57 Dec 15 '14 at 08:17
  • You could get that kind of results with this approach using solid angles, but it's going to be quite hard... – hjhjhj57 Dec 15 '14 at 08:23