circumscribing sphere of tetrahedron

Question

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?

Answer 1

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.