# Tetrahedron from its inscribed sphere

I'm facing a geometrical problem:

Given a sphere $S$, I want to calculate the vertices of the tetrahedron $T$ whose inscribed sphere is $S$. In other words I want to calculate a tetrahedron from its inscribed sphere.

If anyone knows the solution, don't hesitate to share with me.

• Given any sphere $S$, there are essentially $8$ degree of freedom for constructing a tetrahedron with $S$ as its inscribed sphere. What sort of tetrahedron you want? – achille hui May 18 '15 at 19:45
• For any $u > 2$, let $v > 2$ be the root of the equation $\frac{1}{u^2-1} + \frac{1}{v^2-1} = 1$, then the four vertices $$\left( -u, 0, \pm \frac{u+v}{\sqrt{v^2-1}}\right), \left( v, \pm \frac{u+v}{\sqrt{u^2-1}}, 0 \right)$$ is one family of orthogonal tetrahedra with the unit sphere inscribed inside it. In particular, when $u = v = \sqrt{3}$, you get the regular tetrahedron. – achille hui May 18 '15 at 21:29
• oops, in the above comment, the constraint should be $u, v > \sqrt{2}$. sorry about that. – achille hui May 19 '15 at 8:54
The radius of the circumsphere is three times the radius of the inscibed sphere. Hence one tetrahedron with the sphere of radius $1$ around the origin would be given by the vertices $(0,0,3)$, $(\sqrt 8,0,-1)$, $(-\sqrt 2,\sqrt 6,-1)$, $(-\sqrt 2,-\sqrt 6,-1)$.
If the radius of inscribed sphere is $r$ then the edge length of the tetrahedron is $2r\sqrt{6}$ & radius of circumscribed sphere is $3r$
Then in general form, for inscribed sphere with a radius $r$ centered at the origin, the vertices of the tetrahedron are $(0, 0, 3r)$,$\left(2r\sqrt{2},0, -r\right)$, $\left(-r\sqrt{2},r\sqrt{6}, -r\right)$, $\left(-r\sqrt{2}, -r\sqrt{6},-r\right)$