Maximum volume ball in tetrahedron

Consider the problem : "Find the maximum volume of the three dimensionnal ball inscribed to a tetraedron."

The tetraedron have the following summit (just to have a concrete case): $(0,0,0), (0,\pi,0),(\pi,0,0),(\pi,\pi,\pi)$.

Geometrically, from the optimization point of view, the problem can be formulated as

\begin{align} &\max_{a,b,c,r\in \mathbb{R}}\qquad r^3\\ &r-|c|<0\\ &r-|a-b|/\sqrt{2}<0\\ &r-|b-\pi|<0\\ &r-|a-c|/\sqrt{2}\\ &a-\pi<0\\ &b-\pi<0\\ &c-\pi<0\\ &a>0\\ &b>0\\ &c>0\\ &r>0\\ \end{align}

($r$ stands for the radius of the ball of center $(a,b,c)$). The optimal volume is obtained for $r\simeq 0.65$. Now, I have difficulty in formulating the problem with a ball of dimension $4$ in a tetraedron of dimension $4$ because I don't have the geometric point of view and I don't know how to write the constraints.

• Smilia, usually "circumscribed" is reserved for the sphere going around something. I will look it up, for a sphere tangent to the faces of the tetrahedron, i would probably say inscribed – Will Jagy Jul 6 '18 at 19:17
• – Will Jagy Jul 6 '18 at 19:19
• What is a tetrahedron of dimension 4? – MPW Jul 6 '18 at 19:22
• @MPW pretty sure it is just a simplex of full dimension; I also think that the max volume problem for inscribed sphere is the same as the max radius.... trying to think of a compass and straightedge type of construction, don't have it yet – Will Jagy Jul 6 '18 at 19:25
• @MPW naively it is a tetraedron whose volume varies to make the fourth dimension, but it doesn't help. As I said I am looking for a generalization that I don't even know how to formulate it. – Smilia Jul 6 '18 at 19:29

In $\mathbb R^3,$ for each pair of planes bounding the tetrahedron, draw the plane that bisects the angle between the two planes. The six new planes meet in the center of the inscribed sphere.
There is a little linear algebra/geometry involved in describing these bisecting hyperplanes... In $\mathbb R^n,$ a plane containing a face of the simplex is an $(n-1)$-plane. Two of them meet in an $(n-2)$-plane, call it $P.$ The orthogonal complement of $P$ (through a favorite point) is a $2$-plane, call it $O.$ The intersection of $O$ with the two bounding hyperplanes is just two intersecting lines; we construct the angle bisector between these two lines, and take the sum of that with $P,$ getting us back up to a bisecting $(n-1)$-plane.