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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 $r\simeq 0.65$.

solution

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.

Ethan Bolker
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prolea
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  • 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 06 '18 at 19:17
  • https://en.wikipedia.org/wiki/Circumscribed_sphere ...... https://en.wikipedia.org/wiki/Inscribed_sphere – Will Jagy Jul 06 '18 at 19:19
  • What is a tetrahedron of dimension 4? – MPW Jul 06 '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 06 '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. – prolea Jul 06 '18 at 19:29
  • If I understand the problem I think I know how to solve it. You can assume one vertex is at the origin. Then a linear map will convert the tetrahedron to a regular tetrahedron (or to the one whose other vertices are the unit coordinate vectors). This will scale the volume by the determinant of the linear map. Now make the calculation for the nice tetrahedron. Should work in any dimension. – Ethan Bolker Jul 06 '18 at 19:45

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Alright, still not sure what the problem is; however...

To find the circle inscribed in a triangle in the plane, draw the angle bisectors through each vertex. These meet in the center of the inscribed circle.

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.

In higher dimension, for each pair of hyperplanes bounding the simplex, construct the hyperplane bisecting that angle; these new hyperplanes 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.

Will Jagy
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