It is a mere geometrical quest, however within some $D+2$ dimensional space, and I have some trouble figuring it out.
Consider you have 3 all $D$ dimensional polytopes $P_i$, each with a subsumed to be existant circumradius $r_i=r(P_i)$, circumcenters $C_i=c(P_i)$, and with mutual pairwise offset $h_{ij}=dist(C_i,C_j)$. That is those 3 polytopes are arranged in a triangle with side lengths $h_{ij}$. Moreover this arrangement shall take place within perpendicular space, i.e. those 3 polytopes all are situated orthogonal to the 2D displacement space. So we indeed are considering a $D+2$ dimensional configuration.
Now I'm interested into the $D+2$ dimensional convex hull polytope $P$. That one is, what I'm calling the $trigonic$ of the given 3 polytopes $P_i$.
What I would like to know then is the circumradius $R=r(P)$. Surely it ought be possible to calculate it by means of the so far given measures. I.e. I'm looking for a formula
$$R=r(P) = f(r_1,r_2,r_3;h_{12},h_{13},h_{23})$$
Could anyone provide that for me? Possibly with some hints on how to obtain it?
--- rk
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Edit: Note that this problem well can be reduced to 3 dimensions:
Just sketch the triangle, given by the side lengths $h_{ij}$ onto the xy-plane. Then place peaks of respective heights $r_i$ on the verious vertices. Now the radius $R$ of the sphere, which is running through those tips is requested, meeting the additional constraint, that the center of that also lives somewhere in this xy-plane.
In fact, this reduced setup is nothing but the projection of the $D$ dimensional perp space onto a 1 dimensional one. The additional constraint comes from the fact that the peaks actually are 1 dimensional circles, i.e. the opposite point each on the other side of the xy-plane likewise should lie on the searched for sphere.
Might be that this reduced 3D setup makes my quest broader accessible.
--- rk
The reduction of the problem to three “peaks” sounds promising. Intuitively it seems correct.
Assuming the problem really can be reduced this way, the three “peaks” form a triangle in some plane. Find the circumcircle of that triangle. From the center of that circle, project a line perpendicular to the plane of the circle. Where that line intersects the $x,y$ plane is the center of your sphere.