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Question:
is there a theorem that guarantees that
$\mathcal{P}\subset\mathbb{E}^n$ is finite set of points in a Euclidean space and all radii of the $(n-1)$-spheres that are defined by the $n$-simplices with $n$ corners in $\mathcal{P}$ are equal $\implies$
the centers of the circumspheres are identical.

Calculating the radius of a simplexes circumsphere is possible by means of sidelengths , cf e.g. https://westy31.home.xs4all.nl/Circumsphere/ncircumsphere.htm.

I want to utilize the above criterion for being co-spheric in a graph-theoretic algorithm and want to be sure that no "exotic" exception is lurking somewhere.

Edit:

the not so "exotic" counter example of @fedja

enter image description here

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    $\begingroup$ A tiny point of vocabulary : corners -> vertices $\endgroup$
    – Jean Marie Becker
    Commented Feb 14, 2021 at 9:46
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    $\begingroup$ @JeanMarieBecker I use corners to hint the geometric interpretation whereas I tend to reserve vertex for graph theoretic interpretations. In geometry vertices need not be corners; e.g. the vertex of a parabola as depicted here $\endgroup$
    – Manfred Weis
    Commented Feb 14, 2021 at 10:15
  • $\begingroup$ Take four noncoplanar points of your group. They determine a sphere who's radius and center can be determined geometrically. Think of one of these points asthe North Pole of the sphere and consider the stereographic projection onto the tangent plane to the South pole. You are now left to decide if the projection of the remaining remaining points are coplanar. That is an easier proposition. $\endgroup$
    – Liviu Nicolaescu
    Commented Feb 14, 2021 at 13:13
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    $\begingroup$ Three vertices of an equilateral triangle on the plane together with its center give a sure counterexample but I wouldn't call this one "exotic" :-). It looks like the statement may, indeed, hold if the points are sufficiently many though... $\endgroup$
    – fedja
    Commented Feb 14, 2021 at 20:30
  • $\begingroup$ @fedja in your example there are four simplices of which the three that contain the centerpoint have equal radius of circum circle, but the equilateral triangle's radius is different or did I miss the point? Maybe I have to edit my question. $\endgroup$
    – Manfred Weis
    Commented Feb 15, 2021 at 3:48

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