I was assuming a nonequilateral triangle and a nonregular simplex. Jim On Sun, Nov 6, 2016 at 5:56 PM, Dan Asimov <dasimov@earthlink.net> wrote:
The center of mass has to be invariant under symmetries, and both mass distributions have symmetry group equal to the symmetric group on the vertices.
Taking the n-simplex to be the convex hull of the standard basis of R^(n+1):
{(x_j) in R^(n+1) | Sum x_j = 1, 0 <= x_j <= 1}
, there is only one point invariant under its symmetries, i.e., permutations of coordinates, namely x_j = 1/(n+1) all j.
—Dan
----- From: Bill Gosper <billgosper@gmail.com> Sent: Nov 6, 2016 12:41 PM
but is it obvious why the center of mass of such a simplex is the center of mass of unit masses at its vertices? -----
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