6 Nov
2016
6 Nov
'16
3:57 p.m.
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? -----