Ah, but I think Dan was right after all, if one interprets "symmetries" to include all linear transformations! If I had the time I think I could write a paragraph argument showing that the center of mass should behave well under linear transformations. Then one can transform irregular simplices so that they become regular, and then apply Dan's isometry argument. Jim On Sun, Nov 6, 2016 at 6:49 PM, James Propp <jamespropp@gmail.com> wrote:

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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