How to calculate the centroid of a Polytope?

Question

Given a polytope is divided into simplexes, is it correct to calculate the centroid of the polytope as the average sum of its simplex centroid coordinates

Answer 1

No, it would be the weighted average, weighted by each simplex's volume
$$\frac{\sum_i V_i\vec{c_i}}{\sum_iV_i}$$ The volume of a simplex is the absolute value of a determinant. Given the $i$th vertex has vertices the $n+1$ points $\vec{x_j}\in\mathbb{R}^n,j=0..n$ let $M$ be the matrix whose $j$th column is $x_j-x_0$. $$c_i=\frac1{n+1}\sum_j\vec{x_j}\\ V_i=\frac1{n!}\text{abs det}M$$

Answer 2

No, it isn't true in general.

Take a look at the following picture illustrating it in dimension 2:

with a pentagon partitioned into 3 simplexes (triangles). Taking the partial centers of masses $F,G,H$ and then the center of mass $I$ of triangle $FGH$ doesn't give the center of mass $J$ of the pentagon.

Why that ? Because

$$\frac13(F+G+H)=\frac13\left(\frac{(A+B+C)}{3}+\frac{(A+C+D)}{3}+\frac{(A+D+E)}{3}\right)$$

$$=\frac{3A+B+2C+2D+E}{9} \ne \frac{A+B+C+D+E}{5}$$