## § Barycentric subdivision: edge length decreases

For a edge $E$, subdiving the edges into two at the center produces two edges both $1/2$ the original length. Given a triangle $T$, we wish to prove that subdividing the triangle by joining the barycenter to the vertices reduces edge length by $2/3$ of the maximum length. More generally, we wish to show that the edge length decreases to $n/(n+1)$ of the largest length for an $n$ dimensional figure. The barycenter is at the location $b \equiv 1/n \sum_i v_i$. The distance from a vertex $v_k$ is $||v_j - b|| = ||v_k - (1/n \sum_i v_i)|| = ||1/n(\sum_i v_k - v_i)||$. By Cauchy Schwarz, we have that $||v_j - b|| \leq 1/n ||\sum_i v_k - v_i||$. One of the terms, where $k = i$ will be zero, and the other (n-1) terms are at most $l$, the length of the longest edge. This gives $||v_j - b|| \leq (n-1)l/n$, hence the edge length decreases by a factor of $(n-1)/n$.