Solution
= Solution
For every nonempty simplex $\sigma$ of $K$, let $b_\sigma$ be its barycentre. The first <barycentric subdivision> $K'$ has the $b_\sigma$ as vertices, and
$$
[b_{\sigma_0},\ldots,b_{\sigma_q}]
$$
is a simplex exactly when, after reordering, $\sigma_0\subsetneq\cdots\subsetneq\sigma_q$. Its realization is the same <polyhedron> as that of $K$. The <iterated barycentric subdivision> is defined by
$$
\boxed{K^{(0)}=K,
\qquad K^{(r+1)}=(K^{(r)})'.}
$$