Ratio of Volume of Standard Simplex to its Dual In the two-dimensional case, the ratio of the area of an equilateral triangle whose vertices are the mid-points of the edges of another equilateral triangle is 1/4th.  For a tetrahedron and its dual, the ratio is known to be 1/27.
In the general case, is it $\frac{1}{n^{n}}$?  By general case, I mean the standard simplex $S$ in $\mathbb{R}^n$ and the simplex whose vertices are the Euclidean barycenters of the faces of $S$.
 A: I think the key is to take the $n$th power of the ratio of the side lengths.
In what follows, I'll view the $n$-simplex as living in $\mathbb{R}^{n + 1}$ with vertices the standard basis vectors.
First, the outer simplex always has side length $\sqrt{2}$.
To get the side length of the inner simplex, we can compute the distance between the midpoints of two different faces. Since these midpoints occur at the average of $n - 1$ of the vertices, they have one coordinate 0 and the other $n$ equal to $1 / n$. Therefore the distance between two of them is $\sqrt{\frac{1}{n^2} + \frac{1}{n^2}} = \frac{\sqrt{2}}{n}$.
So, the ratio of the side lengths is
\begin{align*}
\frac{\frac{\sqrt{2}}{n}}{\sqrt{2}} = \frac 1 n ,
\end{align*}
and the $n$-volumes have ratio $1 / n^n$ as you claimed.
A: The barycenter of an $n$-simplex is the average of all the vertices: $O=\frac{1}{n+1}(V_1+\ldots+V_{n+1})$. The barycenter of the face opposite $V_i$ is the average of all the vertices except $V_i$: $W_i=\frac{1}{n}[(V_1+\ldots+V_{n+1})-V_i]=\frac{n+1}{n}O-\frac{1}{n} V_i$. So $O=\frac{n}{n+1}W_i+\frac{1}{n+1}V_i$. This means that $O$ lies on the segment $V_iW_i$ and divides it in ratio $n:1$ (this is a generalization of the fact that the point of intersection of the medians divides each median in the ratio of $2:1$). Therefore the "dual" simplex is obtained from the original one by a similarity transform centered at $O$ and with ratio $-\frac{1}{n}$. Since we are in $n$ dimensions, this multiplies the volume by $\frac{1}{n^n}$.
In fact this is true for $n$-volume any $n$-simplex in arbitrary $\mathbb{R}^m$, with same proof.
