Radius of inscribed sphere of n-simplex.

Question

I want to calculate the radius of inscribed sphere of $n$-simplex, where the side length of $n$-simplex is 1. For example, when $n=2$, the 2-simplex is equilateral triangle with side length is 1. And the radius of inscribed sphere is $\frac{1}{2\sqrt 5}$. But when $n\ge 4$, I don't know how to calculate it.

Answer 1

A $n$-simplex can be embedded in $\mathbb{R}^{n+1}$ by choosing $(1,0,0,\ldots),(0,1,0,\ldots),\ldots$ as vertices.
By this way the centroid of the simplex lies at $\left(\frac{1}{n+1},\frac{1}{n+1},\ldots,\frac{1}{n+1}\right)$, the centroid of the (hyper-)face opposite to $(1,0,0,\ldots)$ lies at $\left(0,\frac{1}{n},\ldots,\frac{1}{n}\right)$ and the edge length is $\sqrt{2}$, while the inradius equals

$$ \sqrt{\frac{1}{(n+1)^2}+n\cdot\frac{1}{n^2 (n+1)^2}} =\frac{1}{\sqrt{n(n+1)}}.$$ By scaling, a regular $n$-simplex with unit edge length has an inradius equal to $\frac{1}{\sqrt{2n(n+1)}}$.
Similarly, the circumradius equals $\sqrt{\frac{n}{2(n+1)}}$ and $\frac{R}{r}=n$.