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Let $r_n$ denote the inradius of a regular $n$-simplex $\triangle^n$ in $\mathbb{R}^n$, and $a$ denote the uniform edge length.

It is well-known that

$r_1 = a \frac{1}{2} \\ r_2 = a \frac{1}{6} \sqrt{3} \\ r_3 = a \frac{1}{12} \sqrt{6}$

But how to generalize $r_n$ to arbitrary dimensions?

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As we see in Height of $n$-simplex, the altitude (height) of a regular $n$-simplex of edge length $a$ is

$$h_n = a\sqrt{\frac{n+1}{2n}}.$$

The inradius $r_n$ is the distance from the centroid of the regular $n$-simplex to the center of a hyperface, so $r_n = \frac 1{n+1} h_n.$ Therefore

$$r_n = \frac a{n+1} \sqrt{\frac{n+1}{2n}} = \frac{a}{\sqrt{2n(n+1)}}.$$

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You might check the proof of

Cartesian coordinates for vertices of a regular 16-simplex?

which gives, via an inductive midpoint construction, the inradii as $h_n$ for a unit length regular $n$-simplex.

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