Height of $n$-simplex

Question

$n$-simplex is a generalization of triangle or tetrahedron (with $n + 1$ vertices). The problem is to find its height.

I kindly ask to check my solution. I am not fluent with $n$-dimensional space yet, and can make a mistake.

$h^2 + r_0^2 = 1$, $h$ is height, $r_0$ is the radius of the circle, described around the $n-1$-simplex (which is the side of our simplex).

$r_0 = \sqrt{\frac{n(n-1)}{2n^2}}$ (I am pretty sure in it, it is easy to calculate).

So $h = \sqrt{1 - \frac{n(n-1)}{2n^2}} = \sqrt{\frac{n^2 + n}{2n^2}}$.

Still not sure I generalized it correctly, because all the time I used tetrahedron to imagine the problem.

Answer 1

Your question didn't state this explicitely, but I assume you're referring to a regular simplex of unit edge length. (A comment indicates as much).

I like coordinates, so I'd use coordinates to verify your result. Consider the standard simplex. Its corners are the unit vectors of a cartesian coordinate system of dimension $n+1$, and therefore its edge length is $\sqrt2$. It's height would be the minimal distance between $e_1=(1,0,0,\ldots,0)$ and a point on the opposite face, i.e. some $p=(0,a_2,a_3,\ldots,a_{n+1})$ with $a_2+a_3+\dots+a_{n+1}=1$. For reasons of symmetry, the point with minimal distance has to be in the center of that opposite face, i.e. $a_2=a_3=\dots=a_{n+1}=\frac{1}{n}$. So the height would be

$$\sqrt{1+n\left(\frac{1}{n}\right)^2} =\sqrt{1+\frac{1}{n}} =\sqrt{\frac{n+1}{n}}$$

If your simplex has edge length $1$ instead of $\sqrt2$, then you have to scale everything down by that factor, so your final height would be

$$h=\sqrt{\frac{n+1}{2n}}$$

which is equivalent to the $\sqrt{\frac{n^2+n}{2n^2}}$ you got.

Answer 2

You might check the proof of

Cartesian coordinates for vertices of a regular 16-simplex?

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