Existence of the inscribed hypersphere of a simplex

Question

Letting $\textbf{T}=[u_0,...,u_n]$ be a $n$-simplex of $\mathbb{R}^n$, how does one prove the existence of the inscribed hypersphere ?

Looking at the possible duplicates, people only seem to ask what is either the radius or the volume, and only require computations. But how to actually prove it exists in the first place ?

The queston is : how to prove the existence (not uniqueness, by the way) of a vector $x\in\overset{\circ}{\textbf{T}}$ such that $x$ is equidistant to all faces $f_i=[u_0,...,u_{i-1},u_{i+1},...,u_n]$ ?

Answer 1

Let $M\in\mathbb{N}, M \geq 1$ and let $u_0, u_1, u_2, ..., u_M \in \mathbb{R}^{M+1}$ form an affinely independent set. Let us denote the convex hull of all $u_i$ by $S$. Then $S$ is a Simplex. Without loss of generality we may assume $u_0=0$. From this together with the affine independency it follows that $u_1, ... , u_n$ are linearly independant.

Let us denote the usual inner product of $x,y\in\mathbb{R}^{M+1}$ by $<x,y>$ and define $||x||:=\sqrt{<x,x>}$. Now the simplex has the one Face which is contained in the hyperplane which contains $u_1, u_2, ..., u_M$. Note that this hyperplane does not contain the zero vector. Let us denote a normal vector to that hyperplane by $k$. Then $||k||=1$ and there is a $c\in \mathbb{R}, c \neq 0$ such that it holds $<k,u_i>=c$ for all $i=1,..M$. We can choose $k$ such that $c>0$.

All other Faces contain the zero vector. Let us denote $F_i$ as the Face which is contained in the hyperplane spanned by $0, u_1, u_2, ..., u_{i-1}, u_{i+1}, ... u_M.$ Let us denote a normal vector to that hyperplane by $n_i$. So $||n_i||=1$ and $<n_i,u_j>=0 \quad j=1,..., i-1,i+1,..,M$. We can choose $n_i$ such that $<n_i,u_j> >0$.

Then the radius of the inscribed ball in $S$ is the real number $$r := \frac{c}{1+\sum_{i=1}^M\frac{c}{<u_i,n_i>}}$$

For $i=1,...,M$ denote $\theta_i$ by $$\theta_i:= \frac{\frac{c}{<u_i,n_i>}}{1+\sum_{j=1}^M \frac{c}{<u_j,n_j>}}$$ Note that $\theta_i>0$ and $\sum_{i=1}^M\theta_i < 1$. Let us define $\theta_0:=1-\sum_{i=1}^M\theta_i$, so that $\sum_{i=0}^M\theta_i = 1$

Then the center of the inscribed ball in $S$ is at $m\in \mathbb{R}^{M+1}$ with $$m:=\sum_{i=0}^M\theta_i u_i $$

Answer 2

Here's a geometric argument that convinces me.

Temporarily remove one of the $n+1$ facets. Position the simplex with the opposite vertex "down" and drop a small (hyper)sphere through the opening at the top. It will come to rest tangent to the other $n$ facets. Now consider the plane containing the omitted face and move it down parallel to itself until it's tangent to the sphere. Finally, scale the small simplex up to the original size.

Answer 3

In case it might be sufficient to show the inscribed hypersphere of the unit simplex in $\mathbb{R}^n$ :

Let us denote $e_1, e_2, ... e_n \in\mathbb{R}^n$ with $e_1 :=(1,0,0,...0),e_2 :=(0,1,0,...0), ..., e_n=(0,0, ...0,1)$ and $d \in\mathbb{R}^n$ with with $d :=(1,1,1,...1)$. Let us denote $||x||_2:=\sqrt{\sum_{i=1}^n x_i^2}$ for all $x\in \mathbb{R}^n$.

Then the convex hull of $0, e_1, e_2, ... e_n$ is denoted as the unit simplex in $\mathbb{R}^n$.

Furthermore lets define $\lambda := \frac{1}{n+\sqrt{n}}$. Then the inscribed hypersphere of the unit simplex in $\mathbb{R}^n$ is the sphere centered at $\lambda d$ and radius $\lambda$, or more precisely: All $x\in \mathbb{R}^n$ which satisfy the equation $$ ||x-\lambda d||_2 = \lambda$$ form the inscribed hypersphere .