Simplex with edges of length at least s having smallest circumradius

Question

Is it true that of all $k$-simplices with edge lengths greater than or equal to some parameter $s$, the regular simplex with edge lengths $s$ has the smallest circumradius?

Please supply a proof or reference if possible. Thank you.

Answer 1

Yes, this is true. It is easier to work with an equivalent formulation: among all $k$-simplices contained in the unit ball $B=\{x:\|x\|\le 1\}$, the regular simplex maximizes $\min \ell(e)$ (that is, the minimum of lengths of edges).

Step 1. It suffices to consider simplices with all vertices on the sphere $\partial B$. Indeed, suppose $v_k$ is in the interior of $B$. Let $H$ be the hyperplane passing through $\{v_j:j\ne k\}$. Let $L$ be the line passing through $v_k$ and perpendicular to $H$. If we move $v_k$ along $L$ away from $H$, none of the edges get shorter. We can keep moving $v_k$ until it hits $\partial B$.

Step 2. Since $\|v_k-v_j\|=\cos^{-1}(v_k\cdot v_j)$, we can just as well talk about the maximal dot product $v_k\cdot v_j$ instead of minimal edge length. Let $\mu=\max\{v_k\cdot v_j: k\ne j\}$.

Step 3. Since $$0\le \|v_0+\dots +v_{k}\|^2=(k+1)+\sum_{k\ne j}v_k\cdot v_j \le (k+1)+\mu k(k+1)\tag{a}$$ we have $\mu\ge -1/k$.

Step 4. For the regular simplex equality holds throughout in (a); thus, this simplex achieves $\mu=-1/k$. $\quad \Box$