Let $M$ be an $n$-dimensional Alexandrov space with curvature $\geqslant k$. A point $p\in M$ is said to be an $(n,\delta)$ strained point if there are $n$ pairs of points $a_i, b_i$ such that $$ \tilde{\angle}a_i p b_i >\pi-\delta, \quad\tilde{\angle}a_i p a_j >\frac{\pi}{2}-\delta, $$ $$ \tilde{\angle}a_i pb_j>\frac{\pi}{2}-\delta,\quad \tilde{\angle}b_ipb_j>\frac{\pi}{2}-\delta. $$ We know if $\delta<c(n)$, then the map $\varphi(q)=(|a_1q|,...,|a_nq|)$ is a bi-Lipschitz homeomorphism between a certain neighborhood of $p$ and a domain in $\mathbb{R}^n$.

I want to know how big is $c(n)$? In the paper "A.D. Alexandrov space...", Theorem 5.4 says that if $\delta<\frac{1}{2n}$, then $\varphi$ is a $\frac{(1-2n\delta)}{\sqrt{n}}$-open Lipschitz map. So is the sharp constant $c(n)\leqslant \frac{c_1}{n}$? Can someone give an example that if $\delta>\frac{c_2}{n}$, then a neighborhood of $p$ is not homeomorphic to a domain in $\mathbb{R}^n$, $n\geqslant 3$.