# Neighborhood of $(n,\delta)$-strained point in Alexandrov space homeomorphic to $\mathbb{R}^n$, how big is $\delta$?

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 $$\delta, then the map $$\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 $$\mathbb{R}^n$$, $$n\geqslant 3$$.

Find the minimal value $$\delta_n$$ such that $$\mathbb{S}^{n-1}$$ contains $$2\cdot(n+1)$$ points on the distance at least $$\tfrac\pi2-\delta_n$$ from each other.
I guess it is not hard to solve, if the oprimal configuration is centrally symmetric, then $$\delta_n$$ is the bound you are looking for.
On the other hand, there is another version of strainers where the sharp bound is obvious. The condition is $$\tilde\measuredangle(p\,{}^{a_i}_{a_j})>\tfrac\pi2+\delta$$ for a point array $$a_0,\dots,a_n$$. In this case the distance map $$M\to \mathbb{R}^n$$ defined by $$x\mapsto (|a_1-x|,\dots,|a_n-x|)$$ is bi-Lipschitz in a neighborhood of $$p$$ (note that we did not use $$a_0$$).
This is not true if $$\delta\le 0$$, even for $$n$$-dimensional Euclidean space.