We are given a convex shape $S$ in the $d$-dimensional Euclidean space, whose boundary is formed by portions of $2d$ different spheres, one portion per sphere. The radius of each sphere is the same, $a+1$, and the spheres centers are $\pm a\textbf{e}\_j$ for $1\le j\le d$, where $a\ge 0$. Hence, for $a>0$ each sphere passes through one and only one of the $2d$ points in the set $P:=\{\pm\textbf{e}\_j | 1\le j\le d\}$.


Finally, for each point $\mathbf{p}\in P$, the portion of the sphere $\mathcal{S}\_{\mathbf{p}}$ passing through $\mathbf{p}$ consists of all points of $\mathcal{S}\_{\mathbf{p}}$ whose distance from $\mathbf{p}$ is smaller or equal to the distance between $\mathbf{p}$ and any point of the other $2d-1$ spheres.




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**Questions:** How can we provide a tight lower bound of $a$ in terms of $d$ to have that $\frac{V(S)}{2^d}$ is a constant bounded away from $0$ when $d\to\infty$?



Volume of a shape whose boundary consists of portions of spheres symmetrically placed about the origin in $d\gg 1$ dimensions
If we set $a = \beta d^2$ for $\beta > 0$ then $$\frac{V(S)}{2^d} \to \exp(-1/(6\beta)).$$


The idea is to choose $x \in [-1,1]^d$ uniformly at random. Then $\frac{V(S)}{2^d}$ is just the probability that $x \in S$.


Let $S\_j^+$ (respectively $S\_j^-$) denote the ball centered at $e\_j a$ (respectively $-e\_j a$) of radius $a+1$. For a point $x \in [-1,1]^d$, note that $x \in S\_j^+$ if and only if $$(x\_j - a)^2 + \sum\_{i \neq j} x\_i^2 \leq (a + 1)^2$$
i.e. if and only if $$-2x\_j a + \|x \|\_2^2 \leq 2a + 1\,.$$


Arguing the same for $S\_j^{-}$ we have that $x \in S\_j^+ \cap S\_j^-$ if and only if $$ \|x\|\_2^2 \leq 2a(1 - |x\_j|) + 1\,.$$


This implies that $x \in S = \bigcap\_j (S\_j^+ \cap S\_j^-)$ if and only if $$\|x \|\_2^2 \leq 2a (1 - \|x \|\_\infty) + 1$$
i.e. if $$\frac{\|x\|\_2^2}{d} \leq 2\frac{a}{d^2} \left (d (1 - \|x \|\_{\infty}) \right) + \frac{1}{d}.$$


As $d \to \infty$ we have $\|x\|\_2^2/d \to 1/3$, while $d(1 - \|x \|\_\infty)$ converges to an exponential random variable $Y$ of mean $1$. Thus we have $$\frac{V(S)}{2^d} \to P(1/3 \leq 2 \beta Y) = \exp(-1/(6\beta)).$$


