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There is a simple question where a unit square contains four circles of radius $\frac14$ with a fifth inner circle, and the aim is to find that the radius of the inner circle is $r_2=\frac{\sqrt{2}-1}{4} \approx 0.10355$. Consider the line segment from a vertex passing through the centre of an outer circle to the centre of the square.

circles in box

Going up a dimension, with a unit cube with eight spheres of radius $\frac14$ and a ninth inner sphere, it is not much harder to find the radius of the inner sphere is $r_3=\frac{\sqrt{3}-1}{4} \approx 0.18301$. And the simple geometry looking at the distance from a vertex of the cube to its centre generalises further: with a unit hypercube with $2^n$ hyperspheres of radius $\frac14$ and an inner hypersphere, the radius of the inner hypersphere is $r_n=\frac{\sqrt{n}-1}{4}$.

For $n=4$ this gives $r_4=\frac14$ so the inner hypersphere is then the same size as the sixteen outer hyperspheres. For $n=9$ it gives $r_9=\frac12$ so the inner hypersphere touches the hypercube.

Counterintuitively, for $n=10$ this gives $r_{10}\approx 0.54057$ so the inner hypersphere goes outside the hypercube which bounds the outer hyperspheres which bound the inner hypersphere, and similarly for larger $n$.

The $n$-volume of the hypercube is $1$. The $n$-volume of the inner $n$-ball is $\frac{\left(\sqrt{\pi}\frac{\sqrt{n}-1}{4}\right)^n}{\Gamma\left(\frac n2+1\right)}$: it is $0$ for $n=1$, about $0.033688$ for $n=2$, then decreasing as $n$ increases to $264$ where it is about $10^{-5}$, and then increases with $n$, exceeding $1$ when $n\ge 1206$ and exceeding $2$ when $n \ge 1244$.

For what dimension $n$ is more than half of the inner $n$-ball outside the hypercube?

Calculating the volumes of the $2n$ hyperspherical caps may not lead to the answer, as it seems that for large enough $n$ (possibly $n\ge 15\gt 9+\sqrt{32} \approx 14.657$ and certainly $n\ge 17$) the hyperspherical caps may overlap each other.

Simulation using the R code below suggests $n\ge 43$ will work, but this is not an analytic answer.

pointinnball <- function(n, r=1){
  x <- rnorm(n)
  rmsx <- sqrt(sum(x^2))
  x * r * runif(1)^(1/n) / rmsx
  }
innerradius <- function(n){
  (sqrt(n) - 1) / 4
  }
pointoutsidecube <- function(n){
  max(abs(pointinnball(n, r=innerradius(n)))) > 1/2
  }

set.seed(2025)
mean(replicate(10^6, pointoutsidecube(n=42)))
# 0.494222
mean(replicate(10^6, pointoutsidecube(n=43)))  
# 0.512103
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    $\begingroup$ It must be a nightmare to compute a close form of the volume! Nice problem though $\endgroup$ Commented Aug 28 at 15:11
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    $\begingroup$ related intersection of hypercube and hypersphere $\endgroup$ Commented Aug 30 at 16:12
  • $\begingroup$ @user953715 following that link and links from it, I came across an expression in (2.6) in arxiv.org/pdf/1804.07861 (Peter J. Forrester, "Comment on 'Sum of squares of uniform random variables' by I. Weissman") for the intersection of a hypercube and $n$-ball but my attempts to use it to answer this question failed miserably. $\endgroup$ Commented Sep 4 at 13:25
  • $\begingroup$ @Henry Unfortunately, it seems the user had deleted their account recently. $\endgroup$ Commented Sep 4 at 13:27
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    $\begingroup$ @Soheil It was still a helpful comment including the diagram in the linked question of overlapping spherical caps, and especially the links from LeoDucas and E. Postlethwaite, even if I failed when trying to implement some of the referenced expressions. $\endgroup$ Commented Sep 5 at 22:32

1 Answer 1

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Here's some progress: I can show that this holds for $n\geq 98$.

We will scale everything up by $2$. Let $V_n$ be the volume of an $n$-ball of radius $(\sqrt n-1)/2$, and let $W_n$ be the volume of the intersection of said $n$-ball with the hypercube $[-1,1]^n$. We are concerned with when $W_n/V_n<1/2$.

Note that $W_n/2^n$ is the probability that, if $x_1,\ldots,x_n\sim\operatorname{Unif}([-1,1])$ are chosen independently, then $x_1^2+\cdots+x_n^2\leq (\sqrt n-1)^2/4=(n-2\sqrt n+1)/4$. Let $D$ be the distribution of each $x_i^2$, and let $y_i=x_i^2$ so that $y_1,\ldots,y_n\sim D$. Note that $\mathbb E[y_i]=1/3$. Write $$\delta=\delta_n=1-\frac{\left(\frac{\sqrt n-1}2\right)^2}{\frac n3},$$ so that $$2^{-n}W_n=\Pr\big[y_1+\cdots+y_n<(1-\delta)\mathbb E(y_1+\cdots+y_n)\big].$$ We now run a Chernoff-bound calculation manually. Let $\lambda=\lambda_n>0$ be a parameter to choose later. We have \begin{align*} \Pr\big[y_1+\cdots+y_n<(1-\delta)\mathbb E(y_1+\cdots+y_n)\big] &\leq\frac{\mathbb E(e^{-\lambda(y_1+\cdots+y_n)})}{e^{-\lambda(1-\delta)n/3}}\\ &=e^{\lambda(1-\delta)n/3}\left(\mathbb E_{y\sim D}(e^{-\lambda y})\right)^n\\ &=\left(e^{\lambda(1-\delta)/3}\int_0^1e^{-\lambda x^2}dx\right)^n. \end{align*} Therefore, for any $\lambda>0$, $$\frac{W_n}{V_n}\leq\left(\frac n2\right)!\cdot\left(\frac{2}{\sqrt\pi\cdot\frac{\sqrt n-1}2}e^{\lambda(1-\delta)/3}\int_0^1e^{-\lambda x^2}dx\right)^n.$$ This is strong enough to prove that $W_n/V_n<1/2$ when $n=98$ by taking $\lambda=1.83$. This computation is laid out in this Desmos snapshot.

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    $\begingroup$ The same R simulation suggests that with $n=98$ about $95.6\%$ of the volume of the inner $n$-ball may be outside the hypercube. $\endgroup$ Commented Sep 3 at 7:39

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