Let there be $m$ points in $\mathbb R^n$. Let $D$ be the longest distance between two of these points and let $d$ be the smallest. What is the smallest possible value of $\frac{D}{d}$ for each value of $n$ and $m$, and which configurations reach it?


It is clear when $n$ is $1$ the solution is reached when the points are evenly spaced in a line.


For $n=2$ I think the answer may be the regular polygons. For higher dimensions maybe it is reached by setting all the points in a sphere and trying to spread them out as best possible?



minimum ratio between the shortest and longest distances between $m$ points in $\mathbb R^n$
This is only a partial answer. Though it hints that the situation is already complicated enough even asymptotically for $m$ large.


In the following we define $\lambda(n, m):=\min D/d$, where minimum is taken over all possible sets of $m$ points in $\mathbb{R}^n$.


Turns out, asymptotically, for $m$ large we have $\lambda(2,m) = (12/\pi^2)^{1/4}m^{1/2}$, realised by a part of the hexagonal lattice. Yet, for fixed $m$ the exact answer is tricky to find, seems like it's only known for $m$ up to $8$ with $\lambda(2, 7)=2$ and $\lambda(2, 8)=\frac{1}{2\sin{(\pi/14)}}$. See <https://core.ac.uk/download/pdf/82075638.pdf> for the proof of the latter equality and a short (though not self-contained) argument for the asymptotic behaviour of $\lambda(2, m)$.


One can also see (and rigorously proof) that for fixed $n$, the asymptotic behaviour of $\lambda(n, m)$ is given by $\lambda(n, m) \sim \Delta\_n^{-1/n}\cdot m^{1/n}$, where $\Delta\_n$ if the sphere packing constant in dimension $n$. So the problem for higher dimensions is tricky...


