Given points $A_0,A_1,\ldots,A_n$ in the plane, let $m$ denote the minimum distance among any two points. What is the minimum value of $$\dfrac{|A_0A_1|\cdot|A_0A_2|\cdot\ldots\cdot|A_0A_n|}{m^n}?$$

Here $|AB|$ denotes the Euclidean distance between points $A$ and $B$.

[Source: Based on Chinese competition problem]

  • $\begingroup$ I'm pretty sure by $|A_0 A_1|$ etc. you mean the distance between the two points but you should make that clear. $\endgroup$ – user2566092 Oct 24 '14 at 19:43
  • $\begingroup$ @user2566092 Yes, thanks. $\endgroup$ – simmons Oct 24 '14 at 19:51
  • $\begingroup$ The simple answer is $1$ (unless any two of the points are coincident, in which case the $m=0$ and the ratio does not exist) but I can see there's probably a better solution because for $n>6,m$ is likely $<|A_0A_k|$. $\endgroup$ – Mark Hurd Oct 28 '14 at 1:54
  • $\begingroup$ A natural configuration is to construct a triangular lattice with $A_0$ at the origin, and then to take the $n$ nearest neighbors to $A_0$ as the other $n$ points. This is clearly optimal for $n\leq 6$. Is there any $n$ for which this is provably suboptimal? $\endgroup$ – dshin Jan 24 '16 at 0:57
  • $\begingroup$ I guess this Chinese competition problem (China high school math competition (Oct 14, 2012) problem 15). $\endgroup$ – Alex Ravsky Nov 7 '17 at 2:55

Lets consider




We know that for any k:


Thus the minimum value for


would be $1$, but we must show that such a configuration can indeed exist on the plane. For this we will just pick three points $A_0,A_1,A_2$ such that they form an equilateral triangle and we get that our expression evaluates to $1$, so we are finished.

In the case where $m=0$ and $2$ points coincied our expression is undefined so it is not at a minimum there either.

  • $\begingroup$ Presumably the $n+1$ points are distinct, so that $m\gt 0$ can be assumed. As the Comments on the Question explain, the ratio $1$ can be attained for $n\le 6$, but not for larger counts $n$. The Original Poster has not been around for more than a year, so there may not be a clarification of the actual scope of the problem forthcoming. $\endgroup$ – hardmath Jun 27 '17 at 15:53

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