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Let $t_{m}$ be the sup of the sum of the pairwise distances between any $2m$ points in the unit disk. Does $t_{m}/m^{2}$ go to $0$ as $m\rightarrow\infty$?

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closed as off-topic by Benoît Kloeckner, Ramiro de la Vega, Ricardo Andrade, Suvrit, Kevin P. Costello Sep 23 '13 at 22:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Benoît Kloeckner, Ramiro de la Vega, Ricardo Andrade, Suvrit, Kevin P. Costello
If this question can be reworded to fit the rules in the help center, please edit the question.

I'm puzzled by how you could think this would go to $0$. You are asking whether the average distance between points must go to $0$ when you have more and more of them... of course not. – Douglas Zare Sep 23 '13 at 22:33
I can't even deduce a precise meaning and you already closed the question? Wasn't it better first to be a bit patient to find out what the question really was? – Włodzimierz Holsztyński Oct 6 '13 at 3:02

No, it need not go to $0$: just put $m$ points at $(0,-1/2)$ and the other $m$ points at $(0,+1/2)$; the sum of the pairwise distances is $t_m=m^2$, so $t_m/m^2=1$ does not vanish as $m\rightarrow\infty$. More generally, if you distribute the points inside the unit disk with some generic density function, there will be some nonzero average pairwise distance $\bar{t}$, and $t_m/m^2\simeq m(2m-1)\bar{t}/m^2$ will remain nonzero as $m\rightarrow\infty$.

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