# $n$ points on plane with sum of squares of L2 norm = 1

Let $p_1,p_2$ be two points on the 2 dimensional plane. $|p_1|$ denotes the $L^2$ norm of $p_1$ and $\delta(p_1,p_2)$ denotes the euclidean distance between $p_1$ and $p_2$. Let $f(p_1,\ldots, p_n) = \min_{i\neq j} \delta(p_i,p_j)$. Given $n$, we want to find $n$ points $p_1,\ldots,p_n$ on the 2 dimensional plane such that $f(p_1,\ldots,p_n)$ is maximized, given the following restriction :

$$\sum_i |p_i|^2 = 1$$

Is there any closed form expression for $p_i$s?

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Write $\mathbb R^{2n}$ elements as $(x_1,y_1,\dots,x_n,y_n)$. Define linear transformations $T_{ij}$ on $\mathbb R^{2n}$ which swaps $x_i$ with $x_j$ and $y_i$ with $y_j$. Then your question is, find an element $p\in \mathbb R^{2n}$ with $|p|=1$ such that $$\min_{i,j} |T_{ij}v-v|$$ is maximized. This in turn is because if $p_i=(x_i,y_i)$ then $|T_{ij}v-v|=2\delta(p_i,p_j)$. Not sure if that helps any, however. –  Thomas Andrews Mar 20 '13 at 19:56
Whoops, where I wrote "$p\in\mathbb R^{2n}$ with $|p|=1$" read $p$ as $v$. –  Thomas Andrews Mar 20 '13 at 20:22
Maximizing the minimum pairwise separation with fixed moment of inertia is equivalent to minimizing the moment of inertia with a fixed minimum pairwise separation. In other words, this is equivalent to finding a minimum-energy configuration of hard discs in a radial potential. For large $n$, it would be surprising if the "core" of such a configuration weren't hexagonally close-packed. –  mjqxxxx Mar 20 '13 at 21:40