Consider $n$ points $\{p\_i\}\_{i=1}^n$ located inside or on a circle with radius $r$ in the plane. The question is: how to place the $n$ points so that the sum of inter-point distances,
$$J=\sum\_{i=1}^n\sum\_{j=1}^n \|p\_i-p\_j\|^a,$$
is maximized? Here, $a$ could be 1 or 2.


The intuition is that the optimal solution is that all the points should distribute evenly on the circle. Any hints about it? Thanks.




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Another relevant problem is: what if these points are located on the surface of a sphere instead of the plane? In particular, the points should be within a bounded area on a sphere and the distance is the length of the shortest arc connecting the two points instead of Euclidean distance.



Maximizing the distance sum of some points inside a circle
Let 0 be the centre of your circle of radius $r$.


If $a=2$, we expand the brackets and write $\sum\_{i,j} (p\_i-p\_j)^2=2n\sum p\_i^2-2(\sum p\_i)^2\leqslant 2nr^2$, with equality if and only if all points lie on a circle and 0 is a barycentre.


If $a=1$, we again may suppose that they lie on circle (since the function $x\to\|p-x\|$ is convex, the sum of such functions is maximized when all points lie on the boundary of the disc they belong to.) So, they form a convex cyclic $n$-gon, without loss of generality their order is $p\_1p\_2\ldots p\_n$. For each $j=1,2,\ldots,\lceil n/2\rceil$ the cyclic sum of $\|p\_i-p\_{i+j}\|$ is maximized for the regular polygon. Indeed, if you denote by $2\varphi\_i$ the angle measure of the arc $p\_ip\_{i+1}$ (not containing $p\_{i+2}$), so that $\sum \varphi\_i=\pi$, you should maximize $\sum \sin (\varphi\_{i}+\varphi\_{i+1}+\ldots+\varphi\_{j-1})$. All arguments of the sine function $\varphi\_{i}+\varphi\_{i+1}+\ldots+\varphi\_{j-1}$ belong to $[0,\pi]$, and their sum is fixed and equals $j\pi$. Since sine is concave on $[0,\pi]$, the sum of sines is maximal when all sums are equal to $\pi j/n$. This is the case for a regular polygon, and already for $j=1$ this holds only for regular polygons.


