# Maximizing the distance sum of some points inside a circle

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.

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.

• For, $a=2$, $J=n\sum_{i}r_i^2 -[S_x^2+S_y^2+S_3^2]$ where, $r_i$ is the distance from the origin and $S_{x,y,z}=\sum_{i} \{x,y,z\}_i$. Hence, it will be maximum iff $(S_x,S_y,S_z)=\vec{0}$ and distances are maximum possible. This happens only when all the points are on the circle and forms a regular $n$-gon. May 9 at 15:45
• @AlapanDas not only, it happens when they are on the circle and 0 is barycentre May 10 at 8:49
• @Fedor Petrov, oh, I see. Thank you for correcting me. I had mistakenly assumed the barycentre at 0 as the points making regular $n$-gon on the sphere. May 10 at 9:09
• I believe the new question (on a sphere) is sufficiently different that it would be better asked in a separate question. May 10 at 11:58

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.