Let $A\_1,A\_2,\ldots, A\_n$ be distinct in the plane. For every $1\le i \le n$, let
$S\_i=\sum\limits\_{j=1}^n d(A\_i,A\_j)$ be the sum of distances from $A\_i$ to all the other points.


Assume that $S\_i=S\_j$ for all $1\le i<j\le n$. Is it true that the points $A\_1,A\_2,\ldots, A\_n$ must be the vertices of a convex $n$-gon?



metric condition forcing convex position
Suppose that the points are not in convex position, then there is a point (WLOG let it be $A\_n$) that is a convex combination of other points:


$$A\_n = \sum\_{i=1}^{n-1} \lambda\_i A\_i\text,$$


where $\lambda\_i\ge0$ and $\sum\_i\lambda\_i=1$. Now consider the function $f\_i(X) = d(A\_i,X)$. This function is convex: its value at a convex combination of points is no more than the same convex combination of the values at the points. It is strictly convex as long as the points are not on a line through $A\_i$. Therefore, you have that


$$S\_n = \sum\_{i=1}^{n-1} d(A\_i,A\_n) < \sum\_{i=1}^{n-1}\sum\_{j=1}^{n-1} \lambda\_j d(A\_i,A\_j) < \sum\_{j=1}^{n-1} \lambda\_j S\_j\text.$$


This is impossible if $S\_i = S\_j$ for all $i,j$, so, by contradiction, the points must be in convex position.


