7
$\begingroup$

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?

$\endgroup$

1 Answer 1

7
$\begingroup$

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.

$\endgroup$
2
  • 2
    $\begingroup$ very nice Yoav! Here is a harder question: do the points have to be concyclic? $\endgroup$ Nov 19, 2015 at 19:16
  • 1
    $\begingroup$ Good question. I don't know. Might be worth another question? $\endgroup$ Nov 20, 2015 at 2:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.