# n points can be equidistant from each other only in dimensions $\ge n-1$?

2 points are from equal distance to each other in dimensions 1,2,3,...

3 points can be equidistant from each other in 2,3,... dimensions

4 points can be equidistant from each other only in dimensions 3,4,...

What is the property of number dimensions that relates the number of points that can be equidistant to all other points?

• mathoverflow.net/questions/30270/… – Martin R May 28 '15 at 12:04
• Let $E^{N}$ denote $N$-dimensional Euclidean space, $n$ a positive integer, and $r$ a positive real number. Assuming the question is: "If $E^{N}$ contains a set of $n$ points, any two of which are at distance $r$, is $n \leq N + 1$?", the answer is "yes". The accepted answer to this question sketches a stronger result. – Andrew D. Hwang May 28 '15 at 12:44

## 1 Answer

One side of this, there is a standard picture. In $\mathbb R^n,$ take the $n$ points $$(1,0,0,\ldots,0),$$ $$(0,1,0,\ldots,0),$$ $$\cdots$$ $$(0,0,0,\ldots,1).$$ These are all at pairwise distance $\sqrt 2$ apart.

At the same time, they lie in the $(n-1)$-dimensional plane $$x_1 + x_2 + \cdots + x_n = 1.$$ If you wish to work at it, you can rotate this into $\mathbb R^{n-1};$ in any case, $n$ points in $\mathbb R^{n-1}.$

If you prefer, you can keep $\mathbb R^n$ and place a point numbered $(n+1)$ at $$(-t,-t,-t, \ldots, -t)$$ for a special value of $t > 0$ that makes all the distances $\sqrt 2.$ I get $$n t^2 + 2 t - 1 = 0$$ or, with $t>0,$ $$t = \frac{\sqrt {n+1} - 1}{n } = \frac{1}{\sqrt {n+1} + 1}$$