How is it proved that there is no configuration of five or more mutually equidistant points in $R^3$?
Is it done by induction? I'm stuck. Help would be appreciated. Well, surely equilateral polyhedron does work.
$\begingroup$
$\endgroup$
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
It should be obvious that the configurations for 3 and 4 points are the equilateral triangle and the equilateral tetrahedron.
Now try to add another point to the tetrahedron configuration. There is only one point that has the same distance to all vertices: the centroid. But the distance of the centroid to a vertex is not equal to the length of the edges. Therefore, there is no such configuration for 5 points.
answered Oct 16, 2014 at 6:30
$\begingroup$
$\endgroup$
2
We can check by induction that the only way to place $n+1$ equidistant points in $\mathbb{R}^n$ is to take the vertices of a regular simplex. Then, there will be no place for $n+2$th point.
-
$\begingroup$ this sounds nice. can you please elaborate it a bit? $\endgroup$ Commented Oct 15, 2014 at 20:05
-
1$\begingroup$ Assume you have $n$ points in $\mathbb{R}^n$. By induction they form a regular $(n-1)$-simplex. Where can the $(n+1)$th point be located? What about the $(n+2)$th point? $\endgroup$– user2097Commented Oct 16, 2014 at 15:35
$\begingroup$
$\endgroup$
2
There may be possibilities is the parameters allow for the fourth dimension. Otherwise, the problem is not configurable
-
1
-