What is the maximum number of points in space that can be equidistant from each other?

Question

What is the maximum number of points in space that can be equidistant from each other?

Answer 1

4

I just don't know how to prove it mathematically. All I know is that the only two dimensional way of doing that is three points in an equilateral triangle - in the third dimension, only one more point can be added.

Answer 2

Humm, interesting question.

I think the answer is probably three, an equlataral triangle.

I think this is because there are no crossed verticies so the logest distance is between the points and no a section across the shape.

Answer 3

4

Tetrahedron

Pyramid with triangular base.

Answer 4

Let X be a set {p1,...,pn} of mutually different equidistant points, with distance a, so d(pi,pj)=a for i != j. Suppose n greater or equal 3, so d(p1,p2)=d(p2,p3)=d(p1,p3), and a small calculation shows that the points will form an equilateral triangle (see Appendix).

If you try to add a fourth point, you can (easily) calculate that only 2 points in three-space exist, which are equidistant to p1-p3. Adding one of them will result in a tetrahedron, and n=4. Adding both will not satisfy d(p4,p5)=a, therefore n=4 is maximum.

Appendix:

Without loss of generality we can consider p1=(0,0,0), p2=(a,0,0) and p3=(x3,y3,0).

Since x3²+y3²=a², and (x3-a)²+y3²=a² we get

x3=a/2 and y3=±sqrt(3)·a/2, again without loss we get p3=(a/2,sqrt(3)·a/2,0), in other words three points will form an equilateral triangle, chosen to be in the x-y plane. If you carry this on, you get p4=(a/2,sqrt(3)·a/4, ±3/4·a), as described earlier.

Answer 5

4 and you can prove it by drawing spheres from the 4 points. The points of intersection of 3 spheres are the only possible answers.