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I found a number of them. Here's how I would go about systematically trying to find them one by one. I suppose this method could be built up to a proof. First

I found a number of them. Here's how I would go about systematically trying to find them one by one. I suppose this method could be built up to a proof. First consider the shape of the outside points: point - no straight line segment - no, after a little work triangle - maybe quadrialteral - yes pentagon and up - no For the case of 4 outside points, consider the possible lengths. aaaa - Yes. Square. Other angles - maybe aaab, b is longer - Maybe aaab, b is shorter - Maybe aabb - Maybe abab - Maybe For the case of 3 outside points, consider the possible lengths. aaa - maybe aab, b is shorter - maybe aab, b is longer - maybe

AstralStorm 7y ago
This is not even close. Obviously you missed a crossed circle and every tesselation of regular polygon has at least one of those. And that's just a start. Metric spaces are a huge area of mathematics. 1) Question did not mention distinct points. Two joined equal length segments with overlapping points qualify. 2) Question did not mention metric used, you assumed euclidean distance. 3) There are many more solutions in multidimensional spaces even just for euclidean metric. Question did not specify 2D euclidean (but explanation sort of assumed that) nor even planar. Have fun finding these in hyperbolic geometry - current model of reality due to General Relativity. Or in quantized space. 4) Do infinite distances count as distinct? Zero distances are more obvious, but what about zero limits with different signs? Mainly etc. Never ask a general question about geometry. ;)
- kangnkodos 7y ago
  I deleted my list of spoiler solutions. But you are correct. I was only thinking in 2D euclidean space. Khaaaaaan! Your other ideas are very creative, and as far as I understand them, correct (except for the idea of four DISTINCT points), but probably outside the problem. How many solutions would there be on a hyperbolic plane? No idea.
- sonofgod 7y ago
  1) is answered in the article Clarifications: The four points must be distinct. That is, no point is allowed to be superimposed on another point.

AstralStorm 7y ago

This is not even close. Obviously you missed a crossed circle and every tesselation of regular polygon has at least one of those. And that's just a start. Metric spaces are a huge area of mathematics. 1) Question did not mention distinct points. Two joined equal length segments with overlapping points qualify. 2) Question did not mention metric used, you assumed euclidean distance. 3) There are many more solutions in multidimensional spaces even just for euclidean metric. Question did not specify 2D euclidean (but explanation sort of assumed that) nor even planar. Have fun finding these in hyperbolic geometry - current model of reality due to General Relativity. Or in quantized space. 4) Do infinite distances count as distinct? Zero distances are more obvious, but what about zero limits with different signs? Mainly etc. Never ask a general question about geometry. ;)

kangnkodos 7y ago
I deleted my list of spoiler solutions. But you are correct. I was only thinking in 2D euclidean space. Khaaaaaan! Your other ideas are very creative, and as far as I understand them, correct (except for the idea of four DISTINCT points), but probably outside the problem. How many solutions would there be on a hyperbolic plane? No idea.
sonofgod 7y ago
1) is answered in the article Clarifications: The four points must be distinct. That is, no point is allowed to be superimposed on another point.