# The Four Points Puzzle

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## 2017/09/26 - Four Points, Two Distances

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At the MOVES conference in New York in August I was lucky enough to spend some time with Peter Winkler, mathematician, puzzle master extraordinaire, and author of "Mathematical Puzzles: A Connoisseur's Collection". As we talked about many things he set me this puzzle:

• Find all configurations of four (distinct) points in the plane that determine exactly two distances.

 It's worth noting that you can't have four points in the plane with just a single distance, because if all the points are equi-distant then they will have to be at the vertices of a tetrahedron, which is most definitely non-planar.
In general, four points in the plane will determine six distances, as there are six pairs, and there is no reason to expect that any of those distances would be repeated. But we can choose to place the points to create duplication, and the challenge here is to find all possible ways to do that and end up with only two distances.

Now, you can probably find one very quickly, and possibly another. But Peter said (and I paraphrase) "Nearly everyone misses at least one, and for each possible solution, it's been missed by at least one person."

Well that, of course, put me on guard, and I found all the solutions. But having said that, I found $n-1$ solutions, then started to prove that I had them all, and it was in that process that I found another.

So apparently I now have them all, and you'd think that would be enough. But no, while I had proved that I've got them all, the proof wasn't at all satisfying. It was a lengthy case-by-case analysis, grinding away until it was complete. But I felt there had to be a better way.

 You can find Josh's solution on line if you search for it, and you can find how many configurations there are. But I've not seen a simple, clear, clean, incisive proof of the result. It's not hard to find them all. The proof is the thing.
Josh Jordan took a completely different approach, and that gave me hope that there were other approaches lurking in corners, and perhaps one (or more) would be cleaner, clearer, and more enlightening.

So there you are. It's not enough to find all $n$ solutions, the real challenge is to find a simple proof that you have them all.

I'd love to hear what you try.

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