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2017/09/26  Four Points, Two Distances
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 equidistant
then they will have to be at the vertices of a tetrahedron, which is
most definitely nonplanar. 
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 $n1$ 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 casebycase 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.
Send us a comment ...
