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2017/10/03 - Four points - the proof
That's reasonable, and lots of people have sent me their solutions. Peter had also said:
Remember, it's the proof that matters.
We have four points, and therefore six pairs. But we only have two distances, so some are short, and some are long. The short ones are all the same as each other, and the long ones are all the same as each other. Obvious, I know, but we're just warming up.
So let's prove a few simple results.
Lemma: No three points are collinear.
If $D$ is the same distance from $A$ and $C$ then that must be the long distance, but then $|BD|$ is neither $|AB|$ nor $|AC|$. So that doesn't work.
So $|AD|$ and $|CD|$ are different lengths. We can compute exactly (up to symmetry) where $D$ must lie, then observe that again, $|BD|$ is neither of our possible lengths, so that doesn't work either.
So there is nowhere to put point $D$, and hence no three points can be collinear.
Observation: Each edge is in exactly two triangles.
Lemma: There are at least two short edges that share a vertex.
We cannot have no short edges, because then we'd have all long edges, and the points would have to be in a tetrahedron, which isn't in a plane.
We cannot have exactly one short edge, because then the other five would all be long, and have to form two equilateral triangles with a shared edge. The remaining edge would have to be longer.
We cannot have $|AB|$ and $|CD|$ as the only short edges, because both $C$ and $D$ would have to be on the perpendicular bisector of the line $AB$, and they would have to be on opposite sides of the line, making $|CD|$ too long. So if there are exactly two short edges they must share a vertex, which is what we need.
Any three or more edges must share at least one vertex.
So there are (at least) two short edges that share a vertex.
Now for the single insight that, I think, make the enumeration of all solutions quite straight-forward. The fact is simple, making certain it is a fact is a little tedious - feel free to skip this section if you're happy with the statement of the lemma.
Lemma: There is a triangle with two short and one long side.
We proceed by the cases of how many long edges there are.
We cannot have no long edges.
If there is exactly one long edge then either triangle it forms will suffice.
If there are exactly two long edges, consider one of them. It is part of exactly two triangles, at most one of which can contain the other long edge, so at least one of which has two short sides.
If there are exactly three long edges then we have three short edges. Take two that share a vertex - $AB$ and $BC$. If $AC$ is short then all distances to $D$ would have to be long, which is not possible. So $AC$ is long, and $ABC$ is our required triangle.
If there are exactly four long edges then the two short edges share a vertex and the remaining edge of that implied triangle is long.
We cannot have exactly five or six long edges.
So we are done.
Finding all solutions ...
Now let's consider the fourth point, $D$. Without loss of generality we can assume that $D$ is on or to the right of the blue line. If not, reflect around that line.
The distances from $D$ to each of $A$, $B$, and $C$ must be short or long, so let's write all the possible combinations of each of those distances.
Cases 1 and 3 are not possible, for then $D$ would be to the left of the blue line, which we have excluded.
So there are 8 combinations, two of which are impossible, and the other six give exactly one solution.
And we are done.
Other solutions ...
Here is the solution from Josh Jordan:
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