Trivia

Subject:
Re: Friday Puzzler -- Four Dots - Proof?

Larry Barrett
Problem: How many ways can 4 dots be placed on a sheet of paper so that there are at most two different distances between pairs of dots.

Place two dots on a sheet of paper with a distance x between them.
Now place a third dot on the paper. This dot must be the same distance, x, from at least one of the first two dots. If not, the three dots would have three different distances between each pair, violating the premise.
Case 1: the third dot is the same distance x from each of the first two. In this case you have formed an equilateral triangle. The fourth dot could be place at the centroid of the triangle, with a distance y between it and each of the other three dots; or it could be placed outside the triangle the same distance x from two of the three dots but distance z to the fourth dot. This forms a rhombus. If it was outside the triangle but distance x from just one of the first three dots, then there would be three separate distances between pairs.

Case 2:The third dot is a distance x from just one of the first two, but is a distance y from the other dot. The third dot could be on the extended line between dots 1 and 2 (case2a), or it could be at an angle 90 degrees from one dot, say dot 1 (case 2b), or it could be at an angle less than 90 degrees to dot 1(case 2c).

For case 2a, I do not believe there is any way to place the fourth dot that will not violate the premise of no more than two distinct distances between pairs of dots.

For case 2b, the third dot is placed at 90 degrees and at a distance x from dot 1 and at a distance y to dot 2. Then the fourth dot would have to be at 90 degrees to both dot 3 and dot 2, and at a distance x to both dot 3 and dot 2, and at a distance y to dot 1, forming a square.

For case 2c, the third dot is place at a distance x from dot 1 and at an angle less than (or greater than) 90 degrees. It will then have a distance y to dot 2. The fourth dot must then be placed so that it has a distance x to dots 2, 3, and 4, forming a rhombus as in case 1.

So it seems to me that there are only 3 solutions to this problem, the ones identified in other replies.

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