Points at unequal distances

Putting n points on a plane, this defines Cn2 = n(n-1)/2 distances.
We care of the case when these distances have exactly two different values.

3 points : There is only one pattern : a triangle. With one distance, only an equilateral triangle.
With two different distances, any isosceles triangle. That is the ratio of distances is undefined.

4 points, this becomes more interesting :
How many different patterns with two values of distances ? Ratio of these values ?
   Solution
"Of course" we can't put 4 equidistant points on a plane !
With 4 points and 3 values for distances, the problem becomes uninteresting, as for the 3 points and two values case...

5 points, with two values ?
   Solution
With three values for distances ?

6 points ? That is put 6 points at two values for all the distances.

Details

 

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