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Source. Theorem 2, p. 66, of Peter Fishburn, "Convex polygons with few intervertex distances," Computational Geometry 5 (1995), no. 2, 65--93, doi:10.1016/0925-7721(94)00020-v, the edition named on the source card.
Read depth. Claims checked: the notation of p. 66, the statement, Fig. 1 (p. 67) and the closing paragraph of Section 3 (p. 81) were read clause by clause on the page images. The proofs of Sections 2 and 3 were read for structure only. Nothing here is independently reviewed.
Statement
Setting (p. 66). For , is the class of all convex -gons in the plane, and means that and are similar (rotation, reflection, translation and uniform rescaling). A subclass of "contains polygons" when there are pairwise dissimilar such that the subclass consists of exactly the similar to some . is the regular -gon and , for , is a regular -gon with vertices deleted, a member of . With vertex set , is the number of distinct intervertex distances of , and .
Theorem 2 (p. 66, quoted). " contains 4 polygons, and contains 3 polygons: see Fig. 1. For every even , contains 2 polygons, and ."
In the corpus's words: by Altman's bound (Theorem 1) a convex -gon with even has at least distinct distances, and for every even it has exactly if and only if it is similar to or to .
The small cases (Fig. 1, p. 67, whose caption reads " is all convex -gons with exactly intervertex distances"). Each polygon is labelled with its multiplicity vector, the multiplicities of its distances in decreasing order without regard to which distance has which (pp. 66, 68):
- : with , drawn as a rhombus whose one diagonal is the only other distance; with ; with ; and with .
- : with ; with ; and with .
- , as an instance of the general case: with and with .
Proof pointer
The case is stated without proof as straightforward (p. 66). is Section 2 (pp. 69--71), a case analysis on where the longest distance sits, using Altman's Lemma 1 and his Lemmas 2 and 3 (p. 69). The case , , is Section 3 (pp. 72--81): Lemma 4 (p. 72) confines the longest distances from a chosen vertex 1 to the vertices and possibly or . When has a single longest segment, Lemma 5 (p. 72) makes every segment between opposite vertices , one of the two longest distances, and further use of Lemmas 1--3 forces equal sides and equal short diagonals on one circle, so (p. 73). Otherwise Lemmas 6--9 (pp. 73--81) place vertices equally spaced on a circle and then every other vertex on the same circle, giving . The paper notes (p. 81) that this argument fails at , where is a third polygon, and that also follows from the classification of in Theorem 3 by deleting a vertex of an octagon in (p. 72).
Dependencies
Theorem 1 and Lemma 1, with Altman's Lemmas 2 and 3 (p. 69).
Bears on
- Problem 132: the problem asks whether every points in the plane have two distances that each occur at least once but between at most pairs. The theorem is stated for the vertices of convex polygons only. Its two polygons for even have every distance occurring at most times: in each distance other than the diameter occurs times and the diameter times, and in each distance occurs times (a count made here, not printed in the paper). The paper says nothing about sets not in convex position or about the growth of the number of such distances.