---

Source. Theorem 3, p. 67, 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 statement, the remark after it (p. 67), Fig. 2 (p. 68), Fig. 10 (p. 84) and the closing sentences of Sections 4 and 5 (pp. 84, 92) were read clause by clause on the page images. The proofs were read for structure only. Nothing here is independently reviewed.

Statement

Notation as on Theorem 2: $M_n(t)$ is the class of convex $n$-gons with exactly $t$ distinct intervertex distances, "contains $N$ polygons" counts similarity classes, and $R_n-k$ is a regular $n$-gon with $k$ vertices deleted, its "dissimilar versions" coming from deleting different combinations of vertices (p. 66).

Theorem 3 (p. 67, quoted). "$M_3(2)$ is the class of all nonequilateral isosceles triangles. $M_5(3)$ contains the 15 pentagons shown in Fig. 2. $M_7(4)$ contains 5 polygons, namely $R_8-1$ and the four dissimilar versions of $R_9-2$."

For odd $n$ these are the classes one above Altman's minimum $(n-1)/2$ (Theorem 1).

The pentagons (Fig. 2, p. 68), labelled (5.1)--(5.15), with their multiplicity vectors (multiplicities in decreasing order, not matched to the distances): one with $(6,3,1)$, one with $(6,2,2)$, one with $(5,4,1)$, four with $(5,3,2)$, four with $(4,4,2)$ and four with $(4,3,3)$. The figure also names a construction for each but (5.11), such as $R_4+1$, $A_6-1$ or $R_7-2$.

The heptagons (Fig. 10, p. 84): $R_8-1$ with multiplicity vector $(6,6,6,3)$, and four versions of $R_9-2$, each with $(6,5,5,5)$.

A suggestion, not a theorem (p. 67). The paper writes that the result for $n=7$ "suggests" that for odd $n\geqslant9$, $M_n((n+1)/2)$ contains $(n+3)/2$ polygons, $R_{n+1}-1$ and the $(n+1)/2$ dissimilar versions of $R_{n+2}-2$. An added-in-proof note (p. 93) states that the case $n=9$ is verified in a separate paper of Erdős and Fishburn, then in press in Geometriae Dedicata.

Proof pointer

$M_3(2)$ is not argued. $M_5(3)$ is Section 4 (pp. 81--84): a pentagon is built by adding a fifth vertex to a member of $M_4(2)$ where possible, which gives eight of the fifteen, and the remaining cases, where no quadrilateral of the pentagon has two distances, are settled by case analysis on the longest segments with the plane facts collected as Lemma 0 (pp. 68--69) and Altman's Lemma 1 and Lemmas 2 and 3. $M_7(4)$ is Section 5 (pp. 84--92), split by the position of the longest segments into three parts and many cases, using the classifications of $M_6(3)$ (Theorem 2) and $M_5(3)$ after deleting vertices.

Dependencies

Theorem 2 (the cases $M_4(2)$ and $M_6(3)$) and Lemma 1.

Bears on

• Problem 132: the printed multiplicity vectors show that each of the fifteen pentagons has at least two distances occurring at most $5$ times, and each of the five heptagons has all four distances occurring at most $7$ times (a reading of the figures made here). These are convex examples only; the paper makes no statement about the problem.