A note on Erdős's mysterious remark

TL;DR

The paper studies Erdős's problem of a 6-point planar set $S$ with every triple forming an isosceles triangle, providing an alternative algebraic-geometry proof that the only solution is the regular pentagon with its center. It encodes isosceles constraints as polynomial relations, uses the Rabinowitsch trick to enforce nondegeneracy, and applies elimination to derive all candidate configurations; by reducing to a 5-point case and enumerating possibilities with a computer algebra system, it shows that no 6th point can extend a valid 5-point template. It then confirms Erdős's remark #91 in the plane case by characterizing all 5-point minimizing sets and verifying the uniqueness of the regular pentagon up to similarity, aided by a partial computer-assisted analysis. Overall, the work demonstrates how computational algebraic geometry can resolve classical discrete-geometry questions and clarifies the structure of extremal distance-sets in the plane.

Abstract

We give an alternative proof of the statement, by using elimination from algebraic geometry, that the only set $S\subset\mathbb{R}^2$, $\left|S\right|=6$ such that all subsets that form a triangle are isosceles triangles, is the regular pentagon with its center. Our proof can be extended to answer some related questions raised by Erdős.

Figures (3)

• Figure 1: A geometric realization of the obtained 33 points.
• Figure 2: Configurations with a 5 shape.
• Figure 3: Configurations that form a pentagram with its center.