Four equidistant neighbours yield six isosceles base pairs
ProvedProblem97.iCountAt_ge_six_of_K4convex-geometrydouble-countingequidistant-neighbourserdos-97-96-countingisosceles-triangleslean-formalization
Let be a finite planar point set and a point. If at least four members of lie at one common positive distance from , then at least six unordered two-point subsets of are equidistant from . In the notation of the formalization, the local isosceles count satisfies .
Preamble
/- Generated theorem stub from Erdos9796Proof.P97.IsoscelesCount by Stage 2 proof cut; source SHA-256 207564b11a38e9270192c8e5de3202b113b7f89c7ece5af3bc12954928d3013c -/ import Definitions.Def_Erdos9796Counting_Adapter import Definitions.Def_Erdos9796Counting_Foundation import Definitions.Def_Erdos9796Counting_IsoscelesCount import Mathlib.Algebra.BigOperators.Group.Finset.Basic import Mathlib.Algebra.Order.BigOperators.Group.Finset import Mathlib.Data.Finset.Powerset open Problem97 /-! # Isosceles count for Erdős Problem 97 (Dumitrescu lower bound, Milestone 3) Defines the per-vertex and total isosceles count of a finite point set, in the Dumitrescu 2006 / Nivasch–Pach–Pinchasi–Zerbib 2013 convention (equilaterals counted three times), and proves the easy lower bound: Per-vertex `K4` ⇒ each vertex contributes `≥ C(4,2) = 6` isosceles pairs. Summing: `6 · |A| ≤ I(A)`. The matching upper bound `I(A) ≤ (11·|A|²−18·|A|)/12` for convex point sets is Dumitrescu 2006 eq. (5), still open (`p97-isosceles-count-upper-bound`). References: doc slug `p97-isosceles-obstruction`. -/ set_option linter.style.openClassical false open scoped EuclideanGeometry open Finset Classical
Formal statement
theorem Problem97.iCountAt_ge_six_of_K4 (A : Finset ℝ²) (p : ℝ²)
(hp : HasNEquidistantPointsAt 4 A p) : 6 ≤ iCountAt A p := by sorrySource