If $\mathbb{R}^2$ is finitely coloured then must there exist some colour class which contains the
vertices of a rectangle of every area?
Graham, "On Partitions of 𝔼ⁿ", Journal of Combinatorial Theory, Series A 28, 89-91 (1980).
(See "Concluding Remarks" on page 96.)
Solved (with answer `False`, as formalised below) in:
Vjekoslav Kovač, "Coloring and density theorems for configurations of a given volume", 2023
https://arxiv.org/abs/2309.09973
In fact, Kovač's colouring is even Jordan measurable (the topological boundary of each
monochromatic region is Lebesgue measurable and has measure zero).
Actions
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theorem erdos_189 :
answer(False) ↔ Erdos189For
(fun a b c d ↦
line[ℝ, a, b].direction ⟂ line[ℝ, b, c].direction ∧
line[ℝ, b, c].direction ⟂ line[ℝ, c, d].direction ∧
line[ℝ, c, d].direction ⟂ line[ℝ, d, a].direction)
(fun a b c d ↦ dist a b * dist b c)
