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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 [Gr80] has shown that this is true if we replace rectangle by right-angled triangle. The same question can be asked for parallelograms. It is not true for rhombuses.

This is false; Kovač [Ko23] provides an explicit (and elegantly simple) colouring using 25 colours such that no colour class contains the vertices of a rectangle of area $1$. The question for parallelograms remains open.

Additional thanks to: Ryan Alweiss, Vjekoslav Kovac