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Is there a nice proof for the following fact?

In a plane, there does not exist a square such that its vertices are at a rational distance from each vertex of some equilateral triangle.

What if we replace the square with a cyclic quadrilateral?

Well, it seems the following idea works for the first part of the question: It is well known that there is no equilateral triangle in the Cartesian plane with rational coordinates. Let $(x,y)$ be at rational distance from the square with vertices, $(0,0), (0,a), (a,a), (a,0)$. Then it is easily seen that $2a(x-a)$ is rational. Hence if the coordinate system is scaled by a factor of $a$ and the origin is shifted to $(a,a)$, the coordinates become rational.

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1 Answer 1

It turns out to be false for general cyclic quadrilateral. Let the triangle have unit side length. Take point $P$ on circumcircle so that $PA=\frac{3}{7}; PB=\frac{5}{7}$ (it would lie on a circle, thats easy to prove from cosine rule). By Pompieu theorem, $PC=\frac{8}{7}$. Now take points $Q,R,S$ with the same but permitted distances, and we have a quadrilateral satisfy yours conditions.

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