Rational distance from vertices of an equilateral triangle

Question

A colleague in my department posed the following question...

Let $A=(0,0)$, $B=(1,0)$, and $C=(1/2,\sqrt{3}/2)$. Then $\Delta ABC$ is an equilateral triangle with sides of length 1. Let $B_{\epsilon}({\bf x}) = \{ {\bf y} \in \mathbb{R}^2 \;|\; \mbox{distance}({\bf y},{\bf x})<\epsilon \}$ be an epsilon ball centered at ${\bf x}$ using the usual Euclidean distance.

Given a point ${\bf x} \in \mathbb{R}^2$ and $\epsilon>0$, is there a point $P \in B_{\epsilon}({\bf x})$ such that the distances from $P$ to the vertices of $\Delta ABC$ are all rational?

It seems like it ought to be true. It is if you just demand 2 of the distances to be rational or if you allow $P$ to live in a 3D-ball (i.e. if we spill out of $\mathbb{R}^2$ into $\mathbb{R}^3$).

Thanks for your thoughts!

Answer 1

Yes. In fact, more is true. T. G. Berry ( "Points at rational distance from the vertices of a triangle", Acta Arith. 62 (1992), no. 4, 391–398) proved:

Theorem. Let $ABC$ be a triangle such that the length of at least one side is rational and the squares of the lengths of all sides are rational. Then the set of points $P$ whose distances $PA$, $PB$, $PC$ to the vertices of the triangle are rational is dense in the plane of the triangle.

Berry says that the case where all the sides of the triangle are rational was previously proved by J. H. J. Almering ("Rational quadrilaterals," Indag. Mat. 25 (1963), 192–199).