# Rational distance from vertices of an equilateral triangle

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 $\mathbb{R}^3$).

• There seems to be a typo: is your ball centered at $P$ or $x$? – Joonas Ilmavirta Sep 5 '14 at 21:02
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.