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!



Rational distance from vertices of an equilateral triangle
Yes. In fact, more is true. T. G. Berry (["Points at rational distance from the vertices of a triangle"](http://matwbn.icm.edu.pl/ksiazki/aa/aa62/aa6246.pdf), *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).


