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Let us assume that the radius of the circumcircle of the triangle $AXY$ is equal to $\sqrt{2014}$. In the triangle $AXY$ the angle opposite to the side $XY$ is $60^{\circ }$ because the triangle $ABC$ is equilateral. If $R$ is the radius of the circumcircle of the triangle $AXY$, then $$|XY|=2R\sin 60^{\circ }=\sqrt{2014}\cdot \sqrt{3}.$$ Let $|AX|=m$, $|AY|=n$, where $m$ and $n$ are positive integers. Applying the cosine rule to the triangle $AXY$ we get $m^2+n^2-2mn\cos 60^{\circ }=|XY{|}^2$, which leads to $$m^2+n^2-mn=2014\cdot 3.$$ Let us first notice that the right-hand side of the equation is even. If $m$ and $n$ are of different parity or if they are both odd, then the left-hand side is odd, which is impossible. The only remaining case is when $m$ and $n$ are both even. Then $m^2+n^2-mn$ is divisible by $4$, but $2014\cdot 3$ is not, so we get another impossible case.

Therefore, the initial assumption is wrong, which means that $R$ can not be equal to $\sqrt{2014}$.