Croatian Mathematical Competitions

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 $\sqrt{2014}$.