Bulgaria competition problems, 2022

From the circumscribed circle of $\triangle ADE$ we derive (using inscribed angles and their corresponding arcs) $MD = ME$ and $\angle BDM = 180^\circ - \angle ADM = \angle AEM$, which combined with $AE = BD$ gives rise to $\triangle AEM \cong \triangle BDM$ - thus $AM = MB$, meaning that $M$ is the intersection point of the segment bisector of $AB$ and the angle bisector of $\triangle BAC$. Analogously, $N$ is the intersection point of the segment bisector of $AB$ and the angle bisector of $\triangle ABC$. In particular, we deduce that $MN$ is the segment bisector of $AB$. Let $MN$ intersect $AB$ at its midpoint $K$. Since $AB^2 + AC^2 = BC^2$, it follows that $\angle BAC = 90^\circ$. Now $BC = 2AB$ gives rise to $\angle ABC = 60^\circ$. Hence, $\angle MAK = 45^\circ$ and $\angle NBK = 30^\circ$. In conclusion, from $\triangle AMK$ we deduce $MK = AK = \frac{1}{2}$, while from $\triangle BNK$ we deduce $NK = \frac{1}{2\sqrt{3}}$ (indeed, if $NK = y$, then $BN = 2y$ since $\angle NBK = 30^\circ$ and $y^2 + (\frac{1}{2})^2 = 4y^2$, i.e. $y = \frac{1}{2\sqrt{3}}$). Finally, $MN = MK - NK = \frac{3-\sqrt{3}}{6}$.