The distances from a certain point inside a regular hexagon to three of its consecutive vertices are equal to $1, 1$ and $2$, respectively. Determine the length of this hexagon's side.

(Mikhail Evdokimov)
1. Let \( A, B, C \) be three consecutive vertices of the regular hexagon, with \( AB = BC = a \). Let \( M \) be a point inside the hexagon such that \( MA = MB = 1 \) and \( MC = 2 \). We need to determine the side length \( a \) of the hexagon.

2. Consider the perpendicular from \( M \) to \( AB \), meeting \( AB \) at \( N \). Since \( M \) is equidistant from \( A \) and \( B \), \( \angle MAB = \angle MBA = \alpha \) where \( \alpha \in (0^\circ, 90^\circ) \).

3. Since \( AB = a \) and \( N \) is the midpoint of \( AB \), we have \( AN = BN = \frac{a}{2} \).

4. In the triangle \( MAN \), using the definition of cosine, we have:
   \[
   AN = AM \cos \alpha \implies \frac{a}{2} = 1 \cdot \cos \alpha \implies a = 2 \cos \alpha
   \]

5. Next, we use the Law of Cosines in triangle \( MBC \). The angle \( \angle ABC = 120^\circ \) and \( \angle MBC = 120^\circ - \alpha \). Applying the Law of Cosines:
   \[
   MC^2 = MB^2 + BC^2 - 2 \cdot MB \cdot BC \cdot \cos(\angle MBC)
   \]
   Substituting the known values:
   \[
   2^2 = 1^2 + a^2 - 2 \cdot 1 \cdot a \cdot \cos(120^\circ - \alpha)
   \]
   Simplifying:
   \[
   4 = 1 + a^2 - 2a \cos(120^\circ - \alpha)
   \]

6. Using the angle addition formula for cosine:
   \[
   \cos(120^\circ - \alpha) = \cos 120^\circ \cos \alpha + \sin 120^\circ \sin \alpha = -\frac{1}{2} \cos \alpha + \frac{\sqrt{3}}{2} \sin \alpha
   \]
   Substituting this into the equation:
   \[
   4 = 1 + a^2 - 2a \left( -\frac{1}{2} \cos \alpha + \frac{\sqrt{3}}{2} \sin \alpha \right)
   \]
   Simplifying further:
   \[
   4 = 1 + a^2 + a \cos \alpha - a \sqrt{3} \sin \alpha
   \]

7. Recall that \( a = 2 \cos \alpha \). Substituting \( a \) into the equation:
   \[
   4 = 1 + (2 \cos \alpha)^2 + 2 \cos \alpha \cdot \cos \alpha - 2 \cos \alpha \cdot \sqrt{3} \sin \alpha
   \]
   Simplifying:
   \[
   4 = 1 + 4 \cos^2 \alpha + 2 \cos^2 \alpha - 2 \sqrt{3} \cos \alpha \sin \alpha
   \]
   \[
   4 = 1 + 6 \cos^2 \alpha - 2 \sqrt{3} \cos \alpha \sin \alpha
   \]

8. Using the double-angle identities \( \cos 2\alpha = 2 \cos^2 \alpha - 1 \) and \( \sin 2\alpha = 2 \sin \alpha \cos \alpha \):
   \[
   4 = 1 + 3 (\cos 2\alpha + 1) - \sqrt{3} \sin 2\alpha
   \]
   \[
   4 = 1 + 3 \cos 2\alpha + 3 - \sqrt{3} \sin 2\alpha
   \]
   \[
   4 = 4 + 3 \cos 2\alpha - \sqrt{3} \sin 2\alpha
   \]
   \[
   0 = 3 \cos 2\alpha - \sqrt{3} \sin 2\alpha
   \]

9. Solving for \( \tan 2\alpha \):
   \[
   3 \cos 2\alpha = \sqrt{3} \sin 2\alpha \implies \tan 2\alpha = \frac{\sqrt{3}}{3} = \sqrt{3}
   \]
   \[
   2\alpha = 60^\circ \implies \alpha = 30^\circ
   \]

10. Finally, substituting \( \alpha = 30^\circ \) back into the expression for \( a \):
    \[
    a = 2 \cos 30^\circ = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}
    \]

The final answer is \( \boxed{\sqrt{3}} \).