Junior Balkan Mathematical Olympiad Shortlist

Solution:

Let $K, F, H, Z$ be the projections of point $M$ on the sides of the square.
Then by Pythagorean Theorem we can prove that $M A^2 + M C^2 = M B^2 + M D^2$.
From the given condition we obtain $M A = 1$.
With center $A$ and angle $60^{\circ}$, we rotate $\triangle A M E$, so we construct the triangle $A N B$.

Since $A M = A N$ and $\widehat{M A N} = 60^{\circ}$, it follows that $\triangle A M N$ is equilateral and $M N = 1$. Hence $\triangle B M N$ is right-angled because $B M^2 + M N^2 = B N^2$.
So $m(\widehat{B M A}) = m(\widehat{B M N}) + m(\widehat{A M N}) = 150^{\circ}$.
Applying Pythagorean Generalized Theorem in $\triangle A M B$, we get:
$$A B^2 = A M^2 + B M^2 - 2 A M \cdot B M \cdot \cos 150^{\circ} = 1 + 2 + 2 \sqrt{2} \cdot \sqrt{3} : 2 = 3 + \sqrt{6}$$
We conclude that the area of the square $A B C D$ is $3 + \sqrt{6}$.