Macedonian Mathematical Competitions

$AB$ and $CD$ are parallel so we get that $\angle BAM = \angle AMD$. Because $\angle AMB = \angle AMD$ we conclude that $\angle BAM = \angle AMB$. So we obtain that the triangle $AMB$ is isosceles and $\overline{AB} = \overline{MB}$.

In the right-angled triangle $BCM$ $BM$ is a hypotenuse and two times bigger than $BC$ ($\overline{AB} = 2\overline{BC}$, $\overline{AB} = \overline{MB}$). From that we obtain that $\angle BMC = 30^\circ$ so $\angle AMB + \angle AMD = 150^\circ$. Hence $\angle AMB = \angle AMD = 75^\circ$.

Let $\overline{BC} = b$, $\overline{CD} = 2b$. From the condition $\overline{DM} = 1$ we get $\overline{CD} = \overline{CM} + \overline{MD} = \overline{CM} + 1$ i.e. $2b = \overline{CM} + 1$.

Now we get that $\overline{CM} = \sqrt{\overline{BM}^2 - \overline{BC}^2} = \sqrt{(2b)^2 - b^2} = b\sqrt{3}$.

Or $2b = b\sqrt{3} + 1$ from where we obtain that $b = \frac{1}{2 - \sqrt{3}} = 2 + \sqrt{3}$.

The area of the rectangle is $P = ab = 2b^2 = 2(2+\sqrt{3})^2 = 2(7+4\sqrt{3})$.