Olimpiadi della Matematica

Solution:

The answer is (B). Since the quadrilateral $CDMN$ is cyclic, we know that the angle $\angle CDM$ is equal to the supplement of the angle $\angle MNC$, that is, to the angle $\angle MNB$. On the other hand, the angles $\angle CDM$ and $\angle AMD$ are equal, being alternate interior angles formed by the parallels $AB$ and $CD$ with the transversal $DM$. We therefore deduce that $\angle AMD = \angle MNB$.

The right triangles $ADM$ and $BMN$ thus have an equal acute angle and, by hypothesis, equal hypotenuses $(DM = MN)$, and are therefore congruent. In particular, $AD = MB$ holds, and since $AD = BC$, the triangle $BCM$ is isosceles with respect to the base $CM$ (besides being right-angled at $B$), so that the ratio between its hypotenuse and one of its legs is $\sqrt{2}$. It then holds that $AB / BC = CD / BC = CM / BC = \sqrt{2}$.