Given the square $A B C D$, when it is centrally enlarged from vertex $A$, we obtain the square $A B^{\prime} C^{\prime} D^{\prime}$. The intersection point of the diagonals of the enlarged square is $M$. We know that $M C = B B^{\prime}$. What is the ratio of the enlargement?
If the new square falls on the opposite side of vertex $A$, then $B B^{\prime}=\lambda a+a, M C=a \cdot \sqrt{2}+\lambda a \cdot \frac{\sqrt{2}}{2}$ (where $a$ is the original side length, $\lambda$ is the scaling ratio); if the enlarged square contains the original, then $B B^{\prime}=\lambda a-a, M C$ and if $\lambda \leq 2$, then $a \cdot \sqrt{2}-\lambda a \cdot \frac{\sqrt{2}}{2}$, if $\lambda>2$, then $\lambda a \cdot \frac{\sqrt{2}}{2}-a \cdot \sqrt{2}$; solving the obtained equations yields $\lambda=\sqrt{2}$. (For the case $\lambda>2$, there is no solution.)

Many omitted the case distinction; among those who did, those who drew the two-sided placement or the one-sided placement with $\lambda=2$ as a starting point mostly found the problem unsolvable.