225. Given two squares with respectively parallel sides. Determine the set of points $M$ such that for any point $P$ from the first square, there is a point $Q$ from the second such that the triangle $M P Q$ is equilateral. Let the side of the first square be $a$, and the second be $b$. For what ratio between $a$ and $b$ is the desired set not empty?
225. If (Fig. 61) we rotate the first square around point $M$ by $60^{\circ}$, either clockwise or counterclockwise, it must fit entirely inside the second square. Conversely, for every square located inside the larger one, equal to the smaller one, with sides forming angles of $30^{\circ}$ and $60^{\circ}$ with the sides of the larger square,
![](https://cdn.mathpix.com/cropped/2024_05_21_6282cf4a54d2a9633c48g-125.jpg?height=314&width=604&top_left_y=974&top_left_x=320)

Fig. 61.

there corresponds a point $M$ with the required property. (On the diagram, this square is indicated by a dashed line.) This point will be the center of rotation by $60^{\circ}$, transforming square $A B C D$ into square $A_{1} B_{1} C_{1} D_{1}$, which can be obtained from $O_{1}$ by rotating in the required direction by $60^{\circ}$ around $O$. Consider the extreme positions of squares $A_{1} B_{1} C_{1} D_{1}$ (when two vertices fall on the sides of the larger square). Their centers serve as the vertices of square $K L R N$, the side of which is respectively equal to $b-\frac{1}{2} a(\sqrt{3}+1)$ (the sides of square $K L R N$ are parallel to the sides of the given squares, the center coincides with the center of the larger one). Thus, the desired set consists of the union of two squares, one of which is obtained
from square $K L R N$ by rotating around $O$ by $60^{\circ}$ in one direction, the other - by rotating by $60^{\circ}$ in the opposite direction.

The problem has a solution if $b \geqslant \frac{a}{2}(\sqrt{\overline{3}}+1)$ (points $P$ and $Q$ can be on the boundary of their squares).