86. Given two squares with respectively parallel sides. Determine the geometric locus of points $M$ such that for any point $P$ from the first square, there exists a point $Q$ from the second such that the triangle $M P Q$ is equilateral. Let the sides of the first square be $a$, and the second be $b$. For what ratio between $a$ and $b$ is the sought geometric locus of points non-empty?
86. If (Fig. 24) the first square is rotated around point $M$ by $60^{\circ}$, either clockwise or counterclockwise, it must fit entirely inside the second square. Conversely, for each square located
![](https://cdn.mathpix.com/cropped/2024_05_21_35c23d544adcc44dca41g-131.jpg?height=318&width=604&top_left_y=1130&top_left_x=331)

Fig. 24

inside the larger one, equal to the smaller one, whose sides form angles of $30^{\circ}$ and $60^{\circ}$ with the sides of the larger square, 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 $A_{1} B_{1} C_{1} D_{1}$, and it can be obtained from $O_{1}$ by rotating in the required direction by $60^{\circ}$ around $O$. Consider the extreme positions of the 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 $K L R N$, whose side length is $b-\frac{1}{2} a(\sqrt{3}+1)$ (the sides of square $K L R N$ are parallel to the sides of the given squares, and the center coincides with the center of the larger square). The centers of the other family of squares, forming angles of $30^{\circ}$ and $60^{\circ}$ with the sides of the larger square, also fill the square $K L R N$. Thus, the desired geometric locus of points 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, and the other by rotating by $60^{\circ}$ in the opposite direction.

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