Problem 5. Three square paths with a common center are 1 m apart from each other (see figure). Three ants start simultaneously from the lower left corners of the paths and run at the same speed: Mu and Ra counterclockwise, and Wei clockwise. When Mu reaches the lower right corner of the largest path, the other two, who have not yet completed a full circle, are on the right sides of their paths, and all three are on the same straight line. Find the sides of the squares.

[6 points] (A. V. Shapovalov)
Answer: 4 m, 6 m, 8 m.

Solution: The lengths of the sides of two adjacent paths differ by 2 m (Fig. 3). Therefore, at the moment when Mu reached the corner, Ra had run 2 m along the right side of the path and was at a distance of $2+1=3$ m from the "lower" side of the outer path. Since $\mathrm{Pa}$ is halfway between Mu and Vee, Vee is at twice the distance from this side, 6 m (since the two highlighted right triangles in Fig. 4 are equal in hypotenuse and angles). That is, Vee still has to run $6-1-1=4$ m along the side.

![](https://cdn.mathpix.com/cropped/2024_05_06_031fac0a0a7981eac22cg-10.jpg?height=230&width=379&top_left_y=644&top_left_x=301)

Fig. 3

![](https://cdn.mathpix.com/cropped/2024_05_06_031fac0a0a7981eac22cg-10.jpg?height=351&width=344&top_left_y=565&top_left_x=807)

Fig. 4

But if Vee had run counterclockwise, he would have run the entire lower side and another 4 m along the right side (since this side is 4 m shorter than the side of the outer square), i.e., he would have ended up at the same point. Since Vee ends up at the same point whether he runs clockwise or counterclockwise, this point is the upper right corner of the square. Therefore, the side of this square is 4 m. Accordingly, the sides of the other two squares are $4+2=6$ m and $6+2=8$ m.