Shapovalov A.V.

Three square paths with a common center are spaced 1 m apart (see fig.). 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 large path, the other two, not yet having 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.

![](https://cdn.mathpix.com/cropped/2024_05_06_0e468145bb05e34a17aag-10.jpg?height=703&width=646&top_left_y=0&top_left_x=706)
The lengths of the sides of two adjacent paths differ by 2 m. 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 Ra is halfway between Mu and Vee, Vee is at twice the distance from this side, 6 m. That is, Vee still has to run $6-1-1=4$ m along the side.

![](https://cdn.mathpix.com/cropped/2024_05_06_0e468145bb05e34a17aag-10.jpg?height=592&width=581&top_left_y=1022&top_left_x=767)

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), which means 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 must be 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.

## Answer

$4 \mathrm{~m}, 6 \mathrm{~m}, 8 \mathrm{~m}$.