Shapovalov A.V.

Three square paths with a common center are spaced 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, 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_1aa9346639cd54e59f2ag-06.jpg?height=683&width=640&top_left_y=1202&top_left_x=709)
The lengths of the sides of two adjacent paths differ by 2 meters. Therefore, at the moment when Mu reached the corner, Ra had run 2 meters along the right side of the path and was at a distance of $2+1=3$ meters from the "lower" side of the outer path. Since Ra is halfway between Mu and Vey, Vey is at twice the distance from this side, which is 6 meters. This means Vey still has to run $6-1-1=4$ meters along the side.

![](https://cdn.mathpix.com/cropped/2024_05_06_1aa9346639cd54e59f2ag-06.jpg?height=586&width=577&top_left_y=2211&top_left_x=755)

But if Vey had run counterclockwise, he would have run the entire lower side and an additional 4 meters along the right side (since this side is 4 meters shorter than the side of the outer square), thus ending up at the same point. Since Vey 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 meters. Accordingly, the sides of the other two squares are $4+2=6$ meters and $6+2=8$ meters.

## Answer

4 meters, 6 meters, 8 meters.