Problem 2. On the side $AB$ of triangle $ABC$, a point $M$ is taken. It starts moving parallel to $BC$ until it intersects with $AC$, then it moves parallel to $AB$ until it intersects with $BC$, and so on. Is it true that after a certain number of such steps, point $M$ will return to its original position? If this is true, what is the minimum number of steps sufficient for the return?
Solution. Let the length of side $AB$ be 1, and let point $M$ be at a distance $a$ from point $B$. From the properties of the parallelogram, the small triangles are equal, so after 3 steps, point $M$ will be at a distance $a$ from point $A$, which is at a distance $1-a$ from point $B$. After another 3 steps, the point will be at a distance $1-(1-a)=a$ from point $B$, meaning it will return to its initial position.

![](https://cdn.mathpix.com/cropped/2024_05_06_cf8f1bd3b01b92531235g-2.jpg?height=369&width=600&top_left_y=1049&top_left_x=859)

A special case is when $a=1/2$. Then $1-a=a$, and the return will occur after 3 steps.

Answer: Correct.

It takes 3 steps if point $M$ divides side $AB$ in half, and 6 steps in other cases.