9.6. Petya and Misha start on a circular track from the same point in the counterclockwise direction. Both run at constant speeds, with Misha's speed being greater than Petya's. Petya always runs counterclockwise, while Misha can change his direction at any moment, immediately before which he has run half a lap or more in one direction. Show that while Petya runs the first lap, Misha can meet (catch up to or overtake) him three times, not counting the start.
(I. Rubanov)
Solution. Let Misha, after running half a lap, turn around and run back. While he runs half a lap back, he will meet Petya. When Misha reaches the starting point, Petya will not have reached it yet. Therefore, if Misha continues to run in the same direction, he will meet Petya at some point at a distance from the start. Let him run an additional positive distance , less than , and then turn around (he can do this). Then, while Petya covers the remaining distance , Vasya will run . This means he will already have passed the starting point, and therefore, he will have caught up with Petya for the third time before that.
Comment. Any correct description of Misha's actions (possibly including phrases like "a sufficiently small distance" and the like) -7 points.