2. A cruciform labyrinth represents the intersection of two segments, each 2 kilometers long, with a common midpoint. A nearsighted policeman on a defective motorcycle is chasing a gangster. The policeman's speed is 20 times that of the gangster, but he can only see 1 meter (in all directions), and his motorcycle can turn (including making a U-turn) no more than 60 times. Will the policeman be able to catch the gangster?
2. Solution. Let's conduct the reasoning in the general case. Let the lengths of all corridors be $a$, and the speed of the policeman be $n$ times the speed of the cyclist. We will drive around all the corridors (this will take five turns), and return to the center. If the gangster is not caught, it means he has moved from one corridor to another, and consequently, he has moved away from the center by no more than $\frac{6 a}{n}$. Now, let's choose $x$ such that by driving around distances $x$ from the center, we would definitely catch the gangster if he does not pass through the center. For this, it is sufficient to satisfy the inequality $\frac{6 a}{n}+\frac{7 x}{n} \leqslant x$, from which $x \geqslant \frac{6 a}{n-7}$. Thus, each time the length of the corridor that needs to be driven around is multiplied by $\frac{6}{n-7}$. In our case, $a=1000$, and $\frac{6}{n-7}<\frac{1}{2}$. This means that in 10 such operations, we will either catch the gangster or he will be at a distance from the center not exceeding $1000 \cdot \frac{1}{2^{1} 0}<1$. Therefore, being in the center, the policeman will see him.