In the middle between two parallel streets, identical houses with a side length of $a$ are lined up in a row. The distance between the streets is $3 a$, and the distance between two adjacent houses is $2 a$ (see figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_2434826925a2883e8926g-31.jpg?height=312&width=847&top_left_y=970&top_left_x=617)

One street is patrolled by police officers who move at a distance of $9 a$ from each other at a speed of $v$. By the time the first police officer passes by the middle of a certain house, a gangster appears exactly opposite him on the other street. At what constant speed and in which direction should the gangster move along this street to avoid being seen by any police officer?
Let's draw two lines $x$ and $y$ parallel to the streets through the edges of the houses (the line $x$ closer to the street where the police are standing). Connect the point where the gangster is located with segments to all points where the police are located. Number these segments with integers from $-\infty$ to $+\infty$ in the order in which the police are located on the first street. Let $x_{i}$ and $y_{i}$ be the points where the $i$-th segment intersects the lines $x$ and $y$. Notice that the points $x_{i}$ are spaced at intervals of $6a$, and the points $y_{i}$ are spaced at intervals of $3a$. Therefore, at the initial moment, all these points are located on the walls of the houses, meaning no police can see the gangster. It is clear that all points $x_{i}$ move at the same constant speed. In particular, they are all stationary if the gangster moves towards the police at a speed of $2v$, and in this case, the police will never see the gangster. If the gangster moves towards the police at a speed of $v/2$, then the points $x_{i}$ are stationary, and the police will also not see the gangster. We will prove that the gangster has no other options.

Suppose the points $x_{0}$ and $y_{0}$ are moving (then all other points are also moving). Consider a large time interval $T$. During the time $T$, more than half of the time, the point $x_{0}$ was not on the sides of the houses, and more than half of the time, the point $y_{0}$ was not on the sides of the houses. Then there will be a moment when neither of the points $x_{0}$ and $y_{0}$ was on the sides of the houses. We will prove that at this moment, one of the police can see the gangster. Consider the segment of the line $x$ between the houses where the point $x_{0}$ is located. As stated above, the points $x_{i}$ are spaced at intervals of $6a$, and the points $y_{i}$ are spaced at intervals of $3a$. Therefore, on the segment of the line $y$ located opposite the considered segment of the line $x$, there will be one of the points $y_{i}$. For the same reason, the points $x_{-i}$ and $y_{-i}$ are located on the same interval between the houses. Then the police with number $-i$ can see the gangster.

## Answer

The gangster must move towards the police at a speed of $2v$ or $v/2$.

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