A problem from Moscow Mathematical Olympiad in 1973. goes like this:
At the center of a square stands a cop and at one of the square’s vertices stands a robber. The Rule allows the cop to run anywhere in the square and even digress outside its limits, while the robber can only move along the square’s sides. For each of the following ratios of the cop’s top speed to that of the robber a) 1/2; b) 0.49; c) 0.34; d) 1/3, prove that the cop can run so as to be on the same side as the robber at some moment.
I found the solution which I don't quite understand:
If the speed of the cop is 1/2 v, where v is the speed of the robber, it suffices for the cop to run first directly towards the robber and, when the robber runs along a side of the square, to run directly to that side. If the cop’s speed is smaller then 1/2 v, (s)he must use the fact that if (s)he runs inside a square with the same center but three times as small as the original square, (s)he can stay abreast with the robber, i.e., on the same ray coming out of the center of the square, or even outrun the robber. So the cop should get out beyond the boundary of this square and then rush back to the boundary at an appropriate moment.
Now, the $\frac{1}{2} v$ case is pretty clear, but for the life of me, I don't understand the rest of the solution. How should a cop move in the case of $\frac{1}{3} v$? Is this a lower bound on the speed of the cop for him to catch to robber? If not, what is?
Edit: Finally, some progress! Let $v$ be the robber's speed and $v/n$ cop's speed. Then it is clear that as long as the cop moves in the square which side is $n$ times smaller of the original square, he and robber can be on the same ray. Now, it is optimal for the robber for him to be in the middle of one side of the square so that he has two equally good possibilities at a distance of $a/2$, where $a$ is the side of the original square, to run to. OTOH, the cop wants the robber to be in the corner (and himself in the corresponding corner of the smaller square) because the robber would then have to run a distance of $a$ to save himself. However, the cop would have to run a distance of $\frac{n-1}{2n}a$ to get to the larger square. This implies that as long as $n<3$, the cop would always catch the robber.
Now, the problem is to get them both in the corners of their respective squares. I'm not sure if this is possible?
Can't blame you. Explanation aren't clear. I believe they're trying to say; wherever the robber goes, cop can stay in ray coming out of the center and going through the robber when (s)he reaches side of his/her small square. While this is definitely true, how could (s)he find appropriate moment to run through robber when (s)he is on side of small square if robber runs always upwards (if robber starts from bottom right and the cop's speed is $v\over 3$). If cop wants to stand on ray then (s)he must run always to right, and when (s)he is on the side of small square they are horizontal to each other. Here is a detailed explanation of the above situation:
Assume robber starts at $r_1$. If robber goes upwards, cop, starting from $c_1$ and going through the speed of $v \over 3$, goes right; and if robber goes left, cop goes downwards; making a symmetric situation like above. When robber reaches $r_2$, cop reaches $c_2$. From this situation, cop can't be on the same side with robber if robber always on the move, because if robber goes upwards or downwards (Assuming (s)he never touch corners..) cop must go upwards or downwards to stay on ray, and if (s)he tries to get close to robber at some point, (s)he misses his/her ray, therefore, robber.
Note that if robber make the mistake of coming to $r_3$, cop, being in $c_3$, can reach the robber's side: Since if robber goes left cop runs upwards, and if robber goes downwards cop runs right. If robber goes right after moving left a little, cop simply runs to the right, making a better situation than before. And lastly, if robber stays, cop just runs through the robber. This leads me to consider if there's something wrong with the wording of the problem. Maybe robber can only stays corners and middle of sides, meaning that (s)he can't monkey around, for example, between $r_1$ and $r_2$. If that is the case, once (s)he get to the point $r_2$ his/her only two alternatives $r_1$ and $r_3$ giving him/her a losing situation.
General case:
Assume square has side length $1$, robber's speed $v$ and cop's speed $v\over r$ where $r$ is a positive real number greater than $1$. Cop, using above strategy, can reach middle of his/her small square's side which has a length $1\over r$, when robber reach the middle of side of the big square. From this point, if cop cannot reach robber when (s)he directly run towards it, (s)he can forget about reaching ever at all; because all (s)he can do is stay on ray if robber always runs without touching corners. Therefore, cop must travel a distance of length $\frac{r-1}{2r}$ before robber can travel a distance of length $1\over 2$. Since robber's travel time of this distance is $1\over 2v$, cop's travel time must not be greater than this value; giving an inequality,
$$ \frac{r-1}{2v}\le \frac{1}{2v} $$
which gives $r\le 2$. $\blacksquare$