Difference between revisions of "2002 OIM Problems/Problem 6"
| Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
| − | A police officer tries to catch a thief on a <math>2001 \times 2001</math> board. They play alternately. Each player, on his turn, must move one square in one of the three senses <math>\downarrow</math>, <math>\to</math>, <math>\ | + | A police officer tries to catch a thief on a <math>2001 \times 2001</math> board. They play alternately. Each player, on his turn, must move one square in one of the three senses <math>\downarrow</math>, <math>\to</math>, <math>\vwarrow</math>. |
If the policeman is in the bottom right corner square, he can use his turn to move directly to the square in the upper left corner (the thief cannot do this move). | If the policeman is in the bottom right corner square, he can use his turn to move directly to the square in the upper left corner (the thief cannot do this move). | ||
Revision as of 03:49, 14 December 2023
Problem
A police officer tries to catch a thief on a 
board. They play alternately. Each player, on his turn, must move one square in one of the three senses 
, 
, $\vwarrow$ (Error compiling LaTeX. Unknown error_msg).
If the policeman is in the bottom right corner square, he can use his turn to move directly to the square in the upper left corner (the thief cannot do this move).
Initially the policeman is in the central square and the thief is in the neighboring diagonal square top right to the policeman. The policeman starts the game. Show that:
1. The thief manages to move at least 10,000 without being caught.
2. The police officer has a strategy to capture the thief.
Note: The policeman captures the thief when he enters the square where the thief is. If the thief enters the police officer's box, there is no capture.
~translated into English by Tomas Diaz. orders@tomasdiaz.com
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
