Problem:
A thief starts at the point on a number line, and runs either left or right at a constant rate. One hour later, a policeman who moves twice as fast as the thief appears at the point . However, the policeman does not know which direction the thief went and cannot see the thief. Show that the policeman can still catch the thief.
Berkeley Math Circle
Solution
Solution:
Assume for simplicity that the rate at which the thief moves is unit per hour. The police should drive left for an hour, then drive right for four hours.
This works for the following reason: if the thief runs left, he will be caught in one hour by the left-moving police at the position . If the thief runs right, then after the first hour of the police's movement, the thief will be at the position while the officer is at position . Then, by going right for four hours, the police will catch the robber at the position .
Want a route through all this instead of an archive? The track
puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.