At a point on the real line sits a greyhound. On one of the sides a hare runs, away from the hound. The only thing known is that the (maximal) speed of the hare is strictly less than the (maximal) speed of the greyhound (but not their precise ratio). Does the greyhound have a strategy for catching the hare in a finite amount of time? (Dan Schwarz)
Problem
Source: Stars Of Mathematics 2014, Juniors, Problem 4
Tags: combinatorics
09.03.2015 15:38
I don't quite understand the problem. Why can't the hound just run directly towards the hare? Or is the hare invisible to the hound?
09.03.2015 15:45
I think it's supposed that the hound can't see the hare.
09.03.2015 16:07
It says "The only thing known is that the (maximal) speed of the hare is strictly less than the (maximal) speed of the greyhound (but not their precise ratio)." This means the actual road on which the hare runs is unknown.
09.03.2015 17:35
The main idea is that the hound can't go in one direction,so it must switch directions.Now,it is logical that every period that the hound goes in one direction is larger than the previous one as much as possible,so from these we get the strategy. Now,if the total time that the greyhound is chasing the hare is $t$,go $n*t$ in other direction,where $n$ is an arbirtary integer and it grows by $1$ in each move.Let the distance be $s$.In some switching,the greyhound will go in the same way as the hare goes,so their distance will be at most $d=s+(a+b)*t-(a-b)*t*n$,where $a$ is the speed of the greyhound and $b$ is the speed of the hare.Now,for arbirtary large $n$,$d$ will be less than $0$ so we are done.