We have a line and 1390 points around it such that the distance of each point to the line is less than 1 centimeters and the distance between any two points is more than 2 centimeters. prove that there are two points such that their distance is at least 10 meters (1000 centimeters).
Problem
Source: Iran second round 2011
Tags: geometry, analytic geometry, combinatorics proposed, combinatorics
28.04.2011 19:37
Draw circles with the radius 1. No two circle have a common point. so 1389π≤π+4d+4 where d is the longest distance. Done.
28.04.2011 20:27
Actually I got 694√3 on the exam, which is more than 1000 obviously. I will post the solution if needed.
30.12.2011 10:59
ArefS wrote: Draw circles with the radius 1. No two circle have a common point. so 1389π≤π+4d+4 where d is the longest distance. Done. why is this true?
30.12.2011 11:08
Solution of amparvardi's (hey I'm psychic): Assume the line is the x-axis. By PHP, there is a side in which there are at least 695 points in it. Since the width of the area that can be occupied on one side of the line is 1 cm while the distance between two points in on the line is at least 2 cm, then the x-coordinate difference between two adjacent points is at least √3 cm by Pythagorean Theorem. Thus since there are 694 such gaps, the x-distance between the two farthest points on that side is at least 694√3 cm.
31.12.2011 11:02
sahilsharma94 wrote: ArefS wrote: Draw circles with the radius 1. No two circles have a common point. so 1389π≤π+4d+4 where d is the longest distance. Done. why is this true? this was my solution during the exam. we only need to find a figure such that each circle lies in the figure. not hard to show that it is what I've shown below.
Attachments:
01.01.2012 00:21
You can also partition the given 2 by 1000 block into blocks of 1 by sqrt 3 and notice the only way 2 points are in the same square is if they are diagonally opposite then u can consider 4 such blocks making a bigger block and you can easily break this bound of 1000.