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January 31st, 2010, 06:59 AM  #1 
Newbie Joined: Jan 2010 Posts: 1 Thanks: 0  Need Help with geometry question
I thought this question might be an application of the pigeonhole principle, but I wasn't able to make any progress... plz help anyone... Question: In the interior of a unit square there are 61 points. Show that there are 2 points at a distance < 1/5. 
January 31st, 2010, 04:11 PM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Need Help with geometry question
Here's how I see it. For each point, draw a circle of radius 1/10 around it. If any two circles overlap, the points are closer than 1/5 from each other. Let A = pi * (1/10)^2 be the area of of of these circles. Now at most 20 circles could reach outside the square (else two would be closer than 1/5). At most four of these will go over more than one side, else the shared corner will be overlap. So the area "used up" by these 16 circles is at least 4 A/4 + 16 A/2. The 41 remaining circles must be entirely inside the square, so the total area used by all the circles is A + 8A + 41A = 50A. But 50A = 1.57..., which is larger than the square's area. Thus, at least some of the circles overlap. Alternately, consider a square with side length 1.2 (1/10 past the sides of the original square), and require that all circles be inside it. (Clearly, the circles will be in the big square if the points are in the little square.) Now compare 61A to 1.2^2 = 1.44. 

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