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Pigeon hole principle

 
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Oct13-10, 01:55 AM   #1
 

Pigeon hole principle


1. The problem statement, all variables and given/known data
How many points can be placed in an equilateral triangle where each side is of length 2 such that no 2 points are within 1 of each other?



2. Relevant equations
Need to use pigeon hole principle.

3. The attempt at a solution
I know that there are at least 3. Visually, if you sketch the triangle it looks like there will be a 4. However, I don't know how to prove this.
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Oct13-10, 02:39 AM   #2
 
Quote by auk411 View Post
I know that there are at least 3. Visually, if you sketch the triangle it looks like there will be a 4. However, I don't know how to prove this.
divide the triangle in 4 area's that can only contain 1 point each
Oct13-10, 05:48 AM   #3
 
Quote by willem2 View Post
divide the triangle in 4 area's that can only contain 1 point each
Yes. I'm asking how does one show this. I don't know how to show this. Someone who claims to have the right answer told me that we need to add root 3 over 2 plus root 3 over 6 to get about 1.3.

I am completely lost as to how this is an answer.
Oct14-10, 07:08 PM   #4
 

Pigeon hole principle


Quote by auk411 View Post
Yes. I'm asking how does one show this. I don't know how to show this. Someone who claims to have the right answer told me that we need to add root 3 over 2 plus root 3 over 6 to get about 1.3.

I am completely lost as to how this is an answer.
Apparently they computed the distance from a corner to a center of the triangle, but this indeed not the answer.

If you can divide the triangle in 4 pieces, such that the maximum distance between 2 points in a single piece is 1, then there can be only a single point in each piece, so the maximum amount of points is 4. You then only have to give a configuration of 4 points to prove that 4 is indeed possible.
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discrete mathematics, pigeon hole
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