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For some reason this post was edited out. I can only imagine that it was deliberate. I have replaced the original question as best as I could remember and locked the thread to prevent further destruction of it. - bobbym
Thirteen points are arbitrarily placed inside a 1 m square. Prove that there must be at least 2 points that are less than 5 / 12 m apart.
Last edited by Srikantan (2016-06-04 20:56:53)
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Hi;
Have you tried the pigeonhole principle?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Yes but was not able to apply in this and crack it. Please help
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Hi;
I divided the unit square into 12 subsquares.
The diagonal of each subsquare is
So each square is totally covered by a circle of radius
Now that means the greatest distance between any two points in that subsquare is
So after you place the first 12 points in each subsqare you are forced by the pigeonhole principle to put the 13th in an already occupied subsquare which will have a distance less than 5 / 12 m with the point already in there.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Wow thats awesome. Thanks a ton
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Hi;
That is the first one of that type I have ever done. I do not like proofs and usually will not do them so I suggest you check that very well.
I have closed this topic to prevent erasure of the whole thread. If anyone wishes to have it reopened please start a new thread using this problem. Sorry for the inconvenience.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
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Topic closed