IMO 2014/6

A set of lines in the plane is in general position if no two are parallel and no three pass through the same point. A set of lines in general position cuts the plane into regions, some of which have finte area; we call these its finite regions. Prove that for all sufficiently large nn, in any set of nn lines in general position it is possible to colour at least n \sqrt{n} of the lines blue in such a way that none of its finite regions has a completely blue boundary.

Note: Results with n \sqrt{n} replaced by cn c \sqrt{n} will be awarded points depending on the value of the constant cc.

Note by Calvin Lin
7 years ago

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Note: Results with n\sqrt{n} replaced by cnc\sqrt{n} will be awarded points depending on the value of the constant nn.

That single sentence got me scared at the IMO. I didn't even read the problem after that. :)

Shenal Kotuwewatta - 6 years, 9 months ago

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Does anyone have a solution for this one? Based on what I've heard from the Indian IMOers, only one of them attempted it (and got 2). I'll work on this in school.

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There exists an elementary solution using the extremal principle. Surprisingly, the probabilistic method isn't needed for c = 1.

Zi Song Yeoh - 7 years ago

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Only 15 people got full marks for this. I'm not too sure how to proceed with this, and I'm guessing that a probabilistic expectation method is used.

Calvin Lin Staff - 7 years ago

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I've heard that there exists a purely elementary solution: one can simply construct an algorithm which satisfies the problem conditions. However since this is an IMO P6 and I extremely suck at combinatorics, I haven't made much significant progress yet.

And well, a probabilistic approach given by Evan Chen (TWN 2; v_Enhance on AoPS) gives a bound of 16e3n\dfrac{1}{\sqrt[3]{6e}}\sqrt{n} which is unfortunately smaller than n.\sqrt{n}. There should exist a solution using the probabilistic method, though. Also, I have heard that the bound can actually be improved to nln(n).\sqrt{n \ln (n)}.

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