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 \(n\), in any set of \(n\) lines in general position it is possible to colour at least \( \sqrt{n} \) of the lines blue in such a way that none of its finite regions has a completely blue boundary.

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

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TopNewestDoes 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. – Sreejato Bhattacharya · 3 years ago

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– Calvin Lin Staff · 3 years ago

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.Log in to reply

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

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– Zi Song Yeoh · 3 years ago

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

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

That single sentence got me scared at the IMO. I didn't even read the problem after that. :) – Shenal Kotuwewatta · 2 years, 9 months ago

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