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 \(c\).

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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.

Log in to reply

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

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 \(\dfrac{1}{\sqrt[3]{6e}}\sqrt{n}\) 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 \(\sqrt{n \ln (n)}.\)

Log in to reply

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. :)

Log in to reply