Colouring proofs for beginners

Can anybody please explain me with examples, how to solve tough colouring related problems ( as in CRMO-4 question paper).

Please give me some advice on how to solve these kind of problems.

PS: I've already tried Arthur Engel. So please give me important methods and not the basic idea.

Easy Math Editor

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Comments

Hi priyansh sangule, I suggest that you wait for a while because I will soon talk about imo standard coloring problens in my daily elementary techniques in the imo posts.

Anqi Li - 7 years, 6 months ago

Thank you!

Priyansh Sangule - 7 years, 6 months ago

here's the answer- Since there are three red vertices than among the remaining 17 vertices there are nine of them of the same colour, say blue. We can divide the vertices of the regular 20-gon into four disjoint sets such that each set consists of vertices that form a regular pentagon. Since there are nine blue points, at least one of these sets will have three blue points. Since any three points on a pentagon form an isosceles triangle, the statement follows

another solution-

divide the 20-gon in 4 pentagon so that we can have isosceles triangle in each of these pentagon by joining any three vertices. now as there are 3 red vertices than we are left with 17 vertices. now there will be 1 pentagon which will not have any red vertice as by pigeonhole principle and also by pigeonhole principle that pentagon should have 3 vertices with same colour. hence it follows that there are 3 vertices with same colour such that they form an isosceles triangle.

and btw priyansh did u qualified RMO?

aditya dokhale - 7 years, 5 months ago

No Aditya, I couldn't but some of my friends did. And Thanx for the solution !

Priyansh Sangule - 7 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$