Maths Olympiad Prep

Library / /30 of 94

Combinatorics Difficulty 4.6 AIME Prove it United States

Problem:

Prove that, for all n>3n > 3, there exists a graph with chromatic number nn that does not contain any nn-cliques.

Solution

Solution:

We prove the claim by induction on nn. The case n=3n = 3 was addressed in (a).

Let n3n \geq 3 and suppose GG is a graph with chromatic number nn containing no nn-cliques. We produce a graph GG' with chromatic number n+1n+1 containing no (n+1)(n+1)-cliques as follows. Add a vertex vv to GG, and add an edge from vv to each vertex of GG.

To see this graph has chromatic number n+1n+1, observe that any coloring of the vertices of GG' restricts to a valid coloring of the vertices of GG. So at least nn distinct colors must be used among the vertices of GG. In addition, another color must be used for vv. By coloring vv a new color, we have constructed a coloring of GG' having n+1n+1 colors.

Lastly, any (n+1)(n+1)-clique in GG' must have at least nn vertices in GG which form an nn-clique, which is impossible. Therefore, GG' has no (n+1)(n+1)-cliques.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.