Monochromatic vertex disjoint triangles in $K_{6n}$

Question

I'm trying to show that every red/blue-colouring of the edges of $K_{6n}$ contains $n$ vertex-disjoint triangles with all $3n$ edges of the same colour.

My idea is to show this by induction, we just pick a $K_6$ subgraph which has a monochromatic traingle and the remaining $K_{6(n-1)}$ has $n-1$ monochromatic vertex disjoint triangles by induction. Now the issue is combining these two facts, as it mustn't be that the triangles are of the same color. So the proof boils down to finding a $K_6$ subgraph which contains both a red and a blue triangle. Indeed, if such a subgraph wouldn't exist, namely among every choice of 6 vertices there are only red or blue triangles, we can pick a $K_6$ with a blue triangle and 2 red edges, and one with a red triangle and 2 blue edges (if we assume the coloring to be non trivial, ie. there are both red and blue triangles). I do not quite see how to combine these 2 subgraphs to obtain a contradiction.

Answer 1

By induction on $n$, the base case is clear since $R(3)=6$.

Fix any red/blue edge coloring of $K_{6n}$. If it is trivial in the sense that it only contains triangles of one color we are done as we may split the graph into $n$ disjoint copies of $K_6$ which each contain a monochromatic triangle, and these $n$ triangles will all be vertex disjoint of the same color.

So assume there is both a red and a blue triangle, these will use at most 6 vertices. Remove them and apply induction to the remaining graph to find $n-1$ monochromatic vertex disjoint triangles of the same color. Now add the red or blue one to conclude the proof.