Prove $R(2K_3, 2K_3)=10$ My question is how to prove that $R(2K_3, 2K_3)$ is 10.
First, $2K_3$ are two distinct triangles, so the question is to prove that if we colour the edges of $K_{10}$ in two colours, there will be two triangles with the same colour.
The only thing that I accomplished is that there is a 'butterfly', that is two monochromatic triangles sharing a vertex.
 A: First, we must show that coloring $K_9$ is not enough. Here is a counterexample:

Red $K_3$'s are restricted to vertices $\{1,2,3,4,5\}$, so they're out: any two of these share a vertex. A blue $K_3$ can use at most one of the vertices $\{1,2,3,4,5\}$, so it must use two vertices from $\{6,7,8,9\}$; this cannot include $6$, because the edges from $6$ to $\{7,8,9\}$ are red. So any blue $K_3$ includes two vertices from $\{7,8,9\}$, and any two of them share a vertex.

Here's a proof that $K_{10}$ is enough. First, observe that some color $c \in \{\text{red}, \text{blue}\}$ must have the following property: no matter which two vertices of $K_{10}$ you pick, there is a $c$-colored monochromatic $K_3$ not using those two vertices. 
(If not, then we can remove some two vertices and destroy all red $K_3$'s; then remove some two vertices and destroy all blue $K_3$'s; but now we still have a complete graph on $6$ vertices, which we know contains either a red or a blue $K_3$. Contradiction!)
Suppose that red triangles have this property, and assume also that there is (are?) no red $2K_3$. This implies something much stronger: any two red $K_3$'s either intersect only at one vertex, or form part of a larger clique. To see this, suppose that we have red $K_3$'s induced by $\{v_1, v_2, v_3\}$ and $\{v_2, v_3, v_4\}$. Deleting the vertices $v_2$ and $v_3$, we know we must be able to find another red $K_3$. If it intersects both of these, it must include vertices $v_1$ and $v_4$, so the edge $v_1 v_4$ is also red.
So let $\{v_1, v_2, \dots, v_n\}$ be a maximal red clique. By the observation above, any other vertex $v_i$, $i \ge n+1$, can only have a red edge to one of $v_1, \dots, v_n$. We consider two cases for the value of $n$.
Case 1: ${n\le 4}$. Then the coloring of the remaining $6$ vertices must have a monochromatic $K_3$; we assumed there is no red $2K_3$, so it must be blue. Moreover, if the vertices $\{w_1, w_2, w_3\}$ form the blue $K_3$, among the remaining vertices there must be a blue edge $x_1x_2$ (or else they form a red $K_3$). Since there is at most one red edge $x_1 v_i$ and at most one red edge $x_2 v_j$, we get another blue $K_3$ formed by $\{x_1, x_2, v_k\}$ for some $k$, disjoint from the first.
Case 2: $n=5$. If the remaining $5$ vertices do not contain a red $K_3$, they must at the very least have two disjoint blue edges $w_1w_2$ and $x_1x_2$. Each element of $\{w_1, w_2, x_1, x_2\}$ can only have one red edge to a vertex in $\{v_1,\dots,v_5\}$, so there are at least three choices of $v_i$ and at least two choices $v_j \ne v_i$ such that $\{w_1, w_2, v_i\}$ and $\{x_1, x_2, v_j\}$ form disjoint blue $K_3$'s.

A more general result is proved in (Burr, Erdős, Spencer 1975): $$R(mK_3, nK_3) = 3m+2n \qquad \text{for }m \ge n \ge 2.$$ I didn't read it because I wanted to solve this one myself, but it probably has a nicer proof.
