Exact evaluation R(3, 3) = 6

**Step 1 (upper bound $R(3,3) \le 6$).** Pick any vertex $v$ of $K_6$. Its $5$ incident edges are colored red or blue. By the pigeonhole principle ($\lceil 5/2 \rceil = 3$), at least $3$ of these edges have the same color — say edges $vu_1, vu_2, vu_3$ are all red.

**Step 2 (case split on $\{u_1, u_2, u_3\}$).** Look at the $3$ edges inside the triangle $\{u_1, u_2, u_3\}$. If any one of them — say $u_1u_2$ — is red, then $\{v, u_1, u_2\}$ is a red $K_3$. Otherwise all $3$ edges $u_1u_2, u_2u_3, u_3u_1$ are blue, so $\{u_1, u_2, u_3\}$ itself is a blue $K_3$.

**Step 3 (lower bound $R(3,3) > 5$).** Label the vertices of $K_5$ by $\mathbb{Z}/5\mathbb{Z}$. Color edge $ij$ red if $i - j \equiv \pm 1 \pmod 5$ and blue if $i - j \equiv \pm 2 \pmod 5$. Both color classes form a $5$-cycle, triangle-free, proving $R(3,3) > 5$ and hence $R(3,3) = 6$.