IMO 1992 Problem 3

Consider $9$ points in space, no four of which are coplanar. Each pair of points is joined by an edge (that is, a line segment) and each edge is either colored blue or red or left uncolored. Find the smallest value of  $\,n\,$ such that whenever exactly $\,n\,$ edges are colored, the set of colored edges necessarily contains a triangle all of whose edges have the same color.

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$n=33$. 

Rephrase as: what is the $9$-vertex simple graph $G$ with most edges that permits a $2$-edge-coloring without a chromatic triangle? 

First, we show that $G$ does not have $33$ edges. If it does, then $G$ has a vertex $v$ of degree $8$.

If $v$ is incident to $4$ red and $4$ blue edges, say edges $vx,vy,vz,vw$ are red. Then $x,y,z,w$ have at least $2$ absent edges, making a total of at least $4$ absent edges.

If $v$ is incident to $3$ red and $5$ blue edges, say $vx,vy,vz$ are red and $vw,va,vb,vc,vd$ are blue. Then $x,y,z$ have at least $1$ edge absent and $w,a,b,c,d$ at least $3$ edges, making a total of at least $4$ absent edges.

If $v$ is incident to $2$ red and $6$ blue edges, let $S:=\{x\in V(G):vx\text{ is blue}\}$. Since $R(3,3)=6$, at least $3$ edges, forming a triangle, are absent in $G[S]$, or else $G[S]$ has a red triangle. Clearly at least one more edge is absent in $G[S]$ or else it still has a red triangle.

Below is a construction showing that $n=33$ is sharp. The idea is to make $v$ incident to $4$ red and $4$ blue edges, say edges $vx,vy,vz,vw$ are red and edges $va,vb,vc,vd$ are blue. Then draw $4$ blue edges among $x,y,z,w$ and $4$ red edges among $a,b,c,d$, and then color edges between $\{x,y,z,w\}$ and $\{a,b,c,d\}$ without making any chromatic triangle.