KenHungKK wrote: There are $ n$ points on the plane, no three of which are collinear. Each pair of points is joined by a red, yellow or green line. For any three points, the sides of the triangle they form consist of exactly two colours. Show that $ n < 13$. I can certainly show that $ n < 17$. Suppose, to the contrary, that there exists a set of $ n = 17$ points such that any triangle is colored with exactly $ 2$ colors. Consider a given point $ A$, and assume without loss of generality that the color most commonly used in lines connected to $ A$ is red. Then at least $ 6$ points must be joined with a red line to $ A$. Any two of these points must be joined with yellow or green (lest we have a triangle with all red lines). Consider a given point in this subset, $ B$. Assume without loss of generality that the color most commonly used in lines, within this subset, connected to $ B$ is yellow. Then at least $ 3$ points must be joined with a yellow segment to $ B$. Any two of these points must be joined with green (lest we have a triangle with all yellow lines). This gives us a triangle with all green lines, which provides a contradiction.

NoYoo-hooForMe, what you have proved is essentially the Ramsey number $ N(3,3,3,2) \le 17$. (But here is it is equal to 17 due to the additional condition that any triangle will have two same colored edges. I think this problem is quite hard(atleast for me). Can someone give a hint?