Monochromatic triangle similar to a given triangle Given a scalene triangle, $A$ and $B$ play a game. Each move, $A$ chooses a point on the plane, and $B$ colors it red or green. If three points of the same color form a triangle similar to the original triangle, $A$ wins. 
What is the minimum number of moves after which $A$ can always win, regardless of the original triangle?
For 4 moves, $B$ can color two points red and two points green, so that there is no monochromatic triangle. That means the minimum is at least 5. But 5 moves doesn't look sufficient to create so many triangles similar to the original one.
 A: I have a strategy for $A$ using at most $6$ points that works for any nondegenerate triangle.
Choose three points $C, D, E$ such that the triangle $CDE$ is similar to your given triangle. If they all have the same color, player $A$ wins. Otherwise, two of these points have different colors, let's say $C$ is red and $D$ is green.
Next, choose the midpoint $M$ of $CD$. By symmetry, let's say $M$ is red. Finally, choose $P$ and $Q$ such that $PMC \sim CMQ \sim CDE$. Since $C$ and $M$ are both red, either there is a red triangle or both $P$ and $Q$ must be green.
Now if both $P$ and $Q$ are green, then $PDQ \sim CDE$ is a green triangle, so $A$ has won either way.

Proof that $PDQ \sim CDE$:
Choose complex coordinates such that $M = 0$ and $C = 1$ (hence $D = -1$). Since $PMC \sim CMQ$, we have $$P = \frac{P}{1} = \frac{P-M}{C-M} = \frac{C-M}{Q-M} = \frac{1}{Q}.$$
Now we have
$$P(Q+1) = PQ+P = P+1,$$
dividing by $(Q+1)$ we get
$$\frac{P-D}{Q-D} = \frac{P+1}{Q+1} = P = \frac{P-M}{C-M}$$
and thus $PDQ \sim PMC \sim CDE$.
