How to win when placing colored dots on a plane against one opponent? A‘s color is purple, B‘s color is green.
They alternately set a dot in their color on a set 2D-plane.
Whoever manages to

*

*construct a triangle in their own color without a dot in the other color inside of it

*then place a fourth dot inside of said triangle

wins the game.
Is it possible for one of the players to force their win?
 A: I believe the following strategy always allows Player 1 to win in five turns, so long as points on the edges of the triangles don't count as "blocking" the triangle.  Let $x_i$ denote the $i$th point placed by Player 1, and $y_i$ denote the $i$th point placed by Player 2.

*

*Players 1 and 2 each place their first two points $\{x_1, y_1, x_2, y_2\}$.  No triangles yet.


*Player 1 places $x_3$ so that neither $y_1$ nor $y_2$ is in the interior of the triangle $x_1x_2x_3$.  Player 2 must respond by placing $y_3$ in the interior of this triangle, or Player 1 will win on the next turn.


*Player 1 places $x_4$ in the region of intersection of the triangle $x_1x_2x_3$ and $y_1 y_2 y_3$.  This prevents Player 2 from winning on their fourth turn.  It also gives Player 1 two "empty" triangles:  the point $y_3$ is in the interior of precisely one of the triangles $x_1x_2x_4$, $x_1 x_3 x_4$, and $x_2 x_3 x_4$, and the other two triangles do not have any of Player 2's points in their interior.


*Regardless of where Player 2 plays $y_4$, Player 1 will have one "empty" triangle in which to place $x_5$, winning the game.
Note that if one's opponent can block your triangles by blocking the edges (instead of needing to play in the interior of the triangle), this strategy doesn't guarantee Player 1 a win at this point:  Player 2 can place $y_4$ on common edge between the Player 1's two "empty" triangles.  I do not know how (or whether) Player 1 could guarantee a win in such a game.
