Hi, could you please explain what do you mean by a "hexagonal lattice grid'?

chaotic_iak wrote: Alzim and Badril are playing a game on a hexagonal lattice grid with 37 points (4 points a side), all of them uncolored. On his turn, Alzim colors one uncolored point with the color red, and Badril colors two uncolored points with the color blue. The game ends either when there is an equilateral triangle whose vertices are all red, or all points are colored. If the former happens, then Alzim wins, otherwise Badril wins. If Alzim starts the game, does Alzim have a strategy to guarantee victory? can you give me the map of the grid?

I think this is the image, perhaps anyone can give a better image?

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First of all, draw two more different hexagons such that the vertex lies on the $6$ line segments(that is connected to the center of hexagon). Denote the small one with $H_1$, the medium one with $H_2$, and the largest with $H_3$. Assume Alzim colors the center of $H_3$(the hexagonal grid the problem mentioned) on the first turn, we will show his winning strategy. Case 1. Badril colors two vertex of $H_1$ such that there are 3 collinear points on the second turn. On the third turn, Alzim will color one of the vertex of $H_2$ such that there are no blue points connected by the $6$ line segments mentioned earlier, forcing Badril to draw two more points on the vertex of $H_2$ to stop Alzim on the fourth move. Then, on the fifth move Alzim will color the vertex of $H_3$ such that there are $3$ collinear red points connected by the $6$ line segments. Alzim will win because Alzim have more than two choices to form an equilateral triangle. Case 2. Badril doesn't do the coloring as in case $1$. On the third turn, Alzim will color one of the vertex of $H_1$ such that there are no blue points connected by the vertices of $H_1$, forcing Badril to draw two more points on the vertex of $H_1$ to stop Alzim. On the fifth move, Alzim will draw one more point on the $6$ line segments such that there are three consecutives red points connected by the $3$ lines on the diagram that intersect on the center of hexagon. If Alzim only have one way to do this, then he'll win. If he have two ways to do this, one of the way leads to more than two choices to form an equilateral triangle considering the two points Badril colored on the second turn. Either case, Alzim will win. Then, it is proven that Alzim has a winning strategy.