Now cross-posted to MathOverflow (link).
Question: There is an equilateral triangle. Two players alternate turns placing non-overlapping equilateral triangles of the same size that touch the original one; contact needs to exist for at least one point. When a player cannot make such a placement, they lose. Which player has an optimal strategy, and what is it?
This question is a variation on the classic one about placing circles around an initial circle (for which player two wins e.g. using a mirroring strategy). The game is similarly solvable for a square.
The maximum number of equilateral triangles of the same size that can touch a single equilateral triangle of that size is twelve (alternatively phrased, the kissing number of an equilateral triangle is $12$). For this reason, the game ends after finitely many moves; in particular, it will be done within twelve moves. Therefore, one of the two players has an optimal strategy. What is it?
If this has been studied elsewhere – or already posed on MSE! – then please alert me. (My searches produced nothing.)
One of the principal difficulties appears to be a lack of exploitable symmetry. Given the solid black equilateral triangle to begin the game as below, suppose player one (blue) moves as follows:
Initially, one might believe that player two's moves can now be reflected using some sort of mirror symmetry. However, player two (red) can now win the game as follows:
From this arrangement, we see that blue's subsequent moves can be mirrored by red, which means red cannot encounter an impossible placement unless blue did in the previous turn. Thus, we detect one strategy that is not optimal for player one. But I do not see a clear generalization or next step around how to form an optimal strategy for either player.