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A game of triangles and lines

There is a game between two players that I have played and it is as follows:

Arrange \(\frac{n(n+1)}{2}\) points in an equilateral triangle, such as in figure 1, when \(n\) is 4. The players connect points with a single line segment at each turn, and each line segment must pass through at least one point. No two line segments can intersect each other. A player loses when there are no more possible lines to be drawn, like in figure 2. Red is the first player while Blue is the second. Blue has drawn the 6th line and there are no more lines for Red to draw, so Red loses.

Is there a winning strategy for any of the two players for a particular \(n\)?

To clarify, the line segments should be drawn as short as possible. Therefore, line 2 in figure 2 cannot be extended to prevent Red from drawing line 3.

Note by Yu Tse Lee
3 years, 9 months ago

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If people like this game, they should check out this variation from Yong See F. Silas Hundt Staff · 3 years, 9 months ago

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This game appears to be winning for player 2 for n=4. Player 2 must follow symmetry with player 1 as long as possible. This ensures that always an even number of points are left to be 'lined' which results in player 2 winning. Yash Talekar · 3 years, 9 months ago

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@Yash Talekar If player 1 starts with line 3 from the example the symmetry-strategy goes wrong though... Ton De Moree · 3 years, 9 months ago

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@Ton De Moree Well the symmetry strategy does go wrong but then I can use the method for \(n = 2k - 1\) which you yourself have shown in the post below :) Yash Talekar · 3 years, 9 months ago

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@Yash Talekar I believe that n=4 is winning for player 2, but can only think of brute force as a way to prove that it is true. Yu Tse Lee · 3 years, 9 months ago

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Also why isn't there any game theory in the Olympiads? This is one of the most interesting math areas. We should have at least a practice session on game theory on Brilliant as we have one on calculus. Yash Talekar · 3 years, 9 months ago

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We start with \(n=2\). This position is obviously losing.

For \(n=2k-1\) \((k=1,2,3,...)\) we draw a line through the top and the middle point at the bottom and proceed to win by mirroring our opponents moves in that line.

That just leaves us \(n=2k\) \((k=2,3,4,...)\) to consider. At this point I don't see a strategy to win, so perhaps this always loses? Ton De Moree · 3 years, 9 months ago

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@Ton De Moree I see what you are doing for \(n=2k-1\), but that only works if you go first. \(n=2k\) seems pretty interesting! William Cui · 3 years, 9 months ago

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@William Cui "but that only works if you go first"

Ah, I assumed the opponent plays optimally, but that doesn't have to be the case of course. It makes it nearly impossible to come up with strategies though, since the number of possible opening moves increases dramatically with \(n\). Ton De Moree · 3 years, 9 months ago

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I don't get it, you say that the line should be as short as possible but in your figure 2, line no. 1 contradicts that. Yash Talekar · 3 years, 9 months ago

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@Yash Talekar No, basically if one wants to connect one point, he cannot extend the line so far that it almost touches the second point; if he wants to connect four points, it is allowed to connect all four, as long as the line does not extend past that. William Cui · 3 years, 9 months ago

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