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# 10-gon game

## Question

There is a coin at each vertex of a regular 10-gon. Alice and Bob take turns removing one coin, with Alice going first. A coin can be removed only if there is an acute-angled triangle between the coin one wants to remove and two other coins [the center of the shape is the center of the angle]. A player who cannot move loses. (Note: A 90 angle is not acute.)
Who has a winning strategy?

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Didn't even think about this yet, but I'm just gonna go ahead and guess Bob since Alice's first turn doesn't really count because of symmetry, so it's less likely that she's be able to control the game.

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Ok, how about this? Bob just removes the coin opposite Alice's every turn. This guarantees that Bob always has a move since the board will always be symmetrical (every coin always has a coin opposite of it, and the board seen from the point of view of any coin is the same as it is seen from the coin opposite, except flipped) and removing a coin never affects whether its opposite can be removed (since a triangle formed by a coin and its opposite and any other coin is always right angled).

Edited by gavinksong
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Ok, how about this? Bob just removes the coin opposite Alice's every turn. This guarantees that Bob always has a move since the board will always be symmetrical (every coin always has a coin opposite of it, and the board seen from the point of view of any coin is the same as it is seen from the coin opposite, except flipped) and removing a coin never affects whether its opposite can be removed (since a triangle formed by a coin and its opposite and any other coin is always right angled).

remember right angled is not considered acute in the game's directions

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Ok, how about this? Bob just removes the coin opposite Alice's every turn. This guarantees that Bob always has a move since the board will always be symmetrical (every coin always has a coin opposite of it, and the board seen from the point of view of any coin is the same as it is seen from the coin opposite, except flipped) and removing a coin never affects whether its opposite can be removed (since a triangle formed by a coin and its opposite and any other coin is always right angled).

remember right angled is not considered acute in the game's directions

Right. So if coin A can be removed, it means that you're able to make an acute triangle with that coin and two others. Since the board is symmetrical, you should also be able to make an acute triangle with the coin opposite that coin (coin B) and the coins opposite the two others. Thus, you would be able to remove coin B as well. Removing coin A won't change that fact. The only way it would is if the vertices of the triangle contained both coin A and coin B, but if it did, you wouldn't be able to remove coin A in the first place since such a triangle would be right angled (not acute). In other words, as long as Bob sticks to his strategy, if Alice has a move then Bob has a move.

Edited by gavinksong
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remember right angled is not considered acute in the game's directions

I don't see why 90° should be an issue. There are no coins that make 90° -- 10 sides for 360°; angle between any 2 vertices is a multiple of 36° and will never be 90°.

I agree with gavin on the solution.

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remember right angled is not considered acute in the game's directions

I don't see why 90° should be an issue. There are no coins that make 90° -- 10 sides for 360°; angle between any 2 vertices is a multiple of 36° and will never be 90°.

I agree with gavin on the solution.

The problem is talking about acute triangles (and right triangles and obtuse triangles) formed by three vertices, not angles formed by only two vertices and the center point. I actually got this confused at first too. So any triangle whose base is the diameter of a circle and whose remaining vertex lies on the circle is a right triangle. If you imagine the 10-gon as being circumscribed by a circle, you get a right triangle whenever two of its vertices are opposite each other.

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