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2020 AMC 12B Problem 14

Problem 14 of 25IntermediateCombinatoricsProblem-Solving Techniques

Bela and Jenn play the following game on the closed interval [0,n][0, n] of the real number line, where nn is a fixed integer greater than 4.4. They take turns playing, with Bela going first. At his first turn, Bela chooses any real number in the interval [0,n].[0, n]. Thereafter, the player whose turn it is chooses a real number that is more than one unit away from all numbers previously chosen by either player. A player unable to choose such a number loses. Using optimal strategy, which player will win the game?

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Solution

Bela first plays the midpoint n2.\tfrac{n}{2}. This choice makes the configuration symmetric about the center of the interval. Thereafter, whenever Jenn picks a number x,x, Bela responds with its mirror image n−x.n - x. Since Bela has already chosen n2,\tfrac n2, Jenn’s legal move satisfies ∣x−n2∣>1.\left|x-\tfrac n2\right|>1. Therefore ∣x−(n−x)∣=2∣x−n2∣>2,|x-(n-x)|=2\left|x-\tfrac n2\right|>2, so Bela’s response is far enough from Jenn’s new point. Symmetry shows that it is also far enough from every earlier point. Thus Bela always has a move whenever Jenn does, so Jenn is the first to be stuck. Bela always wins. Thus, the correct answer is A.
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Tagged: combinatorial game · symmetry

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