Question

Bela and Jenn play the following game on the closed interval $[0, n]$ of the real number line, where $n$ is a fixed integer greater than $4.$ They take turns playing, with Bela going first. At his first turn, Bela chooses any real number in the interval $[0, n].$ 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?

Bela will always win.

Jenn will always win.

Bela will win if and only if $n$ is odd.

Jenn will win if and only if $n$ is odd.

Jenn will win if and only if $n \gt 8.$

Answer: A

Concepts:combinatorial game symmetry

Difficulty rating: 1500

Solution:

Bela first plays the midpoint $\tfrac{n}{2}.$ This choice makes the configuration symmetric about the center of the interval.

Thereafter, whenever Jenn picks a number $x,$ Bela responds with its mirror image $n - x.$ Since Bela has already chosen $\tfrac n2,$ Jenn's legal move satisfies $\left|x-\tfrac n2\right|>1.$ Therefore $|x-(n-x)|=2\left|x-\tfrac n2\right|>2,$ 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.

2020 AMC 12B Problem 14

Answer: A

Difficulty rating: 1500

Solution:

Problem 14 in Other Years