Question

Bela 和 Jenn 在实数轴闭区间 $[0, n]$ 上玩如下游戏，其中 $n$ 是大于 $4$ 的固定整数。两人轮流行动，Bela 先手。Bela 第一次可以在区间 $[0, n]$ 中任选一个实数。此后，轮到的玩家必须选择一个与此前任一玩家选过的所有数都相距大于一的实数。无法选择者失败。在最优策略下，谁会获胜？

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 总会获胜。

Bela will always win.

Jenn 总会获胜。

Jenn will always win.

Bela 获胜当且仅当 $n$ 为奇数。

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

Jenn 获胜当且仅当 $n$ 为奇数。

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

Jenn 获胜当且仅当 $n \gt 8$ 。

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

答案：A

知识点：组合游戏对称性

难度评级：1500

解答：

Bela 第一步选择区间中点 $\tfrac{n}{2}$ 这样局面关于区间中心对称。

之后每当 Jenn 选择 $x$ ，Bela 就选择其镜像 $n - x$ 。 $\tfrac n2$ ， $\left|x-\tfrac n2\right|>1$ 。 $|x-(n-x)|=2\left|x-\tfrac n2\right|>2$ ，

因为 Jenn 行动前局面对称且她的选择合法，所以镜像点也合法且不同。因此只要 Jenn 有合法行动，Bela 就也有合法行动，所以 Jenn 会先无路可走。Bela 总会获胜。所以正确答案是 A。

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 第 14 题

答案：A

难度评级：1500

解答：

其他年份的第 14 题