Question

Bela 和 Jenn 在实数轴的闭区间 $[0, n]$ 上玩如下游戏，其中 $n$ 是一个大于 $4$ 的固定整数。他们轮流操作，Bela 先手。Bela 第一次可以在区间 $[0, n]$ 中选择任意实数。此后，轮到的玩家必须选择一个实数，且它与此前任一玩家选择过的所有数的距离都大于一。无法选择这样一个数的玩家输。若双方都采取最优策略，谁会赢？

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 总会赢。

Bela will always win.

Jenn 总会赢。

Jenn will always win.

当且仅当 $n$ 为奇数时 Bela 会赢。

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

当且仅当 $n$ 为奇数时 Jenn 会赢。

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

当且仅当 n>8 时 Bela 会赢。

Jenn will win if and only if n>8.

答案：A

知识点：组合游戏对称性

难度评级：1480

解答：

Bela 首先选择中点 $n/2$ 。之后每当 Jenn 选择一个数 $x$ ，Bela 就选择关于中点的对称点 $n-x$ 。

只要 Jenn 的选择合法，这个对称点也合法：关于 $n/2$ 的反射会保持到此前所选点的距离，而且 Jenn 不能选择 $n/2$ ，因为它已经是 Bela 的第一步。因此 Jenn 的每一步都有对应的 Bela 回应，所以 Jenn 会先无合法步可走。

所以正确答案是 A。

Bela can first choose the midpoint $n/2.$ After that, whenever Jenn chooses a number $x,$ Bela chooses the reflected number $n-x.$

This reflected number is legal whenever Jenn's move is legal: distances from previously chosen numbers are preserved by the reflection about $n/2,$ and Jenn cannot choose $n/2$ because it was Bela's first move. Therefore every Jenn move has a matching Bela response, so Jenn is the first player who can run out of legal moves.

Thus, A is the correct answer.

2020 AMC 10B 第 16 题

答案：A

难度评级：1480

解答：

其他年份的第 16 题