AMC10 2020 B

AMC10 2020 B · Q16

AMC10 2020 B · Q16. It mainly tests Geometry misc, Games (basic).

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 和 Jenn 在实数轴上的闭区间 $[0, n]$ 上玩以下游戏，其中 $n$ 是大于 4 的固定整数。他们轮流玩，Bela 先手。在他的第一回合，Bela 选择区间 $[0, n]$ 中的任意实数。此后，轮到的一方选择一个与之前双方选择的所有数都距离超过 1 个单位的实数。无法选择这样数的玩家输。使用最优策略，谁会赢？

(A) Bela will always win. Bela 总是获胜。

(B) Jenn will always win. Jenn 总是获胜。

(D) Jenn will win if and only if $n$ is odd. Jenn 获胜当且仅当 $n$ 为奇数。

(E) Jenn will win if and only if $n > 8. Jenn 获胜当且仅当 $n > 8$。

Answer

Correct choice: (A)

正确答案：（A）

Solution

Bela has a straightforward strategy that will guarantee winning. At his first turn he chooses the number $\frac{n}{2}$. This splits the playing field into two parts—the numbers less than $\frac{n}{2}$ and the numbers greater than $\frac{n}{2}$. Thereafter, whatever legal move Jenn makes, Bela makes the symmetric move in the other half of the playing field. Specifically, if Jenn chooses $x$, then Bela chooses $n - x$. By symmetry, if Jenn's move is possible, so is Bela's. Thus the first player to be unable to make a legal move will be Jenn, and she will lose. The game will end after a finite number of moves, because of the restriction that the chosen numbers are at least one unit apart.

Bela 有一个简单的策略保证获胜。在他的第一回合，他选择数字 $\frac{n}{2}$。这将游戏区域分成两部分——小于 $\frac{n}{2}$ 的数和大于 $\frac{n}{2}$ 的数。此后，无论 Jenn 做什么合法移动，Bela 在另一半游戏区域做对称移动。具体来说，如果 Jenn 选择 $x$，则 Bela 选择 $n - x$。由对称性，如果 Jenn 的移动是可能的，那么 Bela 的也是。因此，最先无法合法移动的玩家将是 Jenn，她会输。由于选择的数至少相距 1 个单位，游戏将在有限步后结束。

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