Question

坐标平面中的正方形 $ABCD$ 的顶点为 $A(1, 1)$ $B(-1, 1)$ $C(-1, -1)$ 和 $D(1, -1)$ 考虑以下四个变换：

$L$ ，绕原点逆时针旋转 $90^\circ$ ；

$R$ ，绕原点顺时针旋转 $90^\circ$ ；

$H$ ，关于 $x$ -轴反射；

$V$ ，关于 $y$ -轴反射。

每个变换都会把正方形映到自身，但标记顶点的位置会改变。例如，先做 $R$ 再做 $V$ ，会把顶点 $A$ 从 $(1, 1)$ 送到 $(-1, -1)$ ，并把顶点 $B$ 从 $(-1, 1)$ 送回原位。从 $\{L, R, H, V\}$ 中选出长度为 $20$ 的变换序列，有多少个会把所有标记顶点都送回原位？（例如， $R, R, V, H$ 是一个长度为 $4$ 的有效序列。）

Square $ABCD$ in the coordinate plane has vertices at the points $A(1, 1),$ $B(-1, 1),$ $C(-1, -1),$ and $D(1, -1).$ Consider the following four transformations:

$L,$ a rotation of $90^\circ$ counterclockwise around the origin;

$R,$ a rotation of $90^\circ$ clockwise around the origin;

$H,$ a reflection across the $x$ -axis; and

$V,$ a reflection across the $y$ -axis.

Each of these transformations maps the square onto itself, but the positions of the labeled vertices will change. For example, applying $R$ and then $V$ would send the vertex $A$ at $(1, 1)$ to $(-1, -1)$ and would send the vertex $B$ at $(-1, 1)$ to itself. How many sequences of $20$ transformations chosen from $\{L, R, H, V\}$ will send all of the labeled vertices back to their original positions? (For example, $R, R, V, H$ is one sequence of $4$ transformations that will send the vertices back to their original positions.)

$2^{37}$

$3 \cdot 2^{36}$

$2^{38}$

$3 \cdot 2^{37}$

$2^{39}$

答案：C

知识点：变换分类讨论

难度评级：2000

解答：

将顶点依次标为 $0,1,2,3$ 。每个允许的变换都形如 $j\mapsto \varepsilon j+\delta\pmod 4,$ 其中 $\varepsilon\in\{1,-1\}$ 且 $\delta\in\{1,-1\}.$ 这四种选择恰好给出两个四分之一周旋转和题中所述的两个反射。

在复合时， $\delta$ 的奇偶性在每一步都会改变。因此， $19$ 个允许变换的复合仍有奇数的 $\delta,$ 所以它仍是四个允许变换之一。它的逆变换也被允许。因此，前 $19$ 步的每个序列都恰有一种最后一步的选择，共得到 $4^{19}=2^{38}$ 个成功序列。

因此，正确答案是 C。

Label the vertices $0,1,2,3$ cyclically. Each allowed transformation has the form $j\mapsto \varepsilon j+\delta\pmod 4,$ where $\varepsilon\in\{1,-1\}$ and $\delta\in\{1,-1\}.$ These four choices give exactly the two quarter-turns and the two stated reflections.

Under composition, the parity of $\delta$ changes at every move. Thus a composition of $19$ allowed transformations again has odd $\delta,$ and so is one of the four allowed transformations. Its inverse is also allowed. Consequently every sequence of the first $19$ moves has exactly one choice for the final move, giving $4^{19}=2^{38}$ successful sequences.

Thus, the correct answer is C.

2020 AMC 12B 第 19 题

答案：C

难度评级：2000

解答：

其他年份的第 19 题