AMC10 2020 B

AMC10 2020 B · Q23

AMC10 2020 B · Q23. It mainly tests Basic counting (rules of product/sum), Transformations.

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: $\bullet$ $L$, a rotation of $90^\circ$ counterclockwise around the origin; $\bullet$ $R$, a rotation of $90^\circ$ clockwise around the origin; $\bullet$ $H$, a reflection across the $x$-axis; and $\bullet$ $V$, a reflection across the $y$-axis. Each of these transformations maps the square onto itself, but the positions of the labeled vertices will change. How many sequences of 20 transformations chosen from $\{L, R, H, V\}$ will send all of the labeled vertices back to their original positions?

坐标平面上的正方形 $ABCD$ 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：\bullet$ $L$，绕原点逆时针旋转 $90^\circ$；\bullet$ $R$，绕原点顺时针旋转 $90^\circ$；\bullet$ $H$，关于 $x$ 轴反射；\bullet$ $V$，关于 $y$ 轴反射。这些变换都将正方形映射到自身，但标记顶点的位置会改变。从 $\{L, R, H, V\}$ 中选择 20 个变换的序列有多少个能使所有标记顶点回到原始位置？

(A) 2^{37} 2^{37}

(B) 3 \cdot 2^{36} 3 \cdot 2^{36}

(D) 3 \cdot 2^{37} 3 \cdot 2^{37}

(E) 2^{39} 2^{39}

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Label the sides of the square $p$, $q$, $r$, and $s$, with $p$ being the side that starts as the upper horizontal side, $q$ being the side that starts as the left vertical side, $r$ being the side that starts as the bottom horizontal side, $s$ being the side that starts as the right vertical side. The position of the square after a sequence of transformations can be represented by a list of these four letters, starting with the side that is in the top horizontal position and moving counterclockwise. Thus $L$ takes $pqrs$ to $spqr$, $R$ takes $pqrs$ to $qrsp$, $H$ takes $pqrs$ to $psrq$, and $V$ takes $pqrs$ to $rqps$. Let $E=\{pqrs, qpsr, rspq, srqp\}$ and let $O=\{psrq, qrsp, rqps, spqr\}$. Each of the given transformations $L$, $R$, $H$, and $V$ will map each element of $E$ to a different element of $O$, and vice versa. Because $19$ is an odd number, any sequence of $19$ transformations will map the square to a position in set $O$, and exactly one of the given transformations will map this position back to position $pqrs$. Therefore there are $4^{19}=2^{38}$ possible sequences that will return the square to its original position.

答案（C）：将正方形的四条边标记为 $p$、$q$、$r$、$s$，其中 $p$ 为起初位于上方的水平边，$q$ 为起初位于左侧的竖直边，$r$ 为起初位于下方的水平边，$s$ 为起初位于右侧的竖直边。经过一系列变换后，正方形的位置可以用这四个字母的排列来表示：从处于顶部水平位置的边开始，按逆时针方向依次记录。于是 $L$ 将 $pqrs$ 变为 $spqr$，$R$ 将 $pqrs$ 变为 $qrsp$，$H$ 将 $pqrs$ 变为 $psrq$，$V$ 将 $pqrs$ 变为 $rqps$。令 $E=\{pqrs, qpsr, rspq, srqp\}$，令 $O=\{psrq, qrsp, rqps, spqr\}$。给定的变换 $L$、$R$、$H$、$V$ 会把 $E$ 中的每个元素映射到 $O$ 中不同的一个元素，反之亦然。由于 $19$ 是奇数，任意由 $19$ 次变换组成的序列都会把正方形映射到集合 $O$ 中的某个位置，并且恰好有一种给定变换会把该位置映射回 $pqrs$。因此，能使正方形回到原始位置的序列共有 $4^{19}=2^{38}$ 种。

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