AMC12 2022 A

AMC12 2022 A · Q18

AMC12 2022 A · Q18. It mainly tests Transformations.

Let $T_k$ be the transformation of the coordinate plane that first rotates the plane $k$ degrees counterclockwise around the origin and then reflects the plane across the $y$-axis. What is the least positive integer $n$ such that performing the sequence of transformations $T_1, T_2, T_3, \cdots, T_n$ returns the point $(1,0)$ back to itself?

设 $T_k$ 是坐标平面的变换，先绕原点逆时针旋转 $k$ 度，然后关于 $y$ 轴反射平面。执行变换序列 $T_1, T_2, T_3, \cdots, T_n$ 后，使得点 $(1,0)$ 返回自身的最小正整数 $n$ 是多少？

(A) 359 359

(B) 360 360

(D) 720 720

(E) 721 721

Answer

Correct choice: (A)

正确答案：（A）

Solution

Let $P=(r,\theta)$ be a point in polar coordinates, where $\theta$ is in degrees. Rotating $P$ by $k^{\circ}$ counterclockwise around the origin gives the transformation $(r,\theta)\rightarrow(r,\theta+k^{\circ}).$ Reflecting $P$ across the $y$-axis gives the transformation $(r,\theta)\rightarrow(r,180^{\circ}-\theta).$ Note that \begin{align*} T_k(P)&=(r,180^{\circ}-\theta-k^{\circ}), \\ T_{k+1}(T_k(P)) &= (r,\theta -1^{\circ}). \end{align*} We start with $(1,0^{\circ})$ in polar coordinates. For the sequence of transformations $T_1, T_2, T_3, \cdots, T_k,$ it follows that - After $T_1,$ we have $(1,179^{\circ}).$ - After $T_2,$ we have $(1,-1^{\circ}).$ - After $T_3,$ we have $(1,178^{\circ}).$ - After $T_4,$ we have $(1,-2^{\circ}).$ - After $T_5,$ we have $(1,177^{\circ}).$ - After $T_6,$ we have $(1,-3^{\circ}).$ - ... - After $T_{2k-1},$ we have $(1,180^{\circ}-k^{\circ}).$ - After $T_{2k},$ we have $(1,-k^{\circ}).$ The least such positive integer $k$ is $180.$ Therefore, the least such positive integer $n$ is $2k-1=\boxed{\textbf{(A) } 359}.$

设点 $P=(r,\theta)$ 用极坐标表示，其中 $\theta$ 以度为单位。绕原点逆时针旋转 $k^{\circ}$ 给出变换 $(r,\theta)\rightarrow(r,\theta+k^{\circ})$。关于 $y$ 轴反射给出 $(r,\theta)\rightarrow(r,180^{\circ}-\theta)$。注意 \begin{align*} T_k(P)&=(r,180^{\circ}-\theta-k^{\circ}), \\ T_{k+1}(T_k(P)) &= (r,\theta -1^{\circ}). \end{align*} 我们从 $(1,0^{\circ})$ 开始。对于变换序列 $T_1, T_2, T_3, \cdots, T_k$，有 - 经过 $T_1$ 后，$(1,179^{\circ})$。 - 经过 $T_2$ 后，$(1,-1^{\circ})$。 - 经过 $T_3$ 后，$(1,178^{\circ})$。 - 经过 $T_4$ 后，$(1,-2^{\circ})$。 - 经过 $T_5$ 后，$(1,177^{\circ})$。 - 经过 $T_6$ 后，$(1,-3^{\circ})$。 - ... - 经过 $T_{2k-1}$ 后，$(1,180^{\circ}-k^{\circ})$。 - 经过 $T_{2k}$ 后，$(1,-k^{\circ})$。最小这样的正整数 $k$ 是 $180$。因此，最小正整数 $n$ 是 $2k-1=\boxed{\textbf{(A) } 359}$。

Topics

GE.011 Transformations