AMC10 2008 A

AMC10 2008 A · Q19

AMC10 2008 A · Q19. It mainly tests Area & perimeter, Transformations.

Rectangle $PQRS$ lies in a plane with $PQ = RS = 2$ and $QR = SP = 6$. The rectangle is rotated 90° clockwise about $R$, then rotated 90° clockwise about the point that $S$ moved to after the first rotation. What is the length of the path traveled by point $P$?

矩形 $PQRS$ 位于平面内，$PQ = RS = 2$，$QR = SP = 6$。矩形先绕 $R$ 顺时针旋转 $90^\circ$，然后绕第一次旋转后 $S$ 移动到的点顺时针旋转 $90^\circ$。求点 $P$ 所经过路径的长度。

(A) $(2\sqrt{3} + \sqrt{5})\pi$ $(2\sqrt{3} + \sqrt{5})\pi$

(B) $6\pi$ $6\pi$

(D) $(\sqrt{3} + 2\sqrt{5})\pi$ $(\sqrt{3} + 2\sqrt{5})\pi$

(E) $2\sqrt{10}\pi$ $2\sqrt{10}\pi$

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Let $P'$ and $S'$ denote the positions of $P$ and $S$, respectively, after the rotation about $R$, and let $P''$ denote the final position of $P$. In the rotation that moves $P$ to position $P'$, the point $P$ rotates $90^\circ$ on a circle with center $R$ and radius $PR=\sqrt{2^2+6^2}=2\sqrt{10}$. The length of the arc traced by $P$ is $(1/4)(2\pi\cdot 2\sqrt{10})=\pi\sqrt{10}$. Next, $P'$ rotates to $P''$ through a $90^\circ$ arc on a circle with center $S'$ and radius $S'P'=6$. The length of this arc is $\frac{1}{4}(2\pi\cdot 6)=3\pi$. The total distance traveled by $P$ is $$ \pi\sqrt{10}+3\pi=(3+\sqrt{10})\pi. $$

答案（C）：设$P'$和$S'$分别表示绕$R$旋转后$P$和$S$的位置，并设$P''$表示$P$的最终位置。在将$P$移动到$P'$的旋转中，点$P$以$R$为圆心、半径$PR=\sqrt{2^2+6^2}=2\sqrt{10}$的圆旋转$90^\circ$。$P$所走弧长为$(1/4)(2\pi\cdot 2\sqrt{10})=\pi\sqrt{10}$。接着，$P'$以$S'$为圆心、半径$S'P'=6$的圆再转过$90^\circ$到$P''$。该弧长为$\frac{1}{4}(2\pi\cdot 6)=3\pi$。因此$P$走过的总路程为 $$ \pi\sqrt{10}+3\pi=(3+\sqrt{10})\pi. $$

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