AMC10 2016 A

AMC10 2016 A · Q16

AMC10 2016 A · Q16. It mainly tests Coordinate geometry, Transformations.

A triangle with vertices $A(0,2)$, $B(-3,2)$, and $C(-3,0)$ is reflected about the $x$-axis; then the image $\triangle A'B'C'$ is rotated counterclockwise around the origin by $90^\circ$ to produce $\triangle A''B''C''$. Which of the following transformations will return $\triangle A''B''C''$ to $\triangle ABC$?

一个三角形的顶点为 $A(0,2)$、$B(-3,2)$、$C(-3,0)$，先关于 $x$ 轴对称；然后将所得图形 $\triangle A'B'C'$ 围绕原点逆时针旋转 $90^\circ$，得到 $\triangle A''B''C''$。以下哪一种变换可以将 $\triangle A''B''C''$ 还原为 $\triangle ABC$？

(A) counterclockwise rotation around the origin by $90^\circ$ 绕原点逆时针旋转 90°

(B) clockwise rotation around the origin by $90^\circ$ 绕原点顺时针旋转 90°

(D) reflection about the line $y = x$ 关于直线 $y = x$ 反射

(E) reflection about the $y$-axis 关于 $y$ 轴反射

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): After reflection about the $x$-axis, the coordinates of the image are $A'(0,-2)$, $B'(-3,-2)$, and $C'(-3,0)$. The counterclockwise $90^\circ$-rotation around the origin maps this triangle to the triangle with vertices $A''(2,0)$, $B''(2,-3)$, and $C''(0,-3)$. Notice that the final image can be mapped to the original triangle by interchanging the $x$- and $y$-coordinates, which corresponds to a reflection about the line $y=x$.

答案（D）：关于$x$轴对称后，图像的坐标为$A'(0,-2)$、$B'(-3,-2)$和$C'(-3,0)$。以原点为中心逆时针旋转$90^\circ$会将该三角形映射到顶点为$A''(2,0)$、$B''(2,-3)$和$C''(0,-3)$的三角形。注意，最终图像可以通过交换$x$与$y$坐标映射回原三角形，这对应于关于直线$y=x$的对称。

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