AMC10 2008 B

AMC10 2008 B · Q14

AMC10 2008 B · Q14. It mainly tests Triangles (properties), Transformations.

Triangle OAB has O = (0, 0), B = (5, 0), and A in the first quadrant. In addition, $\angle ABO = 90^\circ$ and $\angle AOB = 30^\circ$. Suppose that OA is rotated $90^\circ$ counterclockwise about O. What are the coordinates of the image of A?

三角形 OAB 有 O = (0, 0)，B = (5, 0)，且 A 在第一象限。此外，$\angle ABO = 90^\circ$ 且 $\angle AOB = 30^\circ$。假设 OA 绕 O 逆时针旋转 $90^\circ$。A 的像的坐标是什么？

(A) $\left(-\frac{10}{3}\sqrt{3}, 5\right)$ $\left(-\frac{10}{3}\sqrt{3}, 5\right)$

(B) $\left(-\frac{5}{3}\sqrt{3}, 5\right)$ $\left(-\frac{5}{3}\sqrt{3}, 5\right)$

(D) $\left(\frac{5}{3}\sqrt{3}, 5\right)$ $\left(\frac{5}{3}\sqrt{3}, 5\right)$

(E) $\left(\frac{10}{3}\sqrt{3}, 5\right)$ $\left(\frac{10}{3}\sqrt{3}, 5\right)$

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Because $\triangle OAB$ is a $30-60-90^\circ$ triangle, we have $BA=\frac{5\sqrt{3}}{3}$. Let $A'$ and $B'$ be the images of $A$ and $B$, respectively, under the rotation. Then $B'=(0,5)$, $B'A'$ is horizontal, and $B'A'=BA=\frac{5\sqrt{3}}{3}$. Hence $A'$ is in the second quadrant and $$ A'=\left(-\frac{5}{3}\sqrt{3},5\right). $$

答案（B）：因为 $\triangle OAB$ 是一个 $30-60-90^\circ$ 三角形，所以 $BA=\frac{5\sqrt{3}}{3}$。设 $A'$ 和 $B'$ 分别是点 $A$ 和点 $B$ 在该旋转下的对应像。则 $B'=(0,5)$，线段 $B'A'$ 是水平的，并且 $B'A'=BA=\frac{5\sqrt{3}}{3}$。因此 $A'$ 在第二象限，且 $$ A'=\left(-\frac{5}{3}\sqrt{3},5\right). $$

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