AMC12 2009 B

AMC12 2009 B · Q23

AMC12 2009 B · Q23. It mainly tests Complex numbers (rare), Geometric probability (basic).

A region $S$ in the complex plane is defined by \[S = \{x + iy: - 1\le x\le1, - 1\le y\le1\}.\] A complex number $z = x + iy$ is chosen uniformly at random from $S$. What is the probability that $\left(\frac34 + \frac34i\right)z$ is also in $S$?

复平面中的区域 $S$ 定义为 \[S = \{x + iy: - 1\le x\le1, - 1\le y\le1\}.\] 从 $S$ 中均匀随机选取一个复数 $z = x + iy$。求 $\left(\frac34 + \frac34i\right)z$ 也在 $S$ 中的概率。

(A) \frac{1}{2} \frac{1}{2}

(B) \frac{2}{3} \frac{2}{3}

(D) \frac{7}{9} \frac{7}{9}

(E) \frac{7}{8} \frac{7}{8}

Answer

Correct choice: (D)

正确答案：（D）

Solution

First, turn $\frac34 + \frac34i$ into polar form as $\frac{3\sqrt{2}}{4}e^{\frac{\pi}{4}i}$. Restated using geometric probabilities, we are trying to find the portion of a square enlarged by a factor of $\frac{3\sqrt{2}}{4}$ and rotated $45$ degrees that lies within the original square. This skips all the absolute values required before. Finish with the symmetry method stated above.

先将 $\frac34 + \frac34i$ 写成极坐标形式：$\frac{3\sqrt{2}}{4}e^{\frac{\pi}{4}i}$。用几何概率重新表述：我们要找的是把正方形按比例 $\frac{3\sqrt{2}}{4}$ 放大并旋转 $45$ 度后，落在原正方形内部的部分所占比例。这样可以避免先前所需的所有绝对值计算。最后用上面所述的对称方法完成。

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