Question

一只可忽略其大小的小虫从坐标平面的原点出发。它先向右移动 $1$ 个单位到达 $(1,0)$ 。然后逆时针转 $90^\circ$ ，再移动 $\frac12$ 个单位到达 $(1,\frac12)$ 。此后它继续以同样方式移动，每次逆时针转 $90^\circ$ ，并且移动距离是前一次的一半。它最接近下列哪个点？

A bug (of negligible size) starts at the origin on the coordinate plane. First it moves $1$ unit right to $(1,0).$ Then it makes a $90^\circ$ turn counterclockwise and travels $\frac12$ a unit to $(1,\frac12).$ If it continues in this fashion, each time making a $90^\circ$ turn counterclockwise and traveling half as far as in the previous move, to which of the following points will it come closest?

$(\frac23,\frac23)$

$(\frac45,\frac25)$

$(\frac23,\frac45)$

$(\frac23,\frac13)$

$(\frac25,\frac45)$

答案：B

知识点：等比数列复数求和

难度评级：1830

小提示：

用乘以 $i$ 表示逆时针旋转 $90^\circ$

Represent a $90^\circ$ counterclockwise turn by multiplication by $i$

大提示：

各段位移向量构成公比为 $\frac{i}{2}$ 的无穷等比级数

The displacement vectors form an infinite geometric series with ratio $\frac{i}{2}$

解答：

把位移向量表示为复数，依次为 $1,\ \frac i2,\ \left(\frac i2\right)^2,\ldots\text{。}$ 它们的和是 $\frac1{1-\frac{i}{2}} =\frac{1+\frac{i}{2}}{1+\frac{1}{4}} =\frac45+\frac25i\text{。}$ 因此小虫趋近于 $(\frac{4}{5},\frac{2}{5})$ 。

因此，正确答案是 B。

As complex numbers, the successive displacement vectors are $1,\ \frac i2,\ \left(\frac i2\right)^2,\ldots.$ Their sum is $\frac1{1-\frac{i}{2}} =\frac{1+\frac{i}{2}}{1+\frac{1}{4}} =\frac45+\frac25i.$ Thus the bug approaches $(\frac{4}{5},\frac{2}{5}).$

Therefore, the correct answer is B.

1980 AMC 12 第 13 题

答案：B

难度评级：1830

小提示：

大提示：

解答：

其他年份的第 13 题