1967 AMC 12 第 20 题

Question

在边长为

m

的正方形中内切一个圆，再在该圆中内接一个正方形，再在后一个正方形中内切一个圆，如此继续。若

S_n

是最前面

n

个这样内切圆的面积之和，则当

n

无限增大时，

S_n

趋近于：

A circle is inscribed in a square of side

m,

then a square is inscribed in that circle, then a circle is inscribed in the latter square, and so on. If

S_n

is the sum of the areas of the first

n

circles so inscribed, then, as

n

grows beyond all bounds,

S_n

approaches:

\dfrac{\pi m^2}{2}

\dfrac{3\pi m^2}{8}

\dfrac{\pi m^2}{3}

\dfrac{\pi m^2}{4}

\dfrac{\pi m^2}{8}

第一个圆的面积为

\frac{\pi m^2}{4}

The first circle has area

\frac{\pi m^2}{4}

每个后继圆的面积都是前一个圆面积的一半

Each successive circle has half the preceding circle’s area

第一个圆的面积为

\pi(\frac{m}{2})^2=\frac{\pi m^2}{4}

。每个内接正方形使下一个圆的半径平方，从而也使其面积缩小为原来的

\frac{1}{2}

。因此极限和为

\begin{aligned} &\frac{\pi m^2}{4} \left(1+\frac12+\frac14+\cdots\right)\\ &\qquad=\frac{\pi m^2}{2} \end{aligned}\text{。}

因此，正确答案是 A。

The first circle has area

\pi(\frac{m}{2})^2=\frac{\pi m^2}{4}.

Each inscribed square reduces the next circle’s squared radius, and hence its area, by a factor of

\frac{1}{2}.

Thus the limiting sum is

\begin{aligned} &\frac{\pi m^2}{4} \left(1+\frac12+\frac14+\cdots\right)\\ &\qquad=\frac{\pi m^2}{2}. \end{aligned}

Therefore, the correct answer is A.

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