1971 AMC 12 第 35 题

Question

一个半径递减的无限圆序列中，每个圆都与下一个圆外切，并与一个给定直角的两边相切。第一个圆的面积与序列中其余所有圆的面积之和的比为：

Each circle in an infinite sequence with decreasing radii is tangent externally to the one following it and to both sides of a given right angle. The ratio of the area of the first circle to the sum of the areas of all the other circles in the sequence is:

(4+3\sqrt2):4

9\sqrt2:2

(16+12\sqrt2):1

(2+2\sqrt2):1

(3+2\sqrt2):1

答案：C

知识点：相切圆等比数列面积比根式

难度评级：2380

小提示：

各圆圆心都在角平分线上；求相邻两圆半径之比

The centers lie on the angle bisector; find the ratio of consecutive radii

大提示：

半径之比为

3-2\sqrt2

，所以面积之比是它的平方

The radius ratio is

3-2\sqrt2

, so the area ratio is its square

解答：

设相邻两圆半径为

r\gt r'

，则它们的圆心位于角平分线上，距顶点分别为

r\sqrt2

和

r'\sqrt2

。外切条件给出

\sqrt2(r-r')=r+r'\text{，}

所以

\frac{r'}{r}=3-2\sqrt2

。相邻两圆的面积比为

q=(3-2\sqrt2)^2\text{。}

因此，第一个圆的面积除以后续所有圆的面积之和为

\begin{aligned} \frac{1-q}{q} &=\frac1q-1\\ &=(3+2\sqrt2)^2-1\\ &=16+12\sqrt2\text{。} \end{aligned}

因此，正确答案为 C。

If consecutive radii are

r\gt r',

their centers lie on the angle bisector at distances

r\sqrt2

and

r'\sqrt2

from the vertex. External tangency gives

\sqrt2(r-r')=r+r',

so

\frac{r'}{r}=3-2\sqrt2.

The ratio of successive areas is

q=(3-2\sqrt2)^2.

Therefore the first area divided by the sum of all later areas is

\begin{aligned} \frac{1-q}{q} &=\frac1q-1\\ &=(3+2\sqrt2)^2-1\\ &=16+12\sqrt2. \end{aligned}

Therefore, the correct answer is C.

1971 AMC 12 第 35 题

答案：C

难度评级：2380

小提示：

大提示：

解答：

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