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：

其他年份的第 35 题