Question

在上半平面中构造一组圆，所有圆都与 $x$ -轴相切，构造分层如下。第 $L_0$ 层包含两个半径分别为 $70^2$ 和 $73^2$ 且外切的圆。对 $k \ge 1$ 将 $\bigcup_{j=0}^{k-1} L_j$ 中的圆按它们与 $x$ -轴的切点顺序排列。对这个顺序中每一对相邻的圆，构造一个与这一对圆都外切的新圆。第 $L_k$ 层由这样构造出的 $2^{k-1}$ 个圆组成。令 $S = \bigcup_{j=0}^{6} L_j$ ，对每个圆 $C$ ，用 $r(C)$ 表示它的半径。求 $\sum_{C \in S} \dfrac{1}{\sqrt{r(C)}}?$

A collection of circles in the upper half-plane, all tangent to the $x$ -axis, is constructed in layers as follows. Layer $L_0$ consists of two circles of radii $70^2$ and $73^2$ that are externally tangent. For $k \ge 1,$ the circles in $\bigcup_{j=0}^{k-1} L_j$ are ordered according to their points of tangency with the $x$ -axis. For every pair of consecutive circles in this order, a new circle is constructed externally tangent to each of the two circles in the pair. Layer $L_k$ consists of the $2^{k-1}$ circles constructed in this way. Let $S = \bigcup_{j=0}^{6} L_j,$ and for every circle $C$ denote by $r(C)$ its radius. What is $\sum_{C \in S} \dfrac{1}{\sqrt{r(C)}}?$

$\dfrac{286}{35}$

$\dfrac{583}{70}$

$\dfrac{715}{73}$

$\dfrac{143}{14}$

$\dfrac{1573}{146}$

答案：D

知识点：相切圆等比数列数学归纳法

难度评级：2650

解答：

如果半径为 $r$ 的圆与 $x$ -轴相切，并嵌在两个半径为 $r_1$ 和 $r_2$ 、也与该轴相切且彼此外切的圆之间，那么 $\dfrac{1}{\sqrt{r}} = \dfrac{1}{\sqrt{r_1}} + \dfrac{1}{\sqrt{r_2}}.$

令 $x = \dfrac{1}{\sqrt{70^2}} + \dfrac{1}{\sqrt{73^2}}$ $= \dfrac{1}{70} + \dfrac{1}{73}$ ，这是 $L_0$ 上的和。 $L_1$ 中唯一的圆也贡献 $x$ 。对 $k \ge 2$ ，每个新圆贡献它两个相邻圆的和；除 $L_0$ 的两个圆外，每个较早的圆都被计算两次。因此， $L_k$ 上的和为 $3^{k-1}x$ 。

因此 $\begin{gathered} \sum_{C \in S} \dfrac{1}{\sqrt{r(C)}} = x \\ {}+ \sum_{k=1}^{6} 3^{k-1}x \\ = x\left(1 + \dfrac{3^6 - 1}{2}\right) \\ = x\cdot\dfrac{3^6 + 1}{2} \\ = 365x. \end{gathered}$

因为 $x = \dfrac{1}{70} + \dfrac{1}{73}$ $= \dfrac{143}{70\cdot 73}$ $= \dfrac{143}{5110}$ ，所以总和为 $365\cdot\dfrac{143}{5110} = \dfrac{143}{14}$ 。

因此，正确答案是 D。

If a circle of radius $r$ is tangent to the $x$ -axis and nestled in the crevice between two circles of radii $r_1$ and $r_2$ that are also tangent to the axis and to each other, then $\dfrac{1}{\sqrt{r}} = \dfrac{1}{\sqrt{r_1}} + \dfrac{1}{\sqrt{r_2}}.$

Let $x = \dfrac{1}{\sqrt{70^2}} + \dfrac{1}{\sqrt{73^2}}$ $= \dfrac{1}{70} + \dfrac{1}{73},$ which is the sum over $L_0.$ The single circle of $L_1$ also contributes $x.$ For $k \ge 2,$ each new circle contributes the sum of its two neighbors, and every earlier circle is counted twice except the two circles of $L_0;$ this yields a sum of $3^{k-1}x$ over $L_k.$

Therefore $\begin{gathered} \sum_{C \in S} \dfrac{1}{\sqrt{r(C)}} = x \\ {}+ \sum_{k=1}^{6} 3^{k-1}x \\ = x\left(1 + \dfrac{3^6 - 1}{2}\right) \\ = x\cdot\dfrac{3^6 + 1}{2} \\ = 365x. \end{gathered}$

Since $x = \dfrac{1}{70} + \dfrac{1}{73}$ $= \dfrac{143}{70\cdot 73}$ $= \dfrac{143}{5110},$ the sum is $365\cdot\dfrac{143}{5110} = \dfrac{143}{14}.$

Thus, the correct answer is D.

2015 AMC 12A 第 25 题

答案：D

难度评级：2650

解答：

其他年份的第 25 题