2015 AMC 12A 第 25 题

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)}}\text{？}

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

小提示：

对夹在两个同切于一条直线的圆之间的圆，有

\dfrac{1}{\sqrt{r}} = \dfrac{1}{\sqrt{r_1}} + \dfrac{1}{\sqrt{r_2}}

。

For a circle nestled between two circles tangent to the same line,

\dfrac{1}{\sqrt{r}} = \dfrac{1}{\sqrt{r_1}} + \dfrac{1}{\sqrt{r_2}}

大提示：

令

x = \dfrac{1}{70} + \dfrac{1}{73}

；各层的和为

x

，

x

，

3x

，

9x, \dots

。

Let

x = \dfrac{1}{70} + \dfrac{1}{73};

the layer sums are

x,

x,

3x,

9x, \dots

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}}\text{。}

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}\text{。}

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

小提示：

大提示：

解答：

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