2015 AMC 12A Problem 25

Question

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}

Answer: D

Concepts:tangent circles geometric sequence induction

Difficulty rating: 2650

Small hint:

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}}

Big hint:

Let

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

the layer sums are

x,

x,

3x,

9x, \dots

Solution:

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.