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 $\sum_{C \in S} \dfrac{1}{\sqrt{r(C)}}?$

$\dfrac{286}{35}$

$\dfrac{583}{70}$

$\dfrac{715}{73}$

$\dfrac{143}{14}$

$\dfrac{1573}{146}$

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

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 Problem 25

Solution:

Problem 25 in Other Years