# 2015 AMC 12A Problems/Problem 25

## Problem

A collection of circles in the upper half-plane, all tangent to the -axis, is constructed in layers as follows. Layer consists of two circles of radii and that are externally tangent. For , the circles in are ordered according to their points of tangency with the -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 consists of the circles constructed in this way. Let , and for every circle denote by its radius. What is

## Solution

Let us start with the two circles in and the circle in . Let the larger circle in be named circle with radius and the smaller be named circle with radius . Also let the single circle in be named circle with radius . Draw radii , , and perpendicular to the x-axis. Drop altitudes and from the center of to these radii and , respectively, and drop altitude from the center of to radius perpendicular to the x-axis. Connect the centers of circles , , and with their radii, and utilize the Pythagorean Theorem. We attain the following equations.

We see that , , and . Since , we have that . Divide this equation by , and this equation becomes the well-known relation of Descartes's Circle Theorem We can apply this relationship recursively with the circles in layers .

Here, let denote the sum of the reciprocals of the square roots of all circles in layer . The notation in the problem asks us to find the sum of the reciprocals of the square roots of the radii in each circle in this collection, which is . We already have that . Then, . Additionally, , and . Now, we notice that because , which is a power of Hence, our desired sum is . This simplifies to .

Note that the circles in this question are known as Ford circles.