Difference between revisions of "2015 AMC 12A Problems/Problem 25"
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− | <math> \textbf{(A)}\ \frac{286}{35} \qquad\textbf{(B)}\ \frac{583}{70} \qquad\textbf{(C)}\ \frac{715}{73} \qquad\textbf{(D) | + | <math> \textbf{(A)}\ \frac{286}{35} \qquad\textbf{(B)}\ \frac{583}{70} \qquad\textbf{(C)}\ \frac{715}{73} \qquad\textbf{(D)}\ \frac{143}{14} \qquad\textbf{(E)}\ \frac{1573}{146}</math> |
==Solution== | ==Solution== |
Revision as of 17:09, 9 March 2015
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 .
See Also
2015 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Problem |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |