2014 AMC 12B Problems/Problem 25

Revision as of 18:04, 20 February 2014 by Kevin38017 (talk | contribs) (Solution)

Problem

Find the sum of all the positive solutions of

$2\cos2x \left(\cos2x - \cos{\left( \frac{2014\pi^2}{x} \right) } \right) = \cos4x - 1$

$\textbf{(A)}\ \pi \qquad\textbf{(B)}\ 35\qquad\textbf{(C)}\ 1008\pi \qquad\textbf{(D)}}\ 1080 \pi \qquad\textbf{(E)}\ 1800\pi$ (Error compiling LaTeX. Unknown error_msg)

Solution

Rewrite $\cos{4x} - 1$ as $2\cos^2{2x} - 2$. Now let $a = \cos{2x}$, and let $b = \cos{\left( \frac{2014\pi^2}{x} \right) }$. We have \[2a(a - b) = 2a^2 - 2\] \[ab = 1\] Notice that either $a = 1$ and $b = 1$ or $a = -1$ and $b = -1$. For the first case, $x = 1