Difference between revisions of "2014 AMC 12B Problems/Problem 25"

Revision as of 17:11, 20 February 2014

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, $a = 1$ only when $x = k\pi$ and $k$ is an integer. $b = 1$ when $\frac{2014\pi^2}{k\pi}$ is an even multiple of $\pi$ , and since $2014 = 2*19*53$ , $b =1$ only when $k$ is an odd divisor of $2014$ . This gives us the possible values for $x$ : $x= \pi, 19\pi, 53\pi, 1007\pi$ For the case where

@@ Line 10: / Line 10: @@
 <cmath>2a(a - b) = 2a^2 - 2</cmath>
 <cmath>ab = 1</cmath>
-Notice that either <math>a = 1</math> and <math>b = 1</math> or <math>a = -1</math> and <math>b = -1</math>.  For the first case, $x = 1
+Notice that either <math>a = 1</math> and <math>b = 1</math> or <math>a = -1</math> and <math>b = -1</math>.  For the first case, <math>a = 1</math> only when <math>x = k\pi</math> and <math>k</math> is an integer.  <math>b = 1</math> when <math>\frac{2014\pi^2}{k\pi}</math> is an even multiple of <math>\pi</math>, and since <math>2014 = 2*19*53</math>, <math>b =1</math> only when <math>k</math> is an odd divisor of <math>2014</math>.  This gives us the possible values for <math>x</math>:
+<cmath>x= \pi, 19\pi, 53\pi, 1007\pi</cmath>
+For the case where

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Difference between revisions of "2014 AMC 12B Problems/Problem 25"

Revision as of 17:11, 20 February 2014

Problem

Solution