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

Revision as of 17:04, 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, $x = 1

@@ Line 7: / Line 7: @@
 ==Solution==
-Rewrite <math>\cos{4x} - 1</math> as <math>2\cos^2{2x} - 2</math>.  Now let <math>x = \cos{2x}</math>, and let <math>y = \cos{\left( \frac{2014\pi^2}{x} \right) }</math>.  We have
+Rewrite <math>\cos{4x} - 1</math> as <math>2\cos^2{2x} - 2</math>.  Now let <math>a = \cos{2x}</math>, and let <math>b = \cos{\left( \frac{2014\pi^2}{x} \right) }</math>.  We have
-<cmath>2x(x - y) = 2x^2 - 2</cmath>
+<cmath>2a(a - b) = 2a^2 - 2</cmath>
-<cmath>xy = 1</cmath>
+<cmath>ab = 1</cmath>
-Notice that <math>x = 1</math> and <math>y = 1</math> or <math>x = -1</math> and <math>y = -1</math>.
+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

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

Revision as of 17:04, 20 February 2014

Problem

Solution