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

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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>:
 
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>
 
<cmath>x= \pi, 19\pi, 53\pi, 1007\pi</cmath>
−
For the case where
+
For the case where <math>a = -1</math>, <math>\cos{2x} = -1</math>, so <math>x = \frac{m}{2\pi}</math>, where m is odd.  <math>\frac{2014\pi^2}{\frac{m}{2\pi}}</math> must also be an odd multiple of \pi in order for <math>b</math> to equal <math>-1</math>, so \frac{4028}{m} must be odd.  We can quickly see that dividing an even number by an odd number will never yield an odd number, so there are no possible values for <math>m</math>, and therefore no cases where <math>a = -1</math> and <math>b = -1</math>. Therefore, the sum of all our possible values for <math>x</math> is
 +
<cmath>\pi + 19\pi + 53\pi + 1007\pi = \boxed{\textbf{(D)}\ 1080 \pi}</cmath>
 +
 
 +
(Solution by kevin38017)

Revision as of 17:17, 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 $a = -1$, $\cos{2x} = -1$, so $x = \frac{m}{2\pi}$, where m is odd. $\frac{2014\pi^2}{\frac{m}{2\pi}}$ must also be an odd multiple of \pi in order for $b$ to equal $-1$, so \frac{4028}{m} must be odd. We can quickly see that dividing an even number by an odd number will never yield an odd number, so there are no possible values for $m$, and therefore no cases where $a = -1$ and $b = -1$. Therefore, the sum of all our possible values for $x$ is \[\pi + 19\pi + 53\pi + 1007\pi = \boxed{\textbf{(D)}\ 1080 \pi}\]

(Solution by kevin38017)

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