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

(→‎Problem)
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==Solution==
 
==Solution==
−
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  
+
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>2a(a - b) = 2a^2 - 2</cmath>
 
<cmath>2a(a - b) = 2a^2 - 2</cmath>
−
<cmath>ab = 1</cmath>
+
 
 +
Therefore, <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, <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 these 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 these 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>
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<cmath>\pi + 19\pi + 53\pi + 1007\pi = \boxed{\textbf{(D)}\ 1080 \pi}</cmath>
 
<cmath>\pi + 19\pi + 53\pi + 1007\pi = \boxed{\textbf{(D)}\ 1080 \pi}</cmath>
  
 +
(Minor edits for clarification done by ~hastapasta)
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2014|ab=B|num-b=24|after=Last Question}}
 
{{AMC12 box|year=2014|ab=B|num-b=24|after=Last Question}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 13:46, 7 April 2022

Problem

Find the sum of all the positive solutions of

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

$\textbf{(A)}\ \pi \qquad\textbf{(B)}\ 810\pi \qquad\textbf{(C)}\ 1008\pi \qquad\textbf{(D)}\ 1080 \pi \qquad\textbf{(E)}\ 1800\pi$

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: \[2a(a - b) = 2a^2 - 2\]

Therefore, \[ab = 1\].

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 these possible values for $x$: \[x= \pi, 19\pi, 53\pi, 1007\pi\] For the case where $a = -1$, $\cos{2x} = -1$, so $x = \frac{m\pi}{2}$, where m is odd. $\frac{2014\pi^2}{\frac{m\pi}{2}}$ 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 \[\pi + 19\pi + 53\pi + 1007\pi = \boxed{\textbf{(D)}\ 1080 \pi}\]

(Minor edits for clarification done by ~hastapasta)

See also

2014 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Question
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

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