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

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<cmath>2c - \frac{2014}{c} \equiv 0 \mod 2</cmath>
 
<cmath>2c - \frac{2014}{c} \equiv 0 \mod 2</cmath>
  
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In order for this to be satisfied, <math>\frac{2014}{c}</math> must be an even integer. Factoring <math>2014 = 2 \cdot 19 \cdot 53</math>, we see that our only positive valid <math>c = 19, 53, 1007</math>. Our answer is just <math>\pi(19+53+1007) = 1080\pi \implies \textbf{(D)}</math>.  
+
In order for this to be satisfied, <math>\frac{2014}{c}</math> must be an even integer. Factoring <math>2014 = 2 \cdot 19 \cdot 53</math>, we see that <math>c</math> must be a positive odd integer that divides <math>2014</math>. Our only positive valid <math>c</math> are <math>c = 1, 19, 53, 1007</math>. Our answer is just <math>\pi(1+19+53+1007) = 1080\pi \implies \textbf{(D)}</math>.  
  
  
 
-FIREDRAGONMATH16
 
-FIREDRAGONMATH16
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==Solution 3==
 +
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Rewriting <math>\cos{4x} - 1</math> as <math>2\cos^2{2x} - 2</math> and transposing <math>2\cos{2x}</math> from the LHS to the RHS, we get,
 +
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<cmath>\cos{2x} - \cos{\left(\frac{2014\pi^2}{x}\right)} = 1 - \frac{1}{\cos{2x}}</cmath>
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<cmath>\implies \underbrace{\cos{2x} + \frac{1}{\cos{2x}}}_{\text{LHS}} = \underbrace{1 + \cos{\left(\frac{2014\pi^2}{x}\right)}}_{\text{RHS}}</cmath>
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 +
By the [https://artofproblemsolving.com/wiki/index.php/AM-GM_Inequality AM-GM Inequality],
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<cmath>\cos{2x} + \frac{1}{\cos{2x}} \in (-\infty, -2] \cup [2, \infty)</cmath>
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 +
Also, because of the range of <math>\cos</math>,
 +
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<cmath>1 + \cos{\left(\frac{2014\pi^2}{x}\right)} \in [0, 2]</cmath>
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Hence, <math>\text{LHS} = \text{RHS} = 2</math>, and we get (<math>m, n \in \mathbb{Z}</math>),
 +
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* <math>\cos{2x} = 1 \implies x = m\pi \hspace{75pt} \cdots\cdots (1)</math>
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* <math>\cos{\left(\frac{2014\pi^2}{x}\right)} = 1 \implies x = \frac{1007\pi}{n} \hspace{22pt} \cdots\cdots (2)</math>
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 +
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From <math>(1)</math> and <math>(2)</math>,
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<cmath>m = \frac{1007}{n}</cmath>
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<cmath>\implies m \in \{1, 19, 53, 1007\}</cmath>
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<cmath>\implies x \in \{1\pi, 19\pi, 53\pi, 1007\pi\}</cmath>
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Therefore, sum of values of <math>x</math> is <cmath>\pi + 19\pi + 53\pi + 1007\pi = \boxed{\textbf{(D)}\ 1080\pi}</cmath>.
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~ [https://artofproblemsolving.com/wiki/index.php/User:Plusone plusone]
  
 
== 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}}

Latest revision as of 09:08, 17 August 2023

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 1

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 \[\pi + 19\pi + 53\pi + 1007\pi = \boxed{\textbf{(D)}\ 1080 \pi}\]

Solution 2

Very similar to the solution above, re-write the expression using $\cos 4x = 2 \cos^2 2x - 1$:

\[2\cos2x \left(\cos2x - \cos{\left( \frac{2014\pi^2}{x} \right) } \right) = 2 \cos^2 2x - 2\]

Now, expand the LHS and cancel terms:

\[2 \cos^2 2x - 2 \cos 2x \cos{\left( \frac{2014\pi^2}{x} \right) } = 2 \cos^2 2x - 2\]

\[\cos 2x \cos{\left( \frac{2014\pi^2}{x} \right) } = 1\]

Now we use product-to-sum identities to get:

\[\frac{1}{2} \left(\cos{2x + \left( \frac{2014\pi^2}{x} \right) } + \cos{2x + \left( \frac{2014\pi^2}{x} \right) } \right) = 1\]

\[\cos{2x + \left( \frac{2014\pi^2}{x} \right) } + \cos{2x + \left( \frac{2014\pi^2}{x} \right) } = 1\]

Notice that for any $\theta$, $\max{\cos \theta} = 1$. This is achieved when $\theta = 0$, or equivalently

\[2x - \frac{2014\pi^2}{x} \equiv 0 \mod 2\pi\]

We can cleverly assume $x=c\pi$ for some real $c$. Then, we must have

\[2c - \frac{2014}{c} \equiv 0 \mod 2\]

In order for this to be satisfied, $\frac{2014}{c}$ must be an even integer. Factoring $2014 = 2 \cdot 19 \cdot 53$, we see that $c$ must be a positive odd integer that divides $2014$. Our only positive valid $c$ are $c = 1, 19, 53, 1007$. Our answer is just $\pi(1+19+53+1007) = 1080\pi \implies \textbf{(D)}$.


-FIREDRAGONMATH16

Solution 3

Rewriting $\cos{4x} - 1$ as $2\cos^2{2x} - 2$ and transposing $2\cos{2x}$ from the LHS to the RHS, we get,

\[\cos{2x} - \cos{\left(\frac{2014\pi^2}{x}\right)} = 1 - \frac{1}{\cos{2x}}\] \[\implies \underbrace{\cos{2x} + \frac{1}{\cos{2x}}}_{\text{LHS}} = \underbrace{1 + \cos{\left(\frac{2014\pi^2}{x}\right)}}_{\text{RHS}}\]

By the AM-GM Inequality,

\[\cos{2x} + \frac{1}{\cos{2x}} \in (-\infty, -2] \cup [2, \infty)\]

Also, because of the range of $\cos$,

\[1 + \cos{\left(\frac{2014\pi^2}{x}\right)} \in [0, 2]\]

Hence, $\text{LHS} = \text{RHS} = 2$, and we get ($m, n \in \mathbb{Z}$),

  • $\cos{2x} = 1 \implies x = m\pi \hspace{75pt} \cdots\cdots (1)$
  • $\cos{\left(\frac{2014\pi^2}{x}\right)} = 1 \implies x = \frac{1007\pi}{n} \hspace{22pt} \cdots\cdots (2)$


From $(1)$ and $(2)$, \[m = \frac{1007}{n}\] \[\implies m \in \{1, 19, 53, 1007\}\] \[\implies x \in \{1\pi, 19\pi, 53\pi, 1007\pi\}\]

Therefore, sum of values of $x$ is \[\pi + 19\pi + 53\pi + 1007\pi = \boxed{\textbf{(D)}\ 1080\pi}\].

~ plusone

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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