2014 AMC 12B Problems/Problem 25

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

2014 AMC 12B (Problems • Answer Key • Resources)
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2014 AMC 12B Problems/Problem 25

Contents

Problem

Solution 1

Solution 2

Solution 3

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