Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2014 AMC 12B Problems/Problem 25"
(→‎Solution 2)
Line 25: Line 25:
 
Now, expand the LHS and cancel terms:
 
Now, expand the LHS and cancel terms:
  
−
<cmath>2 \cos^2 2x - 2 \cos 2x \cos{\left( \frac{2014\pi^2}{x} \right) } \right) = 2 \cos^2 2x - 2</cmath>
+
<cmath>2 \cos^2 2x - 2 \cos 2x \cos{\left( \frac{2014\pi^2}{x} \right) } = 2 \cos^2 2x - 2</cmath>
  
−
<cmath> \cos 2x \cos{\left( \frac{2014\pi^2}{x} \right) } \right) = 1</cmath>
+
<cmath> \cos 2x \cos{\left( \frac{2014\pi^2}{x} \right) } = 1</cmath>
  
 
Now we use product-to-sum identities to get:
 
Now we use product-to-sum identities to get:

Revision as of 16:16, 26 September 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 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: 

\[\frac{1}{2} \left(\cos{2x + \left( \frac{2014\pi^2}{x} \right) } \right) + \cos{2x + \left( \frac{2014\pi^2}{x} \right) } \right) \right) = 1\] (Error compiling LaTeX. Unknown error_msg)
\[\cos{2x + \left( \frac{2014\pi^2}{x} \right) } \right) + \cos{2x + \left( \frac{2014\pi^2}{x} \right) } \right) = 1\] (Error compiling LaTeX. Unknown error_msg)

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 our only positive valid $c = 19, 53, 1007$. Our answer is just $\pi(19+53+1007) = 1080\pi \implies \textbf{(D)}$.


-FIREDRAGONMATH16

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_12B_Problems/Problem_25&oldid=178460"