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 "2006 AIME I Problems/Problem 12"

m
m (→‎Solution)
 
(16 intermediate revisions by 12 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
Find the sum of the values of <math> x </math> such that <math> \cos^3 3x+ \cos^3 5x = 8 \cos^3 4x \cos^3 x, </math> where <math> x </math> is measured in degrees and <math> 100< x< 200. </math>
+
Find the sum of the values of <math> x </math> such that <math> \cos^3 3x+ \cos^3 5x = 8 \cos^3 4x \cos^3 x </math>, where <math> x </math> is measured in degrees and <math> 100< x< 200. </math>
−
 
 
−
 
 
−
 
 
−
 
 
  
 
== Solution ==
 
== Solution ==
 +
Observe that <math>2\cos 4x\cos x = \cos 5x + \cos 3x</math> by the sum-to-product formulas. Defining <math>a = \cos 3x</math> and <math>b = \cos 5x</math>, we have <math>a^3 + b^3 = (a+b)^3 \rightarrow ab(a+b) = 0</math>. But <math>a+b = 2\cos 4x\cos x</math>, so we require <math>\cos x = 0</math>, <math>\cos 3x = 0</math>, <math>\cos 4x = 0</math>, or <math>\cos 5x = 0</math>.
  
 +
Hence we see by careful analysis of the cases that the solution set is <math>A = \{150, 126, 162, 198, 112.5, 157.5\}</math> and thus <math>\sum_{x \in A} x = \boxed{906}</math>.
  
 +
== See also ==
 +
{{AIME box|year=2006|n=I|num-b=11|num-a=13}}
  
−
 
+
[[Category:Intermediate Trigonometry Problems]]
−
== See also ==
+
{{MAA Notice}}
−
* [[2006 AIME I Problems]]
 

Latest revision as of 17:54, 13 December 2017

Problem

Find the sum of the values of $x$ such that $\cos^3 3x+ \cos^3 5x = 8 \cos^3 4x \cos^3 x$, where $x$ is measured in degrees and $100< x< 200.$

Solution

Observe that $2\cos 4x\cos x = \cos 5x + \cos 3x$ by the sum-to-product formulas. Defining $a = \cos 3x$ and $b = \cos 5x$, we have $a^3 + b^3 = (a+b)^3 \rightarrow ab(a+b) = 0$. But $a+b = 2\cos 4x\cos x$, so we require $\cos x = 0$, $\cos 3x = 0$, $\cos 4x = 0$, or $\cos 5x = 0$.

Hence we see by careful analysis of the cases that the solution set is $A = \{150, 126, 162, 198, 112.5, 157.5\}$ and thus $\sum_{x \in A} x = \boxed{906}$.

See also

2006 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME 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=2006_AIME_I_Problems/Problem_12&oldid=88895"