Difference between revisions of "2006 AIME A Problems/Problem 12"

(Solution)
(See also)
Line 10: Line 10:
  
 
== See also ==
 
== See also ==
*[[2006 AIME II Problems/Problem 11 | Previous problem]]
+
{{AIME box|year=2006|n=II|num-b=11|num-a=13}}
*[[2006 AIME II Problems/Problem 13 | Next problem]]
 
*[[2006 AIME II Problems]]
 
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Revision as of 14:57, 25 September 2007

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 $a^3 + b^3 = (a+b)^3 \Leftrightarrow ab(a+b) = 0$. 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 the solution set is $A = \{150, 112, 144, 176, 112.5, 157.5\}$ and thus $\sum_{x \in A} x = 852$.

See also

2006 AIME II (ProblemsAnswer KeyResources)
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