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

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== Solution ==
 
== Solution ==
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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 \Leftrightarrow 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>.
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Hence the solution set is <math>A = \{150, 112, 144, 176, 112.5, 157.5\}</math> and thus <math>\sum_{x \in A} x = 852</math>.
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== See also ==
 
== See also ==
 
*[[2006 AIME II Problems/Problem 11 | Previous problem]]
 
*[[2006 AIME II Problems/Problem 11 | Previous problem]]

Revision as of 14:45, 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