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| − | == Problem ==
| + | #REDIRECT [[2006 AIME I Problems/Problem 12]] |
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| − | 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>
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| − | == 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 ==
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| − | *[[2006 AIME II Problems/Problem 11 | Previous problem]]
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| − | *[[2006 AIME II Problems/Problem 13 | Next problem]]
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| − | *[[2006 AIME II Problems]]
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| − | [[Category:Intermediate Geometry Problems]]
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