Difference between revisions of "2006 AIME A Problems/Problem 12"
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== 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 \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 such that where is measured in degrees and
Solution
Observe that by the sum-to-product formulas. Defining and , we have . But , so we require , , , or .
Hence the solution set is and thus .