Difference between revisions of "2006 AIME I Problems/Problem 12"
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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 \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>. | + | 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>. | 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>. | ||
Revision as of 17:51, 13 December 2017
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 we see by careful analysis of the cases that the solution set is 
and thus 
.
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. 
