Difference between revisions of "2006 AIME I Problems/Problem 12"
(→Solution) |
m (cat) |
||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | + | 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> | |
− | |||
== 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>. |
+ | |||
+ | 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>. | ||
== See also == | == See also == | ||
{{AIME box|year=2006|n=I|num-b=11|num-a=13}} | {{AIME box|year=2006|n=I|num-b=11|num-a=13}} | ||
+ | |||
+ | [[Category:Intermediate Trigonometry Problems]] |
Revision as of 19:35, 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
.
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 |