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

Latest revision as of 17:54, 13 December 2017

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 $a^3 + b^3 = (a+b)^3 \rightarrow ab(a+b) = 0$ . 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 we see by careful analysis of the cases that the solution set is $A = \{150, 126, 162, 198, 112.5, 157.5\}$ and thus $\sum_{x \in A} x = \boxed{906}$ .

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 == 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 ==
-<math> \cos^3 3x+ \cos^3 5x = 8 \cos^3 4x \cos^3 x </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>.
-<math> \cos^3 (4x-x)+ \cos^3 (4x+x) = 8 \cos^3 4x \cos^3 x </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>.
-Using the sum and difference formulas for the cosine function:
+== See also ==
+{{AIME box|year=2006|n=I|num-b=11|num-a=13}}
-<math> ( \cos 4x \cos x + \sin 4x \sin x )^3 + ( \cos 4x \cos x - \sin 4x \sin x )^3 = 8 \cos^3 4x \cos^3 x </math>
-Expanding the expression:
+[[Category:Intermediate Trigonometry Problems]]
+{{MAA Notice}}
-<math> ( \cos^3 4x \cos^3 x + 3 \cos^2 4x \cos^2 x \sin 4x \sin x + 3 \cos 4x \cos x \sin^2 4x \sin^2 x + \sin^3 4x \sin^3 x )</math>
-<math>+</math>
-<math>( \cos^3 4x \cos^3 x - 3 \cos^2 4x \cos^2 x \sin 4x \sin x + 3 \cos 4x \cos x \sin^2 4x \sin^2 x - \sin^3 4x \sin^3 x )</math>
-<math>=</math>
-<math> 8 \cos^3 4x \cos^3 x </math>
-Combining like terms:
-<math> 2 \cos^3 4x \cos^3 x + 6 \cos 4x \cos x \sin^2 4x \sin^2 x = 8 \cos^3 4x \cos^3 x</math>
-<math> -6 \cos^3 4x \cos^3 x + 6 \cos 4x \cos x \sin^2 4x \sin^2 x = 0 </math>
-Factoring <math> -6 \cos 4x \cos x </math>:
-<math> ( -6 \cos 4x \cos x ) ( \cos^2 4x \cos^2 x - \sin^2 4x \sin^2 x )= 0 </math>
-Using the difference of squares factorization:
-<math> ( -6 \cos 4x \cos x ) ( \cos 4x \cos x + \sin 4x \sin x ) ( \cos 4x \cos x - \sin 4x \sin x )= 0 </math>
-Using the sum and difference formulas for cosine in reverse:
-<math> ( -6 \cos 4x \cos x ) (\cos (4x-x)) ( \cos (4x+x))= 0 </math>
-<math> -6 \cos 4x \cos x \cos 3x \cos 5x = 0 </math>
-Setting each non-constant factor equal to 0:
-<math> \cos x = 0 </math>
-<math> x = 90, 270, .... </math>
-<math> \cos 3x = 0 </math>
-<math> 3x = 90, 270, 450, 630, .... </math>
-<math> x = 30, 90, 150, 210, .... </math>
-<math> \cos 4x = 0 </math>
-<math> 4x = 90, 270, 450, 630, 810, .... </math>
-<math> x = 22.5, 67.5, 112.5, 157.5, 202.5, .... </math>
-<math> \cos 5x = 0 </math>
-<math> 5x = 90, 270, 450, 630, 810, 990, 1170, .... </math>
-<math> x = 18, 54, 90, 126, 162, 198, 234, .... </math>
-So the sum of the values of <math>x</math> where <math> 100< x< 200 </math> is:
-<math> 112.5 + 126 + 150 + 157.5 + 162 + 198 = 906 </math>
-== See also ==
-{{AIME box|year=2006|n=I|num-b=4|num-a=6}}

2006 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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Difference between revisions of "2006 AIME I Problems/Problem 12"

Latest revision as of 17:54, 13 December 2017

Problem

Solution

See also