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

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== Problem ==
 
== Problem ==
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[[Equilateral triangle | Equilateral]] <math> \triangle ABC </math> is inscribed in a [[circle]] of [[radius]] 2. Extend <math> \overline{AB} </math> through <math> B </math> to point <math> D </math> so that <math> AD=13, </math> and extend <math> \overline{AC} </math> through <math> C </math> to point <math> E </math> so that <math> AE = 11. </math> Through <math> D, </math> draw a line <math> l_1 </math> [[parallel]] to <math> \overline{AE}, </math> and through <math> E, </math> draw a line <math> l_2 </math> parallel to <math> \overline{AD}. </math> Let <math> F </math> be the [[intersection]] of <math> l_1 </math> and <math> l_2. </math> Let <math> G </math> be the point on the circle that is [[collinear]] with <math> A </math> and <math> F </math> and distinct from <math> A. </math> Given that the [[area]] of <math> \triangle CBG </math> can be expressed in the form <math> \frac{p\sqrt{q}}{r}, </math>  where <math> p, q, </math> and <math> r </math> are [[positive integer]]s, <math> p </math> and <math> r</math>  are [[relatively prime]], and <math> q </math> is not [[divisibility | divisible]] by the [[perfect square | square]] of any [[prime]], find <math> p+q+r. </math>
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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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{{image needed}}
 
  
 
== Solution ==
 
== Solution ==
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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 \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>.
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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>.
  
 
== 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}}
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[[Category:Intermediate Trigonometry Problems]]
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{{MAA Notice}}

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 $\sum_{x \in A} x = \boxed{906}$.

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

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