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Difference between revisions of "1997 AIME Problems/Problem 11"

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== Problem ==
 
== Problem ==
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Let <math>x=\frac{\sum_{n=1}^{44} \cos n^\circ}{\sum_{n=1}^{44} \sin n^\circ}</math>. What is the greatest integer that does not exceed <math>100x</math>?
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Let <math>x=\frac{\sum_{n=1}^{44} \cos n^\circ}{\sum_{n=1}^{44} \sin n^\circ}</math>. What is the [[greatest integer function|greatest integer]] that does not exceed <math>100x</math>?
  
 
== Solution ==
 
== Solution ==
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{{solution}}
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Manipulating the [[numerator]],
 +
 
 +
<cmath>\begin{eqnarray*}
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\sum_{n=1}^{44} \cos n &=& \sum_{n=1}^{44} \cos n + \sum_{n=1}^{44} \sin n - \sum_{n=1}^{44} \sin n\\
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&=& \sum_{n=1}^{44} \sin n + \sin(90-n) - \sum_{n=1}^{44} \sin n\\
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\end{eqnarray*}</cmath>
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Using the identity <math>\sin a + \sin b = 2\sin \frac{a+b}2 \cos \frac{a-b}{2}</math> <math>\Longrightarrow \sin x + \sin (90-x) = 2 \sin 45 \cos (45-x) = \sqrt{2} \cos (45-x)</math>, that first [[summation]] reduces to
 +
 
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<cmath>\begin{eqnarray*}
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\sum_{n=1}^{44} \cos n &=& \sqrt{2}\sum_{n=1}^{44} \cos(45-n) - \sum_{n=1}^{44} \sin n\\
 +
&=& \sqrt{2}\sum_{n=1}^{44} \cos n - \sum_{n=1}^{44} \sin n\\
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(\sqrt{2} - 1)\sum_{n=1}^{44} \cos n &=& \sum_{n=1}^{44} \sin n\\
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\end{eqnarray*}</cmath>
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This is the [[ratio]] we are looking for! This reduces is <math>\frac{1}{\sqrt{2} - 1} = \sqrt{2} + 1</math>, and <math>\lfloor 100(\sqrt{2} + 1)\rfloor = \boxed{241}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1997|num-b=10|num-a=12}}
 
{{AIME box|year=1997|num-b=10|num-a=12}}
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 +
[[Category:Intermediate Trigonometry Problems]]

Revision as of 19:16, 23 November 2007

Problem

Let $x=\frac{\sum_{n=1}^{44} \cos n^\circ}{\sum_{n=1}^{44} \sin n^\circ}$. What is the greatest integer that does not exceed $100x$?

Solution

Manipulating the numerator,

\begin{eqnarray*} \sum_{n=1}^{44} \cos n &=& \sum_{n=1}^{44} \cos n + \sum_{n=1}^{44} \sin n - \sum_{n=1}^{44} \sin n\\ &=& \sum_{n=1}^{44} \sin n + \sin(90-n) - \sum_{n=1}^{44} \sin n\\ \end{eqnarray*}

Using the identity $\sin a + \sin b = 2\sin \frac{a+b}2 \cos \frac{a-b}{2}$ $\Longrightarrow \sin x + \sin (90-x) = 2 \sin 45 \cos (45-x) = \sqrt{2} \cos (45-x)$, that first summation reduces to

\begin{eqnarray*} \sum_{n=1}^{44} \cos n &=& \sqrt{2}\sum_{n=1}^{44} \cos(45-n) - \sum_{n=1}^{44} \sin n\\ &=& \sqrt{2}\sum_{n=1}^{44} \cos n - \sum_{n=1}^{44} \sin n\\ (\sqrt{2} - 1)\sum_{n=1}^{44} \cos n &=& \sum_{n=1}^{44} \sin n\\ \end{eqnarray*}

This is the ratio we are looking for! This reduces is $\frac{1}{\sqrt{2} - 1} = \sqrt{2} + 1$, and $\lfloor 100(\sqrt{2} + 1)\rfloor = \boxed{241}$.

See also

1997 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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