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

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(more intuitive solution, revised version of seraph22's, changed my own sol to be more direct)
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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>?
 
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>?
  
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__TOC__
 
== Solution ==
 
== Solution ==
Manipulating the [[numerator]],
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=== Solution 1 ===
 +
<cmath>\begin{eqnarray*} x &=& \frac {\sum_{n = 1}^{44} \cos n^\circ}{\sum_{n = 1}^{44} \sin n^\circ} = \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\sin 1 + \sin 2 + \dots + \sin 44}\
 +
&=& \frac {\cos (45 - 1) + \cos(45 - 2) + \dots + \cos(45 - 44)}{\sin 1 + \sin 2 + \dots + \sin 44}\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 + \cos x</math> <math>= \sin x + \sin (90-x)</math> <math>= 2 \sin 45 \cos (45-x)</math> <math>= \sqrt{2} \cos (45-x)</math>, that [[summation]] reduces to
 +
 +
<cmath>\begin{eqnarray*}x &=& \left(\frac {1}{\sqrt {2}}\right)\left(\frac {(\cos 1 + \cos2 + \dots + \cos44) + (\sin1 + \sin2 + \dots + \sin44)}{\sin1 + \sin2 + \dots + \sin44}\right)\
 +
&=& \left(\frac {1}{\sqrt {2}}\right)\left(1 + \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\sin 1 + \sin 2 + \dots + \sin 44}\right)
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</cmath>
 +
 +
That fraction is <math>x</math>! Therefore,
 
<cmath>\begin{eqnarray*}
 
<cmath>\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\  
+
x &=& \left(\frac {1}{\sqrt {2}}\right)\left(1 + x\right)\
&=& \sum_{n=1}^{44} \sin n + \sin(90-n) - \sum_{n=1}^{44} \sin n\
+
\frac {1}{\sqrt {2}} &=& x\left(\frac {\sqrt {2} - 1}{\sqrt {2}}\right)\
 +
x &=& \frac {1}{\sqrt {2} - 1} = 1 + \sqrt {2}\\
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\lfloor 100x \rfloor &=& \lfloor 100(1 + \sqrt {2}) \rfloor = \boxed{241}\
 
\end{eqnarray*}</cmath>
 
\end{eqnarray*}</cmath>
  
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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=== Solution 2 ===
 +
A slight variant of the above solution, note that  
  
 
<cmath>\begin{eqnarray*}
 
<cmath>\begin{eqnarray*}
\sum_{n=1}^{44} \cos n &=& \sqrt{2}\sum_{n=1}^{44} \cos(45-n) - \sum_{n=1}^{44} \sin n\
+
\sum_{n=1}^{44} \cos n + \sum_{n=1}^{44} \sin n &=& \sum_{n=1}^{44} \sin n + \sin(90-n)\
&=& \sqrt{2}\sum_{n=1}^{44} \cos n - \sum_{n=1}^{44} \sin n\
+
&=& \sqrt{2}\sum_{n=1}^{44} \cos(45-n) = \sqrt{2}\sum_{n=1}^{44} \cos n\
(\sqrt{2} - 1)\sum_{n=1}^{44} \cos n &=& \sum_{n=1}^{44} \sin n\
+
\sum_{n=1}^{44} \sin n &=& (\sqrt{2}-1)\sum_{n=1}^{44} \cos n</cmath>
\end{eqnarray*}</cmath>
 
  
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>.
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This is the [[ratio]] we are looking for. <math>x</math> reduces to <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 ==

Revision as of 18:55, 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

Solution 1

\begin{eqnarray*} x &=& \frac {\sum_{n = 1}^{44} \cos n^\circ}{\sum_{n = 1}^{44} \sin n^\circ} = \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\sin 1 + \sin 2 + \dots + \sin 44}\\ &=& \frac {\cos (45 - 1) + \cos(45 - 2) + \dots + \cos(45 - 44)}{\sin 1 + \sin 2 + \dots + \sin 44}\end{eqnarray*}

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

\begin{eqnarray*}x &=& \left(\frac {1}{\sqrt {2}}\right)\left(\frac {(\cos 1 + \cos2 + \dots + \cos44) + (\sin1 + \sin2 + \dots + \sin44)}{\sin1 + \sin2 + \dots + \sin44}\right)\\
&=& \left(\frac {1}{\sqrt {2}}\right)\left(1 + \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\sin 1 + \sin 2 + \dots + \sin 44}\right) (Error compiling LaTeX. Unknown error_msg)

That fraction is $x$! Therefore, \begin{eqnarray*} x &=& \left(\frac {1}{\sqrt {2}}\right)\left(1 + x\right)\\ \frac {1}{\sqrt {2}} &=& x\left(\frac {\sqrt {2} - 1}{\sqrt {2}}\right)\\ x &=& \frac {1}{\sqrt {2} - 1} = 1 + \sqrt {2}\\ \lfloor 100x \rfloor &=& \lfloor 100(1 + \sqrt {2}) \rfloor = \boxed{241}\\ \end{eqnarray*}

Solution 2

A slight variant of the above solution, note that

\begin{eqnarray*}
\sum_{n=1}^{44} \cos n + \sum_{n=1}^{44} \sin n &=& \sum_{n=1}^{44} \sin n + \sin(90-n)\\
&=& \sqrt{2}\sum_{n=1}^{44} \cos(45-n) = \sqrt{2}\sum_{n=1}^{44} \cos n\\
\sum_{n=1}^{44} \sin n &=& (\sqrt{2}-1)\sum_{n=1}^{44} \cos n (Error compiling LaTeX. Unknown error_msg)

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

See also

1997 AIME (ProblemsAnswer KeyResources)
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