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

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 $\lfloor 100(\sqrt{2} + 1)\rfloor = \boxed{241}$ .

@@ Line 2: / Line 2: @@
 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>?
+__TOC__
 == Solution ==
-Manipulating the [[numerator]],
+=== 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>
+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)
+</cmath>
+That fraction is <math>x</math>! Therefore,
 <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}\\
+\lfloor 100x \rfloor &=& \lfloor 100(1 + \sqrt {2}) \rfloor = \boxed{241}\\
 \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
+=== Solution 2 ===
+A slight variant of the above solution, note that
 <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>.
+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 ==

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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