Difference between revisions of "1997 AIME Problems/Problem 11"
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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>? | ||
+ | __TOC__ | ||
== Solution == | == Solution == | ||
− | + | === 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*} | <cmath>\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*}</cmath> | \end{eqnarray*}</cmath> | ||
− | + | === Solution 2 === | |
+ | A slight variant of the above solution, note that | ||
<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 + \sin(90-n)\\ |
− | &=& \sqrt{2}\sum_{n=1}^{44} \cos 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</cmath> | |
− | |||
− | This is the [[ratio]] we are looking for | + | 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 19:55, 23 November 2007
Problem
Let . What is the greatest integer that does not exceed
?
Solution
Solution 1
Using the identity
, 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 ! Therefore,
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. reduces to
, and
.
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 |