Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Problem)
m (→‎Solution)
Line 7: Line 7:
 
=== Solution 1 ===
 
=== 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}\\
 
<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>
+
&=& \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
 
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
Line 13: Line 14:
 
<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)\\
 
<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)
 
&=& \left(\frac {1}{\sqrt {2}}\right)\left(1 + \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\sin 1 + \sin 2 + \dots + \sin 44}\right)
 +
\end{eqnarray*}
 
</cmath>
 
</cmath>
  
Line 29: Line 31:
 
\sum_{n=1}^{44} \cos n + \sum_{n=1}^{44} \sin n &=& \sum_{n=1}^{44} \sin n + \sin(90-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(45-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>
+
\sum_{n=1}^{44} \sin n &=& (\sqrt{2}-1)\sum_{n=1}^{44} \cos n
 +
\end{eqnarray*}</cmath>
  
 
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>.
 
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>.

Revision as of 17:48, 10 March 2015

Problem 11

Let $x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}$. 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) \end{eqnarray*}

This fraction is equivalent to $x$. 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 \end{eqnarray*}

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

Solution 3

Consider the sum $\sum_{n = 1}^{44} \text{cis } n^\circ$. The fraction is given by the real part divided by the imaginary part.

The sum can be written $- 1 + \sum_{n = 0}^{44} \text{cis } n^\circ = - 1 + \frac {\text{cis } 45^\circ - 1}{\text{cis } 1^\circ - 1}$ (by De Moivre's Theorem with geometric series)

$= - 1 + \frac {\frac {\sqrt {2}}{2} - 1 + \frac {i \sqrt {2}}{2}}{\text{cis } 1^\circ - 1} = - 1 + \frac {\left( \frac {\sqrt {2}}{2} - 1 + \frac {i \sqrt {2}}{2} \right) (\text{cis } ( - 1^\circ) - 1)}{(\cos 1^\circ - 1)^2 + \sin^2 1^\circ}$ (after multiplying by complex conjugate)

$= - 1 + \frac {\left( \frac {\sqrt {2}}{2} - 1 \right) (\cos 1^\circ - 1) + \frac {\sqrt {2}}{2}\sin 1^\circ + i\left( \left(1 - \frac {\sqrt {2}}{2} \right) \sin 1^\circ + \frac {\sqrt {2}}{2} (\cos 1^\circ - 1)\right)}{2(1 - \cos 1^\circ)}$

$= - \frac {1}{2} - \frac {\sqrt {2}}{4} - \frac {i\sqrt {2}}{4} + \frac {\sin 1^\circ \left( \frac {\sqrt {2}}{2} + i\left( 1 - \frac {\sqrt {2}}{2} \right) \right)}{2(1 - \cos 1^\circ)}$

Using the tangent half-angle formula, this becomes $\left( - \frac {1}{2} + \frac {\sqrt {2}}{4}[\cot (1/2^\circ) - 1] \right) + i\left( \frac {1}{2}\cot (1/2^\circ) - \frac {\sqrt {2}}{4}[\cot (1/2^\circ) + 1] \right)$.

Dividing the two parts and multiplying each part by 4, the fraction is $\frac { - 2 + \sqrt {2}[\cot (1/2^\circ) - 1]}{2\cot (1/2^\circ) - \sqrt {2}[\cot (1/2^\circ) + 1]}$.

Although an exact value for $\cot (1/2^\circ)$ in terms of radicals will be difficult, this is easily known: it is really large!

So treat it as though it were $\infty$. The fraction is approximated by $\frac {\sqrt {2}}{2 - \sqrt {2}} = \frac {\sqrt {2}(2 + \sqrt {2})}{2} = 1 + \sqrt {2}\Rightarrow \lfloor 100(1+\sqrt2)\rfloor=\boxed{241}$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_11&oldid=68640"