# 1997 AIME Problems/Problem 11

## 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 byProblem 10 Followed byProblem 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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