1997 AIME Problems/Problem 11

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

Manipulating the numerator,

$\begin{eqnarray*} \sum_{n=1}^{44} \cos n &=& \sum_{n=1}^{44} \cos n + \sum_{n=1}^{44} \sin n - \sum_{n=1}^{44} \sin n\\ &=& \sum_{n=1}^{44} \sin n + \sin(90-n) - \sum_{n=1}^{44} \sin n\\ \end{eqnarray*}$

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

$\begin{eqnarray*} \sum_{n=1}^{44} \cos n &=& \sqrt{2}\sum_{n=1}^{44} \cos(45-n) - \sum_{n=1}^{44} \sin n\\ &=& \sqrt{2}\sum_{n=1}^{44} \cos n - \sum_{n=1}^{44} \sin n\\ (\sqrt{2} - 1)\sum_{n=1}^{44} \cos n &=& \sum_{n=1}^{44} \sin n\\ \end{eqnarray*}$

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

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
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1997 AIME Problems/Problem 11

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