Difference between revisions of "2000 AIME II Problems/Problem 15"
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== Solution == | == Solution == | ||
− | + | We apply the identity | |
+ | <cmath>\begin{align*} | ||
+ | \frac{1}{\sin n \sin (n+1)} &= \frac{1}{\sin 1} \cdot \frac{\sin (n+1) \cos n - \sin n \cos (n+1)}{\sin n \sin (n+1)} \\ &= \frac{1}{\sin 1} \cdot \left(\frac{\cos n}{\sin n} - \frac{\cos (n+1)}{\sin (n+1)}\right) \\ &= \frac{1}{\sin 1} \cdot \left(\cot n - \cot (n+1)\right). \end{align*}</cmath> | ||
+ | |||
+ | The motivation for this identity arises from the need to decompose those fractions, possibly into [[telescoping series]]. | ||
+ | |||
+ | Thus our summation becomes | ||
+ | |||
+ | <cmath>\sum_{k=23}^{67} \frac{1}{\sin (2k-1) \sin 2k} = \frac{1}{\sin 1} \left(\cot 45 - \cot 46 + \cot 47 - \cdots + \cot 133 - \cot 134 \right).</cmath> | ||
+ | |||
+ | Since <math>\cot (180 - x) = - \cot x</math>, the summation simply reduces to <math>\frac{1}{\sin 1} \cdot \left( \cot 45 - \cot 90 \right) = \frac{1 - 0}{\sin 1} = \frac{1}{\sin 1^{\circ}}</math>. Therefore, the answer is <math>\boxed{001}</math>. | ||
+ | |||
+ | == See also == | ||
{{AIME box|year=2000|n=II|num-b=14|after=Last Question}} | {{AIME box|year=2000|n=II|num-b=14|after=Last Question}} | ||
+ | |||
+ | [[Category:Intermediate Trigonometry Problems]] |
Revision as of 16:49, 30 August 2008
Problem
Find the least positive integer such that
Solution
We apply the identity
The motivation for this identity arises from the need to decompose those fractions, possibly into telescoping series.
Thus our summation becomes
Since , the summation simply reduces to . Therefore, the answer is .
See also
2000 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |