Difference between revisions of "2000 AIME II Problems/Problem 15"
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== Problem == | == Problem == | ||
+ | Find the least positive integer <math>n</math> such that <center><math>\frac 1{\sin 45^\circ\sin 46^\circ}+\frac 1{\sin 47^\circ\sin 48^\circ}+\cdots+\frac 1{\sin 133^\circ\sin 134^\circ}=\frac 1{\sin n^\circ}.</math></center> | ||
− | == Solution == | + | == Solution 1 == |
+ | 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]]. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | We can make an approximation by observing the following points: | ||
+ | |||
+ | The average term is around the 60's which gives <math>\frac{4}{3}</math>. | ||
+ | |||
+ | There are 45 terms, so the approximate sum is 60. | ||
+ | |||
+ | Therefore, the entire thing equals approximately <math>\frac{1}{60}</math>. | ||
+ | |||
+ | Recall that the approximation of <math>\sin(x)</math> in radians is x if x is close to zero. In this case x is close to zero. Converting to radians we see that <math>\sin(1)</math> in degrees is about sin<math>\frac{1}{57}</math> in radians, or is about <math>\frac{1}{57}</math> because of the approximation. What we want is apparently close to that so we make the guess that n is equal to 1 degree. Basically, it boils down to the approximation of <math>\sin(1)=\frac{1}{60}</math> in degrees, convert to radians and use the small angle approximation <math>\sin(x)=x</math>. | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=2000|n=II|num-b=14|after=Last Question}} | |
+ | |||
+ | [[Category:Intermediate Trigonometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 09:37, 26 October 2016
Contents
Problem
Find the least positive integer such that
Solution 1
We apply the identity
The motivation for this identity arises from the need to decompose those fractions, possibly into telescoping.
Thus our summation becomes
Since , the summation simply reduces to . Therefore, the answer is .
Solution 2
We can make an approximation by observing the following points:
The average term is around the 60's which gives .
There are 45 terms, so the approximate sum is 60.
Therefore, the entire thing equals approximately .
Recall that the approximation of in radians is x if x is close to zero. In this case x is close to zero. Converting to radians we see that in degrees is about sin in radians, or is about because of the approximation. What we want is apparently close to that so we make the guess that n is equal to 1 degree. Basically, it boils down to the approximation of in degrees, convert to radians and use the small angle approximation .
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 |
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