2000 AIME II Problems/Problem 15
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
![\[\sum_{k=23}^{67} \frac{1}{\sin (2k-1) \sin 2k} = \frac{1}{\sin 1} \left(\cot 45 - \cot 46 + \cot 47 - \cdots + \cot 133 - \cot 134 \right).\]](http://latex.artofproblemsolving.com/7/9/d/79dfa6a1b147c857e927c3fc52695c5420f19c02.png)
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 \frac{4}{3}.
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 | |
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