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 4/3.
There are 45 terms so the approximate sum is 60.
Therefore the entire thing equals approximately 
.
Recall that the approximation of sinx in radians is x if x is close to zero. In this case x is close to zero. Converting to radians we see that sin1 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 sin1= in degrees, convert to radians and use the small angle approximation sinx=x.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
