Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2000 AIME II Problems/Problem 15"

m (→‎Solution 2)
(→‎Solution 3 (Alternate Finish))
 
(3 intermediate revisions by 2 users not shown)
Line 19: Line 19:
 
We can make an approximation by observing the following points:
 
We can make an approximation by observing the following points:
  
−
The average term is around the 60's which gives \frac{4}{3}.
+
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.
 
There are 45 terms, so the approximate sum is 60.
Line 25: Line 25:
 
Therefore, the entire thing equals approximately <math>\frac{1}{60}</math>.
 
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>.
+
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>.
 +
 
 +
== Solution 3 (Alternate Finish) ==
 +
Let S be the sum of the sequence. We begin the same as in Solution 1 to get
 +
<math>S\sin(1)=\cot(45)-\cot(46)+\cot(47)-\cot(48)+...+\cot(133)-\cot(134)</math>. Observe that this "almost telescopes," if only we had some extra terms. Consider adding the sequence <math>\frac{1}{\sin(46)\sin(47)}+\frac{1}{\sin(48)\sin(49)}+...+\frac{1}{\sin(134)\sin(135)}</math>. By the identity <math>\sin(x)=\sin(180-x)</math>, this sequence is equal to the original one, simply written backwards. By the same logic as before, we may rewrite this sequence as
 +
<math>S\sin(1)=\cot(46)-\cot(47)+\cot(48)-\cot(49)+...+\cot(134)-\cot(135)</math>,
 +
and when we add the two sequences, they telescope to give <math>2S\sin(1)=\cot(45)-\cot(135)=2</math>.
 +
Hence, <math>S=\sin(1)</math>, and our angle is <math>\boxed{001}</math>.
 +
 
 +
~keeper1098
  
 
== See also ==
 
== See also ==

Latest revision as of 20:14, 7 July 2022

Problem

Find the least positive integer $n$ such that

$\frac 1{\sin 45^\circ\sin 46^\circ}+\frac 1{\sin 47^\circ\sin 48^\circ}+\cdots+\frac 1{\sin 133^\circ\sin 134^\circ}=\frac 1{\sin n^\circ}.$

Solution 1

We apply the identity

\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*}

The motivation for this identity arises from the need to decompose those fractions, possibly into telescoping.

Thus our summation becomes

\[\sum_{k=23}^{67} \frac{1}{\sin (2k-1) \sin 2k} = \frac{1}{\sin 1} \left(\cot 45 - \cot 46 + \cot 47 - \cdots + \cot 133 - \cot 134 \right).\]

Since $\cot (180 - x) = - \cot x$, the summation simply reduces to $\frac{1}{\sin 1} \cdot \left( \cot 45 - \cot 90 \right) = \frac{1 - 0}{\sin 1} = \frac{1}{\sin 1^{\circ}}$. Therefore, the answer is $\boxed{001}$.

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 $\frac{1}{60}$.

Recall that the approximation of $\sin(x)$ in radians is x if x is close to zero. In this case x is close to zero. Converting to radians we see that $\sin(1)$ in degrees is about sin$\frac{1}{57}$ in radians, or is about $\frac{1}{57}$ 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 $\sin(1)=\frac{1}{60}$ in degrees, convert to radians and use the small angle approximation $\sin(x)=x$.

Solution 3 (Alternate Finish)

Let S be the sum of the sequence. We begin the same as in Solution 1 to get $S\sin(1)=\cot(45)-\cot(46)+\cot(47)-\cot(48)+...+\cot(133)-\cot(134)$. Observe that this "almost telescopes," if only we had some extra terms. Consider adding the sequence $\frac{1}{\sin(46)\sin(47)}+\frac{1}{\sin(48)\sin(49)}+...+\frac{1}{\sin(134)\sin(135)}$. By the identity $\sin(x)=\sin(180-x)$, this sequence is equal to the original one, simply written backwards. By the same logic as before, we may rewrite this sequence as $S\sin(1)=\cot(46)-\cot(47)+\cot(48)-\cot(49)+...+\cot(134)-\cot(135)$, and when we add the two sequences, they telescope to give $2S\sin(1)=\cot(45)-\cot(135)=2$. Hence, $S=\sin(1)$, and our angle is $\boxed{001}$.

~keeper1098

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_II_Problems/Problem_15&oldid=175714"