Art of Problem Solving
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

1996 USAMO Problems/Problem 1: Difference between revisions

← Older editNewer edit →
Suli (talk | contribs)
editor
445 edits
1=2 (talk | contribs)
4,129 edits
Reconstructed from page template
Line 2: Line 2:
Prove that the average of the numbers <math> n\sin n^{\circ}\; (n = 2,4,6,\ldots,180) </math> is <math>\cot 1^\circ</math>.
Prove that the average of the numbers <math> n\sin n^{\circ}\; (n = 2,4,6,\ldots,180) </math> is <math>\cot 1^\circ</math>.


==Solution 1==
==Solution==
===Solution 1===
First, as <math>180\sin{180^\circ}=0,</math> we omit that term. Now, we multiply by <math>\sin 1^\circ</math> to get, after using product to sum, <math>(\cos 1^\circ-\cos 3^\circ)+2(\cos 3^\circ-\cos5)+\cdots +89(\cos 177^\circ-\cos 179^\circ)</math>.  
First, as <math>180\sin{180^\circ}=0,</math> we omit that term. Now, we multiply by <math>\sin 1^\circ</math> to get, after using product to sum, <math>(\cos 1^\circ-\cos 3^\circ)+2(\cos 3^\circ-\cos5)+\cdots +89(\cos 177^\circ-\cos 179^\circ)</math>.  
This simplifies to <math>\cos 1^\circ+\cos 3^\circ +\cos 5^\circ+\cos 7^\circ+...+\cos 177^\circ-89\cos 179^\circ</math>. Since <math>\cos x=-\cos(180-x),</math> this simplifies to <math>90\cos 1^\circ</math>. We multiplied by <math>\sin 1^\circ</math> in the beginning, so we must divide by it now, and thus the sum is just <math>90\cot 1^\circ</math>, so the average is <math>\cot 1^\circ</math>, as desired.
This simplifies to <math>\cos 1^\circ+\cos 3^\circ +\cos 5^\circ+\cos 7^\circ+...+\cos 177^\circ-89\cos 179^\circ</math>. Since <math>\cos x=-\cos(180-x),</math> this simplifies to <math>90\cos 1^\circ</math>. We multiplied by <math>\sin 1^\circ</math> in the beginning, so we must divide by it now, and thus the sum is just <math>90\cot 1^\circ</math>, so the average is <math>\cot 1^\circ</math>, as desired.
Line 8: Line 9:
<math>\Box</math>
<math>\Box</math>


==Solution 2==
===Solution 2===
Notice that for every <math>n\sin n^\circ</math> there exists a corresponding pair term <math>(180^\circ - n)\sin{180^\circ - n} = (180^\circ - n)\sin n^\circ</math>, for <math>n</math> not <math>90^\circ</math>. Pairing gives the sum of all <math>n\sin n^\circ</math> terms to be <math>90(\sin 2^\circ + \sin 4^\circ + ... + \sin 178^\circ)</math>, and thus the average is <cmath>S = (\sin 2^\circ + \sin 4^\circ + ... + \sin 178^\circ). (*)</cmath> We need to show that <math>S = \cot 1^\circ</math>. Multiplying (*) by <math>2\sin 1^\circ</math> and using sum-to-product and telescoping gives <math>2\sin 1^\circ S = \cos 1^\circ - \cos 179^\circ = 2\cos 1^\circ</math>. Thus, <math>S = \frac{\cos 1^\circ}{\sin 1^\circ} = \cot 1^\circ</math>, as desired.
Notice that for every <math>n\sin n^\circ</math> there exists a corresponding pair term <math>(180^\circ - n)\sin{180^\circ - n} = (180^\circ - n)\sin n^\circ</math>, for <math>n</math> not <math>90^\circ</math>. Pairing gives the sum of all <math>n\sin n^\circ</math> terms to be <math>90(\sin 2^\circ + \sin 4^\circ + ... + \sin 178^\circ)</math>, and thus the average is <cmath>S = (\sin 2^\circ + \sin 4^\circ + ... + \sin 178^\circ). (*)</cmath> We need to show that <math>S = \cot 1^\circ</math>. Multiplying (*) by <math>2\sin 1^\circ</math> and using sum-to-product and telescoping gives <math>2\sin 1^\circ S = \cos 1^\circ - \cos 179^\circ = 2\cos 1^\circ</math>. Thus, <math>S = \frac{\cos 1^\circ}{\sin 1^\circ} = \cot 1^\circ</math>, as desired.


<math>\Box</math>
<math>\Box</math>
== See Also ==
{{USAMO box|year=1996|before=First Question|num-a=2}}
{{MAA Notice}}
{{MAA Notice}}
[[Category:Olympiad Algebra Problems]]

Revision as of 08:23, 20 July 2016

Problem

Prove that the average of the numbers $n\sin n^{\circ}\; (n = 2,4,6,\ldots,180)$ is $\cot 1^\circ$.

Solution

Solution 1

First, as $180\sin{180^\circ}=0,$ we omit that term. Now, we multiply by $\sin 1^\circ$ to get, after using product to sum, $(\cos 1^\circ-\cos 3^\circ)+2(\cos 3^\circ-\cos5)+\cdots +89(\cos 177^\circ-\cos 179^\circ)$. This simplifies to $\cos 1^\circ+\cos 3^\circ +\cos 5^\circ+\cos 7^\circ+...+\cos 177^\circ-89\cos 179^\circ$. Since $\cos x=-\cos(180-x),$ this simplifies to $90\cos 1^\circ$. We multiplied by $\sin 1^\circ$ in the beginning, so we must divide by it now, and thus the sum is just $90\cot 1^\circ$, so the average is $\cot 1^\circ$, as desired.

$\Box$

Solution 2

Notice that for every $n\sin n^\circ$ there exists a corresponding pair term $(180^\circ - n)\sin{180^\circ - n} = (180^\circ - n)\sin n^\circ$, for $n$ not $90^\circ$. Pairing gives the sum of all $n\sin n^\circ$ terms to be $90(\sin 2^\circ + \sin 4^\circ + ... + \sin 178^\circ)$, and thus the average is \[S = (\sin 2^\circ + \sin 4^\circ + ... + \sin 178^\circ). (*)\] We need to show that $S = \cot 1^\circ$. Multiplying (*) by $2\sin 1^\circ$ and using sum-to-product and telescoping gives $2\sin 1^\circ S = \cos 1^\circ - \cos 179^\circ = 2\cos 1^\circ$. Thus, $S = \frac{\cos 1^\circ}{\sin 1^\circ} = \cot 1^\circ$, as desired.

$\Box$

See Also

1996 USAMO (Problems • Resources)
Preceded by
First Question 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1996_USAMO_Problems/Problem_1&oldid=79620"