Difference between revisions of "1996 USAMO Problems/Problem 1"

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==Solution==
 
==Solution==
 
===Solution 1===
 
===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-\cos 5^\circ)+\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.
 
   
 
   

Revision as of 13:30, 11 February 2019

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, $(\cos 1^\circ-\cos 3^\circ)+2(\cos 3^\circ-\cos 5^\circ)+\cdots +89(\cos 177^\circ-\cos 179^\circ)$. 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 (ProblemsResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5 6
All USAMO Problems and Solutions

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