Difference between revisions of "1996 USAMO Problems/Problem 1"
(→Solution) |
|||
| 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== | + | ==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. | ||
| + | <math>\Box</math> | ||
| + | |||
| + | ==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 2^\circ</math> and using sum-to-product and telescoping gives <math>2\sin 2^\circ S = \cos 0^\circ + \cos 2^\circ - \cos 178^\circ - \cos 180^\circ = 2 + 2\cos 2^\circ</math>. Thus, <math>S = \frac{1 + \cos 2^\circ}{\sin 2^\circ} = \cot 1^\circ</math>, as desired. | ||
| + | |||
<math>\Box</math> | <math>\Box</math> | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 22:52, 5 June 2014
Problem
Prove that the average of the numbers 
is 
.
Solution 1
First, as 
we omit that term. Now, we multiply by 
to get, after using product to sum, 
.
This simplifies to 
. Since 
this simplifies to 
. We multiplied by in the beginning, so we must divide by it now, and thus the sum is just 
, so the average is , as desired.
Solution 2
Notice that for every 
there exists a corresponding pair term 
, for 
not 
. Pairing gives the sum of all terms to be 
, and thus the average is ![\[S = (\sin 2^\circ + \sin 4^\circ + ... + \sin 178^\circ). (*)\]](http://latex.artofproblemsolving.com/a/0/8/a081dadc867b44d0b0a481798b4571ed744b03bd.png)
We need to show that 
. Multiplying (*) by 
and using sum-to-product and telescoping gives 
. Thus, 
, as desired.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
