Difference between revisions of "1996 USAMO Problems/Problem 1"
| Line 1: | Line 1: | ||
| − | + | ==Problem== | |
| − | |||
| − | |||
| − | |||
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== | |
| − | |||
| − | |||
| − | |||
| − | |||
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. | ||
Revision as of 20:52, 7 April 2014
Problem
Prove that the average of the numbers 
is 
.
Solution
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.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
