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-\ | + | 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 is .
Solution
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 We need to show that . Multiplying (*) by and using sum-to-product and telescoping gives . Thus, , as desired.
See Also
1996 USAMO (Problems • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.