Difference between revisions of "1996 USAMO Problems/Problem 1"
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===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. | ||
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+ | <math>\Box</math> | ||
+ | |||
+ | ===Solution 3 (Very long and detailed)=== | ||
+ | Make the sum of the numbers equal <math>x</math>. Now, <math>2\sin{2^\circ}+4\sin{4^\circ}+...+180\sin{180^\circ}=x</math> the average of these numbers is <math>\frac{x}{90}</math>. | ||
+ | |||
+ | We know that <math>180\sin{180^\circ}=0</math>, so we can eliminate that term and use the identity, <math>\sin(\theta)=\sin(180-\theta)</math> to get | ||
+ | |||
+ | <math>x=(2\sin{2^\circ}+178\sin{2^\circ})+...+(88\sin{88^\circ}+92\sin{22^\circ})+90\sin{90^\circ}</math> | ||
+ | |||
+ | Or, <math>x=180(\sin{2^\circ}+\sin{4^\circ}+\sin{6^\circ}+...+\sin{88^\circ})+90</math> | ||
+ | |||
+ | Pairing the terms up, using sum-product identity, and simplifying, yields: <math>x=180\sqrt 2 (\sin47^\circ + \sin49^\circ + \ldots + \sin89^\circ)+ 90</math>. | ||
+ | |||
+ | After dividing both sides by <math>90</math>, you get: <math>\frac{x}{90}=2\sqrt 2 (\sin47^\circ + \sin49^\circ + \ldots + \sin89^\circ) + 1=\frac{\cos{1^\circ}}{\sin{1^\circ}}</math>. | ||
+ | |||
+ | Now we have to prove that <math>2\sqrt 2 (\sin47^\circ + \sin49^\circ + \ldots + \sin89^\circ) + 1=\frac{\cos{1^\circ}}{\sin{1^\circ}}</math>. | ||
+ | |||
+ | Multiply both sides by <math>\sin{1^\circ}</math> to get <math>\cos{1^\circ}=2\sqrt{2}(\sin{1^\circ}\sin{47^\circ}+\sin{1^\circ}\sin{49^\circ}+...+\sin{1^\circ}\sin{89^\circ})+\sin{1^\circ}</math>. | ||
+ | |||
+ | After applying product-sum identities, you get <math>\cos{1^\circ}=\sqrt{2}[(\cos{46^\circ}-\cos{48^\circ})+(\cos{48^\circ}-\cos{50^\circ})+...+(\cos{88^\circ}-\cos{90^\circ})]+\sin{1^\circ}</math>. | ||
+ | |||
+ | This is just <math>\cos{1^\circ}=\sqrt{2}(\cos{(1^\circ+45^\circ}))+\sin{1^\circ}</math>. | ||
+ | |||
+ | After applying angle addition formulas, you get: <math>\cos{1^\circ}=\sqrt{2}(\cos{1^\circ}\cos{45^\circ}-\sin{1^\circ}\sin{45^\circ})+\sin{1^\circ}</math>. | ||
+ | |||
+ | Since the cosine and sine of <math>45^\circ</math> are <math>\frac{\sqrt{2}}{2}</math> you can simplify that to: <math>\cos{1^\circ}=(\cos{1^\circ}-\sin{1^\circ})+\sin{1^\circ}</math>. Or, <math>\cos{1^\circ}=\cos{1^\circ}</math>. | ||
+ | |||
+ | Therefore, the average of the numbers <math>n \sin n^\circ</math> (<math>n = 2, 4, 6, \ldots, 180</math>) is <math>\cot 1^\circ</math>. | ||
+ | |||
+ | <cmath>\boxed{\frac{n\sin{n^\circ}(n=2,4,6,\ldots,180)}{90}=\cot{1^\circ}}</cmath> | ||
<math>\Box</math> | <math>\Box</math> |
Revision as of 23:19, 15 August 2017
Contents
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.
Solution 3 (Very long and detailed)
Make the sum of the numbers equal . Now, the average of these numbers is .
We know that , so we can eliminate that term and use the identity, to get
Or,
Pairing the terms up, using sum-product identity, and simplifying, yields: .
After dividing both sides by , you get: .
Now we have to prove that .
Multiply both sides by to get .
After applying product-sum identities, you get .
This is just .
After applying angle addition formulas, you get: .
Since the cosine and sine of are you can simplify that to: . Or, .
Therefore, the average of the numbers () is .
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.