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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>
 
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===Solution 3 (Very long and detailed)===
 
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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>.
 
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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
 
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<math>x=(2\sin{2^\circ}+178\sin{2^\circ})+...+(88\sin{88^\circ}+92\sin{22^\circ})+90\sin{90^\circ}</math>
 
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Or, <math>x=180(\sin{2^\circ}+\sin{4^\circ}+\sin{6^\circ}+...+\sin{88^\circ})+90</math>
 
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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>.
 
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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>.
 
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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>.
 
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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>.
 
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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>.
 
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This is just <math>\cos{1^\circ}=\sqrt{2}(\cos{(1^\circ+45^\circ}))+\sin{1^\circ}</math>.
 
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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>.
 
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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>.
 
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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>.
 
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<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 16:51, 28 January 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, $(\cos 1^\circ-\cos 3^\circ)+2(\cos 3^\circ-\cos5)+\cdots +89(\cos 177^\circ-\cos 179^\circ)$. 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 (Problems • Resources)
Preceded by
First Question 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

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