1996 USAMO Problems/Problem 1

Revision as of 23:19, 15 August 2017 by Onezero (talk | contribs) (Solution)

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.

$\Box$

Solution 3 (Very long and detailed)

Make the sum of the numbers equal $x$. Now, $2\sin{2^\circ}+4\sin{4^\circ}+...+180\sin{180^\circ}=x$ the average of these numbers is $\frac{x}{90}$.

We know that $180\sin{180^\circ}=0$, so we can eliminate that term and use the identity, $\sin(\theta)=\sin(180-\theta)$ to get

$x=(2\sin{2^\circ}+178\sin{2^\circ})+...+(88\sin{88^\circ}+92\sin{22^\circ})+90\sin{90^\circ}$

Or, $x=180(\sin{2^\circ}+\sin{4^\circ}+\sin{6^\circ}+...+\sin{88^\circ})+90$

Pairing the terms up, using sum-product identity, and simplifying, yields: $x=180\sqrt 2 (\sin47^\circ + \sin49^\circ + \ldots + \sin89^\circ)+ 90$.

After dividing both sides by $90$, you get: $\frac{x}{90}=2\sqrt 2 (\sin47^\circ + \sin49^\circ + \ldots + \sin89^\circ) + 1=\frac{\cos{1^\circ}}{\sin{1^\circ}}$.

Now we have to prove that $2\sqrt 2 (\sin47^\circ + \sin49^\circ + \ldots + \sin89^\circ) + 1=\frac{\cos{1^\circ}}{\sin{1^\circ}}$.

Multiply both sides by $\sin{1^\circ}$ to get $\cos{1^\circ}=2\sqrt{2}(\sin{1^\circ}\sin{47^\circ}+\sin{1^\circ}\sin{49^\circ}+...+\sin{1^\circ}\sin{89^\circ})+\sin{1^\circ}$.

After applying product-sum identities, you get $\cos{1^\circ}=\sqrt{2}[(\cos{46^\circ}-\cos{48^\circ})+(\cos{48^\circ}-\cos{50^\circ})+...+(\cos{88^\circ}-\cos{90^\circ})]+\sin{1^\circ}$.

This is just $\cos{1^\circ}=\sqrt{2}(\cos{(1^\circ+45^\circ}))+\sin{1^\circ}$.

After applying angle addition formulas, you get: $\cos{1^\circ}=\sqrt{2}(\cos{1^\circ}\cos{45^\circ}-\sin{1^\circ}\sin{45^\circ})+\sin{1^\circ}$.

Since the cosine and sine of $45^\circ$ are $\frac{\sqrt{2}}{2}$ you can simplify that to: $\cos{1^\circ}=(\cos{1^\circ}-\sin{1^\circ})+\sin{1^\circ}$. Or, $\cos{1^\circ}=\cos{1^\circ}$.

Therefore, the average of the numbers $n \sin n^\circ$ ($n = 2, 4, 6, \ldots, 180$) is $\cot 1^\circ$.

\[\boxed{\frac{n\sin{n^\circ}(n=2,4,6,\ldots,180)}{90}=\cot{1^\circ}}\]

$\Box$

See Also

1996 USAMO (ProblemsResources)
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. AMC logo.png

Invalid username
Login to AoPS