Difference between revisions of "1996 USAMO Problems/Problem 1"

Revision as of 08:29, 20 July 2016

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.

$\Box$

@@ Line 15: / Line 15: @@
 == See Also ==
-{{USAMO box|year=1996|before=First Question|num-a=2}}
+{{USAMO newbox|year=1996|before=First Question|num-a=2}}
 {{MAA Notice}}
 [[Category:Olympiad Algebra Problems]]

1996 USAMO (Problems • Resources)
Preceded by First Question	Followed by Problem 2
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All USAMO Problems and Solutions

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