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

## Problem

Prove that the average of the numbers $n\sin n^{\circ}\; (n = 2,4,6,\ldots,180)$ is $\cot 1^\circ$.

## 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 $2\sin 2^\circ$ and using sum-to-product and telescoping gives $2\sin 2^\circ S = \cos 0^\circ + \cos 2^\circ - \cos 178^\circ - \cos 180^\circ = 2 + 2\cos 2^\circ$. Thus, $S = \frac{1 + \cos 2^\circ}{\sin 2^\circ} = \cot 1^\circ$, as desired. $\Box$ The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 