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

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.

$\Box$ The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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1996 USAMO Problems/Problem 1

Problem

Solution