# 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 | + | 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. |

<math>\Box</math> | <math>\Box</math> | ||

{{MAA Notice}} | {{MAA Notice}} |

## Revision as of 23:55, 5 June 2014

## Problem

Prove that the average of the numbers is .

## Solution 1

First, as we omit that term. Now, we multiply by to get, after using product to sum, . This simplifies to . Since this simplifies to . We multiplied by in the beginning, so we must divide by it now, and thus the sum is just , so the average is , as desired.

## Solution 2

Notice that for every there exists a corresponding pair term , for not . Pairing gives the sum of all terms to be , and thus the average is We need to show that . Multiplying (*) by and using sum-to-product and telescoping gives . Thus, , as desired.

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