1996 USAMO Problems/Problem 1
Prove that the average of the numbers is .
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.
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.
Solution 3 (Very long and detailed)
Make the sum of the numbers equal . Now, the average of these numbers is .
We know that , so we can eliminate that term and use the identity, to get
Pairing the terms up, using sum-product identity, and simplifying, yields: .
After dividing both sides by , you get: .
Now we have to prove that .
Multiply both sides by to get .
After applying product-sum identities, you get .
This is just .
After applying angle addition formulas, you get: .
Since the cosine and sine of are you can simplify that to: . Or, .
Therefore, the average of the numbers () is .
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