1999 AIME Problems/Problem 11
Problem
Given that where angles are measured in degrees, and and are relatively prime positive integers that satisfy find
Solution
Let . We could try to manipulate this sum by wrapping the terms around (since the first half is equal to the second half), but it quickly becomes apparent that this way is difficult to pull off. Instead, we look to telescope the sum. Using the identity , we can rewrite as
This telescopes to Manipulating this to use the identity , we get and our answer is .
Alternate Solution
We note that . We thus have that The desired answer is thus .
- Only if is in radians, which it is not. However, the solution is still viable, so keep reading.
Solution 3
Let . By Euler's Formula, .
The sum we want is thus
We factor the and split into two geometric series to get
However, we note that , so , so our two geometric series are actually the same. We combine the terms and simplify to get
Apply Euler's identity and simplify again to get
Now, we need to figure out how to express this as the tangent of something. We note that .
So, we set the two equal to each other to solve for . Cross multiplying gets . Expanding yields . Simplifying yields . Since is the smallest solution, we have , and the argument of tangent is . The requested sum is .
See also
1999 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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