2013 AIME I Problems/Problem 14
For , let
so that . Then where and are relatively prime positive integers. Find .
Noticing the and in both and we think of the angle addition identities:
With this in mind, we multiply by and by to try and use some angle addition identities. Indeed, we get after adding term-by-term. Similar term-by-term adding yields This is a system of equations; rearrange and rewrite to get and Subtract the two and rearrange to get Then, square both sides and use Pythagorean Identity to get a quadratic in Factor that quadratic and solve for Since we're given is nonpositive. We therefore use the negative solution, and our desired answer is
Use sum to product formulas to rewrite and
Plug in to the previous equation and cancel out the "P" terms to get:
Then use the pythagorean identity to solve for ,
Thus, the following identities follow immediately:
Consider, now, the sum . It follows fairly immediately that:
This follows straight from the geometric series formula and simple simplification. We can now multiply the denominator by it's complex conjugate to find:
Comparing real and imaginary parts, we find:
Squaring this equation and letting :
Clearing denominators and solving for gives sine as .
A bit similar to Solution 3. We use because the progression cycles in . So we could rewrite that as .
Setting complex , we get
The important part is the ratio of the imaginary part to the real part. To cancel out the imaginary part from the denominator, we must add to the numerator to make the denominator a difference (or rather a sum) of squares. The denominator does not matter. Only the numerator, because we are trying to find a PROPORTION of values. So denominators would cancel out.
Setting , we obtain .
Since because , . Adding up, .
Solution 5 (utterly disgusting)
We notice and
We observe that both and can be split into parts, namely the terms which contain the and the terms which contain the
The part of can be expressed as:
Repeating the above process, we find that the part of is the part of is and finally, the part of is
Converting back to trigonometric form, we have Using the double identity and simplifying, we have Factoring the denominator, we have Simplifying We set as , and by the Pythagorean Identity, we have . This factors into , which yields the 2 solutions . As , the latter root is erroneous, and we are left with . Thus, our final answer is .
Follow solution 3, up to the point of using the geometric series formula
Moving everything to the other side, and considering only the imaginary part, we get
We can then write , and , (). Thus, we can substitute and divide out by k.
Since , we get , and thus,
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