2015 AIME I Problems/Problem 13
Contents
Problem
With all angles measured in degrees, the product , where and are integers greater than 1. Find .
Solution 1
Let . Then from the identity we deduce that (taking absolute values and noticing ) But because is the reciprocal of and because , if we let our product be then because is positive in the first and second quadrants. Now, notice that are the roots of Hence, we can write , and so It is easy to see that and that our answer is .
Solution 2
Let
because of the identity
we want
Thus the answer is
Solution 3
Similar to Solution , so we use and we find that: Now we can cancel the sines of the multiples of : So and we can apply the double-angle formula again: Of course, is missing, so we multiply it to both sides: Now isolate the product of the sines: And the product of the squares of the cosecants as asked for by the problem is the square of the inverse of this number: The answer is therefore .
Solution 4
Let .
Then, .
Since , we can multiply both sides by to get .
Using the double-angle identity , we get .
Note that the right-hand side is equal to , which is equal to , again, from using our double-angle identity.
Putting this back into our equation and simplifying gives us .
Using the fact that again, our equation simplifies to , and since , it follows that , which implies . Thus, .
Solution 5
Once we have the tools of complex polynomials there is no need to use the tactical tricks. Everything is so basic (I think).
Recall that the roots of are , we have Let , and take absolute value of both sides, or, Let be even, then, so, Set and we have , -Mathdummy
Solution 6
Recall that Since it is in csc, we can write in sin and then take reciprocal. We can group them by threes, . Thus So we take reciprocal, , the desired answer is leads to answer
~bluesoul
Solution 7
We have
Multiplying by gives
Using gives
Thus, the answer is
See Also
2015 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.