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 5
Recall that
Since it is in csc, we can write in sin and then take reciprocal.
we can group them three by three, having![]()
So we take reciprocal, the expression is , the desired answer is $\2^{89}$ (Error compiling LaTeX. Unknown error_msg) leads to answer
~bluesoul
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.