2013 AIME I Problems/Problem 14
Contents
Problem 14
For , let
and
so that . Then
where
and
are relatively prime positive integers. Find
.
Solution
Solution 1
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
The answer format tells us it's the negative solution, and our desired answer is
Solution 2
Use sum to product formulas to rewrite and
Therefore,
Using
Plug in to the previous equation and cancel out the "P" terms to get: .
Then use the pythagorean identity to solve for ,
Solution 3
Note that
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
.
Solution 4
A bit similar to Solution 3. We use because the progression cycles in
. So we could rewrite that as
.
Similarly, .
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 (lots of room for sillies, I wouldn't recommend it)
We notice and
With these, we just quickly find the sum of the infinite geometric series' in and
.
has 2 parts, the
and the
parts. The
part is:
, which can be turned into:
, which is
. This turns into
. For simplicity, we call the denominator
.
To spare you some reading time and me some ing time, we follow the same process as above, and we find that the
part of
is
, the
part of
is
, and finally, the
part of
is
.
We convert all 4 of these "things" into trig, and we end up getting , we divide by
on both numerator and denominator, and we get
. We use some trig identities and we get
. We factor the denominator into
. We cancel out
on both numerator and denominator to get
. We set
as
, and we just solve a quadratic in terms of
,
, cross multiply and simplify, and we get
. We can actually factor this to get
, which yields the 2 solutions
. Since
, the latter solution is deemed invalid, and we are left with
. Our final answer is
.
See also
2013 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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