Difference between revisions of "2019 AIME I Problems/Problem 8"
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Let <math>x</math> be a real number such that <math>\sin^{10}x+\cos^{10} x = \tfrac{11}{36}</math>. Then <math>\sin^{12}x+\cos^{12} x = \tfrac{m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | Let <math>x</math> be a real number such that <math>\sin^{10}x+\cos^{10} x = \tfrac{11}{36}</math>. Then <math>\sin^{12}x+\cos^{12} x = \tfrac{m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | ||
− | ==Solution== | + | ==Solution 1== |
We can substitute <math>y = \sin^2{x}</math>. Since we know that <math>\cos^2{x}=1-\sin^2{x}</math>, we can do some simplification. | We can substitute <math>y = \sin^2{x}</math>. Since we know that <math>\cos^2{x}=1-\sin^2{x}</math>, we can do some simplification. |
Revision as of 13:18, 15 March 2019
Contents
[hide]Problem 8
Let be a real number such that
. Then
where
and
are relatively prime positive integers. Find
.
Solution 1
We can substitute . Since we know that
, we can do some simplification.
This yields . From this, we can substitute again to get some cancellation through binomials. If we let
, we can simplify the equation to
. After using binomial theorem, this simplifies to
. If we use the quadratic theorem, we obtain that
, so
. By plugging z into
(which is equal to
, we can either use binomial theorem or sum of cubes to simplify, and we end up with 13/54. Therefore, the answer is
.
eric2020, inspired by Tommy2002
Solution 2
First, for simplicity, let and
. Note that
. We then bash the rest of the problem out. Take the tenth power of this expression and get
. Note that we also have
. So, it suffices to compute
. Let
. We have from cubing
that
or
. Next, using
, we get
or
. Solving gives
or
. Clearly
is extraneous, so
. Now note that
, and
. Thus we finally get
, giving
.
-Emathmaster
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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