Difference between revisions of "2019 AIME I Problems/Problem 8"
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Newton sums is basically constructing the powers of the roots of the polynomials instead of deconstructing them which was done in solution 2. Let <math>\sin^2x</math> and <math>\cos^2x</math> be the roots of some polynomial <math>F(a)</math>. Then, <math>F(a)=a^2-a+b</math> for some <math>b=\sin^2x\cdot\cos^2x</math>. | Newton sums is basically constructing the powers of the roots of the polynomials instead of deconstructing them which was done in solution 2. Let <math>\sin^2x</math> and <math>\cos^2x</math> be the roots of some polynomial <math>F(a)</math>. Then, <math>F(a)=a^2-a+b</math> for some <math>b=\sin^2x\cdot\cos^2x</math>. | ||
− | Let <math>S_k=\left(\sin^2x\right)^k+\left(\cos^2x\right)^k</math>. We want to find <math>S_6</math>. Clearly <math>S_1=1</math> and <math>S_2=1-2b</math> Newton sums tells us that <math>S_k-S_{k-1}+bS_{k-2}=0\Rightarrow S_k=S_{k-1}-bS_{k-2}</math> where <math>k\ge 3</math> for our polynomial <math>F(a)</math>. | + | Let <math>S_k=\left(\sin^2x\right)^k+\left(\cos^2x\right)^k</math>. We want to find <math>S_6</math>. Clearly <math>S_1=1</math> and <math>S_2=1-2b</math>. Newton sums tells us that <math>S_k-S_{k-1}+bS_{k-2}=0\Rightarrow S_k=S_{k-1}-bS_{k-2}</math> where <math>k\ge 3</math> for our polynomial <math>F(a)</math>. |
Bashing, we have | Bashing, we have |
Revision as of 14:51, 15 March 2019
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
Solution 3 (Newton Sums)
Newton sums is basically constructing the powers of the roots of the polynomials instead of deconstructing them which was done in solution 2. Let and
be the roots of some polynomial
. Then,
for some
.
Let . We want to find
. Clearly
and
. Newton sums tells us that
where
for our polynomial
.
Bashing, we have
Thus
. Clearly,
so
.
Note . Solving for
, we get
. Finally,
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.