Difference between revisions of "2019 AIME I Problems/Problem 8"
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+ | ==Solution 6== | ||
+ | Let <math>m=\sin^2 x</math> and <math>n=\cos^2 x</math>, then <math>m+n=1</math> and <math>m^5+n^5=\frac{11}{36}</math> | ||
+ | |||
+ | <math>m^6+n^6=(m^5+n^5)(m+n)-mn(m^4+n^4)=(m^5+n^5)-mn(m^4+n^4)</math> | ||
+ | |||
+ | Now factoring <math>m^5+n^5</math> as solution 4 yields <math>m^5+n^5=(m+n)(m^4-m^3n+m^2n^2-mn^3+n^4)</math> | ||
+ | <math>=m^4+n^4-mn(m^2-mn+n^2)=m^4+n^4-mn[(m+n)^2-3mn]=m^4+n^4-mn(1-3mn)</math>. | ||
+ | |||
+ | Since <math>(m+n)^4=m^4+4m^3n+6m^2n^2+4mn^3+n^4</math>, <math>m^4+n^4=(m+n)^4-2mn(2m^2+3mn+2n^2)=1-2mn(2m^2+3mn+2n^2)</math>. | ||
+ | |||
+ | Notice that <math>2m^2+3mn+2n^2</math> can be rewritten as <math>[\sqrt{2}(a+b)]^2-mn=2-mn</math>. Thus,<math>m^4+n^4=1-2mn(2-mn)</math> and <math>m^5+n^5=1-2mn(2-mn)-mn(1-3mn)=1-5mn+5(mn)^2=\frac{11}{36}</math>. As in solution 4, we get <math>mn=\frac{1}{6}</math> and <math>m^4+n^4=1-2*\frac{1}{6}(2-\frac{1}{6})=\frac{7}{18}</math> | ||
+ | |||
+ | Substitute <math>m^4+n^4=\frac{7}{18}</math> and <math>mn=\frac{1}{6}</math>, then | ||
+ | <math>m^6+n^6=\frac{11}{36}-\frac{1}{6}*\frac{7}{18}=\frac{13}{54}</math>, and the desired answer is <math>\boxed{067}</math> | ||
+ | |||
==See Also== | ==See Also== | ||
{{AIME box|year=2019|n=I|num-b=7|num-a=9}} | {{AIME box|year=2019|n=I|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 04:41, 13 August 2020
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 formula, we obtain the 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
. 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 . Let
and
be the roots of some polynomial
. Then, by Vieta,
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,
.
Solution 4
Factor the first equation.
First of all,
because
We group the first, third, and fifth term and second and fourth term. The first group:
The second group:
Add the two together to make
Because this equals
, we have
Let
so we get
Solving the quadratic gives us
Because
, we finally get
.
Now from the second equation,
Plug in
to get
which yields the answer
~ZericHang
Solution 5
Define the recursion
We know that the characteristic equation of
must have 2 roots, so we can recursively define
as
.
is simply the sum of the roots of the characteristic equation, which is
.
is the product of the roots, which is
. This value is not trivial and we have to solve for it.
We know that
,
,
.
Solving the rest of the recursion gives
Solving for in the expression for
gives us
, so
. Since
, we know that the minimum value it can attain is
by AM-GM, so
cannot be
.
Plugging in the value of
into the expression for
, we get
. Our final answer is then
-Natmath
Solution 6
Let and
, then
and
Now factoring as solution 4 yields
.
Since ,
.
Notice that can be rewritten as
. Thus,
and
. As in solution 4, we get
and
Substitute and
, then
, and the desired answer is
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.