Difference between revisions of "2019 AIME I Problems/Problem 8"
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==Solution 2== | ==Solution 2== | ||
− | First, for simplicity, let <math>a=\sin{x}</math> and <math>b=\cos{x}</math>. Note that <math>a^2+b^2=1</math>. We then bash the rest of the problem out. Take the | + | First, for simplicity, let <math>a=\sin{x}</math> and <math>b=\cos{x}</math>. Note that <math>a^2+b^2=1</math>. We then bash the rest of the problem out. Take the fifth power of this expression and get <math>a^{10}+b^{10}+5a^2b^2(a^6+b^6)+10a^4b^4(a^2+b^2)=\frac{11}{36}+5a^2b^2(a^6+b^6)+10a^4b^4=1</math>. Note that we also have <math>\frac{11}{36}=a^{10}+b^{10}=(a^{10}+b^{10})(a^2+b^2)=a^{12}+b^{12}+a^2b^2(a^8+b^8)</math>. So, it suffices to compute <math>a^2b^2(a^8+b^8)</math>. Let <math>y=a^2b^2</math>. We have from cubing <math>a^2+b^2=1</math> that <math>a^6+b^6+3a^2b^2(a^2+b^2)=1</math> or <math>a^6+b^6=1-3y</math>. Next, using <math>\frac{11}{36}+5a^2b^2(a^6+b^6)+10a^4b^4=1</math>, we get <math>a^2b^2(a^6+b^6)+2a^4b^4=\frac{5}{36}</math> or <math>y(1-3y)+2y^2=y-y^2=\frac{5}{36}</math>. Solving gives <math>y=\frac{5}{6}</math> or <math>y=\frac{1}{6}</math>. Clearly <math>y=\frac{5}{6}</math> is extraneous, so <math>y=\frac{1}{6}</math>. Now note that <math>a^4+b^4=(a^2+b^2)^2-2a^2b^2=\frac{2}{3}</math>, and <math>a^8+b^8=(a^4+b^4)^2-2a^4b^4=\frac{4}{9}-\frac{1}{18}=\frac{7}{18}</math>. Thus we finally get <math>a^{12}+b^{12}=\frac{11}{36}-\frac{7}{18}\cdot\frac{1}{6}=\frac{13}{54}</math>, giving <math>\boxed{067}</math>. |
'''- Emathmaster''' | '''- Emathmaster''' | ||
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A quick note: this solution uses recursion and is similar to the solution 7 above. It was from trumpeter, posted in the AoPS Forums, Contest Discussion. | A quick note: this solution uses recursion and is similar to the solution 7 above. It was from trumpeter, posted in the AoPS Forums, Contest Discussion. | ||
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==Video Solution By The Power Of Logic== | ==Video Solution By The Power Of Logic== |
Latest revision as of 12:19, 1 February 2024
Contents
[hide]Problem
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
, so
(observe that either choice of
doesn't matter). Substituting
we get:
Therefore, the answer is .
-eric2020, inspired by Tommy2002
Motivation
The motivation to substitute comes so that after applying the binomial theorem to
a lot of terms will cancel out. Note that all the terms with odd exponents in
will cancel out, while the terms with even exponents will be doubled.
mathboy282
Solution 2
First, for simplicity, let and
. Note that
. We then bash the rest of the problem out. Take the fifth 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
Solution 7 (Official MAA)
Let and let
Then for
Because
and
it follows that
and
Hence
or
and because
the only possible value of
is
Therefore
The requested sum is
Solution 8 (Recursion)
Let for non-negative integers
. Then
and
. In addition,
where
. So we can compute
. But by the sin double angle formula,
, so
. Then
so the answer is
as desired.
A quick note: this solution uses recursion and is similar to the solution 7 above. It was from trumpeter, posted in the AoPS Forums, Contest Discussion.
Video Solution By The Power Of Logic
~ Hayabusa1
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.