Difference between revisions of "2019 AIME I Problems/Problem 8"
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~ZericHang | ~ZericHang | ||
+ | |||
+ | ==Solution 5== | ||
+ | Define the recursion <math>a_n=(\sin^2 x)^n+(\cos^2 x)^n</math> | ||
+ | We know that the characteristic equation of <math>a_n</math> must have 2 roots, so we can recursively define <math>a_n</math> as <math>a_n=p*a_{n-1}+q*a_{n-2}</math>. <math>p</math> is simply the sum of the roots of the characteristic equation, which is <math>\sin^2 x+\cos^2 x=1</math>. <math>q</math> is the product of the roots, which is <math>-(\sin^2 x)(\cos^2 x)</math>. This value is not trivial and we have to solve for it. | ||
+ | We know that <math>a_0=2</math>, <math>a_1=1</math>, <math>a_5=\frac{11}{36}</math>. | ||
+ | Solving the rest of the recursion gives | ||
+ | \begin{align*} | ||
+ | <math>a_2=1+2q</math> | ||
+ | <math>a_3=1+3q</math> | ||
+ | <math>a_4=1+4q+2q^2</math> | ||
+ | <math>a_5=1+5q+5q^2=\frac{11}{36}</math> | ||
+ | <math>a_6=1+6q+9q^2+2q^3</math> | ||
+ | \end{align*} | ||
+ | |||
+ | Solving for <math>q</math> in the expression for <math>a_5</math> gives us <math>q^2+q+\frac{5}{36}=0</math>, so <math>q=-\frac{5}{6}, -\frac{1}{6}</math>. Since <math>q=-(\sin^2 x)(\cos^2 x)</math>, we know that the minimum value it can attain is <math>-\frac{1}{4}</math> by AM-GM, so <math>q</math> cannot be <math>-\frac{5}{6}</math>. | ||
+ | Plugging in the value of <math>q</math> into the expression for <math>a_6</math>, we get <math>a_6=1-1+\frac{1}{4}-\frac{1}{108}=\frac{26}{108}=\frac{13}{54}</math>. Our final answer is then <math>13+54=\boxed{67}</math> | ||
+ | |||
+ | -Natmath | ||
==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 15:04, 11 June 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
\begin{align*}
\end{align*}
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
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.