Difference between revisions of "2019 AIME I Problems/Problem 8"
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\end{align*}</cmath> | \end{align*}</cmath> | ||
Because <math>S(0)=2</math> and <math>S(1)=1,</math> it follows that <math>S(2)=1-2c, S(3)=1-3c,S(4)=2c^2-4c+1,</math> and <math>\tfrac{11}{36}=S(5)=5c^2-5c+1.</math> Hence <math>c=\tfrac16</math> or <math>\tfrac56,</math> and because <math>4c=\sin^2{2x},</math> the only possible value of <math>c</math> is <math>\tfrac16.</math> Therefore <cmath>S(6)=S(5)-cS(4)=\frac{11}{36}-\frac16\left(2\left(\frac16\right)^2-4\left(\frac16\right)+1\right)=\frac{13}{54}.</cmath> The requested sum is <math>13+54=67.</math> | Because <math>S(0)=2</math> and <math>S(1)=1,</math> it follows that <math>S(2)=1-2c, S(3)=1-3c,S(4)=2c^2-4c+1,</math> and <math>\tfrac{11}{36}=S(5)=5c^2-5c+1.</math> Hence <math>c=\tfrac16</math> or <math>\tfrac56,</math> and because <math>4c=\sin^2{2x},</math> the only possible value of <math>c</math> is <math>\tfrac16.</math> Therefore <cmath>S(6)=S(5)-cS(4)=\frac{11}{36}-\frac16\left(2\left(\frac16\right)^2-4\left(\frac16\right)+1\right)=\frac{13}{54}.</cmath> The requested sum is <math>13+54=67.</math> | ||
+ | |||
+ | ==Solution 8 (Recursion)== | ||
+ | Let <math>a_n=\sin^nx+\cos^nx</math> for non-negative integers <math>n</math>. Then <math>a_0=2</math> and <math>a_2=1</math>. In addition,<cmath>a_n=\sin^nx+\cos^nx=\left(\sin^{n-2}x+\cos^{n-2}x\right)\left(\sin^2x+\cos^2x\right)-\sin^2x\cos^2x\left(\sin^{n-4}x+\cos^{n-4}x\right)=a_{n-2}-Xa_{n-4},</cmath>where <math>X=\sin^2x\cos^2x</math>. So we can compute | ||
+ | \begin{align*} | ||
+ | a_4&=1-2X\ | ||
+ | a_6&=1-3X\ | ||
+ | a_8&=1-4X+2X^2\ | ||
+ | a_{10}&=1-5X+5X^2=\frac{11}{36} | ||
+ | \end{align*}so <math>X=\frac{1}{6},\frac{5}{6}</math>. But by the sin double angle formula, <math>\sin^2x\cos^2x=\frac{1}{4}\sin^22x\leq\frac{1}{4}</math>, so <math>X=\frac{1}{6}</math>. Then<cmath>a_{12}=a_{10}-Xa_8=\frac{11}{36}-\frac{1}{6}\cdot\frac{7}{18}=\frac{13}{54}</cmath>so the answer is <math>\boxed{067}</math> 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 North America Math Contest Go Go Go== | ==Video Solution By North America Math Contest Go Go Go== |
Revision as of 21:33, 22 January 2023
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 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
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 North America Math Contest Go Go Go
https://www.youtube.com/watch?v=a3Ps425bYUM
~ Please sub.!
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.