Difference between revisions of "2019 AIME I Problems/Problem 8"
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==Solution 4 (Not Fun)== | ==Solution 4 (Not Fun)== | ||
We let <math>a = \sin^2(x)</math> and <math>b = \cos^2(x)</math>, so we have <math>a+b=1</math> and <math>a^5 + b^5 = \frac{11}{36}</math>. Noticing that <math>ab</math> might be a useful value to find, we let <math>c = ab</math>. Then we can work our way up to find <math>c</math>. | We let <math>a = \sin^2(x)</math> and <math>b = \cos^2(x)</math>, so we have <math>a+b=1</math> and <math>a^5 + b^5 = \frac{11}{36}</math>. Noticing that <math>ab</math> might be a useful value to find, we let <math>c = ab</math>. Then we can work our way up to find <math>c</math>. | ||
− | <cmath>a + b = 1</cmath> | + | <cmath>a+b = 1</cmath> |
− | <cmath>(a+b) | + | <cmath>(a+b)^2 = 1</cmath> |
<cmath>a^2 + 2ab + b^2 = 1</cmath> | <cmath>a^2 + 2ab + b^2 = 1</cmath> | ||
<cmath>a^2 + b^2 = -2c + 1</cmath> | <cmath>a^2 + b^2 = -2c + 1</cmath> | ||
− | <cmath>(a+b)(a^2 + b^2) = -2c + 1</cmath> | + | <cmath>(a+b)\left(a^2 + b^2\right) = -2c + 1</cmath> |
<cmath>a^3 + ab^2 + a^2b + b^3 = -2c + 1</cmath> | <cmath>a^3 + ab^2 + a^2b + b^3 = -2c + 1</cmath> | ||
<cmath>a^3 + b^3 + ab(a + b) = -2c + 1</cmath> | <cmath>a^3 + b^3 + ab(a + b) = -2c + 1</cmath> | ||
<cmath>a^3 + b^3 + c = -2c + 1</cmath> | <cmath>a^3 + b^3 + c = -2c + 1</cmath> | ||
<cmath>a^3 + b^3 = -3c + 1</cmath> | <cmath>a^3 + b^3 = -3c + 1</cmath> | ||
− | <cmath>(a + b)(a^3 + b^3) = -3c + 1</cmath> | + | <cmath>(a + b)\left(a^3 + b^3\right) = -3c + 1</cmath> |
<cmath>a^4 + ab^3 + a^3b + b^4 = -3c + 1</cmath> | <cmath>a^4 + ab^3 + a^3b + b^4 = -3c + 1</cmath> | ||
− | <cmath>a^4 + b^4 + ab(a^2 + b^2) = -3c + 1</cmath> | + | <cmath>a^4 + b^4 + ab\left(a^2 + b^2\right) = -3c + 1</cmath> |
<cmath>a^4 + b^4 + c(-2c + 1) = -3c + 1</cmath> | <cmath>a^4 + b^4 + c(-2c + 1) = -3c + 1</cmath> | ||
<cmath>a^4 + b^4 - 2c^2 + c = -3c + 1</cmath> | <cmath>a^4 + b^4 - 2c^2 + c = -3c + 1</cmath> | ||
<cmath>a^4 + b^4 = 2c^2 - 4c + 1</cmath> | <cmath>a^4 + b^4 = 2c^2 - 4c + 1</cmath> | ||
<cmath>a^5 + ab^4 + a^4b + b^5 = 2c^2 - 4c + 1</cmath> | <cmath>a^5 + ab^4 + a^4b + b^5 = 2c^2 - 4c + 1</cmath> | ||
− | <cmath>a^5 + b^5 + ab(a^3 + b^3) = 2c^2 - 4c + 1</cmath> | + | <cmath>a^5 + b^5 + ab\left(a^3 + b^3\right) = 2c^2 - 4c + 1</cmath> |
<cmath>\frac{11}{36} + c(-3c + 1) = 2c^2 - 4c + 1</cmath> | <cmath>\frac{11}{36} + c(-3c + 1) = 2c^2 - 4c + 1</cmath> | ||
<cmath>\frac{11}{36} = 5c^2 - 5c + 1</cmath> | <cmath>\frac{11}{36} = 5c^2 - 5c + 1</cmath> | ||
<cmath>5c^2 - 5c + \frac{25}{36} = 0</cmath> | <cmath>5c^2 - 5c + \frac{25}{36} = 0</cmath> | ||
− | using quadform you get <math>c = \frac{1}{6}</math> or <math>c = \frac{5}{6}</math>. Since <math>c = \sin^2(x)\cos^2(x) = (\sin(x)\cos(x))^2 = (\frac{\sin(2x)}{2})^2</math>, and since <math>\sin(2x)</math> can't exceed 1, <math>c</math> can't exceed <math>(\frac{1}{2})^2 = \frac{1}{4}</math>. Clearly, <math>c = \frac{1}{6}</math>. And finally, | + | using quadform you get <math>c = \frac{1}{6}</math> or <math>c = \frac{5}{6}</math>. Since <math>c = \sin^2(x)\cos^2(x) = (\sin(x)\cos(x))^2 = \left(\frac{\sin(2x)}{2}\right)^2</math>, and since <math>\sin(2x)</math> can't exceed 1, <math>c</math> can't exceed <math>\left(\frac{1}{2}\right) ^2 = \frac{1}{4}</math>. Clearly, <math>c = \frac{1}{6}</math>. And finally, |
− | <cmath>a^5 + b^5 = (a + b)(a^5 + b^5)</cmath> | + | <cmath>a^5 + b^5 = (a + b)\left(a^5 + b^5\right)</cmath> |
<cmath>a^5 + b^5 = a^6 + ab^5 + a^5b + b^6</cmath> | <cmath>a^5 + b^5 = a^6 + ab^5 + a^5b + b^6</cmath> | ||
− | <cmath>\frac{11}{36} = a^6 + b^6 + c(a^4 + b^4)</cmath> | + | <cmath>\frac{11}{36} = a^6 + b^6 + c\left(a^4 + b^4\right)</cmath> |
looking back to previous results, we see that <math>a^4 + b^4 = 2c^2 - 4c + 1 = \frac{14}{36}</math> (it's easier not to simplify the fraction). | looking back to previous results, we see that <math>a^4 + b^4 = 2c^2 - 4c + 1 = \frac{14}{36}</math> (it's easier not to simplify the fraction). | ||
− | <cmath>\frac{11}{36} = a^6 + b^6 + c(\frac{14}{36}) = a^6 + b^6 + \frac{14}{216}</cmath> | + | <cmath>\frac{11}{36} = a^6 + b^6 + c\left(\frac{14}{36}\right) = a^6 + b^6 + \frac{14}{216}</cmath> |
<cmath>a^6 + b^6 = \frac{11}{36} - \frac{14}{216} = \frac{13}{54}</cmath> | <cmath>a^6 + b^6 = \frac{11}{36} - \frac{14}{216} = \frac{13}{54}</cmath> | ||
which yields the answer <math>\boxed{067}</math>. | which yields the answer <math>\boxed{067}</math>. |
Revision as of 07:45, 28 June 2019
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 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,
.
Solution 4 (Not Fun)
We let and
, so we have
and
. Noticing that
might be a useful value to find, we let
. Then we can work our way up to find
.
using quadform you get
or
. Since
, and since
can't exceed 1,
can't exceed
. Clearly,
. And finally,
looking back to previous results, we see that
(it's easier not to simplify the fraction).
which yields the answer
.
~PCampbell
Solution 5
Factor the first equation.
First of all,
because
We group the 1st, 3rd and 5th term and 2nd and 4th term. The 1st group:
The 2nd 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
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.