Difference between revisions of "2013 AIME I Problems/Problem 5"
m (→Video Solution) |
m (→Solution 1) |
||
Line 4: | Line 4: | ||
== Solutions == | == Solutions == | ||
=== Solution 1 === | === Solution 1 === | ||
− | We note that <math>8x^3 - 3x^2 - 3x - 1 = 9x^3 - x^3 - 3x^2 - 3x - 1 = 9x^3 - (x + 1)^3</math>. Therefore, we have that <math>9x^3 = (x+1)^3</math>, so it follows that <math>x\sqrt[3]{9} = x+1</math>. Solving for <math>x</math> yields <math>\frac{1}{\sqrt[3]{9}-1} = \frac{\sqrt[3]{81}+\sqrt[3]{9}+1}{8}</math>, so the answer is <math>\boxed{ | + | We note that <math>8x^3 - 3x^2 - 3x - 1 = 9x^3 - x^3 - 3x^2 - 3x - 1 = 9x^3 - (x + 1)^3</math>. Therefore, we have that <math>9x^3 = (x+1)^3</math>, so it follows that <math>x\sqrt[3]{9} = x+1</math>. Solving for <math>x</math> yields <math>\frac{1}{\sqrt[3]{9}-1} = \frac{\sqrt[3]{81}+\sqrt[3]{9}+1}{8}</math>, so the answer is <math>\boxed{098}</math>. |
=== Solution 2 === | === Solution 2 === |
Revision as of 17:49, 22 July 2020
Problem
The real root of the equation can be written in the form
, where
,
, and
are positive integers. Find
.
Contents
Solutions
Solution 1
We note that . Therefore, we have that
, so it follows that
. Solving for
yields
, so the answer is
.
Solution 2
Let be the real root of the given polynomial. Now define the cubic polynomial
. Note that
must be a root of
. However we can simplify
as
, so we must have that
. Thus
, and
. We can then multiply the numerator and denominator of
by
to rationalize the denominator, and we therefore have
, and the answer is
.
Solution 3
It is clear that for the algebraic degree of to be
that there exists some cubefree integer
and positive integers
such that
and
(it is possible that
, but then the problem wouldn't ask for both an
and
). Let
be the automorphism over
which sends
and
which sends
(note :
is a cubic root of unity).
Letting be the root, we clearly we have
by Vieta's formulas. Thus it follows
.
Now, note that
is a root of
. Thus
so
. Checking the non-cubicroot dimension part, we get
so it follows that
.
Solution 4
We have Therefore
We have
We will find
so that the equation is equivalent to the original one. Let
Easily,
and
So
.
-JZ
Video Solution
https://www.youtube.com/watch?v=9way8JrtD04&t=555s
See Also
2013 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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.