Difference between revisions of "2013 AIME I Problems/Problem 5"
m |
|||
(11 intermediate revisions by 9 users not shown) | |||
Line 1: | Line 1: | ||
+ | __TOC__ | ||
+ | |||
== Problem == | == Problem == | ||
The real root of the equation <math>8x^3 - 3x^2 - 3x - 1 = 0</math> can be written in the form <math>\frac{\sqrt[3]a + \sqrt[3]b + 1}{c}</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are positive integers. Find <math>a+b+c</math>. | The real root of the equation <math>8x^3 - 3x^2 - 3x - 1 = 0</math> can be written in the form <math>\frac{\sqrt[3]a + \sqrt[3]b + 1}{c}</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are positive integers. Find <math>a+b+c</math>. | ||
− | |||
− | |||
− | |||
− | |||
− | === Solution 2 | + | == Solution 1 == |
− | Let <math>r</math> be the real root of the given [[polynomial]]. Now define the cubic polynomial <math>Q(x)=-x^3-3x^2-3x+8</math>. Note that <math>1/r</math> must be a root of <math>Q</math>. However we can simplify <math>Q</math> as <math>Q(x)=9-(x+1)^3</math>, so we must have that <math>(\frac{1}{r}+1)^3=9</math>. Thus <math>\frac{1}{r}=\sqrt[3]{9}-1</math>, and <math>r=\frac{1}{\sqrt[3]{9}-1}</math>. We can then multiply the numerator and denominator of <math>r</math> by <math>\sqrt[3]{81}+\sqrt[3]{9}+1</math> to rationalize the denominator, and we therefore have <math>r=\frac{\sqrt[3]{81}+\sqrt[3]{9}+1}{8}</math>, and 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{98}</math>. |
+ | |||
+ | == Solution 2 == | ||
+ | Let <math>r</math> be the real root of the given [[polynomial]]. Now define the cubic polynomial <math>Q(x)=-x^3-3x^2-3x+8</math>. Note that <math>1/r</math> must be a root of <math>Q</math>. However we can simplify <math>Q</math> as <math>Q(x)=9-(x+1)^3</math>, so we must have that <math>(\frac{1}{r}+1)^3=9</math>. Thus <math>\frac{1}{r}=\sqrt[3]{9}-1</math>, and <math>r=\frac{1}{\sqrt[3]{9}-1}</math>. We can then multiply the numerator and denominator of <math>r</math> by <math>\sqrt[3]{81}+\sqrt[3]{9}+1</math> to rationalize the denominator, and we therefore have <math>r=\frac{\sqrt[3]{81}+\sqrt[3]{9}+1}{8}</math>, and the answer is <math>\boxed{98}</math>. | ||
− | + | == Solution 3 == | |
It is clear that for the algebraic degree of <math>x</math> to be <math>3</math> that there exists some cubefree integer <math>p</math> and positive integers <math>m,n</math> such that <math>a = m^3p</math> and <math>b = n^3p^2</math> (it is possible that <math>b = n^3p</math>, but then the problem wouldn't ask for both an <math>a</math> and <math>b</math>). Let <math>f_1</math> be the [[automorphism]] over <math>\mathbb{Q}[\sqrt[3]{a}][\omega]</math> which sends <math>\sqrt[3]{a} \to \omega \sqrt[3]{a}</math> and <math>f_2</math> which sends <math>\sqrt[3]{a} \to \omega^2 \sqrt[3]{a}</math> (note : <math>\omega</math> is a cubic [[Roots of unity|root of unity]]). | It is clear that for the algebraic degree of <math>x</math> to be <math>3</math> that there exists some cubefree integer <math>p</math> and positive integers <math>m,n</math> such that <math>a = m^3p</math> and <math>b = n^3p^2</math> (it is possible that <math>b = n^3p</math>, but then the problem wouldn't ask for both an <math>a</math> and <math>b</math>). Let <math>f_1</math> be the [[automorphism]] over <math>\mathbb{Q}[\sqrt[3]{a}][\omega]</math> which sends <math>\sqrt[3]{a} \to \omega \sqrt[3]{a}</math> and <math>f_2</math> which sends <math>\sqrt[3]{a} \to \omega^2 \sqrt[3]{a}</math> (note : <math>\omega</math> is a cubic [[Roots of unity|root of unity]]). | ||
Letting <math>r</math> be the root, we clearly we have <math>r + f_1(r) + f_2(r) = \frac{3}{8}</math> by Vieta's formulas. Thus it follows <math>c=8</math>. | Letting <math>r</math> be the root, we clearly we have <math>r + f_1(r) + f_2(r) = \frac{3}{8}</math> by Vieta's formulas. Thus it follows <math>c=8</math>. | ||
− | Now, note that <math>\sqrt[3]{a} + \sqrt[3]{b} + 1</math> is a root of <math>x^3 - 3x^2 - 24x - 64 = 0</math>. Thus <math>(x-1)^3 = 27x + 63</math> so <math>(\sqrt[3]{a} + \sqrt[3]{b})^3 = 27(\sqrt[3]{a} + \sqrt[3]{b}) + 90</math>. Checking the non-cubicroot dimension part, we get <math>a + b = 90</math> so it follows that <math>a + b + c = \boxed{ | + | Now, note that <math>\sqrt[3]{a} + \sqrt[3]{b} + 1</math> is a root of <math>x^3 - 3x^2 - 24x - 64 = 0</math>. Thus <math>(x-1)^3 = 27x + 63</math> so <math>(\sqrt[3]{a} + \sqrt[3]{b})^3 = 27(\sqrt[3]{a} + \sqrt[3]{b}) + 90</math>. Checking the non-cubicroot dimension part, we get <math>a + b = 90</math> so it follows that <math>a + b + c = \boxed{98}</math>. |
+ | |||
+ | == Solution 4 == | ||
+ | We have <math>cx-1=\sqrt[3]{a}+\sqrt[3]{b}.</math> Therefore <math>(cx-1)^3=(\sqrt[3]{a}+\sqrt[3]{b})^3=a+b+3\sqrt[3]{ab}(\sqrt[3]{a}+\sqrt[3]{b})=a+b+3\sqrt[3]{ab}(cx-1).</math> We have | ||
+ | <cmath>c^3x^3-3c^2x^2-(3c\sqrt[3]{ab}-3c)x-(a+b+1-3\sqrt[3]{ab})=0.</cmath> | ||
+ | We will find <math>a,b,c</math> so that the equation is equivalent to the original one. Let <math>\dfrac{3c^2}{c^3}=\dfrac{3}{8}, \dfrac{3c\sqrt[3]{ab}-3c}{c^3}=\dfrac{3}{8}, \dfrac{a+b+1-3\sqrt[3]{ab}}{c^3}=\dfrac{1}{8}.</math> Easily, <math>c=8, \sqrt[3]{ab}=9,</math> and <math>a+b=90.</math> So <math>a + b + c = 90+8=\boxed{98}</math>. | ||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=9way8JrtD04&t=240s | ||
== See Also == | == See Also == |
Latest revision as of 21:11, 27 August 2023
Problem
The real root of the equation can be written in the form
, where
,
, and
are positive integers. Find
.
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
.
Video Solution
https://www.youtube.com/watch?v=9way8JrtD04&t=240s
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.