Difference between revisions of "2013 AIME I Problems/Problem 5"
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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 | + | 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{098}</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{098}</math>. | ||
Revision as of 22:44, 11 November 2019
Problem
The real root of the equation can be written in the form
, where
,
, and
are positive integers. Find
.
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
.
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.