Difference between revisions of "2013 AIME I Problems/Problem 5"
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=== Solution 3 === | === 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 [[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. 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. Thus it follows <math>c=8</math>. |
Revision as of 12:10, 12 August 2013
Problem
The real root of the equation can be written in the form , where , , and are positive integers. Find .
Contents
[hide]Solutions
Solution 1
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. 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 proceed by using the cubic formula.
Let , , , and . Then let and . Then the real root of is Now note that and Thus and hence the answer is .
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.