Difference between revisions of "2013 AIME I Problems/Problem 5"
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== See Also == | == See Also == |
Revision as of 13:57, 18 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.