2013 AIME I Problems/Problem 5
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 |
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