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
.
Contents
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 automorpism 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
See Also
1983 AIME (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
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