2019 AIME I Problems/Problem 8
Problem 8
Let be a real number such that
. Then
where
and
are relatively prime positive integers. Find
.
Solution 1
We can substitute . Since we know that
, we can do some simplification.
This yields . From this, we can substitute again to get some cancellation through binomials. If we let
, we can simplify the equation to
. After using binomial theorem, this simplifies to
. If we use the quadratic formula, we obtain the that
, so
. By plugging z into
(which is equal to
), we can either use binomial theorem or sum of cubes to simplify, and we end up with
. Therefore, the answer is
.
-eric2020, inspired by Tommy2002
Solution 2
First, for simplicity, let and
. Note that
. We then bash the rest of the problem out. Take the tenth power of this expression and get
. Note that we also have
. So, it suffices to compute
. Let
. We have from cubing
that
or
. Next, using
, we get
or
. Solving gives
or
. Clearly
is extraneous, so
. Now note that
, and
. Thus we finally get
, giving
.
- Emathmaster
Solution 3 (Newton Sums)
Newton sums is basically constructing the powers of the roots of the polynomials instead of deconstructing them which was done in Solution . Let
and
be the roots of some polynomial
. Then, by Vieta,
for some
.
Let . We want to find
. Clearly
and
. Newton sums tells us that
where
for our polynomial
.
Bashing, we have
Thus
. Clearly,
so
.
Note . Solving for
, we get
. Finally,
.
Solution 4
Factor the first equation.
First of all,
because
We group the first, third, and fifth term and second and fourth term. The first group:
The second group:
Add the two together to make
Because this equals
, we have
Let
so we get
Solving the quadratic gives us
Because
, we finally get
.
Now from the second equation,
Plug in
to get
which yields the answer
~ZericHang
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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