Difference between revisions of "2000 AIME II Problems/Problem 13"
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<math>z^2 - 2 = y^2 + \frac{1}{y^2}</math> | <math>z^2 - 2 = y^2 + \frac{1}{y^2}</math> | ||
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and | and | ||
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<math>z^3 - 3z = y^3 + \frac{1}{y^3}</math> | <math>z^3 - 3z = y^3 + \frac{1}{y^3}</math> | ||
Revision as of 23:38, 6 October 2019
Problem
The equation has exactly two real roots, one of which is
, where
,
and
are integers,
and
are relatively prime, and
. Find
.
Solution
We may factor the equation as:[1]
Now for real
. Thus the real roots must be the roots of the equation
. By the quadratic formula the roots of this are:
Thus , and so the final answer is
.
^ A well-known technique for dealing with symmetric (or in this case, nearly symmetric) polynomials is to divide through by a power of with half of the polynomial's degree (in this case, divide through by
), and then to use one of the substitutions
. In this case, the substitution
gives
and
, which reduces the polynomial to just
. Then one can backwards solve for
.
Solution 2 (Complex Bash)
It would be really nice if the coefficients were symmetrical. What if we make the substitution, . The the polynomial becomes
It's symmetric! Dividing by and rearranging, we get
Now, if we let , we can get the equations
and
(These come from squaring and subtracting
, then multiplying that result by
and subtracting
)
Plugging this into our polynomial, expanding, and rearranging, we get
Now, we see that the two terms must cancel in order to get this polynomial equal to
, so what squared equals 3? Plugging in
into the polynomial, we see that it works! Is there something else that equals 3 when squared? Trying
, we see that it also works! Great, we use long division on the polynomial by
and we get
.
We know that the other two solutions for z wouldn't result in real solutions for since we have to solve a quadratic with a negative discriminant, then multiply by
. We get that
. Solving for
(using
) we get that
, and multiplying this by
(because
) we get that
for a final answer of
-GrizzyProblemSolver79c
See also
2000 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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.