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> | ||
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We know that the other two solutions for z wouldn't result in real solutions for <math>x</math> since we have to solve a quadratic with a negative discriminant, then multiply by <math>-(\frac{i}{\sqrt{10}})</math>. We get that <math>z = (\frac{i}{-2\sqrt{10}})</math>. Solving for <math>y</math> (using <math>y + \frac{1}{y} = z</math>) we get that <math>y = \frac{-i \pm \sqrt{161}i}{4\sqrt{10}}</math>, and multiplying this by <math>-(\frac{i}{\sqrt{10}})</math> (because <math>x = -(\frac{i}{\sqrt{10}})y</math>) we get that <math>x = \frac{-1 \pm \sqrt{161}}{40}</math> for a final answer of <math>-1 + 161 + 40 = \boxed{200}</math> | We know that the other two solutions for z wouldn't result in real solutions for <math>x</math> since we have to solve a quadratic with a negative discriminant, then multiply by <math>-(\frac{i}{\sqrt{10}})</math>. We get that <math>z = (\frac{i}{-2\sqrt{10}})</math>. Solving for <math>y</math> (using <math>y + \frac{1}{y} = z</math>) we get that <math>y = \frac{-i \pm \sqrt{161}i}{4\sqrt{10}}</math>, and multiplying this by <math>-(\frac{i}{\sqrt{10}})</math> (because <math>x = -(\frac{i}{\sqrt{10}})y</math>) we get that <math>x = \frac{-1 \pm \sqrt{161}}{40}</math> for a final answer of <math>-1 + 161 + 40 = \boxed{200}</math> | ||
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+ | -Grizzy | ||
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+ | ==Video solution== | ||
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+ | https://www.youtube.com/watch?v=mAXDdKX52TM | ||
== See also == | == See also == |
Latest revision as of 00:07, 28 October 2020
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
-Grizzy
Video solution
https://www.youtube.com/watch?v=mAXDdKX52TM
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.