Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2013 AIME II Problems/Problem 12

Revision as of 11:19, 2 January 2016 by Yan justin98 (talk | contribs) (→‎Solution)

Problem 12

Let $S$ be the set of all polynomials of the form $z^3 + az^2 + bz + c$, where $a$, $b$, and $c$ are integers. Find the number of polynomials in $S$ such that each of its roots $z$ satisfies either $|z| = 20$ or $|z| = 13$.

Solution

Every cubic in real coefficients has to have either three real roots or one real and two complex roots which are conjugates. This follows from Vieta's formulas. Factorise the polynomial $(z-r)(z-\omega)(z-\omega^*)$, where $\omega^*$ is the complex conjugate of omega. We know that $r$ is the real root which must be $-20$, $20$, $-13$, or $13$, and it doesn't matter which. $|\omega|=|\omega^*|=20 \ \text{or}\ 13$. Let $\omega=\alpha+i\beta$. Viète tells us that $a=-(r+\omega+\omega^*$), but $\omega+\omega^*=2\Re{(\omega)}$ (i.e., adding the conjugates cancels the imaginary part). a the quadratic coefficient must be an integer so $2\Re{(\omega)}$ is some integer. $|\omega|=|\omega^*|=$20 or 13 so you have a bound on $\Re{(\omega)}$: either $-13\leq\Re{(\omega)}\leq 13$ or $-20\leq\Re{(\omega)}\leq 20$. Don't forget zero! We're not double counting the numbers between $-13$ and $13$ here because there's an imaginary part too -- $\sqrt{\alpha^2+\beta^2}=|\omega|$, and what you get when you solve for beta will depend on what the magnitude was. You have the magnitude so $\Re{(\omega)}$ determines $\omega$ totally (you can solve for the imaginary part) and $\omega$ determines $\omega^*$. Now just count: 4 possibilities for the real root times [(52+1) possibilities if $|\omega|=13$ plus (80+1) possibilities if $|\omega|=20$ = 536. But this is not all, we also have ${4\choose{3}}=4$ ways of constructing a totally real polynomial (all real roots), which gives you $\boxed{540}$.

See Also

2013 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_II_Problems/Problem_12&oldid=74249"