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2013 AIME II Problems/Problem 12

Revision as of 11:50, 2 March 2014 by Nahmid (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 $|z| = 13$.


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 $\Re{(\omega)}$ is some integer over $2$. $|\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 (ProblemsAnswer KeyResources)
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

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