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Difference between revisions of "2013 AIME II Problems/Problem 12"

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Let <math>S</math> be the set of all polynomials of the form <math>z^3 + az^2 + bz + c</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are integers. Find the number of polynomials in <math>S</math> such that each of its roots <math>z</math> satisfies either <math>|z| = 20</math> or <math>|z| = 13</math>.
 
Let <math>S</math> be the set of all polynomials of the form <math>z^3 + az^2 + bz + c</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are integers. Find the number of polynomials in <math>S</math> such that each of its roots <math>z</math> satisfies either <math>|z| = 20</math> or <math>|z| = 13</math>.
  
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==Solution==
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==Solution 1==
  
 
Every cubic with real coefficients has to have either three real roots or one real and two nonreal roots which are conjugates. This follows from [[Vieta's formulas]].  
 
Every cubic with real coefficients has to have either three real roots or one real and two nonreal roots which are conjugates. This follows from [[Vieta's formulas]].  
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Therefore, there are a total of <math>520+20=\boxed{540}</math> polynomials with the desired property.
 
Therefore, there are a total of <math>520+20=\boxed{540}</math> polynomials with the desired property.
  
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==Solution Systematics==
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==Solution 2 (Systematics)==
 
This combinatorics problem involves counting, and casework is most appropriate.
 
This combinatorics problem involves counting, and casework is most appropriate.
 
There are two cases: either all three roots are real, or one is real and there are two imaginary roots.
 
There are two cases: either all three roots are real, or one is real and there are two imaginary roots.
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And <math>520+20=540</math> and we are done.
 
And <math>520+20=540</math> and we are done.
  
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==Comments==
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==Solution 3 (Comments)==
 
If the polynomial has one real root and two complex roots, then it can be factored as <math>(z-r)(z^2+pz+q), </math> where <math>r</math> is real with <math>|r|=13,20</math> and <math>p,q</math> are integers with <math>p^2 <4q.</math> The roots <math>z_1</math> and <math>z_2</math> are conjugates. We have <math>|z_1|^2=|z_2|^2=z_1z_2=q.</math> So <math>q</math> is either <math>20^2</math> or <math>13^2</math>. The only requirement for <math>p</math> is <math>p<\sqrt{4q^2}=2\sqrt{q}.</math> All such quadratic equations are listed as follows:
 
If the polynomial has one real root and two complex roots, then it can be factored as <math>(z-r)(z^2+pz+q), </math> where <math>r</math> is real with <math>|r|=13,20</math> and <math>p,q</math> are integers with <math>p^2 <4q.</math> The roots <math>z_1</math> and <math>z_2</math> are conjugates. We have <math>|z_1|^2=|z_2|^2=z_1z_2=q.</math> So <math>q</math> is either <math>20^2</math> or <math>13^2</math>. The only requirement for <math>p</math> is <math>p<\sqrt{4q^2}=2\sqrt{q}.</math> All such quadratic equations are listed as follows:
  

Revision as of 22:49, 29 August 2021

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 1

Every cubic with real coefficients has to have either three real roots or one real and two nonreal roots which are conjugates. This follows from Vieta's formulas.

  • Case 1: $f(z)=(z-r)(z-\omega)(z-\omega^*)$, where $r\in \mathbb{R}$, $\omega$ is nonreal, and $\omega^*$ is the complex conjugate of omega (note that we may assume that $\Im(\omega)>0$).

The real root $r$ must be one of $-20$, $20$, $-13$, or $13$. By Viète's formulas, $a=-(r+\omega+\omega^*)$, $b=|\omega|^2+r(\omega+\omega^*)$, and $c=-r|\omega|^2$. But $\omega+\omega^*=2\Re{(\omega)}$ (i.e., adding the conjugates cancels the imaginary part). Therefore, to make $a$ an integer, $2\Re{(\omega)}$ must be an integer. Conversely, if $\omega+\omega^*=2\Re{(\omega)}$ is an integer, then $a,b,$ and $c$ are clearly integers. Therefore $2\Re{(\omega)}\in \mathbb{Z}$ is equivalent to the desired property. Let $\omega=\alpha+i\beta$.

  • Subcase 1.1: $|\omega|=20$.

In this case, $\omega$ lies on a circle of radius $20$ in the complex plane. As $\omega$ is nonreal, we see that $\beta\ne 0$. Hence $-20<\Re{(\omega)}< 20$, or rather $-40<2\Re{(\omega)}< 40$. We count $79$ integers in this interval, each of which corresponds to a unique complex number on the circle of radius $20$ with positive imaginary part.

  • Subcase 1.2: $|\omega|=13$.

In this case, $\omega$ lies on a circle of radius $13$ in the complex plane. As $\omega$ is nonreal, we see that $\beta\ne 0$. Hence $-13<\Re{(\omega)}< 13$, or rather $-26<2\Re{(\omega)}< 26$. We count $51$ integers in this interval, each of which corresponds to a unique complex number on the circle of radius $13$ with positive imaginary part.

Therefore, there are $79+51=130$ choices for $\omega$. We also have $4$ choices for $r$, hence there are $4\cdot 130=520$ total polynomials in this case.

  • Case 2: $f(z)=(z-r_1)(z-r_2)(z-r_3)$, where $r_1,r_2,r_3$ are all real.

In this case, there are four possible real roots, namely $\pm 13, \pm20$. Let $p$ be the number of times that $13$ appears among $r_1,r_2,r_3$, and define $q,r,s$ similarly for $-13,20$, and $-20$, respectively. Then $p+q+r+s=3$ because there are three roots. We wish to find the number of ways to choose nonnegative integers $p,q,r,s$ that satisfy that equation. By balls and urns, these can be chosen in $\binom{6}{3}=20$ ways.

Therefore, there are a total of $520+20=\boxed{540}$ polynomials with the desired property.

Solution 2 (Systematics)

This combinatorics problem involves counting, and casework is most appropriate. There are two cases: either all three roots are real, or one is real and there are two imaginary roots.

Case 1: Three roots are of the set ${13, -13, 20, -20}$. By stars and bars, there is $\binom{6}{3}=20$ ways (3 bars between all four possibilities, and then 3 stars that represent the roots themselves).

Case 2: One real root: one of $13, -13, 20, -20$. Then two imaginary roots left; it is well known that because coefficients of the polynomial are integral (and thus not imaginary), these roots are conjugates. Therefore, either both roots have a norm (also called magnitude) of $20$ or $13$. Call the root $a+bi$, where $a$ is not the magnitude of the root; otherwise, it would be case 1. We need integral coefficients: expansion of $(x-(a+bi))(x-(a-bi))=-2ax+x^2+(a^2+b^2)$ tells us that we just need $2a$ to be integral, because $a^2+b^2$ IS the norm of the root! (Note that it is not necessary to multiply by the real root. That won't affect whether or not a coefficient is imaginary.) Therefore, when the norm is $20$, the $a$ term can range from $-19.5, -19, ...., 0, 0.5, ..., 19.5$ or $79$ solutions. When the norm is $13$, the $a$ term has $51$ possibilities from $-12.5, -12, ..., 12.5$. In total that's 130 total ways to choose the imaginary root. Now, multiply by the ways to choose the real root, $4$, and you get $520$ for this case.

And $520+20=540$ and we are done.

Solution 3 (Comments)

If the polynomial has one real root and two complex roots, then it can be factored as $(z-r)(z^2+pz+q),$ where $r$ is real with $|r|=13,20$ and $p,q$ are integers with $p^2 <4q.$ The roots $z_1$ and $z_2$ are conjugates. We have $|z_1|^2=|z_2|^2=z_1z_2=q.$ So $q$ is either $20^2$ or $13^2$. The only requirement for $p$ is $p<\sqrt{4q^2}=2\sqrt{q}.$ All such quadratic equations are listed as follows:

$z^2+pz+20^2,$ where $p=0,\pm1,\pm2,\cdots,\pm 39,$

$z^2+pz+13^2,$ where $p=0,\pm1,\pm2,\cdots,\pm 25$.

Total of 130 equations, multiplied by 4 (the number of cases for real $r$, we have 520 equations, as indicated in the solution.

-JZ

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

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