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

Revision as of 15:39, 24 May 2023

Let $a$ and $b$ be real numbers so that the roots of $z^2 + (10 + ai) z + (27 + bi) = 0$ are complex conjugates. Enter the ordered pair $(a,b).$

Two solutions of $x^4 - 3x^3 + 5x^2 - 27x - 36 = 0$ are pure imaginary. Enter these solutions, separated by commas.

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

Solution 4

There are two cases: either all the roots are real, or one is real and two are imaginary.

Case 1: All roots are real. Then each of the roots is a member of the set $\{-20, 20, -13, 13\}$ . It splits into three sub-cases: either no two are the same, exactly two are the same, or all three are the same.

Sub-case 1.1: No two are the same. This is obviously $\dbinom{4}{3}=4$ .

Sub-case 1.2: Exactly two are the same. There are four ways to choose the root that will repeat twice, and three ways to choose the remaining root. For this sub-case, $4\cdot 3=12$ .

Sub-case 1.3: All three are the same. This is obviously $4$ .

Thus for case one, we have $4+12+4=20$ polynomials in $S$ . We now have case two, which we state below.

Case 2: Two roots are imaginary and one is real. Let these roots be $p-qi$ , $p+qi$ , and $r$ . Then by Vieta's formulas

$-(2p+r)=a$ ;
$p^{2}+q^{2}+2pr=b$ ;
$-\left(p^{2}+q^{2}\right)r=c$ .

Since $a$ , $b$ , $c$ , and $r$ are integers, we have that $p=\frac{1}{2}k$ for some integer $k$ . Case two splits into two sub-cases now:

Sub-case 2.1: $|p-qi|=|p+qi|=13$ . Obviously, $|p|<13$ . The $51$ cases in which $p$ is either $0,\pm\frac{1}{2},\pm\frac{2}{2},\pm\frac{3}{2},\ldots,\pm\frac{25}{2}$ are acceptable. Each can pair with one value of $q$ and four values of $r$ , adding $51\cdot 4=204$ polynomials to $S$ .

Sub-case 2.2: $|p-qi|=|p+qi|=20$ . Obviously, $|p|<20$ . Here, the $79$ cases in which $p$ is either $0,\pm\frac{1}{2},\pm\frac{2}{2},\pm\frac{3}{2},\ldots,\pm\frac{39}{2}$ are acceptable. Again, each can pair with a single value of $q$ as well as four values of $r$ , adding $79\cdot 4=316$ polynomials to $S$ .

Thus for case two, $204+316=520$ polynomials are part of $S$ .

All in all, $20+204+316=\boxed{540}$ polynomials can call $S$ home.

@@ Line 2: / Line 2: @@
 <cmath>z^2 + (10 + ai) z + (27 + bi) = 0</cmath>are complex conjugates. Enter the ordered pair <math>(a,b).</math>
-==Solution 1==
+Two solutions of
+<cmath>x^4 - 3x^3 + 5x^2 - 27x - 36 = 0</cmath>are pure imaginary. Enter these solutions, separated by commas.
-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:  <math>f(z)=(z-r)(z-\omega)(z-\omega^*)</math>, where <math>r\in \mathbb{R}</math>,  <math>\omega</math> is nonreal, and <math>\omega^*</math> is the complex conjugate of omega (note that we may assume that <math>\Im(\omega)>0</math>).
-The real root <math>r</math> must be one of <math>-20</math>, <math>20</math>, <math>-13</math>, or <math>13</math>. By Viète's formulas, <math>a=-(r+\omega+\omega^*)</math>, <math>b=|\omega|^2+r(\omega+\omega^*)</math>, and <math>c=-r|\omega|^2</math>. But <math>\omega+\omega^*=2\Re{(\omega)}</math> (i.e., adding the conjugates cancels the imaginary part). Therefore, to make <math>a</math> an integer, <math>2\Re{(\omega)}</math> must be an integer. Conversely, if <math>\omega+\omega^*=2\Re{(\omega)}</math> is an integer, then <math>a,b,</math> and <math>c</math> are clearly integers. Therefore <math>2\Re{(\omega)}\in \mathbb{Z}</math> is equivalent to the desired property. Let <math>\omega=\alpha+i\beta</math>.
-*Subcase 1.1: <math>|\omega|=20</math>.
-In this case, <math>\omega</math> lies on a circle of radius <math>20</math> in the complex plane. As <math>\omega</math> is nonreal, we see that <math>\beta\ne 0</math>.  Hence <math>-20<\Re{(\omega)}< 20</math>, or rather <math>-40<2\Re{(\omega)}< 40</math>. We count <math>79</math> integers in this interval, each of which corresponds to a unique complex number on the circle of radius <math>20</math> with positive imaginary part.
-*Subcase 1.2: <math>|\omega|=13</math>.
-In this case, <math>\omega</math> lies on a circle of radius <math>13</math> in the complex plane. As <math>\omega</math> is nonreal, we see that <math>\beta\ne 0</math>.  Hence <math>-13<\Re{(\omega)}< 13</math>, or rather <math>-26<2\Re{(\omega)}< 26</math>. We count <math>51</math> integers in this interval, each of which corresponds to a unique complex number on the circle of radius <math>13</math> with positive imaginary part.
-Therefore, there are <math>79+51=130</math> choices for <math>\omega</math>. We also have <math>4</math> choices for <math>r</math>, hence there are <math>4\cdot 130=520</math> total polynomials in this case.
-*Case 2: <math>f(z)=(z-r_1)(z-r_2)(z-r_3)</math>, where <math>r_1,r_2,r_3</math> are all real.
-In this case, there are four possible real roots, namely <math>\pm 13, \pm20</math>. Let <math>p</math> be the number of times that <math>13</math> appears among <math>r_1,r_2,r_3</math>, and define <math>q,r,s</math> similarly for <math>-13,20</math>, and <math>-20</math>, respectively. Then <math>p+q+r+s=3</math> because there are three roots. We wish to find the number of ways to choose nonnegative integers <math>p,q,r,s</math> that satisfy that equation. By balls and urns, these can be chosen in <math>\binom{6}{3}=20</math> ways.
-Therefore, there are a total of <math>520+20=\boxed{540}</math> polynomials with the desired property.
 ==Solution 2 (Systematics)==

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

Page

Toolbox

Search

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

Revision as of 15:39, 24 May 2023

Contents

Solution 2 (Systematics)

Solution 3 (Comments)

Solution 4

See Also