Difference between revisions of "2013 AIME II Problems/Problem 12"
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==Solution== | ==Solution== | ||
− | Every cubic in real coefficients has to have either three real roots or one real and two | + | Every cubic in 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> is 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>a+b+c+d=3</math> because there are three roots. We wish to find the number of ways to choose nonnegative integers <math>a,b,c,d</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. | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2013|n=II|num-b=11|num-a=13}} | {{AIME box|year=2013|n=II|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 00:29, 18 January 2016
Problem 12
Let be the set of all polynomials of the form , where , , and are integers. Find the number of polynomials in such that each of its roots satisfies either or .
Solution
Every cubic in 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: , where , is nonreal, and is the complex conjugate of omega (note that we may assume that ).
The real root must be one of , , , or . By Viète's formulas, , , and . But (i.e., adding the conjugates cancels the imaginary part). Therefore, to make is an integer, must be an integer. Conversely, if is an integer, then and are clearly integers. Therefore is equivalent to the desired property. Let .
- Subcase 1.1: .
In this case, lies on a circle of radius in the complex plane. As is nonreal, we see that . Hence , or rather . We count integers in this interval, each of which corresponds to a unique complex number on the circle of radius with positive imaginary part.
- Subcase 1.2: .
In this case, lies on a circle of radius in the complex plane. As is nonreal, we see that . Hence , or rather . We count integers in this interval, each of which corresponds to a unique complex number on the circle of radius with positive imaginary part.
Therefore, there are choices for . We also have choices for , hence there are total polynomials in this case.
- Case 2: , where are all real.
In this case, there are four possible real roots, namely . Let be the number of times that appears among , and define similarly for , and , respectively. Then because there are three roots. We wish to find the number of ways to choose nonnegative integers that satisfy that equation. By balls and urns, these can be chosen in ways.
Therefore, there are a total of polynomials with the desired property.
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.