Difference between revisions of "2013 AIME II Problems/Problem 12"
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*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>). | *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> | + | 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>. | *Subcase 1.1: <math>|\omega|=20</math>. |
Latest revision as of 18:52, 20 July 2020
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 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: , 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 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.
Solution 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 . By stars and bars, there is ways (3 bars between all four possibilities, and then 3 stars that represent the roots themselves).
Case 2: One real root: one of . 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 or . Call the root , where is not the magnitude of the root; otherwise, it would be case 1. We need integral coefficients: expansion of tells us that we just need to be integral, because 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 , the term can range from or solutions. When the norm is , the term has possibilities from . In total that's 130 total ways to choose the imaginary root. Now, multiply by the ways to choose the real root, , and you get for this case.
And and we are done.
Comments
If the polynomial has one real root and two complex roots, then it can be factored as where is real with and are integers with The roots and are conjugates. We have So is either or . The only requirement for is All such quadratic equations are listed as follows:
where
where .
Total of 130 equations, multiplied by 4 (the number of cases for real , 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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