Difference between revisions of "2013 AIME II Problems/Problem 12"
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+ | ==Problem 12== | ||
+ | |||
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>. | ||
− | Solution | + | ==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: <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)== | ||
+ | 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 <math>{13, -13, 20, -20}</math>. By stars and bars, there is <math>\binom{6}{3}=20</math> ways (3 bars between all four possibilities, and then 3 stars that represent the roots themselves). | ||
+ | |||
+ | Case 2: One real root: one of <math>13, -13, 20, -20</math>. 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 <math>20</math> or <math>13</math>. Call the root <math>a+bi</math>, where <math>a</math> is not the magnitude of the root; otherwise, it would be case 1. We need integral coefficients: expansion of <math>(x-(a+bi))(x-(a-bi))=-2ax+x^2+(a^2+b^2)</math> tells us that we just need <math>2a</math> to be integral, because <math>a^2+b^2</math> 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 <math>20</math>, the <math>a</math> term can range from <math>-19.5, -19, ...., 0, 0.5, ..., 19.5</math> or <math>79</math> solutions. When the norm is <math>13</math>, the <math>a</math> term has <math>51</math> possibilities from <math>-12.5, -12, ..., 12.5</math>. In total that's 130 total ways to choose the imaginary root. Now, multiply by the ways to choose the real root, <math>4</math>, and you get <math>520</math> for this case. | ||
+ | |||
+ | And <math>520+20=540</math> and we are done. | ||
+ | |||
+ | ==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: | ||
+ | |||
+ | <math>z^2+pz+20^2,</math> where <math>p=0,\pm1,\pm2,\cdots,\pm 39,</math> | ||
+ | |||
+ | <math>z^2+pz+13^2,</math> where <math>p=0,\pm1,\pm2,\cdots,\pm 25</math>. | ||
+ | |||
+ | Total of 130 equations, multiplied by 4 (the number of cases for real <math>r</math>, 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. | ||
+ | |||
+ | <b>Case 1:</b> All roots are real. | ||
+ | Then each of the roots is a member of the set <math>\{-20, 20, -13, 13\}</math>. It splits into three sub-cases: either no two are the same, exactly two are the same, or all three are the same. | ||
+ | |||
+ | <b>Sub-case 1.1:</b> No two are the same. | ||
+ | This is obviously <math>\dbinom{4}{3}=4</math>. | ||
+ | |||
+ | <b>Sub-case 1.2:</b> 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, <math>4\cdot 3=12</math>. | ||
+ | |||
+ | <b>Sub-case 1.3:</b> All three are the same. | ||
+ | This is obviously <math>4</math>. | ||
+ | |||
+ | Thus for case one, we have <math>4+12+4=20</math> polynomials in <math>S</math>. We now have case two, which we state below. | ||
+ | |||
+ | <b>Case 2:</b> Two roots are imaginary and one is real. | ||
+ | Let these roots be <math>p-qi</math>, <math>p+qi</math>, and <math>r</math>. Then by Vieta's formulas | ||
+ | * <math>-(2p+r)=a</math>; | ||
+ | * <math>p^{2}+q^{2}+2pr=b</math>; | ||
+ | * <math>-\left(p^{2}+q^{2}\right)r=c</math>. | ||
+ | Since <math>a</math>, <math>b</math>, <math>c</math>, and <math>r</math> are integers, we have that <math>p=\frac{1}{2}k</math> for some integer <math>k</math>. Case two splits into two sub-cases now: | ||
+ | |||
+ | <b>Sub-case 2.1:</b> <math>|p-qi|=|p+qi|=13</math>. | ||
+ | Obviously, <math>|p|<13</math>. The <math>51</math> cases in which <math>p</math> is either <math>0,\pm\frac{1}{2},\pm\frac{2}{2},\pm\frac{3}{2},\ldots,\pm\frac{25}{2}</math> are acceptable. Each can pair with one value of <math>q</math> and four values of <math>r</math>, adding <math>51\cdot 4=204</math> polynomials to <math>S</math>. | ||
+ | |||
+ | <b>Sub-case 2.2:</b> <math>|p-qi|=|p+qi|=20</math>. | ||
+ | Obviously, <math>|p|<20</math>. Here, the <math>79</math> cases in which <math>p</math> is either <math>0,\pm\frac{1}{2},\pm\frac{2}{2},\pm\frac{3}{2},\ldots,\pm\frac{39}{2}</math> are acceptable. Again, each can pair with a single value of <math>q</math> as well as four values of <math>r</math>, adding <math>79\cdot 4=316</math> polynomials to <math>S</math>. | ||
+ | |||
+ | Thus for case two, <math>204+316=520</math> polynomials are part of <math>S</math>. | ||
+ | |||
+ | All in all, <math>20+204+316=\boxed{540}</math> polynomials can call <math>S</math> home. | ||
+ | |||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/-U65hhr1Smw?si=2JfRYL032MhUe276 | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
+ | |||
− | + | ==See Also== | |
− | + | {{AIME box|year=2013|n=II|num-b=11|num-a=13}} | |
− | + | {{MAA Notice}} |
Latest revision as of 17:41, 20 June 2024
Contents
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 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:
, 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 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 . 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.
Solution 3 (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
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 . 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 .
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, .
Sub-case 1.3: All three are the same.
This is obviously .
Thus for case one, we have polynomials in
. We now have case two, which we state below.
Case 2: Two roots are imaginary and one is real.
Let these roots be ,
, and
. Then by Vieta's formulas
;
;
.
Since ,
,
, and
are integers, we have that
for some integer
. Case two splits into two sub-cases now:
Sub-case 2.1: .
Obviously,
. The
cases in which
is either
are acceptable. Each can pair with one value of
and four values of
, adding
polynomials to
.
Sub-case 2.2: .
Obviously,
. Here, the
cases in which
is either
are acceptable. Again, each can pair with a single value of
as well as four values of
, adding
polynomials to
.
Thus for case two, polynomials are part of
.
All in all, polynomials can call
home.
Video Solution
https://youtu.be/-U65hhr1Smw?si=2JfRYL032MhUe276
~MathProblemSolvingSkills.com
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.