Difference between revisions of "2013 AIME II Problems/Problem 12"
(→Solution) |
Yan justin98 (talk | contribs) m (→Solution) |
||
| Line 5: | Line 5: | ||
==Solution== | ==Solution== | ||
| − | Every cubic in real coefficients has to have either three real roots or one real and two complex roots which are conjugates. This follows from [[Vieta's formulas]]. Factorise the polynomial <math>(z-r)(z-\omega)(z-\omega^*)</math>, where <math>\omega^*</math> is the complex conjugate of omega. We know that <math>r</math> is the real root which must be <math>-20</math>, <math>20</math>, <math>-13</math>, or <math>13</math>, and it doesn't matter which. <math>|\omega|=|\omega^*|=20 \ \text{or}\ 13</math>. Let <math>\omega=\alpha+i\beta</math>. Viète tells us that <math>a=-(r+\omega+\omega^*</math>), but <math>\omega+\omega^*=2\Re{(\omega)}</math> (i.e., adding the conjugates cancels the imaginary part). a the quadratic coefficient must be an integer so <math>\Re{(\omega)}</math> is some integer over <math>2</math>. <math>|\omega|=|\omega^*|=</math>20 or 13 so you have a bound on <math>\Re{(\omega)}</math>: either <math>-13\leq\Re{(\omega)}\leq 13</math> or <math>-20\leq\Re{(\omega)}\leq 20</math>. Don't forget zero! We're not double counting the numbers between <math>-13</math> and <math>13</math> here because there's an imaginary part too -- <math>\sqrt{\alpha^2+\beta^2}=|\omega|</math>, and what you get when you solve for beta will depend on what the magnitude was. | + | Every cubic in real coefficients has to have either three real roots or one real and two complex roots which are conjugates. This follows from [[Vieta's formulas]]. Factorise the polynomial <math>(z-r)(z-\omega)(z-\omega^*)</math>, where <math>\omega^*</math> is the complex conjugate of omega. We know that <math>r</math> is the real root which must be <math>-20</math>, <math>20</math>, <math>-13</math>, or <math>13</math>, and it doesn't matter which. <math>|\omega|=|\omega^*|=20 \ \text{or}\ 13</math>. Let <math>\omega=\alpha+i\beta</math>. Viète tells us that <math>a=-(r+\omega+\omega^*</math>), but <math>\omega+\omega^*=2\Re{(\omega)}</math> (i.e., adding the conjugates cancels the imaginary part). a the quadratic coefficient must be an integer so <math>2\Re{(\omega)}</math> is some integer over <math>2</math>. <math>|\omega|=|\omega^*|=</math>20 or 13 so you have a bound on <math>\Re{(\omega)}</math>: either <math>-13\leq\Re{(\omega)}\leq 13</math> or <math>-20\leq\Re{(\omega)}\leq 20</math>. Don't forget zero! We're not double counting the numbers between <math>-13</math> and <math>13</math> here because there's an imaginary part too -- <math>\sqrt{\alpha^2+\beta^2}=|\omega|</math>, and what you get when you solve for beta will depend on what the magnitude was. |
You have the magnitude so <math>\Re{(\omega)}</math> determines <math>\omega</math> totally (you can solve for the imaginary part) and <math>\omega</math> determines <math>\omega^*</math>. | You have the magnitude so <math>\Re{(\omega)}</math> determines <math>\omega</math> totally (you can solve for the imaginary part) and <math>\omega</math> determines <math>\omega^*</math>. | ||
Now just count: 4 possibilities for the real root times [(52+1) possibilities if <math>|\omega|=13</math> plus (80+1) possibilities if <math>|\omega|=20</math> = 536. But this is not all, we also have <math>{4\choose{3}}=4</math> ways of constructing a totally real polynomial (all real roots), which gives you <math>\boxed{540}</math>. | Now just count: 4 possibilities for the real root times [(52+1) possibilities if <math>|\omega|=13</math> plus (80+1) possibilities if <math>|\omega|=20</math> = 536. But this is not all, we also have <math>{4\choose{3}}=4</math> ways of constructing a totally real polynomial (all real roots), which gives you <math>\boxed{540}</math>. | ||
Revision as of 11:16, 2 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 complex roots which are conjugates. This follows from Vieta's formulas. Factorise the polynomial 
, where 
is the complex conjugate of omega. We know that 
is the real root which must be 
, 
, 
, or 
, and it doesn't matter which. 
. Let 
. Viète tells us that 
), but 
(i.e., adding the conjugates cancels the imaginary part). a the quadratic coefficient must be an integer so 
is some integer over 
. 
20 or 13 so you have a bound on 
: either 
or 
. Don't forget zero! We're not double counting the numbers between and here because there's an imaginary part too -- 
, and what you get when you solve for beta will depend on what the magnitude was.
You have the magnitude so determines 
totally (you can solve for the imaginary part) and determines .
Now just count: 4 possibilities for the real root times [(52+1) possibilities if 
plus (80+1) possibilities if 
= 536. But this is not all, we also have 
ways of constructing a totally real polynomial (all real roots), which gives you 
.
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. 
