Difference between revisions of "2013 AIME II Problems/Problem 12"
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− | 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 | + | 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. |
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 17:00, 6 April 2013
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 |