Difference between revisions of "2021 AIME II Problems/Problem 4"
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There are real numbers <math>a, b, c,</math> and <math>d</math> such that <math>-20</math> is a root of <math>x^3 + ax + b</math> and <math>-21</math> is a root of <math>x^3 + cx^2 + d.</math> These two polynomials share a complex root <math>m + \sqrt{n} \cdot i,</math> where <math>m</math> and <math>n</math> are positive integers and <math>i = \sqrt{-1}.</math> Find <math>m+n.</math> | There are real numbers <math>a, b, c,</math> and <math>d</math> such that <math>-20</math> is a root of <math>x^3 + ax + b</math> and <math>-21</math> is a root of <math>x^3 + cx^2 + d.</math> These two polynomials share a complex root <math>m + \sqrt{n} \cdot i,</math> where <math>m</math> and <math>n</math> are positive integers and <math>i = \sqrt{-1}.</math> Find <math>m+n.</math> | ||
− | ==Solution 1== | + | ==Solution 1 (Complex Conjugate Root Theorem)== |
By the <b>Complex Conjugate Root Theorem</b>, the imaginary roots for each of <math>x^3+ax+b</math> and <math>x^3+cx^2+d</math> are a pair of complex conjugates. Let <math>z=m+\sqrt{n}\cdot i</math> and <math>\overline{z}=m-\sqrt{n}\cdot i.</math> It follows that the roots of <math>x^3+ax+b</math> are <math>-20,z,\overline{z},</math> and the roots of <math>x^3+cx^2+d</math> are <math>-21,z,\overline{z}.</math> | By the <b>Complex Conjugate Root Theorem</b>, the imaginary roots for each of <math>x^3+ax+b</math> and <math>x^3+cx^2+d</math> are a pair of complex conjugates. Let <math>z=m+\sqrt{n}\cdot i</math> and <math>\overline{z}=m-\sqrt{n}\cdot i.</math> It follows that the roots of <math>x^3+ax+b</math> are <math>-20,z,\overline{z},</math> and the roots of <math>x^3+cx^2+d</math> are <math>-21,z,\overline{z}.</math> | ||
Revision as of 04:05, 23 March 2021
Contents
Problem
There are real numbers and
such that
is a root of
and
is a root of
These two polynomials share a complex root
where
and
are positive integers and
Find
Solution 1 (Complex Conjugate Root Theorem)
By the Complex Conjugate Root Theorem, the imaginary roots for each of and
are a pair of complex conjugates. Let
and
It follows that the roots of
are
and the roots of
are
By Vieta's Formulas on we have
from which
By Vieta's Formulas on we have
from which
Finally, we get
by
and
~MRENTHUSIASM
Solution 2 (Somewhat Bashy)
, hence
Also, , hence
satisfies both
we can put it in both equations and equate to 0.
In the first equation, we get
Simplifying this further, we get
Hence, and
In the second equation, we get
Simplifying this further, we get
Hence, and
Comparing (1) and (2),
and
;
Substituting these in gives,
This simplifies to
Hence,
Consider case of :
Also,
(because c = 1)
Also,
Also, Equation (2) gives
Solving (4) and (5) simultaneously gives
[AIME can not have more than one answer, so we can stop here also 😁... Not suitable for Subjective exam]
Hence,
-Arnav Nigam
See also
2021 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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