Difference between revisions of "2021 AIME II Problems/Problem 4"
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-Arnav Nigam | -Arnav Nigam | ||
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start off by applying vieta's and you will find that <math>a=m^2+n-40m</math> <math>b=20m^2+20n</math> <math>c=21-2m</math> and <math>d=21m^2+21n</math>. After that, we have to use the fact that <math>-20</math> and <math>-21</math> are roots of <math>x^3+ax+b</math> and <math>x^3+cx^2+d</math>, respectively. Since we know that if you substitute the root of a function back into the function, the output is zero, therefore <math>(-20)^3-20(a)+b=0</math> and <math>(-21)^3+c*(-21)^2+d=0</math> and you can set these two equations equal to each other while also substituting the values of <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> above to give you <math>21m^2+21n-1682m+8000=0</math>, then you can rearrange the equation into <math>21n = -21m^2+1682m-8000</math>. With this property, we know that <math>-21m^2+1682m-8000</math> is divisible by <math>21</math> therefore that means <math>1682m-8000=0(mod 21)</math> which results in <math>2m-20=0(mod 21)</math> which finally gives us m=10 mod 21. We can test the first obvious value of <math>m</math> which is <math>10</math> and we see that this works as we get <math>m=10</math> and <math>n=320</math>. That means your answer will be <math>m + n = 10 + 320 = \boxed{330}</math> | start off by applying vieta's and you will find that <math>a=m^2+n-40m</math> <math>b=20m^2+20n</math> <math>c=21-2m</math> and <math>d=21m^2+21n</math>. After that, we have to use the fact that <math>-20</math> and <math>-21</math> are roots of <math>x^3+ax+b</math> and <math>x^3+cx^2+d</math>, respectively. Since we know that if you substitute the root of a function back into the function, the output is zero, therefore <math>(-20)^3-20(a)+b=0</math> and <math>(-21)^3+c*(-21)^2+d=0</math> and you can set these two equations equal to each other while also substituting the values of <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> above to give you <math>21m^2+21n-1682m+8000=0</math>, then you can rearrange the equation into <math>21n = -21m^2+1682m-8000</math>. With this property, we know that <math>-21m^2+1682m-8000</math> is divisible by <math>21</math> therefore that means <math>1682m-8000=0(mod 21)</math> which results in <math>2m-20=0(mod 21)</math> which finally gives us m=10 mod 21. We can test the first obvious value of <math>m</math> which is <math>10</math> and we see that this works as we get <math>m=10</math> and <math>n=320</math>. That means your answer will be <math>m + n = 10 + 320 = \boxed{330}</math> | ||
-Jske25 | -Jske25 | ||
+ | |||
==See also== | ==See also== | ||
{{AIME box|year=2021|n=II|num-b=3|num-a=5}} | {{AIME box|year=2021|n=II|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:01, 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
Solution 3 (Heavy Calculation Solution)
start off by applying vieta's and you will find that
and
. After that, we have to use the fact that
and
are roots of
and
, respectively. Since we know that if you substitute the root of a function back into the function, the output is zero, therefore
and
and you can set these two equations equal to each other while also substituting the values of
,
,
, and
above to give you
, then you can rearrange the equation into
. With this property, we know that
is divisible by
therefore that means
which results in
which finally gives us m=10 mod 21. We can test the first obvious value of
which is
and we see that this works as we get
and
. That means your answer will be
-Jske25
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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