Difference between revisions of "2021 AIME II Problems/Problem 4"
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==Solution 1 (Complex Conjugate Root Theorem and Vieta's Formulas)== | ==Solution 1 (Complex Conjugate Root Theorem and Vieta's Formulas)== | ||
− | By the | + | By the Complex Conjugate Root Theorem, the imaginary roots for each of <math>x^3+ax+b</math> and <math>x^3+cx^2+d</math> are 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> |
We know that | We know that |
Revision as of 02:08, 2 August 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 and Vieta's Formulas)
By the Complex Conjugate Root Theorem, the imaginary roots for each of and
are complex conjugates. Let
and
It follows that the roots of
are
and the roots of
are
We know that
Applying Vieta's Formulas to
we have
Substituting
into this equation, we get
Applying Vieta's Formulas to we have
or
Substituting
and
into this equation, we get
Finally, the answer is
~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
Solution 4 (Synthetic Division)
We note that and
for some polynomials
and
.
Through synthetic division (ignoring the remainder as we can set and
to constant values such that the remainder is zero),
, and
.
By the complex conjugate root theorem, we know that and
share the same roots, and they share the same leading coefficient, so
.
Therefore, and
. Solving the system of equations, we get
and
, so
.
Finally, by the quadratic formula, we have roots of , so our final answer is
-faefeyfa
Solution 5 (Fast and Easy)
We plug -20 into the equation obtaining , likewise, plugging -21 into the second equation gets
.
Both equations must have 3 solutions exactly, so the other two solutions must be and
.
By Vieta's, the sum of the roots in the first equation is , so
must be
.
Next, using Vieta's theorem on the second equation, you get: However, since we know that the sum of the roots with complex numbers are 20, we can factor out the terms with -21, so
Given that is
, then
is equal to
.
Therefore, the answer to the equation is
Video Solution
https://www.youtube.com/watch?v=sYRWWQayNyQ
Video Solution by TheCALT
https://www.youtube.com/watch?v=HJ0EldshLuE
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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