2021 AIME II Problems/Problem 4
Contents
- 1 Problem
- 2 Solution 1 (Complex Conjugate Root Theorem and Vieta's Formulas)
- 3 Solution 2 (Somewhat Bashy)
- 4 Solution 3 (Heavy Calculation Solution)
- 5 Solution 4 (Synthetic Division)
- 6 Solution 5 (Fast and Easy)
- 7 Solution 6 (solution by integralarefun)
- 8 Video Solution
- 9 Video Solution by TheCALT
- 10 See Also
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
Solution 6 (solution by integralarefun)
Since is a common root and all the coefficients are real, must be a common root, too.
Now that we know all three roots of both polynomials, we can match coefficients(or more specifically, the zero coefficients).
First, however, the product of the two common roots is:
Now, let's equate the two forms of both the polynomials: Now we can match the zero coefficients. Thus, .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.