2021 AIME II Problems/Problem 4
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[hide]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
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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