Difference between revisions of "2021 AIME II Problems/Problem 4"
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==Solution 3 (Somewhat Bashy)== | ==Solution 3 (Somewhat Bashy)== |
Revision as of 03:59, 23 March 2021
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
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 3 (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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