Difference between revisions of "2021 AIME II Problems/Problem 4"
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Simplifying this further, we get <math>(m^{3} - 3mn + am + b) + i(3m^{2} \sqrt{n} - n\sqrt{n} + a\sqrt{n}) = 0</math> | Simplifying this further, we get <math>(m^{3} - 3mn + am + b) + i(3m^{2} \sqrt{n} - n\sqrt{n} + a\sqrt{n}) = 0</math> | ||
− | Hence, <math>m^{3} - 3mn + am + b = 0</math> and <math>3m^{2} \sqrt{n} - n\sqrt{n} + a\sqrt{n} = 0 \Rightarrow 3m^{2} - n + a = 0 | + | Hence, <math>m^{3} - 3mn + am + b = 0</math> and <math>3m^{2} \sqrt{n} - n\sqrt{n} + a\sqrt{n} = 0 \Rightarrow 3m^{2} - n + a = 0 \rightarrow (1)</math> |
In the second equation, we get | In the second equation, we get | ||
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Simplifying this further, we get <math>(m^{3} + m^{2}c - nc - 3mn + d) + i(3m^{2} \sqrt{n} - n\sqrt{n} + 2mc\sqrt{n}) = 0</math> | Simplifying this further, we get <math>(m^{3} + m^{2}c - nc - 3mn + d) + i(3m^{2} \sqrt{n} - n\sqrt{n} + 2mc\sqrt{n}) = 0</math> | ||
− | Hence, <math>m^{3} + m^{2}c - nc - 3mn + d = 0</math> and <math>3m^{2} \sqrt{n} - n\sqrt{n} + 2mc\sqrt{n} = 0 \Rightarrow 3m^{2} - n + 2mc = 0 | + | Hence, <math>m^{3} + m^{2}c - nc - 3mn + d = 0</math> and <math>3m^{2} \sqrt{n} - n\sqrt{n} + 2mc\sqrt{n} = 0 \Rightarrow 3m^{2} - n + 2mc = 0 \rightarrow (2)</math> |
-Arnav Nigam | -Arnav Nigam |
Revision as of 02:40, 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
Conjugate root theorem
Solution in progress
~JimY
Solution 2
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
-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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.