Difference between revisions of "2021 AIME II Problems/Problem 4"
Arnigam2007 (talk | contribs) (→Solution 3 (Somewhat Bashy)) |
Arnigam2007 (talk | contribs) (→Solution 3 (Somewhat Bashy)) |
||
Line 22: | Line 22: | ||
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 | ||
Line 28: | Line 28: | ||
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 01:40, 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.