Difference between revisions of "2021 AIME II Problems/Problem 4"
Arnigam2007 (talk | contribs) (→Solution 3) |
Arnigam2007 (talk | contribs) (→Solution 3 (Somewhat Bashy)) |
||
| Line 12: | Line 12: | ||
==Solution 2== | ==Solution 2== | ||
==Solution 3 (Somewhat Bashy)== | ==Solution 3 (Somewhat Bashy)== | ||
| + | <math>(-20)^{3} + (-20)a + b = 0</math>, hence <math>-20a + b = 8000</math> | ||
| + | |||
| + | Also, <math>(-21)^{3} + c(-21)^{2} + d = 0</math>, hence <math>441c + d = 9261</math> | ||
| + | |||
| + | <math>m + i \sqrt{n}</math> | ||
==See also== | ==See also== | ||
{{AIME box|year=2021|n=II|num-b=3|num-a=5}} | {{AIME box|year=2021|n=II|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 01:25, 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 
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. 
