Difference between revisions of "2020 AIME I Problems/Problem 11"
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== Problem == | == Problem == | ||
| − | == Solution == | + | == Solution ==Either <math>f(2)=f(4)</math> or not. If it is, note that Vieta's forces <math>a = -6</math>. Then, <math>b</math> can be anything. However, <math>c</math> can also be anything, as we can set the root of <math>g</math> (not equal to <math>f(2) = f(4)</math>) to any integer, producing a possible integer value of <math>d</math>. Therefore there are <math>21^2 = 441</math> in this case. If it isn't, then <math>f(2),f(4)</math> are the roots of <math>g</math>. This means by Vieta's, that: |
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| + | <cmath>f(2)+f(4) = -c \in [-10,10]</cmath> | ||
| + | <cmath>20 + 6a + 2b \in [-10,10]</cmath> | ||
| + | <cmath>3a + b \in [-15,5].</cmath> | ||
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| + | Solving these inequalities while considering that <math>a \neq -6</math> to prevent <math>f(2) = f(4)</math>, we obtain <math>69</math> possible tuples and adding gives <math>441+69=\boxed{510}</math>. ~awang11 | ||
==See Also== | ==See Also== | ||
Revision as of 17:23, 12 March 2020
Note: Please do not post problems here until after the AIME.
Problem
== Solution ==Either 
or not. If it is, note that Vieta's forces 
. Then, 
can be anything. However, 
can also be anything, as we can set the root of 
(not equal to 
) to any integer, producing a possible integer value of 
. Therefore there are 
in this case. If it isn't, then 
are the roots of . This means by Vieta's, that:
![\[f(2)+f(4) = -c \in [-10,10]\]](http://latex.artofproblemsolving.com/6/a/3/6a37efd5cf28128ffc86292858c5295324219916.png)
![\[20 + 6a + 2b \in [-10,10]\]](http://latex.artofproblemsolving.com/6/a/2/6a21e367c68e23d262ef4c634a2cac27eadf144c.png)
![\[3a + b \in [-15,5].\]](http://latex.artofproblemsolving.com/9/5/5/95555d558a85629834280abcce4f6c78c0a6cc4c.png)
Solving these inequalities while considering that 
to prevent , we obtain 
possible tuples and adding gives 
. ~awang11
See Also
| 2020 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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. 
