Difference between revisions of "2020 AIME I Problems/Problem 11"

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== Problem ==
 
== Problem ==
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For integers <math>a,b,c</math> and <math>d,</math> let <math>f(x)=x^2+ax+b</math> and <math>g(x)=x^2+cx+d.</math> Find the number of ordered triples <math>(a,b,c)</math> of integers with absolute values not exceeding <math>10</math> for which there is an integer <math>d</math> such that <math>g(f(2))=g(f(4))=0.</math>
  
 
== Solution ==
 
== Solution ==

Revision as of 16:36, 12 March 2020

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Problem

For integers $a,b,c$ and $d,$ let $f(x)=x^2+ax+b$ and $g(x)=x^2+cx+d.$ Find the number of ordered triples $(a,b,c)$ of integers with absolute values not exceeding $10$ for which there is an integer $d$ such that $g(f(2))=g(f(4))=0.$

Solution

Either $f(2)=f(4)$ or not. If it is, note that Vieta's forces $a = -6$. Then, $b$ can be anything. However, $c$ can also be anything, as we can set the root of $g$ (not equal to $f(2) = f(4)$) to any integer, producing a possible integer value of $d$. Therefore there are $21^2 = 441$ in this case. If it isn't, then $f(2),f(4)$ are the roots of $g$. This means by Vieta's, that:

\[f(2)+f(4) = -c \in [-10,10]\] \[20 + 6a + 2b \in [-10,10]\] \[3a + b \in [-15,5].\]

Solving these inequalities while considering that $a \neq -6$ to prevent $f(2) = f(4)$, we obtain $69$ possible tuples and adding gives $441+69=\boxed{510}$. ~awang11

See Also

2020 AIME I (ProblemsAnswer KeyResources)
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

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