2020 AIME I Problems/Problem 11

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

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
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2020 AIME I Problems/Problem 11

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