2020 AIME I Problems/Problem 11

Revision as of 15:43, 22 May 2020 by Oreochocolate (talk | contribs) (Notes For *: For bold, italic, underlining, or Asymptote, instead of using brackets, like usual, use <> instead, but only works here.)


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 1 (Strategic Casework)

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

Solution 2 (Bash)

Define $h(x)=x^2+cx$. Since $g(f(2))=g(f(4))=0$, we know $h(f(2))=h(f(4))=-d$. Plugging in $f(x)$ into $h(x)$, we get $h(f(x))=x^4+2ax^3+(2b+a^2+c)x^2+(2ab+ac)x+(b^2+bc)$. Setting $h(f(2))=h(f(4))$, \[16+16a+8b+4a^2+4ab+b^2+4c+2ac+bc=256+128a+32b+16a^2+8ab+b^2+16c+4ac+bc\]. Simplifying and cancelling terms, \[240+112a+24b+12a^2+4ab+12c+2ac=0\] \[120+56a+12b+6a^2+2ab+6c+ac=0\] \[6a^2+2ab+ac+56a+12b+6c+120=0\] \[6a^2+2ab+ac+20a+36a+12b+6c+120=0\] \[a(6a+2b+c+20)+6(6a+2b+c+20)=0\] \[(a+6)(6a+2b+c+20)=0\]

Therefore, either $a+6=0$ or $6a+2b+c=-20$. The first case is easy: $a=-6$ and there are $441$ tuples in that case. In the second case, we simply perform casework on even values of $c$, to get $77$ tuples, subtracting the $8$ tuples in both cases we get $441+77-8=\boxed{510}$.


Notes For *

In case anyone is confused by this (as I initially was). In the case where $f(2)=f(4)$, this does not mean that g has a double root of $f(2)=f(4)=c$, ONLY that $c$ is one of the roots of g. So basically since $a=-6$ in this case, $f(2)=f(4)=b-8$, and we have $21$ choices for b and we still can ensure c is an integer with absolute value less than or equal to 10 simply by having another integer root of g that when added to $b-8$ ensures this, and of course an integer multiplied by an integer is an integer so $d$ will still be an integer. In other words, you have can have $b$ and $c$ be any integer with absolute value less than or equal to 10 with $d$ still being an integer. Now refer back to the 1st solution. ~First

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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