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

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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
 
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
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== Solution 2 (Bash) ==
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Define <math>h(x)=x^2+cx</math>. Since <math>g(f(2))=g(f(4))=0</math>, we know <math>h(f(2))=h(f(4))=-d</math>. Plugging in <math>f(x)</math> into <math>h(x)</math>, we get <math>h(f(x))=x^4+2ax^3+(2b+a^2+c)x^2+(2ab+ac)x+(b^2+bc)</math>. Setting <math>h(f(2))=h(f(4))</math>, <cmath>16+16a+8b+4a^2+4ab+b^2+4c+2ac+bc=256+128a+32b+16a^2+8ab+b^2+16c+4ac+bc</cmath>. Simplifying and cancelling terms, <cmath>240+112a+24b+12a^2+4ab+12c+2ac=0</cmath> <cmath>120+56a+12b+6a^2+2ab+6c+ac=0</cmath> <cmath>6a^2+2ab+ac+56a+12b+6c+120=0</cmath> <cmath>6a^2+2ab+ac+20a+36a+12b+6c+120=0</cmath> <cmath>a(6a+2b+c+20)+6(6a+2b+c+20)=0</cmath> <cmath>(a+6)(6a+2b+c+20)=0</cmath>
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Therefore, either <math>a+6=0</math> or <math>6a+2b+c=-20</math>. The first case is easy: <math>a=-6</math> and there are <math>441</math> tuples in that case. In the second case, we simply perform casework on even values of <math>c</math>, to get <math>77</math> tuples, subtracting the <math>8</math> tuples in both cases we get <math>441+77-8=\boxed{510}</math>.
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-EZmath2006
  
 
==See Also==
 
==See Also==

Revision as of 21:45, 12 March 2020

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

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 $441+77-8=\boxed{510}$.

-EZmath2006

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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