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

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== Solution 1 (Strategic Casework)==
 
== Solution 1 (Strategic Casework)==
−
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:
+
There can be two different cases for this problem, either <math>f(2)=f(4)</math> or not. If it is, note that by Vieta's formulas <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:
  
 
<cmath>f(2)+f(4) = -c \in [-10,10]</cmath>
 
<cmath>f(2)+f(4) = -c \in [-10,10]</cmath>
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-EZmath2006
 
-EZmath2006
 +
 +
==Solution 3 (Official MAA)==
 +
For a given ordered triple <math>(a, b, c)</math>, the value of <math>g(f(4)) - g(f(2))</math> is uniquely determined, and a value of <math>d</math> can be found to give <math>g(f(2))</math> any prescribed integer value. Hence the required condition can be satisfied provided that <math>a</math>, <math>b</math>, and <math>c</math> are chosen so that
 +
<cmath>\begin{align*}
 +
0 &= g(f(4)) - g(f(2)) = \big(f(4)\big)^{\!2} - \big(f(2)\big)^{\!2} + c\big(f(4) - f(2)\big)\\
 +
&= \big(f(4) - f(2)\big)\big(f(4) + f(2) + c\big)\\
 +
&= (12 + 2a)(20 + 6a + 2b + c).
 +
\end{align*}</cmath>
 +
First suppose that <math>12 + 2a = 0</math>, so <math>a = -6</math>. In this case there are <math>21</math> choices for each of <math>b</math> and <math>c</math> with <math>-10 \le b \le 10</math> and <math>-10 \le c \le 10</math>, so this case accounts for <math>21^2 = 441</math> ordered triples.
 +
 +
Next suppose that <math>a \ne -6</math> and <math>20 + 6a + 2b + c = 0</math>, so <math>c = -6a - 2b - 20</math>. Because
 +
<math>-10 \le c \le 10</math>, it follows that <math>-30 \le 6a + 2b \le -10</math>, and because <math>-10\le b \le10</math>, it follows that <math>-8 \le a \le 1</math>. Then
 +
<math>-15 - 3a \le b \le -5 - 3a</math>. The number of ordered triples for various values of <math>a</math> are presented in the following table.
 +
 +
<cmath>\begin{tabular}{|c|c|c|r|}
 +
\hline
 +
a  & b                  & c & triples \\ \hline
 +
-8 & \{9,10\}          & -6a - 2b - 20    &            2              \\
 +
-7 & \{6, 7, 8, 9, 10\} &  -6a - 2b - 20  &        5                  \\
 +
-6 & \{-10, \ldots, 10\} &  \{-10, \ldots, 10\}  &                  441        \\
 +
-5 & \{0, \ldots, 10\}      &  -6a - 2b - 20  &    11                      \\
 +
-4 & \{-3, \ldots, 7\}        &  -6a - 2b - 20  &  11                      \\
 +
-3 & \{-6, \ldots, 4\}        &  -6a - 2b - 20  &  11                      \\
 +
-2 & \{-9, \ldots, 1\}        &  -6a - 2b - 20  &  11                        \\
 +
-1 & \{-10, \ldots, -2\}  &  -6a - 2b - 20  &          9                \\
 +
0  & \{-10, \ldots, -5\}        &  -6a - 2b - 20  &      6                    \\
 +
1  & \{-10, -9, -8\}        &  -6a - 2b - 20  &      3                    \\
 +
\hline
 +
Total & & & 510\\
 +
\hline
 +
\end{tabular}</cmath>
 +
The total number of ordered triples that satisfy the required condition is <math>510</math>.
  
 
== Notes For * ==
 
== Notes For * ==
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~First
 
~First
  
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=ftqYFzzWKv8&list=PLLCzevlMcsWN9y8YI4KNPZlhdsjPTlRrb&index=8 ~ MathEx
 
==See Also==
 
==See Also==
  
 
{{AIME box|year=2020|n=I|num-b=10|num-a=12}}
 
{{AIME box|year=2020|n=I|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:33, 18 January 2024

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 1 (Strategic Casework)

There can be two different cases for this problem, either $f(2)=f(4)$ or not. If it is, note that by Vieta's formulas $a = -6$. 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

Solution 3 (Official MAA)

For a given ordered triple $(a, b, c)$, the value of $g(f(4)) - g(f(2))$ is uniquely determined, and a value of $d$ can be found to give $g(f(2))$ any prescribed integer value. Hence the required condition can be satisfied provided that $a$, $b$, and $c$ are chosen so that \begin{align*} 0 &= g(f(4)) - g(f(2)) = \big(f(4)\big)^{\!2} - \big(f(2)\big)^{\!2} + c\big(f(4) - f(2)\big)\\ &= \big(f(4) - f(2)\big)\big(f(4) + f(2) + c\big)\\ &= (12 + 2a)(20 + 6a + 2b + c). \end{align*} First suppose that $12 + 2a = 0$, so $a = -6$. In this case there are $21$ choices for each of $b$ and $c$ with $-10 \le b \le 10$ and $-10 \le c \le 10$, so this case accounts for $21^2 = 441$ ordered triples.

Next suppose that $a \ne -6$ and $20 + 6a + 2b + c = 0$, so $c = -6a - 2b - 20$. Because $-10 \le c \le 10$, it follows that $-30 \le 6a + 2b \le -10$, and because $-10\le b \le10$, it follows that $-8 \le a \le 1$. Then $-15 - 3a \le b \le -5 - 3a$. The number of ordered triples for various values of $a$ are presented in the following table.

\[\begin{tabular}{|c|c|c|r|} \hline a & b & c & triples \\ \hline -8 & \{9,10\} & -6a - 2b - 20 & 2 \\ -7 & \{6, 7, 8, 9, 10\} & -6a - 2b - 20 & 5 \\ -6 & \{-10, \ldots, 10\} & \{-10, \ldots, 10\} & 441 \\ -5 & \{0, \ldots, 10\} & -6a - 2b - 20 & 11 \\ -4 & \{-3, \ldots, 7\} & -6a - 2b - 20 & 11 \\ -3 & \{-6, \ldots, 4\} & -6a - 2b - 20 & 11 \\ -2 & \{-9, \ldots, 1\} & -6a - 2b - 20 & 11 \\ -1 & \{-10, \ldots, -2\} & -6a - 2b - 20 & 9 \\ 0 & \{-10, \ldots, -5\} & -6a - 2b - 20 & 6 \\ 1 & \{-10, -9, -8\} & -6a - 2b - 20 & 3 \\ \hline Total & & & 510\\ \hline \end{tabular}\] The total number of ordered triples that satisfy the required condition is $510$.

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 $f(2)=f(4)=k$, ONLY that $k$ 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

Video Solution

https://www.youtube.com/watch?v=ftqYFzzWKv8&list=PLLCzevlMcsWN9y8YI4KNPZlhdsjPTlRrb&index=8 ~ MathEx

See Also

2020 AIME I (Problems • Answer Key • Resources)
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Problem 10 		Followed by
Problem 12
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All AIME Problems and Solutions

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