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

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

@@ Line 1: / Line 1: @@
-Note: Please do not post problems here until after the AIME.
 == Problem ==
+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 1 (Strategic Casework)==
+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>20 + 6a + 2b \in [-10,10]</cmath>
+<cmath>3a + b \in [-15,-5].</cmath>
+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
+== Solution 2 (Bash) ==
+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>
+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>.
+-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*}
+&= 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                \\
+  & \{-10, \ldots, -5\}        &   -6a - 2b - 20  &      6                     \\
+  & \{-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 * ==
+In case anyone is confused by this (as I initially was). In the case where <math>f(2)=f(4)</math>, this does not mean that g has a double root of <math>f(2)=f(4)=k</math>, ONLY that <math>k</math> is one of the roots of g. So basically since <math>a=-6</math> in this case, <math>f(2)=f(4)=b-8</math>, and we have <math>21</math> choices for b and we <i>still can</i> 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 <math>b-8</math> ensures this, and of course an integer multiplied by an integer is an integer so <math>d</math> will still be an integer. In other words, you have can have <math>b</math> and <math>c</math> be any integer with absolute value less than or equal to 10 with <math>d</math> 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==
 {{AIME box|year=2020|n=I|num-b=10|num-a=12}}
 {{MAA Notice}}

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