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

Revision as of 00:01, 25 February 2021

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)

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

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 19: / Line 19: @@
 -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
+\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*}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.
+\begin{tabular}{|c|c|c|r|}
+\hline
+<math>a</math>  & <math>b</math>                  & <math>c</math> & triples \\ \hline
+<math>-8</math> & <math>\{9,10\}</math>           & <math>-6a - 2b - 20</math>    &             <math>2</math>              \\
+<math>-7</math> & <math>\{6, 7, 8, 9, 10\}</math> &   <math>-6a - 2b - 20</math>  &         <math>5</math>                  \\
+<math>-6</math> & <math>\{-10, \ldots, 10\}</math> &  <math>\{-10, \ldots, 10\}</math>   &                  <math>441 </math>       \\
+<math>-5</math> & <math>\{0, \ldots, 10\}</math>      &  <math>-6a - 2b - 20</math>   &    <math>11</math>                       \\
+<math>-4</math> & <math>\{-3, \ldots, 7\}</math>        &  <math>-6a - 2b - 20</math>   &   <math>11</math>                        \\
+<math>-3</math> & <math>\{-6, \ldots, 4\}</math>        &  <math>-6a - 2b - 20</math>   &   <math>11</math>                        \\
+<math>-2</math> & <math>\{-9, \ldots, 1\}</math>          &  <math>-6a - 2b - 20</math>   &   <math>11</math>                        \\
+<math>-1</math> & <math>\{-10, \ldots, -2\}</math>   &  <math>-6a - 2b - 20</math>   &           <math>9</math>                \\
+<math>0</math>  & <math>\{-10, \ldots, -5\}</math>        &   <math>-6a - 2b - 20</math>  &      <math>6</math>                     \\
+<math>1</math>  & <math>\{-10, -9, -8\}</math>        &  <math>-6a - 2b - 20</math>   &      <math>3</math>                     \\
+\hline
+Total & & & 510\\
+\hline
+\end{tabular}The total number of ordered triples that satisfy the required condition is <math>510</math>.
 == Notes For * ==

2020 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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