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

Revision as of 01:05, 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 33: / Line 33: @@
 <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.
-<math>\begin{tabular}{|c|c|c|r|}
+<cmath>\begin{tabular}{|c|c|c|r|}
 \hline
-</math>a<math>  & </math>b<math>                  & </math>c<math> & triples \\ \hline
+a  & b                  & c & triples \\ \hline
-</math>-8<math> & </math>\{9,10\}<math>           & </math>-6a - 2b - 20<math>    &             </math>2<math>              \\
+-8 & \{9,10\}           & -6a - 2b - 20    &             2              \\
-</math>-7<math> & </math>\{6, 7, 8, 9, 10\}<math> &   </math>-6a - 2b - 20<math>  &         </math>5<math>                  \\
+-7 & \{6, 7, 8, 9, 10\} &   -6a - 2b - 20  &         5                  \\
-</math>-6<math> & </math>\{-10, \ldots, 10\}<math> &  </math>\{-10, \ldots, 10\}<math>   &                  </math>441 <math>       \\
+-6 & \{-10, \ldots, 10\} &  \{-10, \ldots, 10\}   &                  441        \\
-</math>-5<math> & </math>\{0, \ldots, 10\}<math>      &  </math>-6a - 2b - 20<math>   &    </math>11<math>                       \\
+-5 & \{0, \ldots, 10\}      &  -6a - 2b - 20   &    11                       \\
-</math>-4<math> & </math>\{-3, \ldots, 7\}<math>        &  </math>-6a - 2b - 20<math>   &   </math>11<math>                        \\
+-4 & \{-3, \ldots, 7\}        &  -6a - 2b - 20   &   11                       \\
-</math>-3<math> & </math>\{-6, \ldots, 4\}<math>        &  </math>-6a - 2b - 20<math>   &   </math>11<math>                        \\
+-3 & \{-6, \ldots, 4\}        &  -6a - 2b - 20   &   11                       \\
-</math>-2<math> & </math>\{-9, \ldots, 1\}<math>          &  </math>-6a - 2b - 20<math>   &   </math>11<math>                        \\
+-2 & \{-9, \ldots, 1\}         &  -6a - 2b - 20   &   11                        \\
-</math>-1<math> & </math>\{-10, \ldots, -2\}<math>   &  </math>-6a - 2b - 20<math>   &           </math>9<math>                \\
+-1 & \{-10, \ldots, -2\}   &  -6a - 2b - 20   &           9                \\
-</math>0<math>  & </math>\{-10, \ldots, -5\}<math>        &   </math>-6a - 2b - 20<math>  &      </math>6<math>                     \\
+  & \{-10, \ldots, -5\}        &   -6a - 2b - 20  &      6                     \\
-</math>1<math>  & </math>\{-10, -9, -8\}<math>        &  </math>-6a - 2b - 20<math>   &      </math>3<math>                     \\
+  & \{-10, -9, -8\}        &  -6a - 2b - 20   &      3                     \\
 \hline
 Total & & & 510\\
 \hline
-\end{tabular}</math>
+\end{tabular}</cmath>
 The total number of ordered triples that satisfy the required condition is <math>510</math>.

2020 AIME I (Problems • Answer Key • Resources)
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

Page

Toolbox

Search