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

(Solution 3 (Official MAA))
(Solution 3 (Official MAA))
 
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<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>-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
 
\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>                     \\
+
0  & \{-10, \ldots, -5\}        &  -6a - 2b - 20  &      6                    \\
</math>1<math> & </math>\{-10, -9, -8\}<math>       &  </math>-6a - 2b - 20<math>   &      </math>3<math>                     \\
+
1  & \{-10, -9, -8\}        &  -6a - 2b - 20  &      3                    \\
 
\hline
 
\hline
 
Total & & & 510\\
 
Total & & & 510\\
 
\hline
 
\hline
\end{tabular}</math>
+
\end{tabular}</cmath>
 
The total number of ordered triples that satisfy the required condition is <math>510</math>.
 
The total number of ordered triples that satisfy the required condition is <math>510</math>.
  

Latest 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

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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