Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2023 AIME I Problems/Problem 9"

(→‎Solution)
(→‎Solution 2)
Line 11: Line 11:
  
 
===Solution 2===
 
===Solution 2===
−
a=-10 and a=-8 give values for b outside the range and a=-6 results in m=2. Therefore shouldn't the answer be 41*8=328?
+
For a=-6, we get m=2 using the quadratic formula -- the polynomial is (m-2)^3=0. Therefore shouldn't it be excluded, giving an answer of 41*8=328?
  
 
==Solution==
 
==Solution==

Revision as of 14:23, 8 February 2023

Problem (Unofficial, please update when official one comes out):

$P(x) = x^3 + ax^2 + bx + c$ is a polynomial with integer coefficients in the range$[-20, -19, -18\cdots 18, 19, 20]$, inclusive. There is exactly one integer $m \neq 2$ such that $P(m) = P(2)$. How many possible values are there for the ordered triple $(a, b, c)$?

Solution

Solution 1

Plugging $2$ into $P(x)$, we get $8+4a+2b+c = m^3+am^2+bm+c$. We can rewrite into $(2-m)(m^2+2m+4+a(2+m)+b)=0$, where $c$ can be any value in the range. Since $m\neq2, m^2+2m+4+a(2+m)+b$ must be $0$. The problem also asks for unique integers, meaning $m$ can only be one value for each polynomial, so the discriminant must be $0$. $m^2+2m+4+a(2+m)+b = m^2+m(2+a)+(2a+b+4)= 0$, and $(2+a)^2-4(2a+b+4)=0$. Rewrite to be $a(a-4)=4(b+3)$. $a$ must be even for $4(b+3)$ to be an integer. $-6<=a<=10$ because $4(20+3) = 92$. There are 9 pairs of $(a,b)$ and 41 integers for $c$, giving \[41\cdot9 = \boxed{369}\]

~chem1kall

Solution 2

For a=-6, we get m=2 using the quadratic formula -- the polynomial is (m-2)^3=0. Therefore shouldn't it be excluded, giving an answer of 41*8=328?

Solution

Define $q \left( x \right) = p \left( x \right) - p \left( 2 \right)$. Hence, for $q \left( x \right)$, beyond having a root 2, it has a unique integer root that is not equal to 2.

We have \begin{align*} q \left( x \right) & = p \left( x \right) - p \left( 2 \right) \\ & = \left( x - 2 \right) \left( x^2 + \left( 2 + a \right) x + 4 + 2a + b \right) . \end{align*}

Thus, the polynomial $x^2 + \left( 2 + a \right) x + 4 + 2a + b$ has a unique integer root and it is not equal to 2.

Following from Vieta' formula, the sum of two roots of this polynomial is $- 2 - a$. Because $a$ is an integer, if a root is an integer, the other root is also an integer. Therefore, the only way to have a unique integer root is that the determinant of this polynomial is 0. Thus, \[ \left( 2 + a \right)^2 = 4 \left( 4 + 2a + b \right) . \hspace{1cm} (1) \]

In addition, because two identical roots are not 2, we have \[ 2 + a \neq - 4 . \]

Equation (1) can be reorganized as \[ 4 b = \left( a - 2 \right)^2 - 16 . \hspace{1cm} (2) \]

Thus, $2 | a$. Denote $d = \frac{a-2}{2}$. Thus, (2) can be written as \[ b = d^2 - 4 . \hspace{1cm} (3) \]

Because $a \in \left\{ -20, -19, -18, \cdots , 18, 19, 20 \right\}$, $2 | a$, and $2 + a \neq -4$, we have $d \in \left\{ - 11, - 10, \cdots, 9 \right\} \backslash \left\{ 4 \right\}$.

Therefore, we have the following feasible solutions for $\left( b, d \right)$: $\left( -4 , 0 \right)$, $\left( -3 , \pm 1 \right)$, $\left( 0 , \pm 2 \right)$, $\left( 5, \pm 3 \right)$, $\left( 12 , 4 \right)$. Thus, the total number of $\left( b, d \right)$ is 8.

Because $c$ can take any value from $\left\{ -20, -19, -18, \cdots , 18, 19, 20 \right\}$, the number of feasible $c$ is 41.

Therefore, the number of $\left( a, b, c \right)$ is $8 \cdot 41 = \boxed{\textbf{(328) }}$.


~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AIME_I_Problems/Problem_9&oldid=188995"