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

(→‎Solution 1)
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===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?
 
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?
 +
 +
==Solution==
 +
 +
Define <math>q \left( x \right) = p \left( x \right) - p \left( 2 \right)</math>.
 +
Hence, for <math>q \left( x \right)</math>, 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) \\
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& = \left( x - 2 \right)
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\left( x^2 + \left( 2 + a \right) x + 4 + 2a + b \right) .
 +
\end{align*}
 +
 +
Thus, the polynomial <math>x^2 + \left( 2 + a \right) x + 4 + 2a + b</math> 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 <math>- 2 - a</math>.
 +
Because <math>a</math> 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, <math>2 | a</math>. Denote <math>d = \frac{a-2}{2}</math>.
 +
Thus, (2) can be written as
 +
\[
 +
b = d^2 - 4 .  \hspace{1cm} (3)
 +
\]
 +
 +
Because <math>a \in \left\{ -20, -19, -18, \cdots , 18, 19, 20 \right\}</math>, <math>2 | a</math>, and <math>2 + a \neq -4</math>, we have <math>d \in \left\{ - 11, - 10, \cdots, 9 \right\} \backslash \left\{ 4 \right\}</math>.
 +
 +
Therefore, we have the following feasible solutions for <math>\left( b, d \right)</math>: <math>\left( -4 , 0 \right)</math>, <math>\left( -3 , \pm 1 \right)</math>, <math>\left( 0 , \pm 2 \right)</math>, <math>\left( 5, \pm 3 \right)</math>, <math>\left( 12 , 4 \right)</math>.
 +
Thus, the total number of <math>\left( b, d \right)</math> is 8.
 +
 +
Because <math>c</math> can take any value from <math>\left\{ -20, -19, -18, \cdots , 18, 19, 20 \right\}</math>, the number of feasible <math>c</math> is 41.
 +
 +
Therefore, the number of <math>\left( a, b, c \right)</math> is <math>8 \cdot 41 = \boxed{\textbf{(328) }}</math>.
 +
 +
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Revision as of 15:22, 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

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?

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)

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