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

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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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==Solution==
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It can be easily noticed that <math>c</math> is independent of the condition <math>P(m) = P(2)</math>, and can thus safely take all <math>41</math> possible values between <math>-20</math> and <math>20</math>.
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There are two possible ways for <math>m\ne2</math> to be the only integer satisfying <math>P(m) = P(2)</math>: <math>P</math> has a double root at <math>2</math> or a double root at <math>m</math>.
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Case 1: <math>P</math> has a double root at <math>2</math>:
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In this case, <math>\frac{dP}{dx}(2) = 0</math>, or <math>12 + 4a + b = 0</math>. Thus <math>a</math> ranges from <math>-8</math> to <math>2</math>. One of these values, <math>(a,b) = (-6,-12)</math> corresponds to a triple root at <math>2</math>, which means <math>m=2</math>. Thus there are <math>10</math> possible values of <math>(a,b)</math>. (It can be verified that <math>m</math> is an integer).
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Case 2: <math>P</math> has a double root at <math>m</math>:
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See the above solution. This yields <math>8</math> possible combinations of <math>a</math> and <math>b</math>.
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Thus, in total we have <math>41*18 = \boxed{738}</math> combinations of <math>(a,b,c)</math>.
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(Checked by computer for all values of <math>-50 < m < 50</math>.)
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-Alex_Z

Revision as of 14:59, 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 $4(20+3) = 92$. However, plugging in $a=-6, b=12$ result in $m=2$. There are 8 pairs of $(a,b)$ and 41 integers for $c$, giving \[41\cdot8 = \boxed{328}\]

~chem1kall

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)

Solution

It can be easily noticed that $c$ is independent of the condition $P(m) = P(2)$, and can thus safely take all $41$ possible values between $-20$ and $20$.

There are two possible ways for $m\ne2$ to be the only integer satisfying $P(m) = P(2)$: $P$ has a double root at $2$ or a double root at $m$.

Case 1: $P$ has a double root at $2$:

In this case, $\frac{dP}{dx}(2) = 0$, or $12 + 4a + b = 0$. Thus $a$ ranges from $-8$ to $2$. One of these values, $(a,b) = (-6,-12)$ corresponds to a triple root at $2$, which means $m=2$. Thus there are $10$ possible values of $(a,b)$. (It can be verified that $m$ is an integer).

Case 2: $P$ has a double root at $m$:

See the above solution. This yields $8$ possible combinations of $a$ and $b$.

Thus, in total we have $41*18 = \boxed{738}$ combinations of $(a,b,c)$.

(Checked by computer for all values of $-50 < m < 50$.)


-Alex_Z

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