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

(Problem (Unofficial, please update when official one comes out):)
(Solution)
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<math>m \neq 2</math> with <math>p(m) = p(2)</math>.
 
<math>m \neq 2</math> with <math>p(m) = p(2)</math>.
  
==Solution==
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Find the number of cubic polynomials <math>p(x) = x^3 + ax^2 + bx + c</math>, where <math>a</math>, <math>b</math>, and <math>c</math>
===Solution 1===
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are integers in <math>{-20, -10, -18, . . . , 18, 19, 20}</math>, such that there is a unique integer
 
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<math>m \neq 2</math> with <math>p(m) = p(2)</math>.
Plugging <math>2</math> into <math>P(x)</math>, we get <math>8+4a+2b+c = m^3+am^2+bm+c</math>. We can rewrite into <math>(2-m)(m^2+2m+4+a(2+m)+b)=0</math>, where <math>c</math> can be any value in the range. Since <math>m\neq2, m^2+2m+4+a(2+m)+b</math> must be <math>0</math>. The problem also asks for unique integers, meaning <math>m</math> can only be one value for each polynomial, so the discriminant must be <math>0</math>. <math>m^2+2m+4+a(2+m)+b = m^2+m(2+a)+(2a+b+4)= 0</math>, and <math>(2+a)^2-4(2a+b+4)=0</math>. Rewrite to be <math>a(a-4)=4(b+3)</math>. <math>a</math> must be even for <math>4(b+3)</math> to be an integer. <math>-6<=a<=10</math> because <math>4(20+3) = 92</math>. However, plugging in <math>a=-6, b=12</math> result in <math>m=2</math>. There are 8 pairs of <math>(a,b)</math> and 41 integers for <math>c</math>, giving <cmath>41\cdot8 = \boxed{328}</cmath>
 
 
 
~chem1kall
 
  
 
==Solution==
 
==Solution==

Revision as of 15:49, 8 February 2023

Find the number of cubic polynomials $p(x) = x^3 + ax^2 + bx + c$, where $a$, $b$, and $c$ are integers in ${−20, −10, −18, . . . , 18, 19, 20}$ (Error compiling LaTeX. Unknown error_msg), such that there is a unique integer $m \neq 2$ with $p(m) = p(2)$.

Find the number of cubic polynomials $p(x) = x^3 + ax^2 + bx + c$, where $a$, $b$, and $c$ are integers in ${-20, -10, -18, . . . , 18, 19, 20}$, such that there is a unique integer $m \neq 2$ with $p(m) = p(2)$.

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 $(a,b,c)$.


-Alex_Z