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Difference between revisions of "2023 AIME I Problems/Problem 9"
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are integers in <math>\{ -20, -19, -18, \dots , 18, 19, 20 \}</math>, such that there is a unique integer
 
are integers in <math>\{ -20, -19, -18, \dots , 18, 19, 20 \}</math>, such that there is a unique integer
 
<math>m \neq 2</math> with <math>p(m) = p(2).</math>
 
<math>m \neq 2</math> with <math>p(m) = p(2).</math>
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==Solution==
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Plugging <math>2</math> and <math>m</math> into <math>P(x)</math> and equating them, we get <math>8+4a+2b+c = m^3+am^2+bm+c</math>. Rearranging, we have <cmath>(m^3-8) + (m^2 - 4)a + (m-2)b = 0.</cmath> Note that the value of <math>c</math> won't matter as it can be anything in the provided range, giving a total of <math>41</math> possible choices for <math>c.</math> So what we just need to do is to just find the number of ordered pairs <math>(a, b)</math> that work, and multiply it by <math>41.</math>
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We can start by first dividing both sides by <math>m-2.</math> (Note that this is valid since <math>m\neq2:</math> <cmath>m^2 + 2m + 4 + (m+2)a + b = 0.</cmath> We can rearrange this so it is a quadratic in <math>m</math>: <cmath>m^2 + (a+2)m + (4 + 2a + b) = 0.</cmath> Remember that <math>m</math> has to be unique and not equal to <math>2.</math> We can split this into two cases: case <math>1</math> being that <math>m</math> has exactly one solution, and it isn't equal to <math>2</math>; case <math>2</math> being that <math>m</math> has two solutions, one being equal to <math>2,</math> but the other is a unique solution not equal to <math>2.</math>
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<math>\textbf{Case 1:}</math>
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There is exactly one solution for <math>m,</math> and that solution is not <math>2.</math> This means that the discriminant of the quadratic equation is <math>0,</math> using that, we have <math>(a+2)^2 = 4(4 + 2a + b),</math> rearranging in a neat way, we have <cmath>(a-2)^2 = 4(4 + b)\Longrightarrow a = 2\pm2sqrt{4+b}.</cmath> Using the fact that <math>4+b</math> must be a perfect square, we can easily see that the values for <math>b</math> can be <math>-4, -3, 0, 5,</math> and <math>12.</math> Also since it's a "<math>\pm</math>" there will usually be <math>2</math> solutions for <math>a</math> for each value of <math>b.</math> The two exceptions for this would be if <math>b = -4</math> and <math>b = 12.</math> For <math>b=-4</math> because it would be a <math>\pm0,</math> which only gives one solution, instead of two. And for <math>b=12</math> because then <math>a = -6</math> and the solution for <math>m</math> would equal to <math>2,</math> and we don't want this. (We can know this by putting the solutions back into the quadratic formula).
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So we have <math>5</math> solutions for <math>b,</math> each of which give <math>2</math> values for <math>a,</math> except for <math>2,</math> which only give one. So in total, there are <math>5*10 - 2 = 8</math> ordered pairs of <math>(a,b)</math> in this case.
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<math>\textbf{Case 2:}</math>
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<math>m</math> has two solutions, but exactly one of them isn't equal to <math>2.</math> This ensures that <math>1</math> of the solutions is equal to <math>2.</math>
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Let <math>r</math> be the other value of <math>m</math> that isn't <math>2.</math> By Vieta: <cmath>r + 2 = -a - 2</cmath> <cmath>2r = 4 + 2a + b.</cmath> From the first equation, we subtract both sides by <math>2</math> and double both sides to get <math>2r = -2a - 8</math> which also equals to <math>4+2a+b</math> from the second equation. Equating both, we have <math>4a + b + 12 = 0.</math> We can easily count that there would be <math>11</math> ordered pairs <math>(a,b)</math> that satisfy that.
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However, there's an outlier case in which <math>r</math> happens to also equal to <math>2,</math> and we don't want that. We can reverse engineer and find out that <math>r=2</math> when <math>(a,b) = (-6, 12),</math> which we overcounted. So we subtract by one and we conclude that there are <math>10</math> ordered pairs of <math>(a,b)</math> that satisfy this case.
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This all shows that there are a total of <math>8+10 = 18</math> amount of ordered pairs <math>(a,b).</math> Multiplying this by <math>41</math> (the amount of values for <math>c</math>) we get <math>18\cdot41=\boxed{738}</math> as our final answer.
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~s214425
  
 
==Solution 1==
 
==Solution 1==

Revision as of 17:23, 8 February 2023

Problem 9

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

Solution

Plugging $2$ and $m$ into $P(x)$ and equating them, we get $8+4a+2b+c = m^3+am^2+bm+c$. Rearranging, we have \[(m^3-8) + (m^2 - 4)a + (m-2)b = 0.\] Note that the value of $c$ won't matter as it can be anything in the provided range, giving a total of $41$ possible choices for $c.$ So what we just need to do is to just find the number of ordered pairs $(a, b)$ that work, and multiply it by $41.$ We can start by first dividing both sides by $m-2.$ (Note that this is valid since $m\neq2:$ \[m^2 + 2m + 4 + (m+2)a + b = 0.\] We can rearrange this so it is a quadratic in $m$: \[m^2 + (a+2)m + (4 + 2a + b) = 0.\] Remember that $m$ has to be unique and not equal to $2.$ We can split this into two cases: case $1$ being that $m$ has exactly one solution, and it isn't equal to $2$; case $2$ being that $m$ has two solutions, one being equal to $2,$ but the other is a unique solution not equal to $2.$


$\textbf{Case 1:}$

There is exactly one solution for $m,$ and that solution is not $2.$ This means that the discriminant of the quadratic equation is $0,$ using that, we have $(a+2)^2 = 4(4 + 2a + b),$ rearranging in a neat way, we have \[(a-2)^2 = 4(4 + b)\Longrightarrow a = 2\pm2sqrt{4+b}.\] Using the fact that $4+b$ must be a perfect square, we can easily see that the values for $b$ can be $-4, -3, 0, 5,$ and $12.$ Also since it's a "$\pm$" there will usually be $2$ solutions for $a$ for each value of $b.$ The two exceptions for this would be if $b = -4$ and $b = 12.$ For $b=-4$ because it would be a $\pm0,$ which only gives one solution, instead of two. And for $b=12$ because then $a = -6$ and the solution for $m$ would equal to $2,$ and we don't want this. (We can know this by putting the solutions back into the quadratic formula).

So we have $5$ solutions for $b,$ each of which give $2$ values for $a,$ except for $2,$ which only give one. So in total, there are $5*10 - 2 = 8$ ordered pairs of $(a,b)$ in this case.


$\textbf{Case 2:}$

$m$ has two solutions, but exactly one of them isn't equal to $2.$ This ensures that $1$ of the solutions is equal to $2.$

Let $r$ be the other value of $m$ that isn't $2.$ By Vieta: \[r + 2 = -a - 2\] \[2r = 4 + 2a + b.\] From the first equation, we subtract both sides by $2$ and double both sides to get $2r = -2a - 8$ which also equals to $4+2a+b$ from the second equation. Equating both, we have $4a + b + 12 = 0.$ We can easily count that there would be $11$ ordered pairs $(a,b)$ that satisfy that.

However, there's an outlier case in which $r$ happens to also equal to $2,$ and we don't want that. We can reverse engineer and find out that $r=2$ when $(a,b) = (-6, 12),$ which we overcounted. So we subtract by one and we conclude that there are $10$ ordered pairs of $(a,b)$ that satisfy this case.


This all shows that there are a total of $8+10 = 18$ amount of ordered pairs $(a,b).$ Multiplying this by $41$ (the amount of values for $c$) we get $18\cdot41=\boxed{738}$ as our final answer.

~s214425

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 2

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 $q(x)=p(x)-p(2)=(x-2)( (x^2+2x+4) + a(x+2) + b)$ 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 3

$p(x)-p(2)$ is a cubic with two integral real roots, therefore it has three real roots, which are all integers. There are exactly two distinct roots, so either $p(x)=p(2)+(x-2)^2(x-m)$ or $p(x)=p(2)+(x-2)(x-m)^2$ with $m\neq 2$. In the first case $p(x)=x^3-(4+m)x^2+(4+4m)x-4m+p(2)$, so $m$ can be $-6,-5,-4,-3,-2,-1,0,1,3,4$ and $-4m+p(2)$ can be any integer from $-20$ to $20$, giving $410$ polynomials. In the second case $p(x)=x^3-(1+2m)x^2+(4m+m^2)x-2m^2+p(2)$, and $m$ can be $-6,-5,-4,-3,-2,-1,0,1$ and $-4m+p(2)$ can be any integer from $-20$ to $20$, giving 328 polynomials. The total is $\boxed{738}$.

~EVIN-

See also

2023 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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