AIME 2023 I · 第 9 题

Solution 1

Plugging $2$ and $m$ into the expression for $p(x)$ 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 we only need to 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 $2,$ 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, we have

$$(a-2)^2 = 4(4 + b)\Longrightarrow a = 2\pm2\sqrt{4+b}.$$ 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. For $b=12$, because then $a = -6$, the solution for $m$ would equal $2,$, and we don't want this. (We can know this by replacing the solutions into the quadratic formula).

So we have $5$ solutions for $b,$ each of which gives $2$ values for $a,$ except for $2$ and $b = -4,$ both of which only give one. So in total, there are $5*2 - 2 = 8$ 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:

$$\begin{aligned} r+2 &= -a-2\\ 2r &= 4+2a+b. \end{aligned}$$ 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 $2,$ 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 is 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 2 (factor the difference)

$p(x)-p(2)$ is a cubic with at least 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 $p(x)=x^3-(4+m)x^2+(4+4m)x-4m+p(2)$, with $|4+4m|\leq 20$ (which entails $|4+m|\leq 20$), so $m$ can be $-6,-5,-4,-3,-2,-1,0,1, (\textbf{not 2}!), 3,4$ and $-4m+p(2)$ can be any integer from $-20$ to $20$, giving $410$ polynomials (since the coefficients are given by linear functions of $m$ and thus are distinct).

In the second case $p(x)=x^3-(2+2m)x^2+(4m+m^2)x-2m^2+p(2)$, 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-

Solution 3 (derivative)

For the equation $p(x) = p(2)$ to have exactly two distinct real roots, the line $y = p(2)$ must be tangent to either the local maximum or the local minimum of the cubic. This occurs when the derivative $p'(x) = 0$ at one of the roots.

Case 1: The repeated root is at $x=2$.

If $x=2$ is a repeated root, then $p'(2) = 0$. Given $p(x) = x^3 + ax^2 + bx + c$, the derivative is $p'(x) = 3x^2 + 2ax + b$. $p'(2) = 3(2)^2 + 2a(2) + b = 12 + 4a + b = 0 \implies b = -4a - 12.$ The constraint $|b| \le 20$ implies: $-20 \le -4a - 12 \le 20 \implies -8 \le a \le 2.$ This gives $11$ possible integers for $a$. However, we must exclude $a = -6$ because it makes $p''(2) = 6(2) + 2(-6) = 0$, creating an inflection point rather than an extremum (which would result in only one distinct root). Thus, there are $11 - 1 = 10$ valid pairs of $(a, b)$. Each pair combined with the $41$ possible values of $c$ gives $10 \times 41 = 410$ polynomials.

Case 2: The repeated root is at $x=m$ ($m \neq 2$).

In this case, the polynomial takes the form $p(x) - p(2) = (x-2)(x-m)^2$. Expanding this: $p(x) - p(2) = x^3 - (2m+2)x^2 + (m^2+4m)x - 2m^2.$ Matching the coefficients with $p(x) = x^3 + ax^2 + bx + c$, we have:

$a = -2m - 2$ $b = m^2 + 4m$

For $a$ and $b$ to be integers, $m$ must be an integer. Testing values of $m$ such that $|a| \le 20$ and $|b| \le 20$: The condition $|a| \le 20$ gives $-11 \le m \le 9$. Within this range, checking $|m^2 + 4m| \le 20$ yields $m \in {-6, -5, -4, -3, -2, -1, 0, 1}$. Note that $m=2$ is excluded. This gives $8$ valid pairs of $(a, b)$. Combined with $41$ values of $c$, we have $8 \times 41 = 328$ polynomials.

The total number of such polynomials is $410 + 328 = \boxed{738}.$

~LI,CHENXI-

Video Solution

https://youtu.be/-Asb_5nTgSg

~MathProblemSolvingSkills.com