2025 AMC 12A Problems/Problem 25

Polynomials $P(x)$ and $Q(x)$ each have degree $3$ and leading coefficient $1$ , and their roots are all elements of $\{1,2,3,4,5\}$ . The function $f(x) = \tfrac{P(x)}{Q(x)}$ has the property that there exist real numbers $a < b < c < d$ such that the set of all real numbers $x$ such that $f(x) \leq 0$ consists of the closed interval $[a,b]$ together with the open interval $(c,d)$ . How many functions $f(x)$ are possible?

Solution 1

Step 1: Sign Chart Analysis

From the given $f(x) \le 0$ on $[a,b] \cup (c,d)$ and $f(x) > 0$ elsewhere, we deduce:

$a$ and $b$ are zeros of $f$ since we transition from $f > 0$ to $f \le 0$ at $a$ and from $f \le 0$ to $f > 0$ at $b$ .
$c$ and $d$ are poles (vertical asymptotes or holes) of $f$ since on $(c,d)$ we have $f \le 0$ , at $c^+$ and $d^-$ the sign is negative, and immediately outside the interval $f > 0$ .

Thus the sign pattern is:

$\textstyle \begin{array}{cccccccccc} & (-\infty, a) & a & (a,b) & b & (b,c) & c & (c,d) & d & (d,\infty) \\ f(x) & + & 0 & - & 0 & + & \text{pole} & - & \text{pole} & + \end{array}$

Step 2: Structure of $f(x)$ According to the given conditions, $ f(x) $ can be expressed as: \[ f(x) = \frac{(x-p_1)(x-p_2)(x-p_3)}{(x-q_1)(x-q_2)(x-q_3)}. \]

Combining the sign analysis from Step 1, we can rewrite this expression as: \[ f(x) = \frac{(x-a)(x-b)(x-p_3)}{(x-c)(x-d)(x-q_3)}. \]

Furthermore, we can break down the expression into two parts: \[ f(x) = \frac{(x-a)(x-b)}{(x-c)(x-d)} \cdot \frac{x-p_3}{x-q_3}. \]

Defining $ f(x) = f_1(x) \cdot f_2(x) $, we have: \[ f_1(x) = \frac{(x-a)(x-b)}{(x-c)(x-d)}, \quad f_2(x) = \frac{x-p_3}{x-q_3}. \]

Step 3: $ p_3 $ and $ q_3 $

$ f_1(x) $ completely satisfies the sign chart for $ f(x) $. We can conclude that $ f_2(x) $ must be positive on every interval; otherwise, it would introduce an extra sign change. This forces the condition $ p_3 = q_3 $, otherwise it would be negative on certain intervals. Let $ p_3 = q_3 = t $.
Furthermore, this means $f(t)$ is undefined, forcing a hole at $x=t$ . Thus, it would contradict the given condition that the set $ \{x : f(x) \le 0\} $ consists of the closed interval $ [a, b] $ and the open interval $ (c, d) $.

Thus, we have cannot have $t = a$ or $b$ , but we can have $t = c$ or $d$ or any number outside $ [a,b] \cup (c,d) $.

Step 4: $a$ , $b$ , $c$ , $d$ , and $t$ Given $a < b < c < d$ and the condition that $\{x: f(x) \leq 0\} = [a,b] \cup (c,d)$ , the number of ways to choose $a,b,c,d$ from $\{1,2,3,4,5\}$ is: $\binom{5}{4} = 5$ The five cases are: $[1,2]\cup (3,4)$ ; $[1,2]\cup (3,5)$ ; $[1,2]\cup (4,5)$ ; $[1,3]\cup (4,5)$ ; $[2,3]\cup (4,5)$