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

We are told that $f(x) = \tfrac{P(x)}{Q(x)}$ , where $P$ and $Q$ are monic cubics whose roots are elements of $\{1,2,3,4,5\}$ . The function satisfies $\{x : f(x) \le 0\} = [a,b] \cup (c,d)$ for some real numbers $a < b < c < d$ .

From the given information, $f(x)$ changes sign only at $a,b,c,d$ . Thus: - $a,b$ are zeros of $f(x)$ . - $c,d$ are poles (values where the denominator is zero).

The sign pattern of $f(x)$ is:

\[(+)\ a\ (−)\ b\ (+)\ c\ (−)\ d\ (+),\] (Error compiling LaTeX. Unknown error_msg)

so $f(x)$ is positive outside $[a,b]\cup(c,d)$ and nonpositive on those intervals.

Step 1. General Form

Because $P$ and $Q$ are monic cubics, we can write $f(x) = \frac{(x-a)(x-b)(x-t)}{(x-c)(x-d)(x-t)}$ for some $t$ . This keeps both numerator and denominator degree 3.

To prevent any additional sign changes, the extra factor $\frac{x-t}{x-t}$ must remain positive, so $t$ must either equal $c$ or $d$ , or lie outside $[a,b]\cup(c,d)$ .

Step 2. Counting Possible $(a,b,c,d,t)$

We must select four distinct values $a<b<c<d$ from $\{1,2,3,4,5\}$ : $\binom{5}{4} = 5$ possible sets: $[1,2]\cup(3,4), \quad [1,2]\cup(3,5), \quad [1,2]\cup(4,5), \quad [1,3]\cup(4,5), \quad [2,3]\cup(4,5).$