2025 AMC 12A Problems/Problem 25

Problem

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 ordered pairs of polynomials $(P, Q)$ are possible?

$\textbf{(A)}~7 \qquad \textbf{(B)}~9 \qquad \textbf{(C)}~11 \qquad \textbf{(D)}~12 \qquad \textbf{(E)}~13 \qquad \textbf{(F)}~8$

Why MAA initially got this wrong

If you look at solution 2, they overcounted the second case, essentially saying that $f$ had $2$ choices because it could equal $c$ or $d$ , so the intended answer was $3 + 5 \cdot 2 = 13.$ This would be correct if the problem was asking for the number of pairs of polynomials. But there is actually only $1$ choice for $f$ because no matter what variable we choose $f$ to equal, both polynomial combinations result in the same function. So the correct answer was $3 + 5 \cdot 1 = 8$ functions.

Desmos graph showing how these functions are exactly the same: https://www.desmos.com/calculator/w6sk67ly1i

~grogg007

Solution 1 (Got Wrong)

None of the answer choices on the official test (which asked for the number of possible functions $f(x)$ ) were correct, but choice E would be correct if this problem asked for the number of pairs of functions $(P(x), Q(x))$ .

Let $R(x) = \frac{P(x)}{Q(x)}$ . Since $R(x) \leq 0$ on $[a, b]$ but not for values slightly less than $a$ or slightly more than $b$ , $P(x) = 0$ at $x = a$ and $x = b$ . Therefore, $P(x) = (x-a)(x-b)(x-r)$ for some $r \in \{1, 2, 3, 4, 5\}$ .

Since $R(x) \leq 0$ on $(c, d)$ but not at $x = c$ or $x = d$ , $R(x)$ is not continuous at $x = c$ or $x = d$ . Therefore, $R(x)$ must be undefined at $x = c$ and $x = d$ , that is, $Q(x) = 0$ at $x = c$ and $x = d$ . So $Q(x) = (x-c)(x-d)(x-s)$ for some $s \in \{1, 2, 3, 4, 5\}$ .

Therefore, $R(x) = \frac{(x-a)(x-b)(x-r)}{(x-c)(x-d)(x-s)}$ . Notice that $\frac{(x-a)(x-b)}{(x-c)(x-d)} \leq 0$ only on $[a, b]$ and $(c, d)$ . Therefore, $\frac{x-r}{x-s}$ must be nonnegative for all $x \notin \{a, b, c, d, r, s\}$ . This only happens if $r = s$ .

Thus $R(x) = \frac{(x-a)(x-b)(x-r)}{(x-c)(x-d)(x-r)}$ , which is the same as $\frac{(x-a)(x-b)}{(x-c)(x-d)}$ except that it is undefined at $x = r$ . Thus $R(x)$ satisfies the desired property as long as $r \notin [a, b] \cup (c, d)$ .

Note that each quintuple $(a, b, c, d, r)$ defines a unique pair of functions $(P(x), Q(x))$ .

If $(a, b, c, d) = (1, 2, 3, 4)$ , $r$ can be $3$ , $4$ , or $5$ .

If $(a, b, c, d) = (1, 2, 3, 5)$ , $r$ can be $3$ or $5$ .

If $(a, b, c, d) = (1, 2, 4, 5)$ , $r$ can be $3$ , $4$ , or $5$ .

If $(a, b, c, d) = (1, 3, 4, 5)$ , $r$ can be $4$ or $5$ .

If $(a, b, c, d) = (2, 3, 4, 5)$ , $r$ can be $1$ , $4$ , or $5$ .

Therefore, there are $\boxed{(\textbf{E})\ 13}$ possible pairs of functions $(P(x), Q(x))$ .

-j314andrews

Solution 2 (credit to Sohil Rathi's video solution, pi_is_3.14 / OmegaLearn)

This is the solution to the original problem, which asked for the number of possible functions, not pairs of polynomials.

First, consider the problem if it were talking about two second degree polynomials. We can see that the function $f(x) = \frac{(x-a)(x-b)}{(x-c)(x-d)}$ by itself satisfies the condition of dropping below the x axis over $[a,b] \cup (c,d).$ Now, we need to add one extra $(x-n)$ term each to the numerator and denominator.

Case 1: $f(x) = \frac{(x-a)(x-b)(x-f)}{(x-c)(x-d)(x-f)}$ for some other value $f$ not equal to $a, b, c,$ or $d.$

Note that $f$ cannot be between $a$ and $b$ or between $c$ and $d$ because this would create a hole in the interval. So the only possibilities are:

$a = 1, b = 2, c = 3, d = 4, f = 5$
$a = 1, b = 2, c = 4, d = 5, f = 3$
$a = 2, b = 3, c = 4, d = 5, f = 1$

This gives us $3$ valid functions so far.

Case 2: $f(x) = \frac{(x-a)(x-b)(x-f)}{(x-c)(x-d)(x-f)}$ where $f$ is $c$ or $d.$

If $f$ equals $c$ or $d$ the resulting function is essentially equivalent to $f(x) = \frac{(x-a)(x-b)}{(x-c)(x-d)}$ because a hole would already exist, so we can safely cancel the $(x-f)$ terms. It doesn't matter whether we choose $f$ to equal $c$ or $d$ because the function will still be the same. There are $\binom{5}{4} = 5$ ways to select the values for $a, b, c,$ and $d$ and $1$ way to choose any variable for $f$ to equal, giving $5 \cdot 1 = 5$ valid functions for this case.

The correct answer is $5 + 3 = \boxed{\textbf{(F) } 8}$ functions.

~grogg007, original solution by OmegaLearn/ Sohil Rathi

i tried my best to show what I think was Sohil's thought process, but please feel free to edit this if you can explain it better :)

Solution 3

Solution 1 by user j314andrews is incomplete; Solution 2 by user grogg007 commits a serious logical error because it only provides functions that satisfy the requirements, without showing that no other functions do. Other places like [1] gives a good solution, but unfortunately it requires calculus. Here I am trying to provide a good solution that does not require calculus.

Let $D$ and $E$ be the set of the real roots of $P(x)$ and $Q(x)$ , respectively. Then, outside of the intervals $[a,b]$ and $(c,d)$ , $f(x)>0$ except when $x \in E$ , where $f(x)$ is undefined.

For any function $g(x)$ , define a real number $t$ as a toggling point for $g(x)$ if there exists a sequence $t_1 < t_2 < ... < t_N$ , with the understanding that $t_0 = -\infty$ and $t_{N+1} = \infty$ , such that

$t = t_k$ for some $1 \leq k \leq N$ ,
$g(x)$ is of the same sign (either always positive or always negative) on each interval $(t_j, t_{j+1})$ , for $j=0, ..., N$ , and
$g(x)$ has opposite signs on $(t_{k-1}, t_k)$ and $(t_k, t_{k+1})$ .

We call these open intervals $(t_j, t_{j+1})$ ( $j=0, ..., N$ ) the bands of $g(x)$ .

The polynomial $P(x)$ has no toggling point outside of $D$ , and $Q(x)$ has no toggling point outside of $E$ . Now in our problem, $f(x)$ has exactly 4 toggling points: $a, b, c, d$ . Our task is to see how these four points are related to the sets $D$ and $E$ .

Lemma. At each toggling point $t$ for $f(x)$ ,

either $P(t)=0$ or $Q(t)=0$ (or both);
if the multiplicities of the root $t$ of $P(x)$ and $Q(x)$ are $m$ and $n$ , respectively, then $m-n$ is odd.

Proof. (1) if both $P(t)$ and $Q(t)$ are nonzero, then $t$ belongs to a band of $P(x)$ and a band of $Q(x)$ . Taking the intersection between these two bands, we get an open interval on which $f(x)$ has the same sign as $f(t)$ . Thus $t$ cannot be a toggling point of $f(x)$ .

(2) If $m$ is odd, there are real numbers $r, s$ , with $r<t<s$ , such that $P(x)$ has the same sign on $(r,t)$ and has the opposite sign on $(t, s)$ . If $m$ is even, there are real numbers $r, s$ , with $r<t<s$ , such that $P(x)$ has the same sign on $(r,t)$ and on $(t,s)$ . Similar statements can be made of $Q(x)$ , having intervals $(u, t)$ and $(t, v)$ so that $Q(t)$ is of the same sign on $(u,t)$ , and of either a different or the same sign on $(t, v)$ , depending on the parity of $n$ . If both $m$ and $n$ are odd or both $m$ and $n$ are even, then $P(t)/Q(t)$ is of the same sign on the interval $(\max\{r, u\}, t)$ and $(t, \min\{s, v\})$ , so $t$ cannot be a toggling point for $f(x)$ .

Solution to the problem. Since $a$ is a toggling point for $f(x)$ , either $P(a)=0$ or $Q(a)=0$ , or both. But since $f(a)$ is defined, we have $Q(a)\neq 0$ , so $P(a)=0$ . The same can be said of $b$ , so $P(b)=0$ . The multiplicities of $a$ and $b$ must be $1$ . Otherwise, point (2) of the Lemma would require the multiplicity of one of them to be at least $3$ , but $P(x)$ is only cubic.

$c$ is a toggling point for $f(x)$ , so either $P(c)=0$ or $Q(c)=0$ , or both. But if $Q(c)\neq 0$ , we have $f(c)=0$ , violating the condition. So $Q(c)=0$ . Similarly, $Q(d)=0$ . Concerning their multiplicities, point (2) of the Lemma gives us three cases, given that $Q(x)$ is only cubic:

(a) $c$ is a single root of $Q(x)$ , $P(c)\neq 0$ , $d$ is a double root of $Q(x)$ and a single root of $P(x)$ ;
(b) switching the roles of $c$ and $d$ ;
(c) both $c$ and $d$ are single roots of $Q(x)$ , and $Q(x)$ has a single root $e$ other than $c$ and $d$ .

Case (a) provides functions in the form of $\frac{(x-a)(x-b)(x-d)}{(x-c)(x-d)^2}$ . There are 5 of them because there are 5 ways to choose $a, b, c, d$ from the set $\{1, 2, 3, 4, 5\}$ .

Case (b) provides functions in the form of $\frac{(x-a)(x-b)(x-c)}{(x-c)^2(x-d)}$ , but they are the same as those provided in case (a). To ensure this, notice that the two expressions have the same domain and they are equal for every $x$ in their domains.

Case (c) requires that $P(e)=0$ , because if $P(e)\neq 0$ then $e$ becomes an extra toggling point for $f(x)$ . So this case provides functions in the form of $\frac{(x-a)(x-b)(x-e)}{(x-c)(x-d)(x-e)}$ . But since $f(e)$ is undefined, we need to make sure that $e$ does not fall into $[a,b]$ or $(c,d)$ . So $e$ can be $1, 3, 5$ . In all these cases, we also have $f(e) = \frac{(e-a)(e-b)}{(e-c)(e-d)} > 0$ , satisfying the conditions. That's 3 more functions.

So the total number of functions is $5 + 3 = \boxed{8}$ .

Lightest (talk) 22:51, 10 November 2025 (EST)

Video Solution 1 by OmegaLearn

https://youtu.be/SPbTyq3Dz_0

2025 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

Page

Toolbox

Search