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 functions $f(x)$ are possible?

$\textbf{(A) } 3 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 7 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 11$

Solution 1

None of the answer choices on the official test 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 positive 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 (credit to Sohil Rathi's video solution, pi_is_3.14 / OmegaLearn)

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 which value $f$ equals, giving $5 \cdot 1 = 5$ valid functions for this case.

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

~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 :)

Video Solution by OmegaLearn

https://youtu.be/SPbTyq3Dz_0

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2025 AMC 12A Problems/Problem 25

Contents

Problem

Solution 1

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

Video Solution by OmegaLearn

See Also