2025 AMC 12A Problems/Problem 25
Polynomials 
and 
each have degree 
and leading coefficient 
, and their roots are all elements of 
. The function 
has the property that there exist real numbers 
such that the set of all real numbers 
such that 
consists of the closed interval ![$[a,b]$](http://latex.artofproblemsolving.com/8/e/c/8ecbd1ba3da8f2adef66a63f2ab32c47e63fa734.png)
together with the open interval 
. How many functions 
are possible?
Solution 1 by Victor Zhang
\documentclass[12pt]{article} \usepackage{amsmath, amssymb, amsthm} \usepackage{enumitem} \begin{document} \section*{Solution}
\subsection*{Step 1: Sign chart analysis}
From the structure 
on ![$[a,b] \cup (c,d)$](http://latex-new.aopstest.com/2/3/e/23ee6de2e2eabfa1fc76b5fa58eb4244b85aa39e.png)
and 
elsewhere,
we deduce:
\begin{itemize}
\item 
and 
are zeros of 
(since we transition from 
to 
at and from 
to at ).
\item 
and 
are poles or holes of (since on we have , at 
and 
the sign is negative, and immediately outside the
interval ).
\end{itemize}
Thus the sign pattern is:\\
\section*{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:
\[
f_1(x) = \frac{(x-a)(x-b)}{(x-c)(x-d)}, \quad f_2(x) = \frac{x-p_3}{x-q_3}.
\]
\section*{Step 3: Analysis of \( p_3 \) and \( q_3 \)}
\begin{itemize}
\item By analyzing the sign changes of \( f_1(x) \), we find that it completely
satisfies the sign chart for \( f(x) \). We can therefore conclude that \( f_2(x) \)
must be positive on every interval; otherwise, it would introduce an extra sign
change. This forces the condition \( p_3 = q_3 \), because if \( p_3 \neq q_3 \),
the factor \( f_2(x) \) would be negative on the interval between \( p_3 \) and \(
q_3 \). Let \( p_3 = q_3 = t \).
\item Furthermore, \( t \) cannot lie within \( [a,b] \cup (c,d) \). If it did, 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 \( (c, d) \).
\item Additionally, \( t \) cannot be equal to \( a \) or \( b \). If it were, that
point would be a hole (a point of discontinuity) in the graph of \( f(x) \), which
contradicts the fact that \( [a, b] \) is a \textbf{closed} interval where \( f(x) \le 0
\). On a closed interval, the function must be defined at the endpoints.
\end{itemize}
In conclusion, \( t \) can only be equal to \( c \) or \( d \), or lie in an interval
outside of \( [a,b] \cup (c,d) \).
\section*{Step 4: The Possible Values of , , , , and 
}
\begin{itemize}
\item Given 
and the condition that ![$\{x: f(x) \leq 0\} = [a,b] \cup (c,d)$](http://latex-new.aopstest.com/2/2/e/22e2f53191235ee4384c6a78fc7e52bee9cc5e46.png)
, the number of ways to choose 
from is:
\[
\binom{5}{4} = 5
\]
The five cases are:
![$[1,2]\cup (3,4)$](http://latex-new.aopstest.com/6/8/6/686861d2fa5b54c7aeb4abab150506d084f1c804.png)
; ![$[1,2]\cup (3,5)$](http://latex-new.aopstest.com/0/8/3/083ed506f39b20d070bc88af133b6d9e55da6f37.png)
; ![$[1,2]\cup (4,5)$](http://latex-new.aopstest.com/9/e/c/9ec56a4c09f23b9f61048b3b9a39b88632bf1cd5.png)
; ![$[1,3]\cup (4,5)$](http://latex-new.aopstest.com/0/3/d/03d29ffc490f096e5edcce3d5c881a2ba8bf3913.png)
;
![$[2,3]\cup (4,5)$](http://latex-new.aopstest.com/f/2/0/f2016005b121947e1204a36ac1ae36b0a319e515.png)
\item \textbf{Case 1:} , can be 3, 4, or 5.\\
This gives us 3 combinations:
\[
f(x) = \frac{(x-1)(x-2)(x-3)}{(x-3)(x-4)(x-3)}
\]
\[
f(x) = \frac{(x-1)(x-2)(x-4)}{(x-3)(x-4)(x-4)}
\]
\[
f(x) = \frac{(x-1)(x-2)(x-5)}{(x-3)(x-4)(x-5)}
\]
\item \textbf{Case 2:} , can be 3 or 5.\\
This gives us 2 combinations:
\[
f(x) = \frac{(x-1)(x-2)(x-3)}{(x-3)(x-5)(x-3)}
\]
\[
f(x) = \frac{(x-1)(x-2)(x-5)}{(x-3)(x-5)(x-5)}
\]
\item \textbf{Case 3:} , can be 3, 4, or 5.\\
This gives us 3 combinations:
\[
f(x) = \frac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-3)}
\]
\[
f(x) = \frac{(x-1)(x-2)(x-4)}{(x-4)(x-5)(x-4)}
\]
\[
f(x) = \frac{(x-1)(x-2)(x-5)}{(x-4)(x-5)(x-5)}
\]
\item \textbf{Case 4:} , can be 4 or 5.\\
This gives us 2 combinations:
\[
f(x) = \frac{(x-1)(x-3)(x-4)}{(x-4)(x-5)(x-4)}
\]
\[
f(x) = \frac{(x-1)(x-3)(x-5)}{(x-4)(x-5)(x-5)}
\]
\item \textbf{Case 5:} , can be 1, 4, or 5.\\
This gives us 3 combinations:
\[
f(x) = \frac{(x-2)(x-3)(x-1)}{(x-4)(x-5)(x-1)}
\]
\[
f(x) = \frac{(x-2)(x-3)(x-4)}{(x-4)(x-5)(x-4)}
\]
\[
f(x) = \frac{(x-2)(x-3)(x-5)}{(x-4)(x-5)(x-5)}
\]
\end{itemize}
\section*{Step 5: Final Answer}
Summing up all possible cases from Step 4 gives us 
possible functions .
Thus the answer is 
.
\end{document}
\section*{Step 4: The possible of a,b,c,d, and t }
\begin{itemize}
\item According to given and \( [a,b] \cup (c,d) \), the ways
of choose a,b,c,d from {1,2,3,4,5}:
![\[\binom{5}{4}=5\]](http://latex-new.aopstest.com/1/a/b/1ab5cf91c0c4e931987a4906daa0bd380259fa7a.png)
List 5 cases as:\\$[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)$\item Case 1:$ (Error compiling LaTeX. Unknown error_msg)[1,2]\cup (3,4)$, t can be 3,4, or 5.\\
Give us 3 combnation:
\[
f(x) = \frac{(x-1)(x-2)(x-3)}{(x-3)(x-4)(x-3)}
\]
\[
f(x) = \frac{(x-1)(x-2)(x-4)}{(x-3)(x-4)(x-4
)}
\]
\[
f(x) = \frac{(x-1)(x-2)(x-5)}{(x-3)(x-4)(x-5)}.
\]
\item Case 2:$ (Error compiling LaTeX. Unknown error_msg)[1,2]\cup (3,5)$, t can be 3, or 5.\\
Give us 2 combnation:
\[
f(x) = \frac{(x-1)(x-2)(x-3)}{(x-3)(x-5)(x-3)}
\]
\[
f(x) = \frac{(x-1)(x-2)(x-5)}{(x-3)(x-4)(x-5
)}
\]
\item Case 3:$ (Error compiling LaTeX. Unknown error_msg)[1,2]\cup (4,5)$, t can be 3,4, or 5.\\
Give us 3 combnation:
\[
f(x) = \frac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-3)}
\]
\[
f(x) = \frac{(x-1)(x-2)(x-4)}{(x-4)(x-5)(x-4
)}
\]
\[
f(x) = \frac{(x-1)(x-2)(x-5)}{(x-4)(x-5)(x-5)}.
\]
\item Case 4:$ (Error compiling LaTeX. Unknown error_msg)[1,3]\cup (4,5)$, t can be 4, or 5.\\
Give us 2 combnation:
\[
f(x) = \frac{(x-1)(x-3)(x-4)}{(x-4)(x-5)(x-4)}
\] \[ f(x) = \frac{(x-1)(x-3)(x-5)}{(x-4)(x-5)(x-5 )} \] \item Case 5:$ (Error compiling LaTeX. Unknown error_msg)[2,3]\cup (4,5)$, t can be 1,4, or 5.\\ Give us 3 combnation: \[ f(x) = \frac{(x-2)(x-3)(x-1)}{(x-4)(x-5)(x-1)} \] \[ f(x) = \frac{(x-2)(x-3)(x-4)}{(x-4)(x-5)(x-4)} \] \[ f(x) = \frac{(x-2)(x-3)(x-5)}{(x-4)(x-5)(x-5)} \]
\section*{Step 5: final answer } \Sum up all possible cases of step 4 give us 13 possible functions f(x).
Thus the answer is$ (Error compiling LaTeX. Unknown error_msg)\boxed{E}$. \end{document}
