Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2025 AMC 12A Problems/Problem 25: Difference between revisions

← Older editNewer edit →
Tiankaizhang (talk | contribs)
editor
42 edits
Tiankaizhang (talk | contribs)
editor
42 edits
Line 2: Line 2:


==Solution 1 ==
==Solution 1 ==
We are told that <imath>f(x) = \tfrac{P(x)}{Q(x)}</imath>, where <imath>P</imath> and <imath>Q</imath> are monic cubics whose roots are elements of <imath>\{1,2,3,4,5\}</imath>. The function satisfies
From the structure <imath>f(x) \le 0</imath> on <imath>[a,b] \cup (c,d)</imath> and <imath>f(x) > 0</imath> elsewhere, we deduce:
<cmath>
\begin{itemize}
\{x : f(x) \le 0\} = [a,b] \cup (c,d)
    \item <imath>a</imath> and <imath>b</imath> are zeros of <imath>f</imath> (since we transition from <imath>f > 0</imath> to <imath>f \le 0</imath> at <imath>a</imath> and from <imath>f \le 0</imath> to <imath>f > 0</imath> at <imath>b</imath>).
</cmath>
    \item <imath>c</imath> and <imath>d</imath> are poles or holes of <imath>f</imath> (since on <imath>(c,d)</imath> we have <imath>f \le 0</imath>, at <imath>c^+</imath> and <imath>d^-</imath> the sign is negative, and immediately outside the interval <imath>f > 0</imath>).
for some real numbers <imath>a < b < c < d</imath>.
\end{itemize}


Thus the sign pattern is:\\


 
<imath>\textstyle
From the given information, <imath>f(x)</imath> changes sign only at <imath>a,b,c,d</imath>. 
\begin{array}{cccccccccc}
Thus:
  & (-\infty, a) & a & (a,b) & b & (b,c) & c & (c,d) & d & (d,\infty) \\
- <imath>a,b</imath> are zeros of <imath>f(x)</imath>.
f(x) & + & 0 & - & 0 & + & \text{pole} & - & \text{pole} & +
- <imath>c,d</imath> are poles (values where the denominator is zero).
\end{array}
 
</imath>
The sign pattern of <imath>f(x)</imath> is:
<cmath>(+) a (−) b (+) c (−) d (+)</cmath>
so <imath>f(x)</imath> is positive outside <imath>[a,b]\cup(c,d)</imath> and nonpositive on those intervals.
 
 
 
Step 1. General Form
Because <imath>P</imath> and <imath>Q</imath> are monic cubics, we can write
<cmath>
f(x) = \frac{(x-a)(x-b)(x-t)}{(x-c)(x-d)(x-t)}
</cmath>
for some <imath>t</imath>.  
This keeps both numerator and denominator degree 3.
 
To prevent any additional sign changes, the extra factor <imath>\frac{x-t}{x-t}</imath> must remain positive, so <imath>t</imath> must either equal <imath>c</imath> or <imath>d</imath>, or lie outside <imath>[a,b]\cup(c,d)</imath>.
 
 
 
Step 2. Counting Possible <imath>(a,b,c,d,t)</imath>
We must select four distinct values <imath>a<b<c<d</imath> from <imath>\{1,2,3,4,5\}</imath>:
<cmath>
\binom{5}{4} = 5
</cmath>
possible sets:
<cmath>
[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).
</cmath>
 
For each case, we count valid <imath>t</imath> values.
 
*Case 1:* <imath>[1,2]\cup(3,4)</imath> 
<imath>t = 3, 4, 5</imath> <imath>\Rightarrow</imath> 3 functions.
 
*Case 2:* <imath>[1,2]\cup(3,5)</imath> 
<imath>t = 3, 5</imath> <imath>\Rightarrow</imath> 2 functions.
 
*Case 3:* <imath>[1,2]\cup(4,5)</imath> 
<imath>t = 3, 4, 5</imath> <imath>\Rightarrow</imath> 3 functions.
 
*Case 4:* <imath>[1,3]\cup(4,5)</imath> 
<imath>t = 4, 5</imath> <imath>\Rightarrow</imath> 2 functions.
 
*Case 5:* <imath>[2,3]\cup(4,5)</imath> 
<imath>t = 1, 4, 5</imath> <imath>\Rightarrow</imath> 3 functions.
 
 
 
Adding up all cases:
<cmath>
3 + 2 + 3 + 2 + 3 = 13.
</cmath>
 
 
 
Therefore, there are <imath>\boxed{13}</imath> possible functions <imath>f(x)</imath>, which corresponds to <imath>\boxed{\textbf{(E)}}</imath>.


- Victor Zhang
- Victor Zhang

Revision as of 02:48, 7 November 2025

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

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

   \item $a$ 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 $b$).
   \item $c$ and $d$ are poles 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 $f > 0$).

\end{itemize}

Thus the sign pattern is:\\

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

- Victor Zhang

Video Solution 1 by OmegaLearn

https://youtu.be/SPbTyq3Dz_0

Retrieved from "https://wiki.aopstest.com/wiki/index.php?title=2025_AMC_12A_Problems/Problem_25&oldid=266785"