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
No edit summary
Grogg007 (talk | contribs)
editor
1,725 edits
Line 1: Line 1:
Polynomials <imath>P(x)</imath> and <imath>Q(x)</imath> each have degree <imath>3</imath> and leading coefficient <imath>1</imath>, and their roots are all elements of <imath>\{1,2,3,4,5\}</imath>. The function <imath>f(x) = \tfrac{P(x)}{Q(x)}</imath> has the property that there exist real numbers <imath>a < b < c < d</imath> such that the set of all real numbers <imath>x</imath> such that <imath>f(x) \leq 0</imath> consists of the closed interval <imath>[a,b]</imath> together with the open interval <imath>(c,d)</imath>. How many functions <imath>f(x)</imath> are possible?
Polynomials <imath>P(x)</imath> and <imath>Q(x)</imath> each have degree <imath>3</imath> and leading coefficient <imath>1</imath>, and their roots are all elements of <imath>\{1,2,3,4,5\}</imath>. The function <imath>f(x) = \tfrac{P(x)}{Q(x)}</imath> has the property that there exist real numbers <imath>a < b < c < d</imath> such that the set of all real numbers <imath>x</imath> such that <imath>f(x) \leq 0</imath> consists of the closed interval <imath>[a,b]</imath> together with the open interval <imath>(c,d)</imath>. How many functions <imath>f(x)</imath> are possible?
==Solution (credit given to Sohil Rathi's video solution, pi_is_3.14)==
First, consider the problem if it were talking about two second degree polynomials. We can see that the function <cmath>f(x) = \frac{(x-a)(x-b)}{(x-c)(x-d)}</cmath> by itself satisfies the condition of dropping below the x axis over <imath>[a,b] \cup (c,d).</imath> Now, we need to add one extra <imath>(x-n)</imath> term each to the numerator and denominator.
Case 1:
<cmath>f(x) = \frac{(x-a)(x-b)(x-f)}{(x-c)(x-d)(x-f)}</cmath> for some other value <imath>f.</imath>
Note that <imath>f</imath> cannot be between <imath>a</imath> and <imath>b</imath> or between <imath>c</imath> and <imath>d</imath> because that would create a hole in that interval. So the only possibilities are:
*<imath>a = 1, b = 2, c = 3, d = 4, f = 5</imath>
*<imath>a = 1, b = 2, c = 4, d = 5, f = 3</imath>
*<imath>a = 2, b = 3, c = 4, d = 5, f = 1</imath>
This gives us <imath>3</imath> valid functions so far.
Case 2:
<cmath>f(x) = \frac{(x-a)(x-b)(x-f)}{(x-c)(x-d)(x-f)}</cmath> where <imath>f</imath> is one of <imath>c</imath> or <imath>d</imath>
If <imath>f</imath> equals <imath>c</imath> or <imath>d</imath> the resulting function is essentially equivalent to <imath>f(x) = \frac{(x-a)(x-b)}{(x-c)(x-d}</imath> because a hole already exists. It doesn't matter whether we choose <imath>f</imath> to equal <imath>c</imath> or <imath>d</imath>, the function will still be the same. There are <imath>\binom{5}{4} = 5</imath> ways to select the values for <imath>a, b, c, and d</imath> and <imath>1</imath> way to choose which value <imath>f</imath> equals, giving <imath>5 \cdot 1 = 5</imath> valid functions for this case.
The correct answer is <imath>5 + 3 = \boxed{8}.</imath>
~[[User:grogg007|grogg007]], original solution by OmegaLearn/ Sohil Rathi


==Video Solution 1 by OmegaLearn==
==Video Solution 1 by OmegaLearn==

Revision as of 01:31, 8 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 (credit given to Sohil Rathi's video solution, pi_is_3.14)

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 $f.$

Note that $f$ cannot be between $a$ and $b$ or between $c$ and $d$ because that would create a hole in that 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 one of $c$ 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 already exists. It doesn't matter whether we choose $f$ to equal $c$ or $d$, 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 $5 + 3 = \boxed{8}.$

~grogg007, original solution by OmegaLearn/ Sohil Rathi

Video Solution 1 by OmegaLearn

https://youtu.be/SPbTyq3Dz_0

See Also

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

These problems are copyrighted © by the Mathematical Association of America.

Retrieved from "https://wiki.aopstest.com/wiki/index.php?title=2025_AMC_12A_Problems/Problem_25&oldid=267451"