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

Revision as of 02:30, 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

We are told that $f(x) = \tfrac{P(x)}{Q(x)}$ , where $P$ and $Q$ are monic cubics whose roots are elements of $\{1,2,3,4,5\}$ . The function satisfies $\{x : f(x) \le 0\} = [a,b] \cup (c,d)$ for some real numbers $a < b < c < d$ .

From the given information, $f(x)$ changes sign only at $a,b,c,d$ . Thus: - $a,b$ are zeros of $f(x)$ . - $c,d$ are poles (values where the denominator is zero).

The sign pattern of $f(x)$ is:

\[(+)\ a\ (−)\ b\ (+)\ c\ (−)\ d\ (+),\] (Error compiling LaTeX. Unknown error_msg)

so $f(x)$ is positive outside $[a,b]\cup(c,d)$ and nonpositive on those intervals.

Step 1. General Form

Because $P$ and $Q$ are monic cubics, we can write $f(x) = \frac{(x-a)(x-b)(x-t)}{(x-c)(x-d)(x-t)}$ for some $t$ . This keeps both numerator and denominator degree 3.

To prevent any additional sign changes, the extra factor $\frac{x-t}{x-t}$ must remain positive, so $t$ must either equal $c$ or $d$ , or lie outside $[a,b]\cup(c,d)$ .

Step 2. Counting Possible $(a,b,c,d,t)$

We must select four distinct values $a<b<c<d$ from $\{1,2,3,4,5\}$ : $\binom{5}{4} = 5$ possible sets: $[1,2]\cup(3,4), \quad [1,2]\cup(3,5), \quad [1,2]\cup(4,5), \quad [1,3]\cup(4,5), \quad [2,3]\cup(4,5).$

For each case, we count valid $t$ values.

Case 1:* $[1,2]\cup(3,4)$

$t = 3, 4, 5$ $\Rightarrow$ 3 functions.

Case 2:* $[1,2]\cup(3,5)$

$t = 3, 5$ $\Rightarrow$ 2 functions.

Case 3:* $[1,2]\cup(4,5)$

$t = 3, 4, 5$ $\Rightarrow$ 3 functions.

Case 4:* $[1,3]\cup(4,5)$

$t = 4, 5$ $\Rightarrow$ 2 functions.

Case 5:* $[2,3]\cup(4,5)$

$t = 1, 4, 5$ $\Rightarrow$ 3 functions.

Adding up all cases: $3 + 2 + 3 + 2 + 3 = 13.$

Therefore, there are $\boxed{13}$ possible functions $f(x)$ , which corresponds to $\boxed{\textbf{(E)}}$ .

- Victor Zhang

Video Solution 1 by OmegaLearn

https://youtu.be/SPbTyq3Dz_0

@@ Line 2: / Line 2: @@
 ==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
+<cmath>
+\{x : f(x) \le 0\} = [a,b] \cup (c,d)
+</cmath>
+for some real numbers <imath>a < b < c < d</imath>.
-We begin by analyzing the sign chart. From the condition <imath>f(x) \le 0</imath> on <imath>[a,b] \cup (c,d)</imath> and <imath>f(x) > 0</imath> elsewhere, we deduce the following:
-- <imath>a</imath> and <imath>b</imath> are zeros of <imath>f</imath>, since we transition from <imath>f>0</imath> to <imath>f\le0</imath> at <imath>a</imath>, and from <imath>f\le0</imath> to <imath>f>0</imath> at <imath>b</imath>.
+From the given information, <imath>f(x)</imath> changes sign only at <imath>a,b,c,d</imath>.
-- <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\le0</imath>, while immediately outside that interval <imath>f>0</imath>.
+Thus:
+- <imath>a,b</imath> are zeros of <imath>f(x)</imath>.
+- <imath>c,d</imath> are poles (values where the denominator is zero).
-Thus, the sign pattern of <imath>f(x)</imath> is
+The sign pattern of <imath>f(x)</imath> is:
 <cmath>
-\begin{array}{cccccccccc}
+(+)\ a\ (−)\ b\ (+)\ c\ (−)\ d\ (+),
- & (-\infty, a) & a & (a,b) & b & (b,c) & c & (c,d) & d & (d,\infty) \\
-f(x) & + & 0 & - & 0 & + & \text{pole} & - & \text{pole} & +
-\end{array}
 </cmath>
+so <imath>f(x)</imath> is positive outside <imath>[a,b]\cup(c,d)</imath> and nonpositive on those intervals.
----
-**Step 2. Structure of <imath>f(x)</imath>**
-According to the given conditions, we can express
+===Step 1. General Form===
-<cmath>
+Because <imath>P</imath> and <imath>Q</imath> are monic cubics, we can write
-f(x) = \frac{(x-p_1)(x-p_2)(x-p_3)}{(x-q_1)(x-q_2)(x-q_3)}.
-</cmath>
-Combining this with the sign analysis, we can write
-<cmath>
-f(x) = \frac{(x-a)(x-b)(x-p_3)}{(x-c)(x-d)(x-q_3)}.
-</cmath>
-We can separate this into two factors:
-<cmath>
-f(x) = \frac{(x-a)(x-b)}{(x-c)(x-d)} \cdot \frac{x-p_3}{x-q_3}.
-</cmath>
-Define
 <cmath>
-f_1(x) = \frac{(x-a)(x-b)}{(x-c)(x-d)}, \qquad f_2(x) = \frac{x-p_3}{x-q_3}.
+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 3. Determining <imath>p_3</imath> and <imath>q_3</imath>**
-By analyzing the sign changes of <imath>f_1(x)</imath>, we see that it already matches the required sign pattern for <imath>f(x)</imath>. Therefore, <imath>f_2(x)</imath> must remain positive on every interval; otherwise, it would introduce extra sign changes.
-This means <imath>p_3 = q_3</imath>, since if <imath>p_3 \ne q_3</imath>, the factor <imath>\frac{x-p_3}{x-q_3}</imath> would change sign between <imath>p_3</imath> and <imath>q_3</imath>.
-Let <imath>p_3 = q_3 = t</imath>.
-Furthermore:
+===Step 2. Counting Possible <imath>(a,b,c,d,t)</imath>===
-- <imath>t</imath> cannot lie within <imath>[a,b] \cup (c,d)</imath>, since that would alter the set <imath>\{x : f(x) \le 0\}</imath>.
+We must select four distinct values <imath>a<b<c<d</imath> from <imath>\{1,2,3,4,5\}</imath>:
-- <imath>t</imath> cannot be equal to <imath>a</imath> or <imath>b</imath>, since that would make those endpoints holes (discontinuities), contradicting that <imath>[a,b]</imath> is a *closed* interval.
-Hence, <imath>t</imath> must either be equal to <imath>c</imath> or <imath>d</imath>, or lie outside <imath>[a,b] \cup (c,d)</imath>.
----
-**Step 4. Possible values of <imath>a,b,c,d,t</imath>**
-Given <imath>a<b<c<d</imath> and <imath>\{x : f(x) \le 0\} = [a,b] \cup (c,d)</imath>, we choose four distinct numbers from <imath>\{1,2,3,4,5\}</imath> for <imath>a,b,c,d</imath>:
 <cmath>
-\binom{5}{4} = 5.
+\binom{5}{4} = 5
 </cmath>
-The five possible cases are:
+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>
+*Case 1:* <imath>[1,2]\cup(3,4)</imath>
-<imath>t</imath> can be <imath>3</imath>, <imath>4</imath>, or <imath>5</imath>, giving 3 functions:
+<imath>t = 3, 4, 5</imath> <imath>\Rightarrow</imath> 3 functions.
-<cmath>
-f(x) = \frac{(x-1)(x-2)(x-3)}{(x-3)(x-4)(x-3)}, \quad
-f(x) = \frac{(x-1)(x-2)(x-4)}{(x-3)(x-4)(x-4)}, \quad
-f(x) = \frac{(x-1)(x-2)(x-5)}{(x-3)(x-4)(x-5)}.
-</cmath>
-**Case 2:** <imath>[1,2]\cup(3,5)</imath>
+*Case 2:* <imath>[1,2]\cup(3,5)</imath>
-<imath>t</imath> can be <imath>3</imath> or <imath>5</imath>, giving 2 functions:
+<imath>t = 3, 5</imath> <imath>\Rightarrow</imath> 2 functions.
-<cmath>
-f(x) = \frac{(x-1)(x-2)(x-3)}{(x-3)(x-5)(x-3)}, \quad
-f(x) = \frac{(x-1)(x-2)(x-5)}{(x-3)(x-5)(x-5)}.
-</cmath>
-**Case 3:** <imath>[1,2]\cup(4,5)</imath>
+*Case 3:* <imath>[1,2]\cup(4,5)</imath>
-<imath>t</imath> can be <imath>3</imath>, <imath>4</imath>, or <imath>5</imath>, giving 3 functions:
+<imath>t = 3, 4, 5</imath> <imath>\Rightarrow</imath> 3 functions.
-<cmath>
-f(x) = \frac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-3)}, \quad
-f(x) = \frac{(x-1)(x-2)(x-4)}{(x-4)(x-5)(x-4)}, \quad
-f(x) = \frac{(x-1)(x-2)(x-5)}{(x-4)(x-5)(x-5)}.
-</cmath>
-**Case 4:** <imath>[1,3]\cup(4,5)</imath>
+*Case 4:* <imath>[1,3]\cup(4,5)</imath>
-<imath>t</imath> can be <imath>4</imath> or <imath>5</imath>, giving 2 functions:
+<imath>t = 4, 5</imath> <imath>\Rightarrow</imath> 2 functions.
-<cmath>
-f(x) = \frac{(x-1)(x-3)(x-4)}{(x-4)(x-5)(x-4)}, \quad
-f(x) = \frac{(x-1)(x-3)(x-5)}{(x-4)(x-5)(x-5)}.
-</cmath>
-**Case 5:** <imath>[2,3]\cup(4,5)</imath>
+*Case 5:* <imath>[2,3]\cup(4,5)</imath>
-<imath>t</imath> can be <imath>1</imath>, <imath>4</imath>, or <imath>5</imath>, giving 3 functions:
+<imath>t = 1, 4, 5</imath> <imath>\Rightarrow</imath> 3 functions.
-<cmath>
-f(x) = \frac{(x-2)(x-3)(x-1)}{(x-4)(x-5)(x-1)}, \quad
-f(x) = \frac{(x-2)(x-3)(x-4)}{(x-4)(x-5)(x-4)}, \quad
-f(x) = \frac{(x-2)(x-3)(x-5)}{(x-4)(x-5)(x-5)}.
-</cmath>
----
-**Step 5. Final Answer**
-Summing all possible cases:
+Adding up all cases:
 <cmath>
 + 2 + 3 + 2 + 3 = 13.
 </cmath>
-Thus, there are <imath>\boxed{13}</imath> possible functions <imath>f(x)</imath>, corresponding to choice <imath>\boxed{E}</imath>.
--Victor Zhang
+Therefore, there are <imath>\boxed{13}</imath> possible functions <imath>f(x)</imath>, which corresponds to <imath>\boxed{\textbf{(E)}}</imath>.
+- Victor Zhang
 ==Video Solution 1 by OmegaLearn==
 https://youtu.be/SPbTyq3Dz_0

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