2025 AMC 12A Problem 25

Attempt Problem 25 of the 2025 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AMC 12A solutions, or check the answer key.

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25.

Polynomials P(x)P(x) and Q(x)Q(x) each have degree 33 and leading coefficient 1,1, and their roots are all elements of {1,2,3,4,5}.\{1, 2, 3, 4, 5\}. The function f(x)=P(x)Q(x)f(x) = \dfrac{P(x)}{Q(x)} has the property that there exist real numbers a<b<c<da \lt b \lt c \lt d such that the set of all real numbers xx such that f(x)0f(x) \le 0 consists of the closed interval [a,b][a, b] together with the open interval (c,d).(c, d). How many functions f(x)f(x) are possible?

77

99

1111

1212

1313

Answer: E
Concepts:functionpolynomialcasework
Difficulty rating: 2540
Small Hint:

The closed endpoints must be zeros of P,P, and the open endpoints must be zeros of QQ

Big Hint:

After those four endpoint factors are fixed, the remaining numerator and denominator factors must cancel; count distinct functions, not polynomial pairs

Solution:

This problem was voided: as written, its answer is not among the choices. Here is the count.

For {f0}=[a,b](c,d),\{f \le 0\} = [a, b] \cup (c, d), the endpoints a,ba, b of the closed interval must be zeros of PP at which Q0,Q\ne0, while c,dc,d must be poles. The required sign pattern is therefore that of g(x)=(xa)(xb)(xc)(xd). g(x)=\frac{(x-a)(x-b)}{(x-c)(x-d)}. The unused third factor of PP must match the unused third factor of Q;Q; otherwise there would be an extra zero, pole, or sign change.

If the common factor is the fifth value not among a,b,c,d,a,b,c,d, it cannot make a hole inside either interval. In the ordered list of five values it may therefore occur before a,a, between bb and c,c, or after d,d, giving 33 functions.

Alternatively, the common factor can equal cc or d.d. For each of the (54)=5\binom54=5 choices of a,b,c,d,a,b,c,d, these two polynomial pairs simplify to the same formula gg and have the same domain (both omit c,dc,d), so they define only one function. This gives 55 more functions, for a literal total of 3+5=8.3+5=8.

Counting the two polynomial pairs separately in each of the last five cases gives 3+25=13,3+2\cdot5=13, the provisional answer (E), but that does not answer the stated question about functions. Thus none of the printed choices is correct.

Problem 24#24
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