2025 AMC 12A Problems/Problem 25: Difference between revisions
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Case 1: | 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> | <cmath>f(x) = \frac{(x-a)(x-b)(x-f)}{(x-c)(x-d)(x-f)}</cmath> for some other value <imath>f</imath> not equal to <imath>a, b, c,</imath> or <imath>d.</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: | 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: | ||
Revision as of 01:32, 8 November 2025
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 (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 ![\[f(x) = \frac{(x-a)(x-b)}{(x-c)(x-d)}\]](http://latex-new.aopstest.com/9/f/9/9f99485766b68d4a3317627793d358303ce09aa1.png)
by itself satisfies the condition of dropping below the x axis over ![$[a,b] \cup (c,d).$](http://latex-new.aopstest.com/2/2/9/229983a6f234f01bb6afd97e58c8c34af0c651fa.png)
Now, we need to add one extra 
term each to the numerator and denominator.
Case 1:
![\[f(x) = \frac{(x-a)(x-b)(x-f)}{(x-c)(x-d)(x-f)}\]](http://latex-new.aopstest.com/a/2/4/a24dd6b5f589db024bf82901fa1a44de4d6d1f4b.png)
for some other value 
not equal to 
or 
Note that cannot be between 
and 
or between 
and 
because that would create a hole in that interval. So the only possibilities are:
This gives us valid functions so far.
Case 2:
where is one of or
If equals or the resulting function is essentially equivalent to 
because a hole already exists. It doesn't matter whether we choose to equal or , the function will still be the same. There are 
ways to select the values for 
and way to choose which value equals, giving 
valid functions for this case.
The correct answer is 
~grogg007, original solution by OmegaLearn/ Sohil Rathi
Video Solution 1 by OmegaLearn
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 | |
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