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2019 AMC 10A Problem 24

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2019 AMC 10A Problem 24
by Yehui Feng - Sunday, November 5, 2023, 3:50 AM

Here is the link to the question and solution: https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_24

I wonder for the first solution, why partial fractions can contradict the condition given. Is it because of that that equation is "polynomial identity"? What does polynomial identity mean? Or there is another reason.

Thank you!

Picture of John Lensmire
Re: 2019 AMC 10A Problem 24
by John Lensmire - Tuesday, November 7, 2023, 10:03 AM

Let's look at a simpler example, and go in reverse (which I think makes things easier to understand).

Suppose we want to find $A$ and $B$ such that$$A(x-1)+B(x-2)=1.$$If we let $x=1$ then we see $B=-1$ and if we let $x=2$ then we see $A=1$. So we have shown$$1\cdot (x-1) - 1\cdot (x-2) = 1.$$Note that this is true for ALL $x$, and the $A$ and $B$ we found were unique.

Now let's divide both sides by $(x-1)(x-2)=x^2-3x+2$, giving that$$\frac{A}{x-2} + \frac{B}{x-1} =\frac{1}{x-2} + \frac{-1}{x-1} = \frac{1}{x^2-3x+2}.$$Of course this equation is undefined if $x=2$ or $x=-1$, but that doesn't change the values of $A$ and $B$.

The same type of argument is going on in #24 from the 2019 AMC 10A. Even if the different versions of the equation have different "domains", the values of the constants are unique regardless.

Note: I can explain more what I think they mean by "polynomial identity" if you like, but my opinion is that it doesn't need to be so complicated.