2021 AMC 12B Problems/Problem 16

Problem

Let $g(x)$ be a polynomial with leading coefficient $1,$ whose three roots are the reciprocals of the three roots of $f(x)=x^3+ax^2+bx+c,$ where $1<a<b<c.$ What is $g(1)$ in terms of $a,b,$ and $c?$

$\textbf{(A) }\frac{1+a+b+c}c \qquad \textbf{(B) }1+a+b+c \qquad \textbf{(C) }\frac{1+a+b+c}{c^2}\qquad \textbf{(D) }\frac{a+b+c}{c^2} \qquad \textbf{(E) }\frac{1+a+b+c}{a+b+c}$

Solution 1

Note that $f(1/x)$ has the same roots as $g(x)$ , if it is multiplied by some monomial so that the leading term is $x^3$ they will be equal. We have $f(1/x) = \frac{1}{x^3} + \frac{a}{x^2}+\frac{b}{x} + c$ so we can see that $g(x) = \frac{x^3}{c}f(1/x)$ Therefore $g(1) = \frac{1}{c}f(1) = \boxed{\textbf{(A) }\frac{1+a+b+c}c}$

Solution 2 (Vieta's bash)

Let the three roots of $f(x)$ be $d$ , $e$ , and $f$ . (Here e does NOT mean 2.7182818...) We know that $a=-(d+e+f)$ , $b=de+ef+df$ , and $c=-def$ , and that $g(1)=1-\frac{1}{d}-\frac{1}{e}-\frac{1}{f}+\frac{1}{de}+\frac{1}{ef}+\frac{1}{df}-\frac{1}{def}$ (Vieta's). This is equal to $\frac{def-de-df-ef+d+e+f-1}{def}$ , which equals $\boxed{(\textbf{A}) \frac{1+a+b+c}{c}}$ . -dstanz5

Solution 3 (Fakesolve)

Because the problem doesn't specify what the coefficients of the polynomial actually are, we can just plug in any arbitrary polynomial that satisfies the constraints. Let's take $f(x) = (x+5)^3 = x^3+15x^2+75x+125$ . Then $f(x)$ has a triple root of $x = -5$ . Then $g(x)$ has a triple root of $-\frac{1}{5}$ , and it's monic, so $g(x) = \left(x+\frac{1}{5}\right)^3 = \frac{125x^3+75x^2+15x+1}{125}$ . We can see that this is $\frac{1+a+b+c}{c}$ , which is answer choice $\boxed{(A)}$ .

-Darren Yao

Video Solution by OmegaLearn (Vieta's Formula)

https://youtu.be/afrGHNo_JcY

~ pi_is_3.14

Video Solution by Hawk Math

https://www.youtube.com/watch?v=p4iCAZRUESs

2021 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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All AMC 12 Problems and Solutions

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