# 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 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

## Solution 4

If we let $p, q,$ and $r$ be the roots of $f(x)$, $f(x) = (x-p)(x-q)(x-r)$ and $g(x) = \left(x-\frac{1}{p}\right)\left(x-\frac{1}{q}\right)\left(x-\frac{1}{r}\right)$. The requested value, $g(1)$, is then $$\left(1-\frac{1}{p}\right)\left(1-\frac{1}{q}\right)\left(1-\frac{1}{r}\right) = \frac{(p-1)(q-1)(r-1)}{pqr}$$ The numerator is $-f(1)$ (using the product form of $f(x)$ ) and the denominator is $-c$, so the answer is $$\frac{f(1)}{c} = \boxed{(\textbf{A}) \frac{1+a+b+c}{c}}$$

- gting

## Solution 5 (Good at Guessing)

The function $g(1) = \text{sum of coefficients}$. If it's $(x-r)(x-s)(x-t)$, then it becomes $(x-\dfrac{1}{r})(x-\dfrac{1}{s})(x-\dfrac{1}{t}).$ So, $-rst$ becomes $-\dfrac{1}{rst}$, so $c$ becomes $\dfrac{1}{c}$. Also, there is a $x^3$ so the answer must include $1$. The only answer having both of these is $A$.

~smellyman

-Extremelysupercooldude (Minor Latex Edits and Grammar)

## Solution 6

It is well known that reversing the order of the coefficients of a polynomial turns each root into its corresponding reciprocal. Thus, a polynomial with the desired roots may be written as $cx^3+bx^2+a+1$. As the problem statement asks for a monic polynomial, our answer is $$\boxed{(\textbf{A}) \frac{1+a+b+c}{c}}$$

## Video Solution (🚀Under 2 min 🚀)

~Education, the Study of Everything

~ pi_is_3.14

## Video Solution by Hawk Math

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 