2021 AMC 12B Problems/Problem 16

Revision as of 09:02, 1 October 2023 by Ap246 (talk | contribs) (Solution 4)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)


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

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 $A$.


-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

Video Solution by OmegaLearn


~ pi_is_3.14

Video Solution by Punxsutawney Phil


Video Solution by OmegaLearn (Vieta's Formula)


Video Solution by Hawk Math


See Also

2021 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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

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