2021 AMC 12B Spring Problem 16

Attempt Problem 16 of the 2021 AMC 12B Spring below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 12B Spring solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

16.

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\lt a\lt b\lt c.

What is

g(1)

in terms of

a,

b,

and

c?

\dfrac{1+a+b+c}{c}

1+a+b+c

\dfrac{1+a+b+c}{c^2}

\dfrac{a+b+c}{c^2}

\dfrac{1+a+b+c}{a+b+c}

Answer: A

Concepts:Vieta’s Formulas polynomial

Difficulty rating: 1720

Small hint:

If

f

has roots

r,s,t,

then

g(1)=\left(1-\tfrac1r\right)\left(1-\tfrac1s\right)\left(1-\tfrac1t\right)

Big hint:

This equals

\dfrac{(r-1)(s-1)(t-1)}{rst};

relate the numerator to

f(1)

and the denominator to

c

Solution:

Let

f

have roots

r,s,t.

Since

g

is monic with roots

\tfrac1r,\tfrac1s,\tfrac1t,

\begin{aligned} g(1) &= \left(1-\tfrac1r\right)\left(1-\tfrac1s\right) \\ &\quad {}\cdot \left(1-\tfrac1t\right) \\ &= \dfrac{(r-1)(s-1)(t-1)}{rst}. \end{aligned}

Now

f(1)=(1-r)(1-s)(1-t)

=1+a+b+c,

so

(r-1)(s-1)(t-1)

=-(1+a+b+c).

Also

rst=-c.

Therefore

g(1)

=\dfrac{-(1+a+b+c)}{-c}

=\dfrac{1+a+b+c}{c}.

Thus, the correct answer is A.

Problem 16 in other years