Difference between revisions of "2021 AMC 12B Problems/Problem 16"

Revision as of 11:56, 12 February 2021

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

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

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
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.

@@ Line 20: / Line 20: @@
 ~ pi_is_3.14
+==Video Solution by Hawk Math==
+https://www.youtube.com/watch?v=p4iCAZRUESs
 ==See Also==
 {{AMC12 box|year=2021|ab=B|num-b=15|num-a=17}}
 {{MAA Notice}}

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