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

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==Solution 2 (Vieta's bash)==
 
==Solution 2 (Vieta's bash)==
 
Let the three roots of f(x) be <math>d</math>, <math>e</math>, and <math>f</math>. (Here e does NOT mean 2.7182818...)
 
Let the three roots of f(x) be <math>d</math>, <math>e</math>, and <math>f</math>. (Here e does NOT mean 2.7182818...)
We know that <math>a=-(d+e+f)</math>, <math>b=de+ef+df</math>, and <math>c=-def</math>, 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} (Veita's). This is equal to \frac{def-de-df-ef+d+e+f-1}{def}, which equals <math>\boxed{(A) \frac{1+a+b+c}{c}}</math>. -dstanz5
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We know that <math>a=-(d+e+f)</math>, <math>b=de+ef+df</math>, and <math>c=-def</math>, and that <math>g(1)=1-\frac{1}{d}-\frac{1}{e}-\frac{1}{f}+\frac{1}{de}+\frac{1}{ef}+\frac{1}{df}-\frac{1}{def}</math> (Vieta's). This is equal to <math>\frac{def-de-df-ef+d+e+f-1}{def}</math>, which equals <math>\boxed{(A) \frac{1+a+b+c}{c}}</math>. -dstanz5
  
 
== Video Solution by OmegaLearn (Vieta's Formula) ==
 
== Video Solution by OmegaLearn (Vieta's Formula) ==

Revision as of 04:31, 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 $\boxed{(A) \frac{1+a+b+c}{c}}$. -dstanz5

Video Solution by OmegaLearn (Vieta's Formula)

https://youtu.be/afrGHNo_JcY

~ pi_is_3.14

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

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