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 <math>f(x)</math> 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 <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{(\textbf{A}) \frac{1+a+b+c}{c}}</math>. -dstanz5 | 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{(\textbf{A}) \frac{1+a+b+c}{c}}</math>. -dstanz5 | ||
Revision as of 03:32, 12 February 2021
Contents
Problem
Let 
be a polynomial with leading coefficient 
whose three roots are the reciprocals of the three roots of 
where 
What is 
in terms of 
and 
Solution
Note that 
has the same roots as , if it is multiplied by some monomial so that the leading term is 
they will be equal. We have
![\[f(1/x) = \frac{1}{x^3} + \frac{a}{x^2}+\frac{b}{x} + c\]](http://latex.artofproblemsolving.com/d/4/4/d445d497c73df12477b4f7a987c8c3d844dd9e27.png)
so we can see that
![\[g(x) = \frac{x^3}{c}f(1/x)\]](http://latex.artofproblemsolving.com/7/0/f/70f8e3089ec7eba045d932a0dff5250f7cc3a43c.png)
Therefore
![\[g(1) = \frac{1}{c}f(1) = \boxed{\textbf{(A) }\frac{1+a+b+c}c}\]](http://latex.artofproblemsolving.com/7/9/9/79989395b4413f6f007ecf21bbe70e43f9bf5984.png)
Solution 2 (Vieta's bash)
Let the three roots of 
be 
, 
, and 
. (Here e does NOT mean 2.7182818...)
We know that 
, 
, and 
, and that 
(Vieta's). This is equal to 
, which equals 
. -dstanz5
Video Solution by OmegaLearn (Vieta's Formula)
~ pi_is_3.14
See Also
| 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. 
