Difference between revisions of "2021 AMC 12B Problems/Problem 16"
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==Solution== | ==Solution== | ||
| − | Note that <math>f(1/x)</math> | + | Note that <math>f(1/x)</math> has the same roots as <math>g(x)</math>, if it is multiplied by some monomial so that the leading term is <math>x^3</math> they will be equal. We have |
<cmath>f(1/x) = \frac{1}{x^3} + \frac{a}{x^2}+\frac{b}{x} + c</cmath> | <cmath>f(1/x) = \frac{1}{x^3} + \frac{a}{x^2}+\frac{b}{x} + c</cmath> | ||
so we can see that | so we can see that | ||
Revision as of 19:45, 11 February 2021
Problem 16
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)
