Difference between revisions of "2021 AMC 12B Problems/Problem 16"
Pi is 3.14 (talk | contribs) (→Solution) |
|||
| Line 11: | Line 11: | ||
Therefore | Therefore | ||
<cmath>g(1) = \frac{1}{c}f(1) = \boxed{\textbf{(A) }\frac{1+a+b+c}c}</cmath> | <cmath>g(1) = \frac{1}{c}f(1) = \boxed{\textbf{(A) }\frac{1+a+b+c}c}</cmath> | ||
| + | |||
| + | |||
| + | == Video Solution by OmegaLearn (Vieta's Formula) == | ||
| + | https://youtu.be/afrGHNo_JcY | ||
| + | |||
| + | ~ pi_is_3.14 | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2021|ab=B|num-b=15|num-a=17}} | {{AMC12 box|year=2021|ab=B|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 22:43, 11 February 2021
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)
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. 
