Difference between revisions of "2021 AMC 12B Problems/Problem 16"
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<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> | ||
+ | ==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...) | ||
+ | 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 | ||
== Video Solution by OmegaLearn (Vieta's Formula) == | == Video Solution by OmegaLearn (Vieta's Formula) == |
Revision as of 22:20, 11 February 2021
Contents
[hide]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 so we can see that Therefore
Solution 2 (Vieta's bash)
Let the three roots of f(x) be , , and . (Here e does NOT mean 2.7182818...) We know that , , and , 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 . -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 |
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