Difference between revisions of "2021 AMC 12B Problems/Problem 16"
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==Solution 5 (good at guessing)== | ==Solution 5 (good at guessing)== | ||
g(1) = sum of coefficients. if its (x-r)(x-s)(x-t), then it becomes (x-1/r)(x-1/s)(x-1/t) | g(1) = sum of coefficients. if its (x-r)(x-s)(x-t), then it becomes (x-1/r)(x-1/s)(x-1/t) | ||
so -rst becomes -1/rst so c becomes 1/c. Also, there is x^3 so the answer must include 1. the only answer having both of these is A. | so -rst becomes -1/rst so c becomes 1/c. Also, there is x^3 so the answer must include 1. the only answer having both of these is A. | ||
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| + | ~smellyman | ||
==Solution 6== | ==Solution 6== | ||
It is well known that reversing the order of the coefficients of a polynomial turns each root into its corresponding reciprocal. Thus, a polynomial with the desired roots may be written as <math>cx^3+bx^2+a+1</math>. As the problem statement asks for a monic polynomial, our answer is <cmath>\frac{f(1)}{c} = \boxed{(\textbf{A}) \frac{1+a+b+c}{c}}</cmath> | It is well known that reversing the order of the coefficients of a polynomial turns each root into its corresponding reciprocal. Thus, a polynomial with the desired roots may be written as <math>cx^3+bx^2+a+1</math>. As the problem statement asks for a monic polynomial, our answer is <cmath>\frac{f(1)}{c} = \boxed{(\textbf{A}) \frac{1+a+b+c}{c}}</cmath> | ||
| − | + | ==Video Solution by Punxsutawney Phil== | |
| + | https://youtube.com/watch?v=vCEJzhDRUoU | ||
| + | |||
| + | == Video Solution by OmegaLearn (Vieta's Formula) == | ||
| + | https://youtu.be/afrGHNo_JcY | ||
==Video Solution by Hawk Math== | ==Video Solution by Hawk Math== | ||
Revision as of 18:36, 29 October 2022
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 1
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
Solution 3 (Fakesolve)
Because the problem doesn't specify what the coefficients of the polynomial actually are, we can just plug in any arbitrary polynomial that satisfies the constraints. Let's take 
. Then has a triple root of 
. Then has a triple root of 
, and it's monic, so 
. We can see that this is 
, which is answer choice 
.
-Darren Yao
Solution 4
If we let 
and 
be the roots of , 
and 
. The requested value, , is then
![\[(1-\frac{1}{p})(1-\frac{1}{q})(1-\frac{1}{r}) = \frac{(p-1)(q-1)(r-1)}{pqr}\]](http://latex.artofproblemsolving.com/c/2/3/c233221075725078a5e44b33da32a70950f46f17.png)
The numerator is 
(using the product form of ) and the denominator is 
, so the answer is
![\[\frac{f(1)}{c} = \boxed{(\textbf{A}) \frac{1+a+b+c}{c}}\]](http://latex.artofproblemsolving.com/9/1/3/91397baa415366721b5e0358b13495ef3eda03b3.png)
- gting
Solution 5 (good at guessing)
g(1) = sum of coefficients. if its (x-r)(x-s)(x-t), then it becomes (x-1/r)(x-1/s)(x-1/t) so -rst becomes -1/rst so c becomes 1/c. Also, there is x^3 so the answer must include 1. the only answer having both of these is A.
~smellyman
Solution 6
It is well known that reversing the order of the coefficients of a polynomial turns each root into its corresponding reciprocal. Thus, a polynomial with the desired roots may be written as 
. As the problem statement asks for a monic polynomial, our answer is
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=vCEJzhDRUoU
Video Solution by OmegaLearn (Vieta's Formula)
Video Solution by Hawk Math
https://www.youtube.com/watch?v=p4iCAZRUESs
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. 
