Difference between revisions of "2021 AMC 12B Problems/Problem 16"
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<math>\textbf{(A) }\frac{1+a+b+c}c \qquad \textbf{(B) }1+a+b+c \qquad \textbf{(C) }\frac{1+a+b+c}{c^2}\qquad \textbf{(D) }\frac{a+b+c}{c^2} \qquad \textbf{(E) }\frac{1+a+b+c}{a+b+c}</math> | <math>\textbf{(A) }\frac{1+a+b+c}c \qquad \textbf{(B) }1+a+b+c \qquad \textbf{(C) }\frac{1+a+b+c}{c^2}\qquad \textbf{(D) }\frac{a+b+c}{c^2} \qquad \textbf{(E) }\frac{1+a+b+c}{a+b+c}</math> | ||
− | ==Solution== | + | ==Solution 1== |
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 | 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> | ||
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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 | ||
+ | |||
+ | ==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 <math>f(x) = (x+5)^3 = x^3+15x^2+75x+125</math>. Then <math>f(x)</math> has a triple root of <math>x = -5</math>. Then <math>g(x)</math> has a triple root of <math>-\frac{1}{5}</math>, and it's monic, so <math>g(x) = \left(x+\frac{1}{5}\right)^3 = \frac{125x^3+75x^2+15x+1}{125}</math>. We can see that this is <math>\frac{1+a+b+c}{c}</math>, which is answer choice <math>\boxed{(A)}</math>. | ||
+ | |||
+ | -Darren Yao | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | If we let <math>p, q, </math> and <math>r</math> be the roots of <math>f(x)</math>, <math>f(x) = (x-p)(x-q)(x-r)</math> and <math>g(x) = \left(x-\frac{1}{p}\right)\left(x-\frac{1}{q}\right)\left(x-\frac{1}{r}\right)</math>. The requested value, <math>g(1)</math>, is then | ||
+ | <cmath>\left(1-\frac{1}{p}\right)\left(1-\frac{1}{q}\right)\left(1-\frac{1}{r}\right) = \frac{(p-1)(q-1)(r-1)}{pqr}</cmath> | ||
+ | The numerator is <math>-f(1)</math> (using the product form of <math>f(x)</math> ) and the denominator is <math>-c</math>, so the answer is | ||
+ | <cmath>\frac{f(1)}{c} = \boxed{(\textbf{A}) \frac{1+a+b+c}{c}}</cmath> | ||
+ | |||
+ | - gting | ||
+ | |||
+ | ==Solution 5 (Good at Guessing)== | ||
+ | The function <math>g(1) = \text{sum of coefficients}</math>. If it's <math>(x-r)(x-s)(x-t)</math>, then it becomes <math>(x-\dfrac{1}{r})(x-\dfrac{1}{s})(x-\dfrac{1}{t}).</math> | ||
+ | So, <math>-rst</math> becomes <math>-\dfrac{1}{rst}</math>, so <math>c</math> becomes <math>\dfrac{1}{c}</math>. Also, there is a <math>x^3</math> so the answer must include <math>1</math>. The only answer having both of these is <math>A</math>. | ||
+ | |||
+ | ~smellyman | ||
+ | |||
+ | -Extremelysupercooldude (Minor Latex Edits and Grammar) | ||
+ | |||
+ | ==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>\boxed{(\textbf{A}) \frac{1+a+b+c}{c}}</cmath> | ||
+ | |||
+ | |||
+ | ==Video Solution (🚀Under 2 min 🚀)== | ||
+ | https://youtu.be/vPw6VxuZvQU | ||
+ | |||
+ | <i>~Education, the Study of Everything</i> | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/M4Ffhp9NLKY?t=923 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution by Punxsutawney Phil== | ||
+ | https://youtube.com/watch?v=vCEJzhDRUoU | ||
== Video Solution by OmegaLearn (Vieta's Formula) == | == Video Solution by OmegaLearn (Vieta's Formula) == | ||
https://youtu.be/afrGHNo_JcY | https://youtu.be/afrGHNo_JcY | ||
− | + | ==Video Solution by Hawk Math== | |
+ | https://www.youtube.com/watch?v=p4iCAZRUESs | ||
==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}} | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 09:02, 1 October 2023
Contents
[hide]- 1 Problem
- 2 Solution 1
- 3 Solution 2 (Vieta's bash)
- 4 Solution 3 (Fakesolve)
- 5 Solution 4
- 6 Solution 5 (Good at Guessing)
- 7 Solution 6
- 8 Video Solution (🚀Under 2 min 🚀)
- 9 Video Solution by OmegaLearn
- 10 Video Solution by Punxsutawney Phil
- 11 Video Solution by OmegaLearn (Vieta's Formula)
- 12 Video Solution by Hawk Math
- 13 See Also
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 so we can see that Therefore
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 The numerator is (using the product form of ) and the denominator is , so the answer is
- gting
Solution 5 (Good at Guessing)
The function . If it's , then it becomes So, becomes , so becomes . Also, there is a so the answer must include . The only answer having both of these is .
~smellyman
-Extremelysupercooldude (Minor Latex Edits and Grammar)
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 (🚀Under 2 min 🚀)
~Education, the Study of Everything
Video Solution by OmegaLearn
https://youtu.be/M4Ffhp9NLKY?t=923
~ pi_is_3.14
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.