Difference between revisions of "2021 AMC 12B Problems/Problem 16"
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==Solution 4== | ==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) = (x-\frac{1}{p})(x-\frac{1}{q})(x-\frac{1}{r})</math>. The requested value, <math>g(1)</math>, is then | + | 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>(1-\frac{1}{p})(1-\frac{1}{q})(1-\frac{1}{r}) = \frac{(p-1)(q-1)(r-1)}{pqr}</cmath> | + | <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 | 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> | <cmath>\frac{f(1)}{c} = \boxed{(\textbf{A}) \frac{1+a+b+c}{c}}</cmath> |
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.