Difference between revisions of "2021 AMC 12B Problems/Problem 16"
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+ | {{duplicate|[[2021 AMC 12B Problems#Problem_16|2021 AMC 12B #16]] and [[2021 AMC 10B Problems#Problem 18|2021 AMC 10B #18]]}} | ||
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==Problem== | ==Problem== | ||
Let <math>g(x)</math> be a polynomial with leading coefficient <math>1,</math> whose three roots are the reciprocals of the three roots of <math>f(x)=x^3+ax^2+bx+c,</math> where <math>1<a<b<c.</math> What is <math>g(1)</math> in terms of <math>a,b,</math> and <math>c?</math> | Let <math>g(x)</math> be a polynomial with leading coefficient <math>1,</math> whose three roots are the reciprocals of the three roots of <math>f(x)=x^3+ax^2+bx+c,</math> where <math>1<a<b<c.</math> What is <math>g(1)</math> in terms of <math>a,b,</math> and <math>c?</math> | ||
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==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}} | ||
− | + | {{AMC10 box|year=2021|ab=B|num-b=17|num-a=19}} | |
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 15:40, 8 March 2021
- The following problem is from both the 2021 AMC 12B #16 and 2021 AMC 10B #18, so both problems redirect to this page.
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 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
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=vCEJzhDRUoU
Video Solution by OmegaLearn (Vieta's Formula)
~ pi_is_3.14
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 |
2021 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.