Difference between revisions of "2021 AMC 12B Problems/Problem 16"
Darren yao (talk | contribs) (Added Solution 3) |
(Nice solution.) |
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+ | ==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 | ||
+ | <cmath>(1-\frac{1}{p})(1-\frac{1}{q})(1-\frac{1}{r}) = \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 | ||
== Video Solution by OmegaLearn (Vieta's Formula) == | == Video Solution by OmegaLearn (Vieta's Formula) == |
Revision as of 05:30, 13 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 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 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.