2021 AMC 12A Problems/Problem 22
Contents
Problem
Suppose that the roots of the polynomial are and , where angles are in radians. What is ?
Solution 1
Part 1: solving for a
is the negation of the sum of roots
The real values of the 7th roots of unity are: and they sum to .
If we subtract 1, and condense identical terms, we get:
Therefore, we have
Part 2: solving for b
is the sum of roots two at a time by Vieta's
We know that
By plugging all the parts in we get:
Which ends up being:
Which was shown in the first part to equal , so
Part 3: solving for c
Notice that
is the negation of the product of roots by Vieta's formulas
Multiply by
Then use sine addition formula backwards:
Finally multiply or .
~Tucker
Solution 2 (Approximation)
Letting the roots be , , and , Vietas gives We use the Taylor series for , to approximate the roots. Taking the sum up to yields a close approximation, so we have Note that these approximations get worse as gets larger, but they will be fine for the purposes of this problem. We then have We further approximate these values to , , and (mostly as this is an AMC problem and will likely use nice fractions). Thus, we have . ~ciceronii
Remark: In order to be more confident in your answer, you can go a few terms further in the Taylor series.
Solution 3 (Only using Product to Sum Identity)
Note sum of roots of unity equal zero, sum of real parts equal zero, and thus which means
By product to sum, so
By product to sum, so
~ ccx09
Solution 4 (Complex Numbers)
Solution 4.1
~MRENTHUSIASM (inspired by Peeyush Pandaya et al)
Solution 4.2
~MRENTHUSIASM (inspired by Peeyush Pandaya et al)
Video Solution by OmegaLearn (Euler's Identity + Vieta's )
~ pi_is_3.14
See also
2021 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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 |
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