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)
Using geometric series, we can show that
Graph of in Desmos, where (the th roots of unity): https://www.desmos.com/calculator/7z54p0mkp1
Geometrically, the imaginary parts of these complex numbers sum to Using the above result, the real parts of these complex numbers sum to too. It follows that from which as it contributes half the real part of Two solutions follow from here:
Solution 4.1 (Function Composition)
We know that are solutions of as they can be verified geometrically or algebraically (by the identity ). Now, let It follows that Rewriting from above in terms of we have It follows that and
~MRENTHUSIASM (inspired by Peeyush Pandaya et al)
Solution 4.2 (Vieta's Formulas--Explains Solution 1 Using De Moivre's Theorem)
Let Since is a th root of unity, Geometrically, it follows that
Recall that (so that ), and let By Vieta's Formulas and the results above, the answer is
~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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