2021 AMC 12A Problems/Problem 22
Contents
- 1 Problem
- 2 Solution 1 (Complex Numbers: Vieta's Formulas)
- 3 Solution 2 (Complex Numbers: Trigonometric Identities)
- 4 Solution 3 (Trigonometric Identities)
- 5 Solution 4 (Product-to-Sum Identity)
- 6 Solution 5 (Approximations)
- 7 Video Solution by OmegaLearn (Euler's Identity + Vieta's Formula)
- 8 See also
Problem
Suppose that the roots of the polynomial are and , where angles are in radians. What is ?
Solution 1 (Complex Numbers: Vieta's Formulas)
Let Since is a th root of unity, we have For all integers note that and It follows that By geometric series, we conclude that Alternatively, recall that the th roots of unity satisfy the equation By Vieta's Formulas, the sum of these seven roots is
As a result, we get Let By Vieta's Formulas, the answer is ~MRENTHUSIASM (inspired by Peeyush Pandaya et al)
Solution 2 (Complex Numbers: Trigonometric Identities)
Let In Solution 1, we conclude that so Since holds for all this sum becomes Note that are roots of as they can be verified either algebraically (by the identity ) or geometrically (by the graph below). Let It follows that Rewriting in terms of we have in which the roots are
Therefore, we obtain from which
~MRENTHUSIASM (inspired by Peeyush Pandaya et al)
Solution 3 (Trigonometric Identities)
We solve for and separately:
- Solve for By Vieta's Formulas, we have
The real parts of the th roots of unity are and they sum to
Note that for all Excluding the other six roots add to from which Therefore, we get
- Solve for By Vieta's Formulas, we have
Note that for all and Therefore, we get
- Solve for By Vieta's Formulas, we have
We multiply both sides by then repeatedly apply the angle addition formula for sine: Therefore, we get
Finally, the answer is
~Tucker (Solution)
~MRENTHUSIASM (Reformatting)
Solution 4 (Product-to-Sum Identity)
Note that the sum of the roots of unity equal zero, so the sum of their real parts equal zero, and We have so
By the Product-to-Sum Identity, we have so
By the Product-to-Sum Identity, we have so
Finally, we get
~ ccx09 (Solution)
~MRENTHUSIASM (Reformatting)
Solution 5 (Approximations)
Letting the roots be and Vieta gives We use the Taylor series 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
Remark
In order to be more confident in your answer, you can go a few terms further in the Taylor series.
~ciceronii (Solution)
~MRENTHUSIASM (Reformatting)
Video Solution by OmegaLearn (Euler's Identity + Vieta's Formula)
~ 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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