Difference between revisions of "2021 AMC 12A Problems/Problem 22"
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===Solution 4.1 (Trigonometric Identities)=== | ===Solution 4.1 (Trigonometric Identities)=== | ||
− | We know that <math>\theta=\frac{2\pi}{7},\frac{4\pi}{7},\frac{6\pi}{7}</math> are roots of <cmath>\cos\theta+\cos(2\theta)+\cos(3\theta)=-\frac12, \ \ \ \ \ (*)</cmath> as they can be verified | + | We know that <math>\theta=\frac{2\pi}{7},\frac{4\pi}{7},\frac{6\pi}{7}</math> are roots of <cmath>\cos\theta+\cos(2\theta)+\cos(3\theta)=-\frac12, \ \ \ \ \ (*)</cmath> as they can be verified algebraically (by the identity <math>\cos\theta=\cos(-\theta)=\cos(2\pi-\theta)</math> for all <math>\theta</math>) or geometrically (by the <b>Remark</b> section). |
Let <math>x=\cos\theta.</math> It follows that | Let <math>x=\cos\theta.</math> It follows that |
Revision as of 00:54, 23 April 2021
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
Alternatively, note that the th roots of unity are for in which By Vieta's Formulas, the sum of these seven roots is
It follows that the real parts of these complex numbers must sum to so we get Since holds for all we can rewrite this as
Two solutions follow from here:
Solution 4.1 (Trigonometric Identities)
We know that are roots of as they can be verified algebraically (by the identity for all ) or geometrically (by the Remark section).
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 4.2 (Vieta's Formulas)
Let Since is a th root of unity, Geometrically (shown in the Remark section), it follows that Recall that (from which ), and let By Vieta's Formulas, the answer is
~MRENTHUSIASM (inspired by Peeyush Pandaya et al)
Remark
Graph of the th roots of unity in Desmos:
- Rectangular Coordinate System: https://www.desmos.com/calculator/rzzk5geric (shown below)
- Polar Coordinate System: https://www.desmos.com/calculator/xta6sfbnot
Geometrically, it is clear that the imaginary parts of these complex numbers sum to
~MRENTHUSIASM
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.