Difference between revisions of "2021 AMC 12A Problems/Problem 22"
MRENTHUSIASM (talk | contribs) (→Solution 1 (Complex Numbers: Vieta's Formulas): Shortened my original solutions considerably. Arranged the solutions based on popularity/easiness. Let me know if you disagree.) |
MRENTHUSIASM (talk | contribs) m (→Solution 2 (Complex Numbers: Trigonometric Identities)) |
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\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}&=-\frac12. | \cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}&=-\frac12. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | Note 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, \hspace{15mm} (\bigstar)</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 graph below). | + | Note 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, \hspace{15mm} (\bigstar)</cmath> as they can be verified either algebraically (by the identity <math>\cos\theta=\cos(-\theta)=\cos(2\pi-\theta)</math> for all <math>\theta</math>) or geometrically (by the graph below). |
<asy> | <asy> | ||
/* Made by MRENTHUSIASM */ | /* Made by MRENTHUSIASM */ |
Revision as of 17:29, 4 September 2021
Contents
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
for all
) 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 sum of roots of unity equal zero, sum of 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 )
~ 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.