Difference between revisions of "2021 AMC 12A Problems/Problem 22"
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===Solution 4.2 (DeMoivre's Theorem)=== | ===Solution 4.2 (DeMoivre's Theorem)=== | ||
Let <math>z=e^{\frac{2\pi i}{7}}.</math> Since <math>z</math> is a <math>7</math>th root of unity, <math>z^7=1.</math> Graphically, it follows that | Let <math>z=e^{\frac{2\pi i}{7}}.</math> Since <math>z</math> is a <math>7</math>th root of unity, <math>z^7=1.</math> Graphically, it follows that | ||
− | < | + | <cmath>\begin{align*} |
− | \cos\frac{2\pi}{7}&=\frac{z+z^6}{2}&=\frac{z+z^{-1}}{2}, \\ | + | \cos{\frac{2\pi}{7}}&=\frac{z+z^6}{2}&=\frac{z+z^{-1}}{2}, \\ |
− | \cos\frac{4\pi}{7}&=\frac{z^2+z^5}{2}&=\frac{z^2+z^{-2}}{2}, \\ | + | \cos{\frac{4\pi}{7}}&=\frac{z^2+z^5}{2}&=\frac{z^2+z^{-2}}{2}, \\ |
− | \cos\frac{7\pi}{7}&=\frac{z^3+z^4}{2}&=\frac{z^3+z^{-3}}{2}. | + | \cos{\frac{7\pi}{7}}&=\frac{z^3+z^4}{2}&=\frac{z^3+z^{-3}}{2}. |
− | \end{align*} | + | \end{align*}</cmath> |
~MRENTHUSIASM (inspired by Peeyush Pandaya et al) | ~MRENTHUSIASM (inspired by Peeyush Pandaya et al) |
Revision as of 05:52, 15 February 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
Desmos graph of where
(the
th roots of unity):
..............................................................................................
Graphically, 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 graphically. 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 (DeMoivre's Theorem)
Let Since
is a
th root of unity,
Graphically, it follows that
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.