Difference between revisions of "2021 AMC 12A Problems/Problem 22"
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~MRENTHUSIASM (Reformatting) | ~MRENTHUSIASM (Reformatting) | ||
− | == Solution 2 ( | + | == Solution 2 (Approximations) == |
− | Letting the roots be <math>p | + | Letting the roots be <math>p=\cos\frac{2\pi}{7},q=\cos\frac{4\pi}{7},</math> and <math>r=\cos\frac{6\pi}{7}.</math> Vieta gives |
− | <cmath>p + q + r = a | + | <cmath>\begin{align*} |
− | + | p + q + r &= a, \\ | |
− | + | pq + qr + rp &= -b, \\ | |
− | We use the Taylor series | + | pqr &= c. |
− | <cmath>\cos x = \sum_{k = 0}^{\infty} (-1)^k \frac{x^{2k}}{(2k)!}</cmath> | + | \end{align*}</cmath> |
− | to approximate the roots. Taking the sum up to <math>k = 3</math> yields a close approximation, so we have | + | We use the Taylor series <cmath>\cos x = \sum_{k = 0}^{\infty} (-1)^k \frac{x^{2k}}{(2k)!}</cmath> |
− | <cmath>\cos\left(\frac{2\pi}{7}\right) \ | + | to approximate the roots. |
− | + | ||
− | + | Taking the sum up to <math>k = 3</math> yields a close approximation, so we have | |
+ | <cmath>\begin{alignat*}{8} | ||
+ | \cos\left(\frac{2\pi}{7}\right) &\approx 1-\frac{\left(\frac{2\pi}{7}\right)^{2}}{2}+\frac{\left(\frac{2\pi}{7}\right)^{4}}{24}-\frac{\left(\frac{2\pi}{7}\right)^{6}}{720} &&\approx 0.623, \\ | ||
+ | \cos\left(\frac{4\pi}{7}\right) &\approx 1-\frac{\left(\frac{4\pi}{7}\right)^{2}}{2}+\frac{\left(\frac{4\pi}{7}\right)^{4}}{24}-\frac{\left(\frac{4\pi}{7}\right)^{6}}{720} &&\approx -0.225, \\ | ||
+ | \cos\left(\frac{6\pi}{7}\right) &\approx 1-\frac{\left(\frac{6\pi}{7}\right)^{2}}{2}+\frac{\left(\frac{6\pi}{7}\right)^{4}}{24}-\frac{\left(\frac{6\pi}{7}\right)^{6}}{720} &&\approx -0.964. | ||
+ | \end{alignat*}</cmath> | ||
Note that these approximations get worse as <math>x</math> gets larger, but they will be fine for the purposes of this problem. We then have | Note that these approximations get worse as <math>x</math> gets larger, but they will be fine for the purposes of this problem. We then have | ||
− | <cmath>p + q + r = a \ | + | <cmath>\begin{alignat*}{8} |
− | + | p + q + r &= a &&\approx -0.56, \\ | |
− | + | pq + qr + rq &= -b &&\approx -0.524, \\ | |
− | We further approximate these values to <math>a \ | + | pqr &= c &&\approx 0.135. |
+ | \end{alignat*}</cmath> | ||
+ | We further approximate these values to <math>a \approx -0.5</math>, <math>b \approx 0.5</math>, and <math>c \approx 0.125</math> (mostly as this is an AMC problem and will likely use nice fractions). Thus, we have <math>abc \approx \boxed{\textbf{(D) } \frac{1}{32}}</math>. | ||
+ | |||
+ | <u><b>Remark</b></u> | ||
+ | |||
+ | In order to be more confident in your answer, you can go a few terms further in the Taylor series. | ||
+ | |||
+ | ~ciceronii (Solution) | ||
− | + | ~MRENTHUSIASM (Reformatting) | |
== Solution 3 (Only Using Product to Sum Identity) == | == Solution 3 (Only Using Product to Sum Identity) == |
Revision as of 04:49, 4 September 2021
Contents
Problem
Suppose that the roots of the polynomial are
and
, where angles are in radians. What is
?
Solution 1 (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 2 (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)
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)
By geometric series, we have
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
or
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, it follows that
Geometrically (shown in the Remark section), we get
Recall that
(from which
), and let
By Vieta's Formulas, the answer is
~MRENTHUSIASM (inspired by Peeyush Pandaya et al)
Remark (Geometric Representations)
Graph of the th roots of unity:
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.