Difference between revisions of "2021 AMC 12A Problems/Problem 22"
m (→Solution) |
(Added approximation solution) |
||
Line 4: | Line 4: | ||
<math>\textbf{(A) }-\frac{3}{49} \qquad \textbf{(B) }-\frac{1}{28} \qquad \textbf{(C) }\frac{^3\sqrt7}{64} \qquad \textbf{(D) }\frac{1}{32}\qquad \textbf{(E) }\frac{1}{28}</math> | <math>\textbf{(A) }-\frac{3}{49} \qquad \textbf{(B) }-\frac{1}{28} \qquad \textbf{(C) }\frac{^3\sqrt7}{64} \qquad \textbf{(D) }\frac{1}{32}\qquad \textbf{(E) }\frac{1}{28}</math> | ||
− | ==Solution== | + | ==Solution 1== |
Part 1: solving for c | Part 1: solving for c | ||
Line 69: | Line 69: | ||
~Tucker | ~Tucker | ||
+ | |||
+ | == Solution 2 (Approximation) == | ||
+ | Letting the roots be <math>p</math>, <math>q</math>, and <math>r</math>, Vietas gives | ||
+ | \begin{align*} | ||
+ | p + q + r &= a \\ | ||
+ | pq + qr + pq &= -b \\ | ||
+ | pqr &= c. | ||
+ | \end{align*} | ||
+ | We use the Taylor series for <math>\cos x</math>, | ||
+ | <cmath>\cos x = \sum_{k = 0}^{\infty} (-1)^k \frac{x^{2k}}{(2k)!}</cmath> | ||
+ | to approximate the roots. Taking the sum up to <math>k = 3</math> yields a close approximation, so we have | ||
+ | <cmath>\cos\left(\frac{2\pi}{7}\right) \simeq 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} \simeq 0.623</cmath> | ||
+ | <cmath>\cos\left(\frac{4\pi}{7}\right) \simeq 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} \simeq -0.225</cmath> | ||
+ | <cmath>\cos\left(\frac{6\pi}{7}\right) \simeq 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} \simeq -0.964.</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 | ||
+ | \begin{align*} | ||
+ | p + q + r &= a \simeq -0.56 \\ | ||
+ | pq + qr + pr &= -b \simeq -0.524 \\ | ||
+ | pqr &= c \simeq 0.135. | ||
+ | \end{align*} | ||
+ | We further approximate these values to <math>a \simeq -0.5</math>, <math>b \simeq 0.5</math>, and <math>c \simeq 0.125</math> (mostly as this is an AMC problem and will likely use nice fractions). Thus, we have <math>abc \simeq \boxed{\textbf{(D) } \frac{1}{32}}</math>. ~ciceronii | ||
== Video Solution by OmegaLearn (Euler's Identity + Vieta's ) == | == Video Solution by OmegaLearn (Euler's Identity + Vieta's ) == |
Revision as of 14:57, 12 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 c
Notice that
is the negation of the product of roots by Vieta's formulas
Multiply by
Then use sine addition formula backwards:
Part 2: starting to solve 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 is shown in the next part to equal , so
Part 3: solving for a and b as the sum of roots
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
Finally multiply or .
~Tucker
Solution 2 (Approximation)
Letting the roots be , , and , Vietas gives \begin{align*}
p + q + r &= a \\ pq + qr + pq &= -b \\ pqr &= c.
\end{align*} 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
\begin{align*}
p + q + r &= a \simeq -0.56 \\ pq + qr + pr &= -b \simeq -0.524 \\ pqr &= c \simeq 0.135.
\end{align*} We further approximate these values to , , and (mostly as this is an AMC problem and will likely use nice fractions). Thus, we have . ~ciceronii
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.