Difference between revisions of "2021 AMC 12A Problems/Problem 13"
MRENTHUSIASM (talk | contribs) m (→Solution 2 (Radians)) |
MRENTHUSIASM (talk | contribs) m (→Solution 2 (Radians)) |
||
| Line 28: | Line 28: | ||
<cmath>\begin{array}{c|ccc|cclclclcc} | <cmath>\begin{array}{c|ccc|cclclclcc} | ||
& & & & & & & & & & & & \\ [-2ex] | & & & & & & & & & & & & \\ [-2ex] | ||
| − | \textbf{Choice} & & \boldsymbol{\theta} & & & & & & \textbf{\ Re}\boldsymbol{(z^5)} & & & & \\ \hline | + | \textbf{Choice} & & \boldsymbol{\theta} & & & & & & \textbf{\ Re}\boldsymbol{(z^5)} & & & & \\ [0.5ex] |
| + | \hline | ||
& & & & & & & & & & & & \\ [-1ex] | & & & & & & & & & & & & \\ [-1ex] | ||
\textbf{(A)} & & \pi & & & &32\cos{(5\pi)}&=&32\cos\pi&=&32(-1)& & \\ [2ex] | \textbf{(A)} & & \pi & & & &32\cos{(5\pi)}&=&32\cos\pi&=&32(-1)& & \\ [2ex] | ||
Revision as of 12:08, 24 March 2021
Contents
Problem
Of the following complex numbers 
, which one has the property that 
has the greatest real part?
Solution 1 (Degrees)
First, 
.
Taking the real part of the 5th power of each we have:
,
which is negative
which is zero
Thus, the answer is 
.
~JHawk0224
Solution 2 (Radians)
For every complex number 
where 
and 
are real numbers and 
its magnitude is 
For each choice, we get that the magnitude is 
Rewriting each choice to the polar form 
we know that by the De Moivre's Theorem, the real part of is ![\[\text{Re}(z^5)=r^5\cos{(5\theta)}.\]](http://latex.artofproblemsolving.com/a/f/1/af127495e241a14498d00545eec2c736d4321c2f.png)
We make a table as follows:
![\[\begin{array}{c|ccc|cclclclcc} & & & & & & & & & & & & \\ [-2ex] \textbf{Choice} & & \boldsymbol{\theta} & & & & & & \textbf{\ Re}\boldsymbol{(z^5)} & & & & \\ [0.5ex] \hline & & & & & & & & & & & & \\ [-1ex] \textbf{(A)} & & \pi & & & &32\cos{(5\pi)}&=&32\cos\pi&=&32(-1)& & \\ [2ex] \textbf{(B)} & & \frac{5\pi}{6} & & & &32\cos{\frac{25\pi}{6}}&=&32\cos{\frac{\pi}{6}}&=&32\left(\frac{\sqrt3}{2}\right)& & \\ [2ex] \textbf{(C)} & & \frac{3\pi}{4} & & & &32\cos{\frac{15\pi}{4}}&=&32\cos{\frac{7\pi}{4}}&=&32\left(\frac{\sqrt2}{2}\right)& & \\ [2ex] \textbf{(D)} & & \frac{2\pi}{3} & & & &32\cos{\frac{10\pi}{3}}&=&32\cos{\frac{4\pi}{3}}&=&32\left(-\frac{1}{2}\right)& & \\ [2ex] \textbf{(E)} & & \frac{\pi}{2} & & & &32\cos{\frac{5\pi}{2}}&=&32\cos{\frac{\pi}{2}}&=&32\left(0\right)& & \\ [1ex] \end{array}\]](http://latex.artofproblemsolving.com/8/d/c/8dc30295a1948263253012ad2f3c524f95a8d7c0.png)
Clearly, the answer is 
~MRENTHUSIASM
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=FD9BE7hpRvg
Video Solution by Hawk Math
https://www.youtube.com/watch?v=AjQARBvdZ20
Video Solution by OmegaLearn (Using Polar Form and de Moivre's Theorem)
~ pi_is_3.14
Solution 5
Video Solution by TheBeautyofMath
https://youtu.be/ySWSHyY9TwI?t=568
~IceMatrix
See Also
| 2021 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 12 |
Followed by Problem 14 |
| 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. 
