Difference between revisions of "2017 AMC 12A Problems/Problem 17"
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<math> \textbf{(A)}\ 0 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 24 </math> | <math> \textbf{(A)}\ 0 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 24 </math> | ||
− | ==Solution== | + | ==Solution 1== |
Note that these <math>z</math> such that <math>z^{24}=1</math> are <math>e^{\frac{ni\pi}{12}}</math> for integer <math>0\leq n<24</math>. So | Note that these <math>z</math> such that <math>z^{24}=1</math> are <math>e^{\frac{ni\pi}{12}}</math> for integer <math>0\leq n<24</math>. So | ||
Line 11: | Line 11: | ||
<math>z^6=e^{\frac{ni\pi}{2}}</math> | <math>z^6=e^{\frac{ni\pi}{2}}</math> | ||
− | This is real | + | This is real if <math>\frac{n}{2}\in \mathbb{Z} \Leftrightarrow (n</math> is even<math>)</math>. Thus, the answer is the number of even <math>0\leq n<24</math> which is <math>\boxed{(D)=\ 12}</math>. |
− | |||
==Solution 2== | ==Solution 2== | ||
<math>z = \sqrt[24]{1} = 1^{\frac{1}{24}}</math> | <math>z = \sqrt[24]{1} = 1^{\frac{1}{24}}</math> | ||
− | By [[Euler's identity]], <math>1 = e^{0 \ | + | By [[Euler's identity]], <math>1 = e^{0 \times i} = \cos (0+2k\pi) + i \sin(0+2k\pi)</math>, where <math>k</math> is an integer. |
+ | |||
+ | Using [[De Moivre's Theorem]], we have <math>z = 1^{\frac{1}{24}} = {\cos (\frac{k\pi}{12}) + i \sin (\frac{k\pi}{12})}</math>, where <math>0 \leq k<24</math> that produce <math>24</math> unique results. | ||
+ | |||
+ | Using De Moivre's Theorem again, we have <math>z^6 = {\cos (\frac{k\pi}{2}) + i \sin (\frac{k\pi}{2})}</math> | ||
− | + | For <math>z^6</math> to be real, <math>\sin(\frac{k\pi}{2})</math> has to equal <math>0</math> to negate the imaginary component. This occurs whenever <math>\frac{k\pi}{2}</math> is an integer multiple of <math>\pi</math>, requiring that <math>k</math> is even. There are exactly <math>\boxed{12}</math> even values of <math>k</math> on the interval <math>0 \leq k<24</math>, so the answer is <math>\boxed{(D)}</math>. | |
− | + | ==Solution 3== | |
+ | From the start, recall from the Fundamental Theorem of Algebra that <math>z^{24} = 1</math> must have <math>24</math> solutions (and these must be distinct since the equation factors into <math>0 = (z-1)(z^{23} + z^{22} + z^{21}... + z + 1)</math>), or notice that the question is simply referring to the 24th roots of unity, of which we know there must be <math>24</math>. Notice that <math>1 = z^{24} = (z^6)^4</math>, so for any solution <math>z</math>, <math>z^6</math> will be one of the 4th roots of unity (<math>1</math>, <math>i</math>, <math>-1</math>, or <math>-i</math>). Then <math>6</math> solutions <math>z</math> will satisfy <math>z^6 = 1</math>, <math>6</math> will satisfy <math>z^6 = -1</math> (and this is further justified by knowledge of the 6th roots of unity), so there must be <math>\boxed{(D) \: 12}</math> such <math>z</math>. | ||
− | + | ==Solution 4 (Quick)== | |
+ | Let <math>a\in\mathbb{R}</math> and <math>a = z^6.</math> We have | ||
+ | <cmath>a^4 = 1 \implies a = 1,-1.</cmath> | ||
+ | <math>z^6 = \pm 1</math> has 6 solutions for <math>1</math> and <math>-1</math> respectively, so <math>6+6=\boxed{(D)\ 12}.</math> <cmath> </cmath> | ||
+ | -svyn | ||
+ | ==Solution 5 (Visual Roots of Unity)== | ||
+ | Because <math>z^{24} = 1</math>, we can plot these points on the Argand plane as a regular 24-gon, as shown: | ||
+ | [[Image:2017AMC12aP17.png]] | ||
+ | These are a graphical representation of all 24 values of z, as stated in the problem. Now, we want <math>z^6</math> to be real. The only 2 cases where this happens are if <math>z^6 = 1</math> or <math>z^6 = -1</math>. Squaring both sides for the latter equation, we get <math>z^{12}=1</math>, which, if one were to square root it, would give us a system of both <math>z^6 = 1</math> and <math>z^6 = -1</math>, just as we desire. We can plot the points for <math>z^{12}=1</math> on an Argand plane again, giving us: | ||
+ | [[Image:2017AMC12aP17-p2.png]] | ||
+ | We note that all of these points are also on the first Argand plane, and counting the points nets us <math>\boxed{(D)\ 12}</math> total values for <math>z</math>. <cmath> </cmath> | ||
+ | -yingkai_0_ <cmath> </cmath> | ||
+ | Credit to Michael Andrejkovics for providing the GeoGebra widget used to make these diagrams! | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2017|ab=A|num-b=16|num-a=18}} | {{AMC12 box|year=2017|ab=A|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 00:42, 19 September 2024
Contents
Problem
There are different complex numbers such that . For how many of these is a real number?
Solution 1
Note that these such that are for integer . So
This is real if is even. Thus, the answer is the number of even which is .
Solution 2
By Euler's identity, , where is an integer.
Using De Moivre's Theorem, we have , where that produce unique results.
Using De Moivre's Theorem again, we have
For to be real, has to equal to negate the imaginary component. This occurs whenever is an integer multiple of , requiring that is even. There are exactly even values of on the interval , so the answer is .
Solution 3
From the start, recall from the Fundamental Theorem of Algebra that must have solutions (and these must be distinct since the equation factors into ), or notice that the question is simply referring to the 24th roots of unity, of which we know there must be . Notice that , so for any solution , will be one of the 4th roots of unity (, , , or ). Then solutions will satisfy , will satisfy (and this is further justified by knowledge of the 6th roots of unity), so there must be such .
Solution 4 (Quick)
Let and We have has 6 solutions for and respectively, so -svyn
Solution 5 (Visual Roots of Unity)
Because , we can plot these points on the Argand plane as a regular 24-gon, as shown: These are a graphical representation of all 24 values of z, as stated in the problem. Now, we want to be real. The only 2 cases where this happens are if or . Squaring both sides for the latter equation, we get , which, if one were to square root it, would give us a system of both and , just as we desire. We can plot the points for on an Argand plane again, giving us: We note that all of these points are also on the first Argand plane, and counting the points nets us total values for . -yingkai_0_ Credit to Michael Andrejkovics for providing the GeoGebra widget used to make these diagrams!
See Also
2017 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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.