Difference between revisions of "2001 AIME II Problems/Problem 14"

Latest revision as of 14:03, 22 July 2018

Problem

There are $2n$ complex numbers that satisfy both $z^{28} - z^{8} - 1 = 0$ and $\mid z \mid = 1$ . These numbers have the form $z_{m} = \cos\theta_{m} + i\sin\theta_{m}$ , where $0\leq\theta_{1} < \theta_{2} < \ldots < \theta_{2n} < 360$ and angles are measured in degrees. Find the value of $\theta_{2} + \theta_{4} + \ldots + \theta_{2n}$ .

Solution

$z$ can be written in the form $\text{cis\,}\theta$ . Rearranging, we find that $\text{cis\,}{28}\theta = \text{cis\,}{8}\theta+1$

Since the real part of $\text{cis\,}{28}\theta$ is one more than the real part of $\text{cis\,} {8}\theta$ and their imaginary parts are equal, it is clear that either $\text{cis\,}{28}\theta = \frac{1}{2}+\frac {\sqrt{3}}{2}i$ and $\text{cis\,} {8}\theta = -\frac{1}{2}+\frac {\sqrt{3}}{2}i$ , or $\text{cis\,}{28}\theta = \frac{1}{2} - \frac{\sqrt{3}}{2}i$ and $\text{cis\,} {8}\theta = -\frac{1}{2}- \frac{\sqrt{3}}{2}i$

Case 1 : $\text{cis\,}{28}\theta = \frac{1}{2}+ \frac{\sqrt{3}}{2}i$ and $\text{cis\,} {8}\theta = -\frac{1}{2}+\frac{\sqrt{3}}{2}i$

Setting up and solving equations, $Z^{28}= \text{cis\,}{60^\circ}$ and $Z^8= \text{cis\,}{120^\circ}$ , we see that the solutions common to both equations have arguments $15^\circ , 105^\circ, 195^\circ,$ and $\ 285^\circ$ . We can figure this out by adding 360 repeatedly to the number 60 to try and see if it will satisfy what we need. We realize that it does not work in the integer values.

Case 2 : $\text{cis\,}{28}\theta = \frac{1}{2} -\frac {\sqrt{3}}{2}i$ and $\text{cis\,} {8}\theta = -\frac {1}{2} -\frac{\sqrt{3}}{2}i$

Again setting up equations ( $Z^{28}= \text{cis\,}{300^\circ}$ and $Z^{8} = \text{cis\,}{240^\circ}$ ) we see that the common solutions have arguments of $75^\circ, 165^\circ, 255^\circ,$ and $345^\circ$

Listing all of these values, we find that $\theta_{2} + \theta_{4} + \ldots + \theta_{2n}$ is equal to $(75 + 165 + 255 + 345) ^\circ$ which is equal to $\boxed {840}$ degrees. We only want the sum of a certain number of theta, not all of it.

@@ Line 3: / Line 3: @@
 __TOC__
-== Solution ==
+==Solution==
-=== Solution 1 ===
-To satisfy  <math>z^{28} - z^{8} - 1 = 0</math>, <math>\text{Im}\,(z^{28})=\text{Im}\,(z^{8})</math> and <math>\text{Re}\,(z^{28})=\text{Re}\,(z^{8})+1</math>.
-Since <math>\mid z \mid = 1</math>, <math>z</math> is on the [[unit circle]] centered at the origin in the [[complex plane]].
+<math>z</math> can be written in the form <math> \text{cis\,}\theta</math>. Rearranging, we find that  <math> \text{cis\,}{28}\theta = \text{cis\,}{8}\theta+1</math>
-Since <math>\text{Im}\,(z^{28})=\text{Im}\,(z^{8})</math>, <math>z^{28}</math> and <math>z^8</math> have the same <math>y</math> coordinate. Since <math>\text{Re}\,(z^{28})=\text{Re}\,(z^{8})+1</math>, <math>z^{28}</math> is <math>1</math> unit to the right of <math>z^{8}</math>. It is easy to see that the only possibilities are <math>(z^{28},z^{8})=(\text{cis}\,(60),\text{cis}\,(120))</math> or <math>(\text{cis}\,{(300)},\text{cis}\,{(240)})</math>.
+Since the real part of <math>\text{cis\,}{28}\theta</math> is one more than the real part of <math>\text{cis\,} {8}\theta</math> and their imaginary parts are equal, it is clear that either <math>\text{cis\,}{28}\theta = \frac{1}{2}+\frac {\sqrt{3}}{2}i</math> and <math>\text{cis\,} {8}\theta =  -\frac{1}{2}+\frac {\sqrt{3}}{2}i</math>, or <math>\text{cis\,}{28}\theta = \frac{1}{2} - \frac{\sqrt{3}}{2}i</math> and  <math>\text{cis\,} {8}\theta =  -\frac{1}{2}- \frac{\sqrt{3}}{2}i</math>
-<center><asy>
+*Case 1  : <math>\text{cis\,}{28}\theta = \frac{1}{2}+ \frac{\sqrt{3}}{2}i</math> and <math>\text{cis\,} {8}\theta =  -\frac{1}{2}+\frac{\sqrt{3}}{2}i</math>
-pathpen = black+linewidth(0.7); pen l = linewidth(0.6);
+Setting up and solving equations, <math>Z^{28}= \text{cis\,}{60^\circ}</math> and <math>Z^8= \text{cis\,}{120^\circ}</math>, we see that the solutions common to both equations have arguments <math>15^\circ , 105^\circ, 195^\circ, </math> and <math>\ 285^\circ</math>. We can figure this out by adding 360 repeatedly to the number 60 to try and see if it will satisfy what we need. We realize that it does not work in the integer values.
-D(unitcircle); D((-1.5,0)--(1.5,0),l,Arrows(5)); D((0,-1.5)--(0,1.5),l,Arrows(5));
-D(D(expi(pi/3))--D(expi(2*pi/3)),EndArrow(3)); D(D(expi(4*pi/3)) -- D(expi(5*pi/3)),BeginArrow(3));
-MP("1",(0.5,0));MP("1",(0,3^.5/2),SE);MP("\mathrm{cis}60",expi(1*pi/3),NE);MP("\mathrm{cis}120",expi(2*pi/3),NW);MP("\mathrm{cis}240",expi(4*pi/3),SW);MP("\mathrm{cis}300",expi(5*pi/3),SE);
-</asy></center>
-For the first possibility:
+*Case 2  : <math>\text{cis\,}{28}\theta = \frac{1}{2} -\frac {\sqrt{3}}{2}i</math> and <math>\text{cis\,} {8}\theta =  -\frac {1}{2} -\frac{\sqrt{3}}{2}i</math>
-<cmath>
+Again setting up equations (<math>Z^{28}= \text{cis\,}{300^\circ}</math> and <math>Z^{8} = \text{cis\,}{240^\circ}</math>) we see that the common solutions have arguments of  <math>75^\circ, 165^\circ, 255^\circ, </math> and <math>345^\circ</math>
-\begin{align*}
-z^{28}=\text{cis}\,(28\theta)=\text{cis}\,(60) \Rightarrow 28\theta \equiv 60 \pmod{360} &\Rightarrow \theta \equiv 15 \pmod{90} \\
-z^{8}=\text{cis}\,(8\theta)=\text{cis}\,(120) \Rightarrow 8\theta \equiv 120 \pmod{360} &\Rightarrow \theta \equiv 15 \pmod{45} \end{align*}</cmath>
-Thus, <math>\theta \equiv 15 \pmod{90}</math>. This yields <math>\theta = 15, 105, 195, 285</math>.
+Listing all of these values, we find that <math>\theta_{2} + \theta_{4} + \ldots + \theta_{2n}</math> is equal to <math>(75 + 165 + 255 + 345) ^\circ</math> which is equal to <math>\boxed {840}</math> degrees. We only want the sum of a certain number of theta, not all of it.
-For the second possibility:
-<cmath> \begin{align*}
-z^{28}=\text{cis}\,(28\theta)=\text{cis}\,(300) \Rightarrow 28\theta \equiv 300 \pmod{360} &\Rightarrow \theta \equiv 75 \pmod{90} \\
-z^{8}=\text{cis}\,(8\theta)=\text{cis}\,(240) \Rightarrow 8\theta \equiv 240 \pmod{360} &\Rightarrow \theta \equiv 30 \pmod{45} \end{align*}</cmath>
-Thus, <math>\theta \equiv 75 \pmod{90}</math>. This yields <math>\theta = 75, 165, 255, 345</math>.
-Therefore <math>(\theta_1,\theta_2,\theta_3,\theta_4,\theta_5,\theta_6,\theta_7,\theta_8)=(15,75,105,165,195,255,285,345)</math> and <math>\theta_2+\theta_4+\theta_6+\theta_8=\boxed{840}</math>.
-=== Solution 2 ===
-Rearrange the given equation as <math>z^8\left(z^{20}-1\right) = 1</math>; the magnitudes of both sides must be equal, so
-<cmath>\left|z^8\left(z^{20}-1\right)\right| = \left|z^{20}-1\right| = \left| 1 \right| = 1</cmath>
-Thus the distance between <math>z^{20} = \text{cis}\, 20\theta </math> and <math>(1,0)</math> on the coordinate plane is <math>1</math>. By the distance formula,
-<cmath>1 = \sqrt{(\cos 20\theta - 1)^2 + \sin ^2 20\theta} = \sqrt{2 - 2 \cos 20\theta} \Longrightarrow \cos 20\theta = \frac 12</cmath>
-And <math>20\theta = 60, 300 + 360n</math>, while <math>z^{20} - 1 = \frac{1}{2} \pm \frac{\sqrt{3}}{2}i - 1 = \text{cis}\,(120,240)</math>. Thus <math>z^8 = \frac{1}{z^{20}-1} = \text{cis}^{-1}\, \{120,240\} = \text{cis}\,\{240,120\}</math>. We thus have <math>20\theta = 60 + 360n</math> and <math>8\theta = 240 + 360n</math> or <math>20\theta = 300 + 360n</math> and <math>8\theta = 120 + 360n</math>. From here, follow the above solution.
 == See also ==
@@ Line 52: / Line 22: @@
 [[Category:Intermediate Algebra Problems]]
+{{MAA Notice}}

2001 AIME II (Problems • Answer Key • Resources)
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Difference between revisions of "2001 AIME II Problems/Problem 14"

Latest revision as of 14:03, 22 July 2018

Problem

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Solution

See also