Difference between revisions of "1984 AIME Problems/Problem 8"

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== Problem ==
 
== Problem ==
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The equation <math>z^6+z^3+1</math> has complex roots with argument <math>\theta</math> between <math>90^\circ</math> and <math>180^\circ</math> in the complex plane. Determine the degree measure of <math>\theta</math>.
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The equation <math>z^6+z^3+1=0</math> has [[complex root]]s with argument <math>\theta</math> between <math>90^\circ</math> and <math>180^\circ</math> in the [[complex plane]]. Determine the degree measure of <math>\theta</math>.
  
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== Solution ==
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== Solution 1 ==
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If <math>r</math> is a root of <math>z^6+z^3+1</math>, then <math>0=(r^3-1)(r^6+r^3+1)=r^9-1</math>. The polynomial <math>x^9-1</math> has all of its roots with absolute value 1 and argument of the form <math>40m^\circ</math> for integer <math>m</math>.
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We shall introduce another factor to make the equation easier to solve. If <math>r</math> is a root of <math>z^6+z^3+1</math>, then <math>0=(r^3-1)(r^6+r^3+1)=r^9-1</math>. The polynomial <math>x^9-1</math> has all of its roots with [[absolute value]] <math>1</math> and argument of the form <math>40m^\circ</math> for integer <math>m</math> (the ninth degree [[roots of unity]]). Now we simply need to find the root within the desired range that satisfies our original equation <math>x^6 + x^3 + 1 = 0</math>.
  
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This reduces <math>\theta</math> to either <math>120^{\circ}</math> or <math>160^{\circ}</math>. But <math>\theta</math> can't be <math>120^{\circ}</math> because if <math>r=\cos 120^\circ +i\sin 120^\circ </math>, then <math>r^6+r^3+1=3</math>. (When we multiplied by <math>r^3 - 1</math> at the beginning, we introduced some extraneous solutions, and the solution with <math>120^\circ</math> was one of them.) This leaves <math>\boxed{\theta=160}</math>.
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== Solution 2 ==
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The substitution <math>y=z^3</math> simplifies the equation to <math>y^2+y+1 = 0</math>. Applying the quadratic formula gives roots <math>y=-\frac{1}{2}\pm \frac{\sqrt{3}i}{2}</math>, which have arguments of <math>120</math> and <math>240,</math> respectively. This means <math>\arg(z) = \frac{120 \; \text{or} \;240}{3} + \frac{360n}{3}</math>, and the only one between 90 and 180 is <math>\boxed{\theta=160}</math>.
  
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This reduces <math>\theta</math> to either 120 or 160. But <math>\theta</math> can't be 120 because if <math>r=\cos 120^\circ +i\sin 120^\circ </math>, then <math>r^3=1</math> and <math>r^6+r^3+1=3</math>, a contradiction. This leaves <math>\theta=160</math>.
 
 
== See also ==
 
== See also ==
 
{{AIME box|year=1984|num-b=7|num-a=9}}
 
{{AIME box|year=1984|num-b=7|num-a=9}}
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* [[AIME Problems and Solutions]]
 
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* [[American Invitational Mathematics Examination]]
 
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* [[Mathematics competition resources]]
 
  
 
[[Category:Intermediate Trigonometry Problems]]
 
[[Category:Intermediate Trigonometry Problems]]
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[[Category:Intermediate Complex Numbers Problems]]
 

Latest revision as of 12:58, 1 June 2020

Problem

The equation $z^6+z^3+1=0$ has complex roots with argument $\theta$ between $90^\circ$ and $180^\circ$ in the complex plane. Determine the degree measure of $\theta$.

Solution 1

We shall introduce another factor to make the equation easier to solve. If $r$ is a root of $z^6+z^3+1$, then $0=(r^3-1)(r^6+r^3+1)=r^9-1$. The polynomial $x^9-1$ has all of its roots with absolute value $1$ and argument of the form $40m^\circ$ for integer $m$ (the ninth degree roots of unity). Now we simply need to find the root within the desired range that satisfies our original equation $x^6 + x^3 + 1 = 0$.

This reduces $\theta$ to either $120^{\circ}$ or $160^{\circ}$. But $\theta$ can't be $120^{\circ}$ because if $r=\cos 120^\circ +i\sin 120^\circ$, then $r^6+r^3+1=3$. (When we multiplied by $r^3 - 1$ at the beginning, we introduced some extraneous solutions, and the solution with $120^\circ$ was one of them.) This leaves $\boxed{\theta=160}$.

Solution 2

The substitution $y=z^3$ simplifies the equation to $y^2+y+1 = 0$. Applying the quadratic formula gives roots $y=-\frac{1}{2}\pm \frac{\sqrt{3}i}{2}$, which have arguments of $120$ and $240,$ respectively. This means $\arg(z) = \frac{120 \; \text{or} \;240}{3} + \frac{360n}{3}$, and the only one between 90 and 180 is $\boxed{\theta=160}$.

See also

1984 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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