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

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== Solution ==
 
== Solution ==
{{solution}}
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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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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 ==
 
* [[1984 AIME Problems/Problem 7 | Previous problem]]
 
* [[1984 AIME Problems/Problem 7 | Previous problem]]
 
* [[1984 AIME Problems/Problem 9 | Next problem]]
 
* [[1984 AIME Problems/Problem 9 | Next problem]]
 
* [[1984 AIME Problems]]
 
* [[1984 AIME Problems]]

Revision as of 20:27, 26 March 2007

Problem

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

Solution

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 $m$.


This reduces $\theta$ to either 120 or 160. But $\theta$ can't be 120 because if $r=\cos 120^\circ +i\sin 120^\circ$, then $r^3=1$ and $r^6+r^3+1=3$, a contradiction. This leaves $\theta=160$.

See also

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