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

Revision as of 08:29, 18 October 2007

Problem

The equation $z^6+z^3+1$ 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 $\theta=160$ .

@@ Line 1: / Line 1: @@
 == Problem ==
-The equation <math>\displaystyle 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 thet complex plane. Determine the degree measure of <math>\theta</math>.
+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 thet complex plane. Determine the degree measure of <math>\theta</math>.
 == Solution ==
@@ Line 12: / Line 12: @@
 * [[American Invitational Mathematics Examination]]
 * [[Mathematics competition resources]]
+[[Category:Trigonometry Problems]]

1984 AIME (Problems • Answer Key • Resources)
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Difference between revisions of "1984 AIME Problems/Problem 8"

Revision as of 08:29, 18 October 2007

Problem

Solution

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