Difference between revisions of "1984 AIME Problems/Problem 8"
(deleted category) |
(→Solution) |
||
| Line 6: | Line 6: | ||
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^3=1</math> and <math>r^6+r^3+1=3</math>, a contradiction. This leaves <math>\boxed{\theta=160}</math>. | 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^3=1</math> and <math>r^6+r^3+1=3</math>, a contradiction. This leaves <math>\boxed{\theta=160}</math>. | ||
| + | |||
| + | Also, | ||
| + | |||
| + | From above, you notice that <math>z^6+z^3+1 = r^9-1/r^3-1</math>. Therefore, the solutions are all of the ninth roots of unity that are not the third roots of unity. After checking, the only angle is <math>\boxed{\theta=160}</math>. | ||
== See also == | == See also == | ||
Revision as of 08:44, 14 March 2010
Problem
The equation 
has complex roots with argument 
between 
and 
in the complex plane. Determine the degree measure of .
Solution
If 
is a root of , then 
. The polynomial 
has all of its roots with absolute value 
and argument of the form 
for integer 
.
This reduces to either 
or 
. But can't be because if 
, then 
and 
, a contradiction. This leaves 
.
Also,
From above, you notice that 
. Therefore, the solutions are all of the ninth roots of unity that are not the third roots of unity. After checking, the only angle is .
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 | ||
