Difference between revisions of "1984 AIME Problems/Problem 8"
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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^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>. | ||
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+ | 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 09:44, 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 |