Difference between revisions of "1984 AIME Problems/Problem 8"
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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>. | 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 Problems and Solutions]] |
| − | * [[ | + | * [[American Invitational Mathematics Examination]] |
| + | * [[Mathematics competition resources]] | ||
Revision as of 13:24, 6 May 2007
Problem
The equation 
has complex roots with argument 
between 
and 
in thet complex plane. Determine the degree measure of .
Solution
If 
is a root of 
, then 
. The polynomial 
has all of its roots with absolute value 1 and argument of the form 
for integer 
.
This reduces to either 120 or 160. But can't be 120 because if 
, then 
and 
, a contradiction. This leaves 
.
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 | ||
