Difference between revisions of "1984 AIME Problems/Problem 8"
(→Solution) |
|||
| Line 2: | Line 2: | ||
The equation <math>z^6+z^3+1</math> has [[complex root]]s with argument <math>\theta</math> between <math>90^\circ</math> and <math>180^\circ</math> in the [[complex plane]]. Determine the degree measure of <math>\theta</math>. | The equation <math>z^6+z^3+1</math> has [[complex root]]s with argument <math>\theta</math> between <math>90^\circ</math> and <math>180^\circ</math> in the [[complex plane]]. Determine the degree measure of <math>\theta</math>. | ||
| − | == Solution == | + | == Solution 1 == |
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]] <math>1</math> and argument of the form <math>40m^\circ</math> for integer <math>m</math>. | 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]] <math>1</math> and argument of the form <math>40m^\circ</math> for integer <math>m</math>. | ||
| Line 11: | Line 11: | ||
From above, you notice that <math>z^6+z^3+1 = \frac {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>. | From above, you notice that <math>z^6+z^3+1 = \frac {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>. | ||
| + | == Solution 2 == | ||
| + | Note that the substitution <math>y=z^3</math> simplifies this to <math>y^2+y+1</math>. Simply applying the quadratic formula gives roots <math>y_{1,2}=\frac{1}{2}\pm \frac{\sqrt{3}i}{2}</math>, which have angles of 120 and 240, respectively. This means <math>arg(z) = \frac{120,240}{3} + \frac{360n}{3}</math>, and the only one between 90 and 180 is <math>\boxed{\theta=160}</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=1984|num-b=7|num-a=9}} | {{AIME box|year=1984|num-b=7|num-a=9}} | ||
[[Category:Intermediate Trigonometry Problems]] | [[Category:Intermediate Trigonometry Problems]] | ||
Revision as of 22:41, 10 February 2011
Contents
Problem
The equation 
has complex roots with argument 
between 
and 
in the complex plane. Determine the degree measure of .
Solution 1
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 .
Solution 2
Note that the substitution 
simplifies this to 
. Simply applying the quadratic formula gives roots 
, which have angles of 120 and 240, respectively. This means 
, and the only one between 90 and 180 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 | ||
