Difference between revisions of "1984 AIME Problems/Problem 8"
Meta Knight (talk | contribs) (→Solution 1) |
Expotential (talk | contribs) (→Solution 2) |
||
Line 12: | Line 12: | ||
== Solution 2 == | == 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>. | + | 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 06:11, 30 June 2012
Contents
Problem
The equation has complex roots with argument between and in the complex plane. Determine the degree measure of .
Solution 1
We shall introduce another factor to make the equation easier to solve. Consider . If is a root of , then . The polynomial has all of its roots with absolute value and argument of the form for integer . Now we simply need to find the root within the desired range that satisfies our original equation .
This reduces to either or . But can't be because if , then . 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 |