Difference between revisions of "1984 AIME Problems/Problem 8"
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== Solution 1 == | == Solution 1 == | ||
− | We shall introduce another factor to make the equation easier to solve. Consider <math>r^3-1</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>. Now we simply need to find the root within the desired range that satisfies our original equation <math>x^6 + x^3 + 1 = 0</math>. | + | We shall introduce another factor to make the equation easier to solve. Consider <math>r^3-1</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> (the ninth degree [[roots of unity]]). Now we simply need to find the root within the desired range that satisfies our original equation <math>x^6 + x^3 + 1 = 0</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^6+r^3+1=3</math>. 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^6+r^3+1=3</math>. This leaves <math>\boxed{\theta=160}</math>. | ||
− | + | Note: From above, notice that <math>z^6+z^3+1 = \frac {r^9-1}{r^3-1}</math>. Therefore, the solutions are all of the ninth degree roots of unity that are not also the third degree roots of unity. Checking, we see that the only angle is <math>\boxed{\theta=160}</math>. | |
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− | From above, | ||
== Solution 2 == | == Solution 2 == |
Revision as of 11:03, 29 June 2013
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 (the ninth degree roots of unity). 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 .
Note: From above, notice that . Therefore, the solutions are all of the ninth degree roots of unity that are not also the third degree roots of unity. Checking, we see that 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 |