Difference between revisions of "1984 AIME Problems/Problem 8"
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== Problem == | == Problem == | ||
− | The equation <math> | + | The equation <math>z^6+z^3+1=0</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 1 and argument of the form <math>40m^\circ</math> for integer <math>m</math>. | + | We shall introduce another factor to make the equation easier to solve. 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>. (When we multiplied by <math>r^3 - 1</math> at the beginning, we introduced some extraneous solutions, and the solution with <math>120^\circ</math> was one of them.) This leaves <math>\boxed{\theta=160}</math>. | ||
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+ | == Solution 2 == | ||
+ | The substitution <math>y=z^3</math> simplifies the equation to <math>y^2+y+1 = 0</math>. Applying the quadratic formula gives roots <math>y=-\frac{1}{2}\pm \frac{\sqrt{3}i}{2}</math>, which have arguments of <math>120</math> and <math>240,</math> respectively. This means <math>\arg(z) = \frac{120 \; \text{or} \;240}{3} + \frac{360n}{3}</math>, and the only one between 90 and 180 is <math>\boxed{\theta=160}</math>. | ||
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== See also == | == See also == | ||
{{AIME box|year=1984|num-b=7|num-a=9}} | {{AIME box|year=1984|num-b=7|num-a=9}} | ||
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− | + | [[Category:Intermediate Trigonometry Problems]] | |
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Latest revision as of 12:58, 1 June 2020
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. 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 . (When we multiplied by at the beginning, we introduced some extraneous solutions, and the solution with was one of them.) This leaves .
Solution 2
The substitution simplifies the equation to . Applying the quadratic formula gives roots , which have arguments of and 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 |