Difference between revisions of "1984 AIME Problems/Problem 8"
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== Solution 2 == | == 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>. | |
== See also == | == See also == |
Revision as of 12:03, 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
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 |