Difference between revisions of "2001 AIME II Problems/Problem 14"
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Z can be written in the form <math> cis\theta</math>. Rearranging, we see that <math> cis{28}\theta</math> = <math>cis{8}\theta</math> <math>+1</math> | Z can be written in the form <math> cis\theta</math>. Rearranging, we see that <math> cis{28}\theta</math> = <math>cis{8}\theta</math> <math>+1</math> | ||
− | Since the real part of <math>cis{28}\theta</math> is one more than the real part of <math>cis {8}\theta</math> and their imaginary parts are equal, it is clear that either <math>cis{28}\theta</math> = <math>{1 | + | Since the real part of <math>cis{28}\theta</math> is one more than the real part of <math>cis {8}\theta</math> and their imaginary parts are equal, it is clear that either <math>cis{28}\theta</math> = <math>\frac{1}{2}+\frac {\sqrt{3}}{2}i</math> and <math>cis {8}\theta</math> = <math>-\frac{1}{2}+\frac {\sqrt{3}}{2}i</math>, or <math>cis{28}\theta</math> = <math>\frac{1}{2} - \frac{\sqrt{3}}{2}i</math> and <math>cis {8}\theta</math> = <math>-\frac{1}{2}- \frac{\sqrt{3}}{2}i</math> |
− | *Case One : <math>cis{28}\theta</math> = <math>{1 | + | *Case One : <math>cis{28}\theta</math> = <math>\frac{1}{2}+ \frac{\sqrt{3}}{2}i</math> and <math>cis {8}\theta</math> = <math>-\frac{1}{2}+\frac{\sqrt{3}}{2}i</math> |
Setting up and solving equations, <math>Z^{28}= cis{60^\circ</math> and <math>Z^8= cis{120^\circ</math>, we see that the only common solutions are <math>15^\circ , 105^\circ, 195^\circ, </math> and <math>\ 285^\circ</math> | Setting up and solving equations, <math>Z^{28}= cis{60^\circ</math> and <math>Z^8= cis{120^\circ</math>, we see that the only common solutions are <math>15^\circ , 105^\circ, 195^\circ, </math> and <math>\ 285^\circ</math> | ||
− | *Case 2 : <math>cis{28}\theta</math> = <math>{1 | + | *Case 2 : <math>cis{28}\theta</math> = <math>\frac{1}{2} -\frac {\sqrt{3}}{2}i</math> and <math>cis {8}\theta</math> = <math>-\frac {1}{2} -\frac{\sqrt{3}}{2}i</math> |
Again setting up equations (<math>Z^{28}= cis{300^\circ</math> and <math>Z^{8} = cis{240^\circ</math>) we see that the only common solutions are <math>75^\circ, 165^\circ, 255^\circ, </math> and <math>345^\circ</math> | Again setting up equations (<math>Z^{28}= cis{300^\circ</math> and <math>Z^{8} = cis{240^\circ</math>) we see that the only common solutions are <math>75^\circ, 165^\circ, 255^\circ, </math> and <math>345^\circ</math> |
Revision as of 09:12, 22 August 2014
Problem
There are complex numbers that satisfy both and . These numbers have the form , where and angles are measured in degrees. Find the value of .
Contents
Solution
Z can be written in the form . Rearranging, we see that =
Since the real part of is one more than the real part of and their imaginary parts are equal, it is clear that either = and = , or = and =
- Case One : = and =
Setting up and solving equations, $Z^{28}= cis{60^\circ$ (Error compiling LaTeX. Unknown error_msg) and $Z^8= cis{120^\circ$ (Error compiling LaTeX. Unknown error_msg), we see that the only common solutions are and
- Case 2 : = and =
Again setting up equations ($Z^{28}= cis{300^\circ$ (Error compiling LaTeX. Unknown error_msg) and $Z^{8} = cis{240^\circ$ (Error compiling LaTeX. Unknown error_msg)) we see that the only common solutions are and
Listing all of these values, it is seen that is equal to which is equal to degrees
See also
2001 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.