Difference between revisions of "2001 AIME II Problems/Problem 14"
Makkahjones (talk | contribs) (→Solution) |
Mathgeek2006 (talk | contribs) m (→Solution) |
||
Line 5: | Line 5: | ||
==Solution== | ==Solution== | ||
− | Z can be written in the form <math> cis\theta</math>. Rearranging, we | + | <math>Z</math> can be written in the form <math> \text{cis\,}\theta</math>. Rearranging, we find that <math> \text{cis\,}{28}\theta = \text{cis\,}{8}\theta+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 | + | Since the real part of <math>\text{cis\,}{28}\theta</math> is one more than the real part of <math>\text{cis\,} {8}\theta</math> and their imaginary parts are equal, it is clear that either <math>\text{cis\,}{28}\theta = \frac{1}{2}+\frac {\sqrt{3}}{2}i</math> and <math>\text{cis\,} {8}\theta = -\frac{1}{2}+\frac {\sqrt{3}}{2}i</math>, or <math>\text{cis\,}{28}\theta = \frac{1}{2} - \frac{\sqrt{3}}{2}i</math> and <math>\text{cis\,} {8}\theta = -\frac{1}{2}- \frac{\sqrt{3}}{2}i</math> |
− | *Case | + | *Case 1 : <math>\text{cis\,}{28}\theta = \frac{1}{2}+ \frac{\sqrt{3}}{2}i</math> and <math>\text{cis\,} {8}\theta = -\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 solutions common to both equations have arguments <math>15^\circ , 105^\circ, 195^\circ, </math> and <math>\ 285^\circ</math> | + | Setting up and solving equations, <math>Z^{28}= \text{cis\,}{60^\circ}</math> and <math>Z^8= \text{cis\,}{120^\circ}</math>, we see that the solutions common to both equations have arguments <math>15^\circ , 105^\circ, 195^\circ, </math> and <math>\ 285^\circ</math> |
− | *Case 2 : <math>cis{28}\theta | + | *Case 2 : <math>\text{cis\,}{28}\theta = \frac{1}{2} -\frac {\sqrt{3}}{2}i</math> and <math>\text{cis\,} {8}\theta = -\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 common solutions have arguments of <math>75^\circ, 165^\circ, 255^\circ, </math> and <math>345^\circ</math> | + | Again setting up equations (<math>Z^{28}= \text{cis\,}{300^\circ}</math> and <math>Z^{8} = \text{cis\,}{240^\circ}</math>) we see that the common solutions have arguments of <math>75^\circ, 165^\circ, 255^\circ, </math> and <math>345^\circ</math> |
− | Listing all of these values, | + | Listing all of these values, we find that <math>\theta_{2} + \theta_{4} + \ldots + \theta_{2n}</math> is equal to <math>(75 + 165 + 255 + 345) ^\circ</math> which is equal to <math>\boxed {840}</math> degrees |
== See also == | == See also == |
Revision as of 18:50, 13 March 2015
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
can be written in the form . Rearranging, we find 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 1 : and
Setting up and solving equations, and , we see that the solutions common to both equations have arguments and
- Case 2 : and
Again setting up equations ( and ) we see that the common solutions have arguments of and
Listing all of these values, we find 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.