Difference between revisions of "2018 AMC 12B Problems/Problem 16"
(Created page with "== Problem == The solutions to the equation <math>(z+6)^8=81</math> are connected in the complex plane to form a convex regular polygon, three of whose vertices are labeled <...") |
|||
| Line 8: | Line 8: | ||
The answer is the same if we consider <math>z^8=81.</math> Now we just need to find the area of the triangle bounded by <math>\sqrt 3i, \sqrt 3,</math> and <math>\frac{\sqrt 3}{\sqrt 2}+\frac{\sqrt 3}{\sqrt 2}i.</math> This is just <math>\boxed{\textbf{B.}}</math> | The answer is the same if we consider <math>z^8=81.</math> Now we just need to find the area of the triangle bounded by <math>\sqrt 3i, \sqrt 3,</math> and <math>\frac{\sqrt 3}{\sqrt 2}+\frac{\sqrt 3}{\sqrt 2}i.</math> This is just <math>\boxed{\textbf{B.}}</math> | ||
| + | |||
| + | |||
| + | == See Also == | ||
| + | |||
| + | {{AMC12 box|year=2018|ab=B|num-b=15|num-a=17}} | ||
| + | {{MAA Notice}} | ||
Revision as of 13:35, 16 February 2018
Problem
The solutions to the equation 
are connected in the complex plane to form a convex regular polygon, three of whose vertices are labeled 
and 
. What is the least possible area of 
Solution
The answer is the same if we consider 
Now we just need to find the area of the triangle bounded by 
and 
This is just 
See Also
| 2018 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 15 |
Followed by Problem 17 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
