Difference between revisions of "2018 AMC 12B Problems/Problem 16"

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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>
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== See Also ==
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{{AMC12 box|year=2018|ab=B|num-b=15|num-a=17}}
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{{MAA Notice}}

Revision as of 14:35, 16 February 2018

Problem

The solutions to the equation $(z+6)^8=81$ are connected in the complex plane to form a convex regular polygon, three of whose vertices are labeled $A,B,$ and $C$. What is the least possible area of $\triangle ABC?$

$\textbf{(A) } \frac{1}{6}\sqrt{6} \qquad \textbf{(B) } \frac{3}{2}\sqrt{2}-\frac{3}{2} \qquad \textbf{(C) } 2\sqrt3-3\sqrt2 \qquad \textbf{(D) } \frac{1}{2}\sqrt{2} \qquad \textbf{(E) } \sqrt 3-1$

Solution

The answer is the same if we consider $z^8=81.$ Now we just need to find the area of the triangle bounded by $\sqrt 3i, \sqrt 3,$ and $\frac{\sqrt 3}{\sqrt 2}+\frac{\sqrt 3}{\sqrt 2}i.$ This is just $\boxed{\textbf{B.}}$


See Also

2018 AMC 12B (ProblemsAnswer KeyResources)
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

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