2018 AMC 12B Problem 16

Problem 16 of 25IntermediateAlgebra Geometry

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?

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Solution

Translating by

6,

the solutions of

z^8=81

are eight points on a circle of radius

81^{\frac{1}{8}}=\sqrt3,

forming a regular octagon. The minimum-area triangle uses three consecutive vertices. Take

A=\left(\tfrac12\sqrt6,\tfrac12\sqrt6\right),

B=(\sqrt3,0),

and

C=\left(\tfrac12\sqrt6,-\tfrac12\sqrt6\right).

Then

AC=\sqrt6

and the height is

\sqrt3-\tfrac12\sqrt6,

so the area is

\begin{gathered} \tfrac12\cdot\sqrt6\left(\sqrt3-\tfrac12\sqrt6\right) \\ =\tfrac{3}{2}\sqrt2-\tfrac{3}{2}. \end{gathered}

Thus, the correct answer is B.