2018 AMC 12B Problem 16

Question

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?

\dfrac{1}{6}\sqrt{6}

\dfrac{3}{2}\sqrt{2}-\dfrac{3}{2}

2\sqrt{3}-2\sqrt{2}

\dfrac{1}{2}\sqrt{2}

\sqrt{3}-1

Shifting by

6

does not change the shape; the roots of

z^8=81

form a regular octagon

The circumradius is

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

and the smallest triangle uses three consecutive vertices

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.

Problem 16 in other years