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 $\triangle ABC?$

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

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

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

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

$\sqrt{3}-1$

Solution:

Translating by $6,$ the solutions of $z^8=81$ are eight points on a circle of radius $81^{1/8}=\sqrt3,$ 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 $\tfrac12\cdot\sqrt6\left(\sqrt3-\tfrac12\sqrt6\right)=\tfrac{3}{2}\sqrt2-\tfrac{3}{2}.$

Thus, the correct answer is B.

2018 AMC 12B Problem 16

Solution:

Problem 16 in Other Years