Some background, the AMC 12 is the first of several USA team Math Olympiad qualifying exams. It consists of 25 problems that can be solved without calculus to be solved in 75 minutes without the use of a calculator.

A past problem I am having trouble with is:

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 $\triangle ABC?$

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

The Art of Problem Solving solution found here is:

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 $\boxed{\textbf{B.}}$

As the answer is written, I don't see where it comes from. Why is the answer the same if we consider $z^8=81$? How do you find the triangle bounds from $z^8=81$? I understand that the bounds are the solutions to the equation, however, I do not understand what I must use to find them analytically. Any help understanding how to solve this would be greatly appreciated.

  • 4
    $\begingroup$ Translation preserves area... $\endgroup$ – Idonknow Feb 3 '19 at 14:59

You don't need complex numbers here.

It's obvious that we need to find $S_{\Delta ABC},$ where $A$, $B$ and $C$ are neighboring vertices of the equilateral octagon with the radius of the circumcircle $$\sqrt[8]{81}=\sqrt3.$$ We see that $$BC=AB=2\sqrt3\sin\measuredangle BCA=2\sqrt3\sin22.5^{\circ}$$ and since $$\measuredangle ABC=180^{\circ}-\frac{360^{\circ}}{8}=135^{\circ},$$ we obtain: $$S_{\Delta ABC}=\frac{1}{2}\left(2\sqrt3\sin22.5^{\circ}\right)^2\sin135^{\circ}=3(1-\cos45^{\circ})\frac{1}{\sqrt2}=\frac{3}{2}(\sqrt2-1).$$

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