# 2011 AMC 12B Problems/Problem 24

## Problem

Let $P(z) = z^8 + \left(4\sqrt{3} + 6\right)z^4 - \left(4\sqrt{3} + 7\right)$. What is the minimum perimeter among all the $8$-sided polygons in the complex plane whose vertices are precisely the zeros of $P(z)$? $\textbf{(A)}\ 4\sqrt{3} + 4 \qquad \textbf{(B)}\ 8\sqrt{2} \qquad \textbf{(C)}\ 3\sqrt{2} + 3\sqrt{6} \qquad \textbf{(D)}\ 4\sqrt{2} + 4\sqrt{3} \qquad \textbf{(E)}\ 4\sqrt{3} + 6$

## Solution

First of all, we need to find all $z$ such that $P(z) = 0$ $P(z) = \left(z^4 - 1\right)\left(z^4 + \left(4\sqrt{3} + 7\right)\right)$

So $z^4 = 1 \implies z = e^{i\frac{n\pi}{2}}$

or $z^4 = - \left(4\sqrt{3} + 7\right)$ $z^2 = \pm i \sqrt{4\sqrt{3} + 7} = e^{i\frac{(2n+1)\pi}{2}} \left(\sqrt{3} + 2\right)$ $z = e^{i\frac{(2n+1)\pi}{4}} \sqrt{\sqrt{3} + 2} = e^{i\frac{(2n+1)\pi}{4}} \left(\frac{\sqrt{3}}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right)$

Now we have a solution at $\frac{n\pi}{4}$ if we look at them in polar coordinate, further more, the 8-gon is symmetric (it is an equilateral octagon) . So we only need to find the side length of one and multiply by $8$.

So answer $= 8 \times$ distance from $1$ to $\left(\frac{\sqrt{3}}{2} + \frac{1}{2}\right)\left(1 + i\right)$

Side length $= \sqrt{\left(\frac{\sqrt{3}}{2} - \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2} + \frac{1}{2}\right)^2} = \sqrt{2\left(\frac{3}{4} + \frac{1}{4}\right)} = \sqrt{2}$

Hence, answer is $8\sqrt{2}$.

## Solution 2

Use the law of cosines. We make $a$ the distance. Now, since the angle does not change the distance from the origin, we can just use the distance. $a^2 = (\frac{\sqrt{3}}{\sqrt{2}} + \frac{1}{\sqrt{2}})^2 + 1^2 -2 \times \Big( \frac{\sqrt{3}}{\sqrt{2}} + \frac{1}{\sqrt{2}} \Big)\times 1 \times \cos \frac{\pi}{4}$, which simplifies to $a^2= 2 + \sqrt3 +1 - 1 - \sqrt3$, or $a^2=2$, or $a=\sqrt2$. Multiply the answer by 8 to get $\boxed{ (B) 8\sqrt2}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 