AMC12 2011 12B, Problem 24

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 $P(z)$?

Solution: 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 $\fbox{8\sqrt{2}}$

Answer: 8\sqrt{2}