Question

Let $P(z)=z^8+(4\sqrt{3}+6)z^4-(4\sqrt{3}+7).$ What is the minimum perimeter among all the $8$ -sided polygons in the complex plane whose vertices are precisely the zeros of $P(z)?$

$4\sqrt{3}+4$

$8\sqrt{2}$

$3\sqrt{2}+3\sqrt{6}$

$4\sqrt{2}+4\sqrt{3}$

$4\sqrt{3}+6$

Solution:

Factoring in $z^4,$ $P(z)=(z^4-1)\big(z^4+(4\sqrt3+7)\big).$ The first factor gives the roots $1,-1,i,-i.$ Since $4\sqrt3+7=(\sqrt3+2)^2$ and $2(\sqrt3+2)=(\sqrt3+1)^2,$ writing $w=\tfrac12(\sqrt3+1)$ the other four roots are $w(\pm1\pm i).$

The eight roots are symmetric about the origin with $4$ -fold symmetry, and every segment joining two of them has length at least $\sqrt2.$ Thus any such polygon has perimeter at least $8\sqrt2,$ and the polygon with vertices $1, w(1+i), i, w(-1+i), -1, w(-1-i), -i, w(1-i)$ achieves it.

Thus, the correct answer is B.

2011 AMC 12B Problem 24

Solution:

Problem 24 in Other Years