2011 AMC 12B Problem 24

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

Treat

P

as a quadratic in

z^4

and factor as

(z^4-1)\big(z^4+(4\sqrt3+7)\big)

Since

4\sqrt3+7=(\sqrt3+2)^2,

the eight roots lie on two squares; the shortest edges have length

\sqrt2

Factoring in

z^4,

\begin{aligned} P(z) &=(z^4-1) \\ &\quad {}\cdot\big(z^4+(4\sqrt3+7)\big). \end{aligned}

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.

Problem 24 in other years