2012 AMC 12B Problem 23

Because $|z_0|=1,$ applying the triangle inequality to the identity $$\begin{aligned} &4z_0^5-(z_0-1)P(z_0) \\ &\quad =z_0^4(4-a)+z_0^3(a-b) \\ &\quad {}+z_0^2(b-c)+z_0(c-d)+d \end{aligned}$$ has left side of absolute value $4.$ The nonnegative coefficient jumps on the right sum to $4,$ so the triangle inequality is an equality and all its nonzero complex summands point in the same direction. If two jumps are nonzero, their quotient shows that $z_0^k$ is a positive real for some $1\le k\le4.$ Since $|z_0|=1,$ this means $z_0^k=1.$ The cases $k=2,3,4$ give, respectively, $a=4,b=c,d=0;$ $a=b=4,c=d=0;$ and a polynomial already in the first family. If no such power exists, equality forces exactly one nonzero jump. The constant jump gives $a=b=c=d=4;$ any other jump again forces $z_0^{5-j}=1$ and returns to the cases just listed. Hence the polynomials are exactly $4z^4+4z^3+bz^2+bz$ for $0\le b\le4,$ together with $4z^4+4z^3+4z^2$ and $4z^4+4z^3+4z^2+4z+4.$ Their values at $1$ are $20,$ $12,$ and $8+2b;$ summing gives $$\begin{gathered} 20+12+\sum_{b=0}^{4}(8+2b) \\ = 32+40+20 \\ = 92. \end{gathered}$$ Thus, the correct answer is B.