2011 AMC 12A Problem 13

Let $I$ be the incenter. Because $\overline{BI}$ bisects $\angle B$ and $MN \parallel BC,$ alternate angles give $\angle MIB = \angle IBC = \angle MBI,$ so $\triangle MBI$ is isosceles with $MB = MI.$ Similarly $NC = NI.$ Therefore the perimeter of $\triangle AMN$ is $$\begin{gathered} AM + MN + NA \\ = AM + (MI + IN) + NA \\ = AM + MB + NC + NA \\ = AB + AC = 12 + 18 = 30. \end{gathered}$$ Thus, the correct answer is B.