2010 AMC 10A Problem 16

Using the Angle Bisector Theorem, we have that $$\dfrac{AB}{3} = \dfrac{BC}{8}$$$$AB = \dfrac{3}{8} BC.$$ For $AB$ and $BC$ to be integers, we must have that $BC$ is a multiple of $8.$ To minimize the perimeter, we can set $BC = 8$ and $AB = 3.$ This, however, makes the triangle degenerate. $BC$ must then be $16$ and $AB = 6.$ Since $AC = AD + DC = 11,$ the perimeter is $$16 + 6 + 11 = 33.$$ Thus, B is the correct answer.