2000 AMC 10 Problem 7

The right angle $\angle ADC = 90^\circ$ is trisected into three $30^\circ$ angles, so $\angle ADP = 30^\circ$ and $\angle ADB = 60^\circ.$ In right triangle $ADP,$ with $AD = 1,$ we get $DP = \dfrac{1}{\cos 30^\circ} = \dfrac{2\sqrt3}{3}$ and $AP = \tan 30^\circ = \dfrac{\sqrt3}{3}.$ In right triangle $ADB,$ with $AD = 1,$ we get $DB = \dfrac{1}{\cos 60^\circ} = 2$ and $AB = \tan 60^\circ = \sqrt3.$ Then $$\begin{aligned} PB = AB - AP &= \sqrt3 - \dfrac{\sqrt3}{3} \\ &= \dfrac{2\sqrt3}{3}. \end{aligned}$$ The perimeter of $\triangle BDP$ is $$\begin{gathered} DP + PB + DB \\ = \dfrac{2\sqrt3}{3} + \dfrac{2\sqrt3}{3} + 2 \\ = 2 + \dfrac{4\sqrt3}{3}. \end{gathered}$$ Thus, the correct answer is B.