+230 In rectangle $ABCD$, $AD=1$, $P$ is on $\overline{AB}$, and $\overline{DB}$ and $\overline{DP}$ trisect $\angle ADC$. Write the perimeter of $\triangle BDP$ in simplest form as: $w + \frac{x \cdot \sqrt{y}}{z}$, where $w, x, y, z$ are nonnegative integers. What is $w, x, y, z$?
+591 We see that APD and DBC are 30-60-90 triangles, so
DC=$\sqrt3$,
AP=1/$\sqrt3$=$\sqrt3$/3
PB=DC-AP=2$\sqrt3$/3
PD=$\sqrt3$/3$\cdot$2=s$\sqrt3$/3
DB=2
PB+DP+DB=$2+\frac{4\cdot\sqrt3}3$
So $w, x, y,$ and $z$ are $\boxed{2,4,3,3}$ respectively.
