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 $w + x + y + z$?

w+x+y+z = 12

When right angle ADC is divided into three equal parts (trisected), each of those parts is 90°/3 = 30°. Thus, ∠CDB = ∠PBD = ∠PDB = 30°.

The sides of a 30°-60°-90° triangle are in the proportion 1 : √3 : 2. Hence BD = 2, and PD = PB = 2/√3 = (2/3)√3.

Then the perimeter of ∆BDP is ...

... perimeter ∆BDP = BD + DP + PB

... = 2 + (2/3)√3 + (2/3)√3

... = 2 + (4√3)/3 . . . . . . . . w=2, x=4, y=3, z=3

w + x + y + z = 2+4+3+3 = 12