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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 $w, x, y, z$?

 Mar 4, 2021
edited by wizzymath  Mar 4, 2021
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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.

 Mar 4, 2021

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