2000 AMC 10 Problems/Problem 7

Problem

In rectangle $ABCD$ , $AD=1$ , $P$ is on $\overline{AB}$ , and $\overline{DB}$ and $\overline{DP}$ trisect $\angle ADC$ . What is the perimeter of $\triangle BDP$ ?

$[asy] draw((0,2)--(3.4,2)--(3.4,0)--(0,0)--cycle); draw((0,0)--(1.3,2)); draw((0,0)--(3.4,2)); dot((0,0)); dot((0,2)); dot((3.4,2)); dot((3.4,0)); dot((1.3,2)); label("$A$",(0,2),NW); label("$B$",(3.4,2),NE); label("$C$",(3.4,0),SE); label("$D$",(0,0),SW); label("$P$",(1.3,2),N); [/asy]$

$\mathrm{(A)}\ 3+\frac{\sqrt{3}}{3} \qquad\mathrm{(B)}\ 2+\frac{4\sqrt{3}}{3} \qquad\mathrm{(C)}\ 2+2\sqrt{2} \qquad\mathrm{(D)}\ \frac{3+3\sqrt{5}}{2} \qquad\mathrm{(E)}\ 2+\frac{5\sqrt{3}}{3}$

Solution

$[asy] draw((0,2)--(3.4,2)--(3.4,0)--(0,0)--cycle); draw((0,0)--(1.3,2)); draw((0,0)--(3.4,2)); dot((0,0)); dot((0,2)); dot((3.4,2)); dot((3.4,0)); dot((1.3,2)); label("$A$",(0,2),NW); label("$B$",(3.4,2),NE); label("$C$",(3.4,0),SE); label("$D$",(0,0),SW); label("$P$",(1.3,2),N); [/asy]$

$AD=1$ .

Since $\angle ADC$ is trisected, $\angle ADP= \angle PDB= \angle BDC=30^\circ$ .

Thus, $PD=\frac{2\sqrt{3}}{3}$

$DB=2$

$BP=\sqrt{3}-\frac{\sqrt{3}}{3}=\frac{2\sqrt{3}}{3}$ .

Adding, $2+\frac{4\sqrt{3}}{3}$ .

$\boxed{\text{B}}$

2000 AMC 10 (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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2000 AMC 10 Problems/Problem 7

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