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# Difference between revisions of "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 $\triangle BDP$?

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

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

## Solution

$[asy] draw((0,2)--(3.4,2)--(3.4,0)--(0,0)--cycle); draw((0,0)--(1.3,2)); draw((0,0)--(3.4,2)); dot((0,0)); dot((0,2)); dot((3.4,2)); dot((3.4,0)); dot((1.3,2)); label("A",(0,2),NW); label("B",(3.4,2),NE); label("C",(3.4,0),SE); label("D",(0,0),SW); label("P",(1.3,2),N); label("1",(0,1),W); label("2",(1.7,1),SE); label("\frac{\sqrt{3}}{3}",(0.65,2),N); label("\frac{2\sqrt{3}}{3}",(0.85,1),NW); label("\frac{2\sqrt{3}}{3}",(2.35,2),N); [/asy]$

$AD=1$.

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, we get $\boxed{\textbf{(B) } 2+\frac{4\sqrt{3}}{3}}$.

## See Also

 2000 AMC 10 (Problems • Answer Key • Resources) Preceded byProblem 6 Followed byProblem 8 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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