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Difference between revisions of "2000 AMC 10 Problems/Problem 7"

m (→‎Problem)
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==Problem==
  
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In rectangle <math>ABCD</math>, <math>AD=1</math>, <math>P</math> is on <math>\overline{AB}</math>, and <math>\overline{DB}</math> and <math>\overline{DP}</math> trisect <math>\angle ADC</math>.  What is the perimeter of <math>\triangle BDP</math>?
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<asy>
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draw((0,2)--(3.4,2)--(3.4,0)--(0,0)--cycle);
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draw((0,0)--(1.3,2));
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draw((0,0)--(3.4,2));
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dot((0,0));
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dot((0,2));
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dot((3.4,2));
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dot((3.4,0));
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dot((1.3,2));
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label("$A$",(0,2),NW);
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label("$B$",(3.4,2),NE);
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label("$C$",(3.4,0),SE);
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label("$D$",(0,0),SW);
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label("$P$",(1.3,2),N);
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</asy>
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<math>\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}</math>
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==Solution==
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<asy>
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draw((0,2)--(3.4,2)--(3.4,0)--(0,0)--cycle);
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draw((0,0)--(1.3,2));
 +
draw((0,0)--(3.4,2));
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dot((0,0));
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dot((0,2));
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dot((3.4,2));
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dot((3.4,0));
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dot((1.3,2));
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label("$A$",(0,2),NW);
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label("$B$",(3.4,2),NE);
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label("$C$",(3.4,0),SE);
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label("$D$",(0,0),SW);
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label("$P$",(1.3,2),N);
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label("$1$",(0,1),W);
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label("$2$",(1.7,1),SE);
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label("$\frac{\sqrt{3}}{3}$",(0.65,2),N);
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label("$\frac{2\sqrt{3}}{3}$",(0.85,1),NW);
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label("$\frac{2\sqrt{3}}{3}$",(2.35,2),N);
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</asy>
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<math>AD=1</math>.
 +
 +
Since <math>\angle ADC</math> is trisected, <math>\angle ADP= \angle PDB= \angle BDC=30^\circ</math>.
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Thus, <math>PD=\frac{2\sqrt{3}}{3}</math>
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<math>DB=2</math>
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<math>BP=\sqrt{3}-\frac{\sqrt{3}}{3}=\frac{2\sqrt{3}}{3}</math>.
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Adding, we get <math>\boxed{\textbf{(B) }  2+\frac{4\sqrt{3}}{3}}</math>.
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==See Also==
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 +
{{AMC10 box|year=2000|num-b=6|num-a=8}}
 +
{{MAA Notice}}

Revision as of 17:12, 6 August 2018

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); 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 by
Problem 6 		Followed by
Problem 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

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