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

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==Problem==
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==Solution==
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<asy>
 
<asy>
 
draw((0,2)--(3.4,2)--(3.4,0)--(0,0)--cycle);
 
draw((0,2)--(3.4,2)--(3.4,0)--(0,0)--cycle);
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<math>AD=1</math>.
 
<math>AD=1</math>.
  
Since <math>\angle ADC</math> is trisected, <math>\angle ADP= \angle PDB= \angle BDC=30</math>.
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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>
  
Thus, <math>PD=\frac{2\sqrt{3}}{3}</math>.
 
 
<math>DB=2</math>
 
<math>DB=2</math>
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<math>BP=\sqrt{3}-\frac{\sqrt{3}}{3}=\frac{2\sqrt{3}}{3}</math>.
 
<math>BP=\sqrt{3}-\frac{\sqrt{3}}{3}=\frac{2\sqrt{3}}{3}</math>.
  
 
Adding, <math>2+\frac{4\sqrt{3}}{3}</math>.
 
Adding, <math>2+\frac{4\sqrt{3}}{3}</math>.
  
B.
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<math>\boxed{\text{B}}</math>
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==See Also==
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{{AMC10 box|year=2000|num-b=6|num-a=8}}

Revision as of 18:35, 8 January 2009

Problem

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

$\boxed{\text{B}}$

See Also

2000 AMC 10 (ProblemsAnswer KeyResources)
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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