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

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==Problem==
 
  
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>?
 
 
<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>
 
 
<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>
 
 
 
<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>
 
 
<math>AD=1</math>.
 
 
Since <math>\angle ADC</math> is trisected, <math>\angle ADP= \angle PDB= \angle BDC=30^\circ</math>.
 
 
Thus, <math>PD=\frac{2\sqrt{3}}{3}</math>
 
 
<math>DB=2</math>
 
 
<math>BP=\sqrt{3}-\frac{\sqrt{3}}{3}=\frac{2\sqrt{3}}{3}</math>.
 
 
Adding, we get <math>2+\frac{4\sqrt{3}}{3}</math>
 
 
<math>\boxed{\text{B}}</math>
 

Revision as of 16:25, 29 April 2018

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