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

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</asy>
 
</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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<math>\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}</math>
  
 
==Solution==
 
==Solution==
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label("$\frac{2\sqrt{3}}{3}$",(0.85,1),NW);
 
label("$\frac{2\sqrt{3}}{3}$",(0.85,1),NW);
 
label("$\frac{2\sqrt{3}}{3}$",(2.35,2),N);
 
label("$\frac{2\sqrt{3}}{3}$",(2.35,2),N);
 +
label("$\sqrt{3}$",(1.7,0),S);
 +
label("$2$",(3,1),W);
 
</asy>
 
</asy>
  
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Adding, we get <math>\boxed{\textbf{(B) }  2+\frac{4\sqrt{3}}{3}}</math>.
 
Adding, we get <math>\boxed{\textbf{(B) }  2+\frac{4\sqrt{3}}{3}}</math>.
 +
 +
== Solution 2 ==
 +
After computing <math>\overline{BP} = \frac{2\sqrt{3}}{3},</math> observe that triangle <math>\triangle BPD</math> is isosceles with <math>\angle DPB = \angle BPD.</math> Therefore, using <math>120 - 30 - 30</math> triangle properties, we see that the perimeter is just <math>(2+ \sqrt{3}) \cdot \frac{2\sqrt{3}}{3} = \boxed{\textbf{(B) }  2+\frac{4\sqrt{3}}{3}}.</math>
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 +
~Sliced_Bread
  
 
==See Also==
 
==See Also==

Latest revision as of 13:44, 7 October 2023

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); label("$\sqrt{3}$",(1.7,0),S); label("$2$",(3,1),W); [/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 $\boxed{\textbf{(B) } 2+\frac{4\sqrt{3}}{3}}$.

Solution 2

After computing $\overline{BP} = \frac{2\sqrt{3}}{3},$ observe that triangle $\triangle BPD$ is isosceles with $\angle DPB = \angle BPD.$ Therefore, using $120 - 30 - 30$ triangle properties, we see that the perimeter is just $(2+ \sqrt{3}) \cdot \frac{2\sqrt{3}}{3} = \boxed{\textbf{(B) } 2+\frac{4\sqrt{3}}{3}}.$

~Sliced_Bread

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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