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

Latest revision as of 23:40, 14 July 2024

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

Video Solution by Daily Dose of Math

https://youtu.be/wITXxUtZj3E?si=KuLEI9SeOFZTw05Y

~Thesmartgreekmathdude

@@ Line 19: / Line 19: @@
 </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>
+<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==
@@ Line 42: / Line 42: @@
 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>
@@ Line 54: / Line 56: @@
 <math>BP=\sqrt{3}-\frac{\sqrt{3}}{3}=\frac{2\sqrt{3}}{3}</math>.
-Adding, we get <math>\boxed{\text{B) }2+\frac{4\sqrt{3}}{3}</math>}.
+Adding, we get <math>\boxed{\textbf{(B) }  2+\frac{4\sqrt{3}}{3}}</math>.
-<math>\boxed{\text{B}}</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>
+~Sliced_Bread
+==Video Solution by Daily Dose of Math==
+https://youtu.be/wITXxUtZj3E?si=KuLEI9SeOFZTw05Y
+~Thesmartgreekmathdude
 ==See Also==
@@ Line 62: / Line 73: @@
 {{AMC10 box|year=2000|num-b=6|num-a=8}}
 {{MAA Notice}}
+[[Category:Introductory Geometry Problems]]

2000 AMC 10 (Problems • Answer Key • Resources)
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