Difference between revisions of "2000 AMC 10 Problems/Problem 7"
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| + | ==Problem== | ||
| + | |||
| + | ==Solution== | ||
| + | |||
<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); | ||
| Line 17: | Line 21: | ||
<math>AD=1</math>. | <math>AD=1</math>. | ||
| − | Since <math>\angle ADC</math> is trisected, <math>\angle ADP= \angle PDB= \angle BDC=30</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>DB=2</math> | ||
| + | |||
<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 | + | <math>\boxed{\text{B}}</math> |
| + | |||
| + | ==See Also== | ||
| + | |||
| + | {{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]](http://latex.artofproblemsolving.com/1/0/a/10ad0d6c91539b40c085057a166894ac1cede67f.png)
.
Since 
is trisected, 
.
Thus, 
.
Adding, 
.
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 | ||
