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 19:35, 8 January 2009
Problem
Solution
.
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 |