Difference between revisions of "2008 AMC 10B Problems/Problem 6"
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==Solution== | ==Solution== | ||
− | Let CD = 1. Then AB = 4(BC+1), and AB+BC = 9 | + | Let <math>\overline{CD} = 1</math>. Then <math>\overline{AB} = 4(\overline{BC}+1)</math>, and <math>\overline{AB}+\overline{BC} = 9\cdot1</math>. From this system of equations we obtain <math>\overline{BC} = 1</math>. Adding <math>\overline{CD}</math> to both sides of the second equation, we obtain <math>\overline{AB}+\overline{BC}+\overline{CD} = 9+1 = 10 = \overline{AD}</math>. Thus, <math>\frac{\overline{BC}}{\overline{AD}} = \frac{1}{10} \implies\text{(C)}</math> |
==See also== | ==See also== | ||
{{AMC10 box|year=2008|ab=B|num-b=5|num-a=7}} | {{AMC10 box|year=2008|ab=B|num-b=5|num-a=7}} |
Revision as of 12:57, 29 October 2009
Problem
Points and lie on . The length of is times the length of , and the length of is times the length of . The length of is what fraction of the length of ?
Solution
Let . Then , and . From this system of equations we obtain . Adding to both sides of the second equation, we obtain . Thus,
See also
2008 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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 |