Difference between revisions of "2008 AMC 10B Problems/Problem 6"
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==Solution== | ==Solution== | ||
| − | Let <math>CD = 1</math>. Then <math>AB = 4(BC + 1)</math> and <math>AB + BC = 9\cdot1</math>. From this system of equations we obtain <math>BC = 1</math>. Adding <math>CD</math> to both sides of the second equation, we obtain <math>AD = AB + BC + CD = 9 + 1 = 10</math>. Thus, <math>\frac{BC}{AD} = \frac{1}{10} \implies\text{(C)}</math> | + | Let <math>CD = 1</math>. Then <math>AB = 4(BC + 1)</math> and <math>AB + BC = 9\cdot1</math>. From this system of equations, we obtain <math>BC = 1</math>. Adding <math>CD</math> to both sides of the second equation, we obtain <math>AD = AB + BC + CD = 9 + 1 = 10</math>. Thus, <math>\frac{BC}{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}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 11:06, 7 June 2021
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
