Difference between revisions of "2008 AMC 10B Problems/Problem 10"

(Problem)
(Solution)
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==Solution==
 
==Solution==
We know that <math>u_6=128</math>, so we plug in <math>n=4</math> to get <math>128=2u_5+u_4</math>. We plug in <math>n=3</math> to get
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Let the center of the circle be O. Draw lines
  
<math>u_5=2u_4+9</math>. Substituting gives
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OA, OB, and OC.
  
<math>128=5u_4+18 \rightarrow u_4=22</math>
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OA=OB=5, since they are both radii.
  
This gives <math>u_5=\frac{128-22}{2}=53</math>.
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OC bisects AB, and AB=6, so letting point D be
  
Answer B is the correct answer
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the intersection of AB and OC, AD=BD=3.
  
NOTE: This is my (BOGTRO) solution, not the official one,
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Also, OD=4. This means that CD=1.
  
and should be ignored in view of a better solution.
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Using the pythagorean theorem,
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 +
<math>AC=\sqrt{3^2+1^2}=\sqrt{10}</math>.
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 +
Answer A is the correct answer.
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2008|ab=B|num-b=9|num-a=11}}
 
{{AMC10 box|year=2008|ab=B|num-b=9|num-a=11}}

Revision as of 17:41, 10 August 2008

Problem

Points A and B are on a circle of radius 5 and AB=6. Point C is the midpoint of the minor arc AB. What is the length of the line segment AC?

Solution

Let the center of the circle be O. Draw lines

OA, OB, and OC.

OA=OB=5, since they are both radii.

OC bisects AB, and AB=6, so letting point D be

the intersection of AB and OC, AD=BD=3.

Also, OD=4. This means that CD=1.

Using the pythagorean theorem,

$AC=\sqrt{3^2+1^2}=\sqrt{10}$.

Answer A is the correct answer.

See also

2008 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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
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