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 03-24-2009 #1 MAS1   Join Date: Dec 2008 Posts: 249 The Lighthouse and the Dog A dog is chained to a point on the outside of a circular lighthouse of radius 1. The length of the dog's chain is Pi. The dog may not go inside the lighthouse. Over exactly how much area is the dog free to roam?
 03-25-2009 #2 Mr. Hui     Join Date: Mar 2005 Posts: 10,609 Drawing a picture seems to be a good start but I could see it might involve math above the middle school level. __________________ Do Math and you can do Anything!
04-04-2009   #3
Sugengz

Join Date: Mar 2009
Posts: 4

Quote:
 Originally Posted by Mr. Hui Drawing a picture seems to be a good start but I could see it might involve math above the middle school level.
I agree...

I just put the drawing... With some comment...

On the left side the semicircle is the area that could cover by the dog with length of rope as it’s restriction… It’s a semicircle with radius of π.

Going to the right… The dog is restricted by the length of the rope and also by the lighthouse.

As the dog move to right with the rope straightened to achieve furthest distance…

If θ in radians... The total length of A to R1
Minor arc AB = r.θ = θ
BR1 = π – θ (The total length of A to R1 = Length of rope = π)

R1 keep on moving at the furthest point exactly at Z.
(since the Arc AZ = r . θ = 1 . π = π)

If we could build an equation of the movement of R1 related to the distance of R1 to line AZ….

Then we could find the area covered by integration of the function from A to Z … Then subtract the result by the area of the light house…
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 02-20-2010 #4 jmw106462     Join Date: Feb 2010 Posts: 388 Put a Pic Up. I Don't Really Know If u type it.
09-13-2010   #5
Sirjon

Join Date: Sep 2010
Posts: 15

Quote:
 Originally Posted by Sugengz I agree... I just put the drawing... With some comment... On the left side the semicircle is the area that could cover by the dog with length of rope as it’s restriction… It’s a semicircle with radius of π. Going to the right… The dog is restricted by the length of the rope and also by the lighthouse. As the dog move to right with the rope straightened to achieve furthest distance… If θ in radians... The total length of A to R1 Minor arc AB = r.θ = θ BR1 = π – θ (The total length of A to R1 = Length of rope = π) R1 keep on moving at the furthest point exactly at Z. (since the Arc AZ = r . θ = 1 . π = π) If we could build an equation of the movement of R1 related to the distance of R1 to line AZ…. Then we could find the area covered by integration of the function from A to Z … Then subtract the result by the area of the light house…
Subtract the area of the smaller circle from the area of the bigger circle?

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