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Math Help - perimeter

  1. #1
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    Wink perimeter

    Two unequal circles (radius x and y, x > y) are touching each other. A rubber band is passed around both of them. What would be the length of the rubber band?
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  2. #2
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    Quote Originally Posted by vipinz View Post
    Two unequal circles (radius x and y, x > y) are touching each other. A rubber band is passed around both of them. What would be the length of the rubber band?
    It would help you to draw a diagram of this situation first...
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  3. #3
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    .....like this?
    Attached Thumbnails Attached Thumbnails perimeter-temp.jpg  
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    Answer

    Very good diagram. But what is the answer.
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    Quote Originally Posted by vipinz View Post
    Very good diagram. But what is the answer.
    any other info on relationship between the two radii...?
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  6. #6
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    There is no other information. The only information available is that the two radii are not equal.
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  7. #7
    Senior Member nikhil's Avatar
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    no other information is needed

    try pythagoras theorem
    use co-axial circles (it will only help in easy visualization)
    answer is
    pi(x+y)+4[xy]^(1/2)
    Last edited by nikhil; November 1st 2009 at 08:51 AM.
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  8. #8
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    Quote Originally Posted by nikhil View Post
    try pythagoras theorem
    use co-axial circles (it will only help in easy visualization)
    answer is
    pi(x+y)+4[xy]^(1/2)
    what do you do with those angles at the centres?
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  9. #9
    Senior Member nikhil's Avatar
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    in the give figure move the line that joins the centres of the circles downwards(or upwards).
    now you will get a right angle triangle whose one side is x+y, other side is x-y and the third side can be found (pythagoras theorem)which is the length of segment of rubber band.by symmetry other other segment of band will also have same length.
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  10. #10
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    Quote Originally Posted by vipinz View Post
    Two unequal circles (radius x and y, x > y) are touching each other. A rubber band is passed around both of them. What would be the length of the rubber band?
    nikhil's formula will work in rough cut situations.
    However, if you need an exact fit, try this:

    Let R be the radius of the larger circle.
    Let r be the radius of the smaller circle.

    Extend the two tangent lines until they intersect.
    Let theta be the angle at the intersection.

     \theta = 2 sin^{-1}\left( \dfrac{R-r}{R+r}\right)

    LengthOfRubberBand=  R(\pi+\theta) + r(\pi -\theta) + 2\sqrt{4Rr}

    Theta in radians.
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  11. #11
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    Nikhil must explain how he got the sum of the lengths of the two arcs as

    pi (x + y). The explanation about the part of the rubber band between the two circles is OK.

    Aidan, I think you are introducing a new variable "theta". Your answer must be only in terms of x and y.

    Vipin
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  12. #12
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    Quote Originally Posted by vipinz View Post
    Nikhil must explain how he got the sum of the lengths of the two arcs as

    pi (x + y). The explanation about the part of the rubber band between the two circles is OK.

    Aidan, I think you are introducing a new variable "theta". Your answer must be only in terms of x and y.

    Vipin
    Sorry, I never saw that condition until now.
    However, it is easily solved:

    Let x be the radius of the larger circle.
    Let y be the radius of the smaller circle.

    Thus:

    LengthOfRubberBand = x[pi + (1/1)(x^1/1) + (1/2)(x^3/3) + (1/2)(3/4) x^5/5) + (1/2)(3/4)(5/6)(x^7/7) + (1/2)(3/4)(5/6)(7/8)(x^9/9) + (1/2)(3/4)(5/6)(7/8)(9/10)(x^11/11) + (1/2)(3/4)(5/6)(7/8)(9/10)(11/12)(x^13/13)] + y(pi - [ (1/1)(((x-y)(x+y))^1/1) + (1/2)(((x-y)(x+y))^3/3) + (1/2)(3/4)(((x-y)(x+y))^5/5) + (1/2)(3/4)(5/6)(((x-y)(x+y))^7/7) + (1/2)(3/4)(5/6)(7/8)(((x-y)(x+y))^9/9) + (1/2)(3/4)(5/6)(7/8)(9/10)(((x-y)(x+y))^11/11) + (1/2)(3/4)(5/6)(7/8)(9/10)(11/12)(((x-y)(x+y))^13/13)] ) + 4 sqrt(xy)

    There all in terms of x and y.

    ok?
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  13. #13
    Senior Member nikhil's Avatar
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    Exclamation oopssss

    actually i did a mistake so let me give the correct solution
    it will be
    pi(x+y)+2arcsin[(x-y)/(x+y)](x-y)+4(xy)^(1/2)
    you might have got the solution but you may ask me how i got it. its easy
    hint: (fit the two circles in a cone then solve)
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