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Math Help - Tricky Circles

  1. #1
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    Tricky Circles

    Hi Guys. I'm hoping there is someone out there who ca help me with question 4 of this problem.

    The rest I have managed to solve, however any advice that anyone may have would be greatly appreciated.

    Thanks in advance
    Dylan Eave
    Attached Thumbnails Attached Thumbnails Tricky Circles-problems.gif  
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  2. #2
    MHF Contributor
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    Line (R+x) is a common tangent to the two outer circles. A tangent is perpenmdicular to a radius, so (R+x) is perpendicular to the straight line (r+r) that connects the centers of the two outer circles.
    And so, the triangle "(R+r)-(r)-(R+x)" is a right triangle, (R+r) being the hypotenuse.
    Hence,
    Let T = theta for less typing,
    tanT = r/(R+x)
    (R+x)*tanT = r
    R+x = r/tanT
    x = r/tanT -R ....(1)

    Since you want x in terms of R only, then we eliminate "r" from (1).
    Back to the same right triangle,
    sinT = r/(R+r)
    (R+r)sinT = r
    R*sinT +r*sinT = r
    R*sinT = r -r*sinT = r(1 -sinT)
    r = R*sinT /(1 -sinT) ....***

    Substitute that into (1),
    x = [R*sinT /(1 -sinT)] /tanT -R
    x = [R*sinT /(1 -sinT)] /[sinT /cosT] -R
    x = [R*sinT /(1 -sinT)] *[cosT /sinT] -R
    x = [R*cosT /(1 -sinT)] -R
    x = R*[cosT/(1-sinT) -1]
    x = R[(cosT -1*(1 -sinT))/(1 -sinT)]
    x = R[(cosT -1 +sinT))/(1 -sinT)]
    x = R(sinT +cosT -1)/(1 -sinT) ......(2)

    ----------
    Since r = r, then (R+x) is a perpendicular bisector of the line (r+r). Then, angle theta is 1/2 of the central angle subtended by the said line (r+r).
    This central angle, which is 2*theta, is also the central angle of the arc of the inner circle between the two points of tangency with the two outer circles.

    ----------
    4.1) When n = 4, (in simplified form).

    Subtended arc = 360/4 = 90 degrees
    So, T = 90/2 = 45 degrees
    Hence,
    x = R(sinT +cosT -1) /(1 -sinT) ......(2)
    x = R[(sin(45deg) +cos(45deg) -1) /(1 -sin(45deg)]
    x = R[(1/sqrt(2) +1/sqrt(2) -1) /(1 -1/sqrt(2))]
    x = R[(1 +1 -sqrt(2)) /sqrt(2)] /(sqrt(2) -1) /sqrt(2)]
    x = R[(2 -sqrt(2)) /(sqrt(2) -1)] ....answer.

    -------------
    4.2) When n = 6 (?), in simplified form,

    Subtended arc = 360/6 = 60 degrees
    So, T = 60/2 = 30 degrees
    Hence,
    x = R(sinT +cosT -1) /(1 -sinT) ......(2)
    x = R[(sin(30deg) +cos(30deg) -1) /(1 -sin(30deg)]
    x = R[(1/2 +sqrt(3)/2 -1) /(1 -1/2)]
    x = R[(1 +sqrt(3) -1*2) /2) /(1/2)]
    x = R(sqrt(3) -1) ....answer.

    -------------
    4.3) When n = 12, correct to two decimal places,

    Subtended arc = 360/12 = 30 degrees
    So, T = 30/2 = 15 degrees
    Hence,
    x = R(sinT +cosT -1) /(1 -sinT) ......(2)
    x = R[(sin(15deg) +cos(15deg) -1) /(1 -sin(15deg)]
    x = R[0.2247 /0.7412]
    x = 0.3032R
    x = 0.30R ....answer.
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  3. #3
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    Apr 2005
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    Wow

    U rock man! This is the first time I've used a math forum of this nature. I'm impressed. Thanks alot! Gonna go try digest that now.

    Dylan
    Last edited by Math Help; April 30th 2005 at 10:25 AM.
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