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Finding radius of circle

  1. Mar 12, 2013 #1
    1. The problem statement, all variables and given/known data
    (see attachment 1)

    In this problem, we have a row of circles placed on a line. All points of tangency are distinct. The circle ##C_n## is uniquely determined.


    2. Relevant equations



    3. The attempt at a solution
    Here's the sketch I drew for the problem.
    2zdwpol.png

    Radius of ##C_n## = ##r_n##
    Radius of ##\Gamma_n## = ##1/2^n##
    Radius of ##\Gamma_{n+1}## = ##1/2^{n+1}##.

    Therefore AB, BC and AC can be easily calculated but how do I calculate ##r_n##. I can't perform the summation until I don't find ##r_n##.

    Any help is appreciated. Thanks!
     

    Attached Files:

  2. jcsd
  3. Mar 12, 2013 #2

    Curious3141

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    Homework Helper

    Try working in a coordinate system with the origin at the point of contact between the largest circle and the line.

    Use Pythagoras theorem to solve for the x-coordinate of the center of the circle of intermediate size.

    Now set up quadratic simult. equations using Pythagoras theorem to solve for the radius of the small circle. The other variable is the x-coordinate of the small circle, which you don't really need.

    It should reduce to a nice expression. The rest becomes quite easy, two geometric sums, etc.
     
  4. Mar 12, 2013 #3
    With centre B? How will I find the x-coordinate, I don't have the distance of B from origin. :(
     
  5. Mar 12, 2013 #4

    ehild

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    Homework Helper
    Gold Member

    You can find some right-angle triangles...

    ehild
     

    Attached Files:

  6. Mar 12, 2013 #5

    Curious3141

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    What ehild said. Red triangle for the first part. Blue and green for the second part (simultaneous quadratics).

    The first equation is easy peasy. The simult. quadratics are a bit icky. But you get a nice expression for ##r_n## in the end.
     
  7. Mar 12, 2013 #6
    I marked the points. (see attachment)

    [tex]AH=\frac{1}{2^n}-\frac{1}{2^{n+1}}[/tex]
    [tex]AB=\frac{1}{2^n}+\frac{1}{2^{n+1}}[/tex]
    [tex]BH^2=\left(\frac{1}{2^n}+\frac{1}{2^{n+1}}\right)^2-\left(\frac{1}{2^n}-\frac{1}{2^{n+1}}\right)^2[/tex]
    [tex]BH=\frac{\sqrt{2}}{2^n}[/tex]

    Calculating EG:
    [tex]AE=\frac{1}{2^n}-r_n[/tex]
    [tex]AG=\frac{1}{2^n}+r_n[/tex]
    Hence
    [tex]EG^2=\left(\frac{1}{2^n}+r_n\right)^2-\left(\frac{1}{2^n}-r_n\right)^2[/tex]
    [tex]EG=2\sqrt{\frac{r_n}{2^n}}[/tex]

    Calculating FG:
    [tex]BF=\frac{1}{2^{n+1}}-r_n[/tex]
    [tex]BG=\frac{1}{2^{n+1}}+r_n[/tex]
    [tex]FG^2=\left(\frac{1}{2^{n+1}}+r_n \right)^2-\left(\frac{1}{2^{n+1}}-r_n \right)^2[/tex]
    [tex]FG=\sqrt{2r_n}{2^n}[/tex]

    Since, EG+FG=BH
    [tex]\sqrt{\frac{2r_n}{2^n}}+2\sqrt{\frac{r_n}{2^n}}= \frac { \sqrt {2} }{ 2^n }[/tex]
    Solving this for ##r_n##
    [tex]r_n=\left(\frac{1}{\sqrt{2}}\right)^{2n}(\sqrt{2}-1)^2[/tex]

    Thanks a lot both of you. If I perform the summation with this expression. I get the right answer. :smile:

    The answer I get is ##3-\sqrt{8}##.
     

    Attached Files:

    Last edited: Mar 12, 2013
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