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Infinite Series Problem Ideas?

  1. Apr 6, 2006 #1
    Hey,

    I'm having a bit of difficulty with a specific problem. I was able to somewhat easily work the problem out, but the few different answers I've tried have been incorrect. I'm beginning to think there's something wrong with my trigonometry or something because I swear the calculus is correct.

    Here's the question:

    In the reallly inaccurate drawing below, there are infinitley many circles approaching the vertices of an equilateral triangle, each circle touching other circles and sides of the triangle. If the triangle has sides of length 1, find the total area occupied by the circles.

    [​IMG]


    Okay, so I've come up with several different answers. The most common of which seems to be 7pi/48, but I'm told that's incorrect. I'm told that something*pi/96 or something is correct, and I'm not sure how I'm wrong here.

    This is what I've done, in a nutshell. I've taken the figure, divided it up, and developed the following information:
    [​IMG]

    [tex]\tan30=\frac{r}{1/2}[/tex]
    [tex]\frac{1}{2}\tan30=r=\frac{\sqrt{3}}{6}[/tex]

    That then, would logically be the radius of the large circle.

    At some point, I don't know how, I somehow determined that something, of the first set of smaller circles outside of the large circle, we'll refer to them as A2, were each 1/3 the radius or area of the large circle, A1.

    This is where I get into a bit of trouble... Assuming that the RADIUS was 1/3 the size of each of these, I came up with the following series to total the area of all 3 sets of these circles, NOT including the large circle:

    [tex]3\pi(\sum_{n = 1}^{\infty} \frac{1}{3^n}*\frac{\sqrt{3}}{6})^2=\frac{\pi}{16}[/tex]

    So then to find the total area, I'd logically add the combined area of all the smaller circles to the area of the large circle which gives:

    [tex]\frac{\pi}{12}+\frac{\pi}{16}=\frac{7\pi}{48}[/tex]

    Can someone explain this? What am I doing wrong? I wrote out the series for this and it seemed to make sense...

    I think there might be something wrong with that 1/3 ratio, I don't know how I got it. How would one go about finding the ratio from the large circle to one of the small circles using the given information??


    Thanks
     
  2. jcsd
  3. Apr 7, 2006 #2

    AKG

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    Do the trigonometry again. The smaller circles will not have radius 1/3 of the previous, they will have radius (1 - tan(30)) the previous. Also, you don't want to square the sum of the radii, you want to sum the squares of the radii.
     
  4. Apr 7, 2006 #3

    Curious3141

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    (sorry, wrong).
     
    Last edited: Apr 7, 2006
  5. Apr 7, 2006 #4

    AKG

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    If I understand what you're saying, Curious, then you're wrong. If the smaller triangle were to have half the side length, then if we were to draw its base in, you would readily see it cuts through the larger triangle, and so does not circumscribe the parent triangle. Also, unless I did the trigonometry wrong, 1 - tan(30) is the correct number, and this number isn't 1/2.
     
  6. Apr 7, 2006 #5

    Curious3141

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    Yes, you're right. I made a mistake.
     
  7. Apr 7, 2006 #6

    Curious3141

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    Actually, now that I have had a bit more time to work it out, the radius of each offspring circle *is* 1/3 that of the parent circle. I'm almost certain this is correct.
     
  8. Apr 7, 2006 #7

    VietDao29

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    No the smaller circle has 1 / 3 the side length, not half as you said. And unless I screw up with my calculation, I get the smaller circle does have radius of 1 / 3 of the larger one. :)
    I just wonder how you can arrive at 1 - tan(30o). Can you show us your approach?
     
  9. Apr 7, 2006 #8

    Curious3141

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    Here's the explanatory diagram. R is the radius of the parent circle, r is that of the offspring circle.

    From triangle OAB,

    sin 30 deg = R/|OA|

    |OA|= 2R

    Since |OQ| = R, |QA| = R

    From triangle QAP,

    tan 30 deg = |QP|/|QA|

    So |QP| = R tan 30

    And finally, from triangle XQP,

    tan 30 = |QX|/|QP|

    So r = |QX| = R (tan 30)^2 = R/3.

    The rest of the problem can be completed with a little geometric series.
     

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  10. Apr 7, 2006 #9

    Curious3141

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    1/3 is right. I've drawn the figure, it should be approved in a while. AKG's method must be wrong.
     
  11. Apr 7, 2006 #10
    Wow. Thank you everyone for all of your help.

    Yeah, I don't quite understand the 1-sin30 answer, I understand the 1/3 answer... Thanks again.

    However, AKG mentioned that I sum the squares of the radius and then take the sum with the squares involved with the series, yes?

    When I do that, I come up with the following:

    [tex] \sum_{n = 1}^{\infty} (\frac{\sqrt{3}}{6*3^n})^2 = \sum_{n = 1}^{\infty} \frac{3}{36*9^n} = \frac{1}{12}\sum_{n = 1}^{\infty} \frac{1}{9^n}=\frac{1}{12}*\frac{1}{8}=\frac{1}{96}[/tex]

    I would then multiply that by 3pi, (if it has already been squared, all it should need is to be multiplied by pi to find the area of one corner's worth of circles, and then multiplied by 3 to find the area of all 3 corners of circles, not including the main circle) Which returns:

    [tex]3\pi(\frac{1}{96})=\frac{3\pi}{96}[/tex]

    To find the total area, I'd add that to the area of the first circle...

    [tex]\frac{\pi}{12}+\frac{3\pi}{96}=\frac{11\pi}{96}[/tex]


    I have no idea if this is correct. Is it?

    Thanks again...
     
    Last edited: Apr 7, 2006
  12. Apr 7, 2006 #11

    Curious3141

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    That's what I got (11pi/96). What I did was conceptually simpler. Since the ratios of the radii between two successive levels of circles is 1/3, the ratio between the areas is 1/9. The sum of all the areas can be expressed in terms of the area A of the largest circle as :

    A(1 + 3*(1/9 + 1/81 + ... + 1/9^n + ...)) = A ( 1 + 3/8) = (11A)/8

    The area of the large circle can be calculated by observing that its radius is [tex]\frac{1}{2\sqrt{3}}[/tex], giving A = [tex]\frac{\pi}{12}[/tex] and the total area = [tex]\frac{11\pi}{96}[/tex]
     
  13. Apr 7, 2006 #12

    AKG

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    Yeah, my calculations were wrong. The smaller circle does have radius 1/3 the one before it, not 1-tan(30). Is there a simple way to see this without calculations?
     
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