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Area of a snowflake

  1. Jun 29, 2011 #1
    http://imageshack.us/photo/my-images/835/mathproblem.jpg/

    I was thinking of an interesting shape (I drew it in paint for help), and if it would be possible to find the area of it. I started to write a series to represent the area but ran into trouble because I think that parts of the snowflake might overlap.

    the formula that I started to work on is incomplete: 1 + 3/4 + [(Σ,∞,n=2) 12(3^(n-2))+...(incomplete)]

    The "12(3^(n-2))" is the number of boxes in that layer, I was going to multiply that by the area of the individual boxes in that layer and then subtract the overlap. But that is where the problems might be.
     
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  3. Jun 29, 2011 #2

    micromass

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    With what rate does the size of the boxes decrease? Your first box is 1x1, what size do the four boxes attached to that have? And what size do the boxes attached to that have?
     
  4. Jun 29, 2011 #3
    opse sorry, when i was calculating it, I used a factor of 1/2 for shrinking the squares. However if that causes the squares to overlap I would like to find out what the minimum rate of size decrease is.
     
  5. Jun 29, 2011 #4

    micromass

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    No, the squares won't overlap. However, I do expect that the entire plane get's filled this way and that the area is equal to 4.

    Let's calculate this. First, let's see how many squares we have to add at each step:

    At step 1, we have 1 square
    At step 2, we add 4 squares
    At step 3, we add [itex]3\cdot 4=12[/itex] squares
    At step 4, we add 324 squares
    At step n, we add 3n-24

    Now let's see what area we add every step:

    At step 1, we add an area of 1.
    At step 2, for each square we add, we will add an area of 1/4. However 1/4 of that area will overlap with our first square. So we only add an area of 1/4-1/16=3/16. We have 4 squares to add so we get 3/4.
    At step 3, for each square we add, we will add an area of 1/16. However, 1/4 of that area will overlap with our original squares. So we only add an area of 1/16-1/64=3/64. We have 12 squares to add, so we get 9/16.
    At step n, we add a square of dimensions 21-n. So we add an area of 22-2n. However, 1/4 of that area will overlap with the original squares. So we only add an area of

    [tex]\frac{1}{2^{2n-2}}-\frac{1}{2^{2n}}=\frac{3}{2^{2n}}[/tex]

    We have 3n-24 squares to add, so we add

    [tex]\frac{3^{n-1}}{2^{2n-2}}[/tex] area.

    So eventually we get the following sum:

    [tex]\sum_{n=1}^{+\infty}{\frac{3^{n-1}}{2^{2n-2}}}=\frac{4}{3}\sum_{n=1}^{+\infty}{\left(\frac{3}{4}\right)^n}[/tex]

    Evaluating this gives an area of 2. Like expected.

    Try to modify this argument if you add a square with a smaller area each time.
     
    Last edited: Jun 29, 2011
  6. Jun 29, 2011 #5
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