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osanseviero

  • one year ago

Get an approximation of pi using a square and triangles inside a circle.

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  1. osanseviero
    • one year ago
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    |dw:1441838480540:dw|

  2. osanseviero
    • one year ago
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    The idea is to first calculate the area of the square and then add triangles.

  3. osanseviero
    • one year ago
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    |dw:1441838520139:dw|

  4. osanseviero
    • one year ago
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    Asquare = (2r)^2 = 2 My problem is with getting the first level of triangles, and then adding triangles to those triangles

  5. phi
    • one year ago
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    what did you pick for r ? how about starting with r= 1?

  6. osanseviero
    • one year ago
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    I'm using r for distance from center to square. The radius is 1.

  7. osanseviero
    • one year ago
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    So with that I can get a height for a triangle 1-r.

  8. phi
    • one year ago
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    ok, but using r for the length of 1/2 the side of the square is confusing (people see "r" and think radius) can I use x instead of r? diameter is 2, and this is the length of the diagonal of the square so its side is 2/sqr(2) = sqr(2) and x= 1/2 of a side = sqr(2)/2

  9. osanseviero
    • one year ago
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    Ok, lets use x instead.

  10. phi
    • one year ago
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    the height of the triangle will be r- x 1 - \sqr(2)/2

  11. osanseviero
    • one year ago
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    Ok, and base would be x*2,right?

  12. phi
    • one year ago
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    yes

  13. osanseviero
    • one year ago
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    Ok. So that would be ((1 - sqrt(2) / 2)(sqrt(2))/2. That times 4 to get total area

  14. phi
    • one year ago
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    yes

  15. osanseviero
    • one year ago
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    Ok, I was good until here. I started to get lost when adding a new layer of triangles

  16. osanseviero
    • one year ago
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    |dw:1441839343074:dw|

  17. osanseviero
    • one year ago
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    So I guess it will pass through the half of the half of the base of the triangle, and then half of the side of the triangle

  18. phi
    • one year ago
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    yes, but now it's not clear what the altitude of the new triangle is. the base of the new triangle is the "hypotenuse" of the old triangle, so we can find it using pythagoras

  19. osanseviero
    • one year ago
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    Ok, give me a minute, I'll try to get it

  20. osanseviero
    • one year ago
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    Ok. The base is pretty simple. The altitude is a little harder. First, I would calculate the distance between the center and the intersection with the square. Then add the intersection with the triangle. For the square, I did: |dw:1441839701196:dw| So a = sqrt((x/2)^2 + x^2)

  21. osanseviero
    • one year ago
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    How to find where the line touches the hypotenuse of the second triangle? We know that the height would be the half.|dw:1441839908971:dw|

  22. osanseviero
    • one year ago
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    Oh...that would be the previous base/2

  23. anonymous
    • one year ago
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    in my mind it looks like this

  24. anonymous
    • one year ago
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    |dw:1441840270963:dw|

  25. phi
    • one year ago
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    this looks like a painful process

  26. osanseviero
    • one year ago
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    It is haha. I think that with this we should get a really close approximation

  27. phi
    • one year ago
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    I know archimedes found the area of triangles formed by an n-gon (n sided polygon) on the inside of the circle and the outside and so the area of the circle is between those two areas. But that process seems easier than adding smaller and smaller triangles

  28. osanseviero
    • one year ago
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    It is easier, I already did that one hehe

  29. osanseviero
    • one year ago
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    Thanks for the help :)

  30. osanseviero
    • one year ago
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    Can you just help me get that last height?

  31. phi
    • one year ago
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    it will take some time

  32. osanseviero
    • one year ago
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    Ok, then don't worry :)

  33. phi
    • one year ago
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    |dw:1441843190370:dw| find "y" using pythagoras and the altitude is 1-y

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