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Math Help - Japanese Geometry

  1. #1
    Member realintegerz's Avatar
    Joined
    Aug 2008
    Posts
    109

    Japanese Geometry

    We have been assigned this math problem (22) and I'm having some trouble solving it.

    Sacred Mathematics: Japanese Temple ... - Google Book Search

    It wants me to show that R=2t, and I know so far that the sides of the square are equal to 4r

    The answer key says to use the Pythagorean theorem on the little circles to show that r= 3/2 t and I'm having problems with that

    This is what I did so far, from diving the square in halves and then connecting the four origins of the small circles inside the square, you then see 4 small triangles. I used the bottom right triangle.

    I got that the hypotenuse is r+t and then the side on the horizontal half line is 2r - t and the side on the vertical half line is r for sure. I tried using the pythagorean theorem but all I came to was that 2r^2 - tr = 2t, for the theorem the sides came up to be r^2 + (2r-t)^2 = (r+t)^2

    Does anyone know what I'm doing wrong here?
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  2. #2
    Member
    Joined
    Jan 2009
    Posts
    94
    You may be making a minor error. 150 years old book should be right !

    I think it should be:
    r^2 + (2r-t)^2 = (r+t)^2
    r^2 = (r+t)^2 - (2r-t)^2
    r^2 = (r+t - 2r+t)(r+t+2r-t)
    r^2 = (2t - r)(3r)
    r = (2t - r)(3)
    r = 6t - 3r
    r=(3/2) t

    Nice question.
    -O
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