Tuesday, December 16, 2008

Archimedes' Book of Lemmas, Proposition #6

Arbelos
Exercise your brain. Archimedes wrote the "Book of Lemmas" more than 2200 years ago. Solve the proposition #6 (high school level) and lift up your geometry skills.


Archimedes' Book of Lemmas #6.
Continue reading at:
gogeometry.com/ArchBooLem06.htm
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1 comment:

  1. PravinJuly 28, 2011 at 11:41 PM

    Let AC = 2a, CB = 2b, DE = 2k .
    Denote the midpoints of AC, CB, AB, DE
    by X, Y, O, Z respectively.
    (These are respectively the centres of the circles o AC, CB,AB, DE as diameters)
    Clearly AO = OB = a + b,
    AX = XC = a, CY = YB = b and so
    XO = b - a, XY = a + b, DE / AB = k / (a + b).
    XZ = a + k, YZ = b + k, OZ = a + b -k
    by tangency considerations
    Also note a /b = 2a / 2b = AC / CB = r.
    Let angle ZXY = x
    Apply Cosine Rule for x in triangles XYZ, XOZ
    2(a + k)(a + b)cos x
    = (a + k)^2 + (a + b)^2 - (b + k)^2
    = 2k(a - b)+ 2a(a + b)
    In the same manner
    2(a + k)(b)cos x
    = (a + k)^2 + b^2 - (a + b - k)^2
    = 2k(2a + b)+ a(a - 2b)
    Dividing respective sides;
    (a + b )/ b
    = [2k(a - b)+ 2a(a + b)]/[2k(2a + b)- 2ab]
    implies
    2bk(a - b)+ 2ab(a + b)
    = 2k(a + b)(2a + b) - 2ab(a + b)
    k(2ab - 2b^2 - 4a^2 - 6ab - 2b^2 ) = - 4ab(a + b)
    k(- 4ab – 4a^2 – 4b^2) = -4ab(a + b)
    k = ab(a + b ) / (a^2 + ab + b^2)
    DE / AB = k / (a + b)
    = ab / (a^2 + ab + b^2)
    = 1 /[(a/b) + 1 + (b/a)]
    = 1 / [r + 1 + (1/r)]
    = r / (r^2 + r +1)

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