Friday, October 5, 2012

Problem 811: Trisecting a Line Segment AB with four Circles and one Line, Radius, Center, Diameter, Chord

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 811.

Online Geometry Problem 811: Trisecting a Line Segment AB with four Circles and one Line, Radius, Center, Diameter, Chord.

3 comments:

  1. Consider ∆DAE and ∆EAF.
    They are both isosceles and ∠DAE=∠EAF.
    Therefore, ∆DAE ~ ∆EAF.

    So AD / AE = AE / AF
    But AD = CB + AB = 3*AB = 3*AE

    Thus
    AB / AF = AE / AF = AD / AE = 3

    Hence, AF = AB / 3.

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  2. Triangles AEF and ADE are isosceles.
    Note that ∠EAF=∠EFA=∠EAD=∠AED
    So triangle AED similar to ADE ( case AA)
    And AF/AE=AE/AD=1/3 => AF=1/3.AE=1/3.AB

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  3. Consider the inversion transform T = I(A, AB^2).

    T(Circle A) = Circle A
    T(Line CAB) = Line CAB

    Now consider circle E.
    T(circle E) would pass through the intersection points of circle A and circle E.
    And since T(circle E) should be a line that does not pass through A,
    so T(circle E) = the common chord of circle A and circle E.

    Since the common chord of circle A and circle E
    is exactly the perpendicular bisector of AE,
    which should pass through point D.

    T(circle E ∩ line CAB) = T(circle E) ∩ line CAB
    Thus, T(F) = D.

    By the definition of inversion,
    AF * AD = AB^2

    But AD = 3*AB,
    hence, AF = AB / 3.

    ReplyDelete