Tuesday, July 19, 2011

Problem 637: Semicircle, Diameter, Perpendicular, Inscribed Circle, Chord, Tangent, Arbelos

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

Click the figure below to see the complete problem 637.

Online Geometry Problem 637: Semicircle, Diameter, Perpendicular, Inscribed Circle, Chord, Tangent, Arbelos.

3 comments:

  1. AG^2=AD^2 - DG^2
    =(2R-r)^2 - r^2
    =4R^2 - 4Rr
    =4R(R-r)
    AE^2=AC.AB (tr AEC and tr ABE are similar)
    =(2R-2r).(2R)
    =4R(R-r)
    AE^2=AG^2,
    AE=AG

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  2. Observe that AEB is a right triangle ( AB is a diameter)
    so AE^2= AC.AB (1) ( Relation in a right triangle)
    In circle center D we also have
    AG^2= AC.AB (2) ( power of point A to circle D)
    Compare (1) and (2) we get AE=AG
    Peter Tran

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  3. Using my proof and the result of Problem 636 AG is a tangent to circle CGB, hence AG^2 = AC.AB

    But AE^2 also = AC.AB since AE is a tangent to circle BCE

    So AE = AG

    Sumith Peiris
    Moratuwa
    Sri Lanka

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