Saturday, February 19, 2011

Problem 579: Quadrilateral, Diagonals, Triangles, Areas

Geometry Problem
Click the figure below to see the complete problem 579.

 Problem 579: Quadrilateral, Diagonals, Triangles, Areas.
Zoom at: Geometry Problem 579

3 comments:

  1. Areas of triangles ABO & CBO are proportionals to bases: S1/S2 = AO/CO
    Areas of triangles ADO & CDO are proportioanls to bases: S3/S4 = CO/AO
    Then we've: S1/S2 = S4/S3 or S1·S3 = S2·S4.

    Migue.

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  2. c.t.e.o has left the following solution of problem 579:

    Draw BG, DH altitude on AC
    AO∙BG = S1, CO∙BG = S2 => S1/S2 = AO/CO
    AO∙DH = S3, CO∙DH = S4 => S4/S3 = AO/CO
    =>
    S1∙S3 = S2∙S4 (Q.E.D.)

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  3. S1=1/2.OA.OB.sin(AOB)
    S2=1/2.OB.OC.sin(BOC)=1/2.OB.OC.sin(180-AOB)=1/2.OB.OC.sin(AOB)
    S3=1/2.OC.OD.sin(COD)=1/2.OC.OD.sin(AOB)
    S4=1/2.OD.OA.sin(AOD)=1/2.OD.OA.sin(180-AOB)=1/2.OD.OA.sin(AOB)

    S1.S3=S2.S4=1/4.sin^2(AOB).OA.OB.OC.OD

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