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

Posted by Antonio Gutierrez at 7:47 AM
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5 comments:

  1. AnonymousFebruary 19, 2011 at 12:30 PM

    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.

    ReplyDelete
  2. Antonio GutierrezFebruary 20, 2011 at 4:10 AM

    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.)

    ReplyDelete
  3. AnonymousMarch 1, 2011 at 5:17 AM

    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

    ReplyDelete
  4. UnknownOctober 11, 2021 at 8:28 AM

    Distance from pt. B to line AC =b
    And from pt D to AC =d
    Area AOB=1/2 . AO. b
    Area COD =1/2 . CO. d
    Area AOB. Area COD
    =1/2.AO.b.1/2.CO.d
    =1/2.AO.d.1/2.CO.b
    =Area AOD. Area BOC

    ReplyDelete
  5. MarcoDecember 26, 2023 at 8:49 PM

    S1:S4=OB:OD=S2:S3
    S1S3=S2S4

    ReplyDelete

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