Thursday, August 10, 2017

Geometry Problem 1341: Isosceles Triangle, 80-20-80 Degrees, Circumcenter, Angle Bisector

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

Geometry Problem 1341: Isosceles Triangle, 80-20-80 Degrees, Circumcenter, Angle Bisector.

8 comments:

  1. Problem 1341
    Suppose the AD crosses the circle at point E, then arcBE=arcEC and <ECB=40, <ECO=50=<OEC (<COE=80).I form the equilateral triangle OCF (F is arcEC ).
    Is < FOE=20, if G is circumcenter of triangle FOE then GO=GE=GF and <GOE=10=<GEO.
    Is triangle OGC=triangle GFC so <GCO=<GCF=60/2=30.But <FCE=<FOE/2=20/2=10
    and <ECG=20.Now triangle DCE=triangle GEC so DC=GE.But triangle OCD=triangle OEG
    (OC=OE,DC=GE,<OCD=10=<OEG) ,soOD=DC=GO=GE.Therefore <ODB=<DOC+<DCO=
    =10+10=20.
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL OF KORYDALLOS PIRAEUS GREECE

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  2. Problem 1341 (sulotion 2)
    Let K point on AB such that KA=KO , then <KAO=10=<KOA. I form the equilateral triangle
    ΟΚL (the point L is located on the right of BC).Then the point K is the circumcenter the
    triangle AOL.So <LAO=<LKO/2=60/2=30=<DAO so the point L belongs to AD.
    Is triangle AOK=triangle COL(OA=OC,OK=OL,<KOA=10=<LOC),then LC=LO=KO=KA and
    <LCO=10,so the point L belongs to BC.
    Therefore the points D,L coincide and <ODB=<OLB=<LOC+<LCO=10+10=20.
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete
    Replies
    1. In order to prove "triangle AOK=triangle COL" above, how do we know that "10=<LOC" since we don't know yet that L is on BC?

      Delete
  3. https://ibb.co/df8rta
    Let AO intersects BC in E. Since AB=BC and OA=OB we have DH is a perpendicular bisector of AE => FA=FE => EF=EB (5). From (3) and (4) and the AAS => triangle ADC=triangel ADF => AC=AF and DC=DF => AD is a perpendicular bisector of FC. Let the perpendicular bisector of BC intersects BC in M and AD in T => TC=TF=TB (6). From (5) and (6) => ET is a perpendicular bisector of FB, i.e. ET _|_ FB. Let TE intersects BF in K => triangle BEK~trianle TEM => D,O,E and T lie on a circle and from (7) => <ODE=<OTE=20.

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  4. Here is my solution...

    https://www.youtube.com/watch?v=d4ZzPnoFogk

    ReplyDelete
    Replies
    1. How did u assume that the circle u drew is tangent to AD?

      Delete
  5. Extend BO to X so that AXC is an equilateral triangle.

    OC bisects <BCX
    So BO/BX = BC/XC = AB/AC =BD/DC since AD bisects <A

    Hence OD//XC

    So < BDO = < BCX = 20

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  6. Problem 1341
    Solution 2
    From Problem 1274, circumradius of ∆ABC, R2 = a2b/(a+b)…(1)

    Now since AD bisects < BAC, CD = ab/(a+b),
    so CD.CB = a2b/(a+b) = R2 from (1).
    Hence CO is tangential at O to circle BDO.
    So < COD = 10 and hence < BDO = 10+10 = 20

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete