Sunday, January 15, 2017

Geometry Problem 1306: Cyclic, Inscribed Quadrilateral, Circle, Tangent, Diagonals, Collinear Points

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

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Geometry Problem 1306: Cyclic, Inscribed Quadrilateral, Circle, Tangent, Diagonals, Collinear Points.

Posted by Antonio Gutierrez at 2:57 PM
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8 comments:

  1. APOSTOLIS MANOLOUDISJanuary 16, 2017 at 12:04 PM

    Problem 1306
    In complete quadrilateral FBECAD the point K is the point Miquel.Then the ΑFCK, ABKD, ABEK and KECD are cyclic.So if x=<FAC=<FKC=<BKE=<BAC=<BDC=<EDC=<EKC.
    Therefore the points F,E and K are collinear.But <BKC=<BKF+<FKC=x+x=2x and <BOC=2.<BAC=2x.So <BKC=<BOC or the BKOC is cyclic.But <OBG=90=<OCG or
    the BOCG is cyclic with <BOG=<GOC=x.Then the BKOCG is cyclic, so <BKG=<BOG=x=<BKF. Therefore the points F,G,E and K are collinear.
    APOSTOLIS MANOLOUDIS KORYDALLOS PIRAEUS GREECE

    ReplyDelete
  2. Sumith PeirisJanuary 17, 2017 at 10:46 AM

    Assume for a moment that F,G,E are not collinear and so let EG extended meet BF at F'. Let circle BGC meet GE extended at P.

    So < GBC = <GCB = < GPB = < BAC = < GPC = < BDC = u say.
    So ABEP and CDPE are concyclic.

    Further let < F'BG = < BCA < BDA = v say. Then since < BGP = < BCP = v+x, (where < ECP = < EDP = x) and so < BF'G = x.
    Hence F'BPD is concyclic and so < BDF' = u.
    But < BDF = u, hence F' and F coincide proving that F,G,E are collinear.

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
    Replies
    1. MotaJanuary 28, 2017 at 1:11 PM

      Beautiful solution. Very clever. But where it has v+x it must be read v-x.

      Delete
    2. Sumith PeirisDecember 3, 2018 at 5:02 PM

      Noted with thanx

      Delete
  3. Sumith PeirisJanuary 17, 2017 at 11:00 AM

    More elegant proof

    Using the same notation as before without using F', < ABP = GEC = < PDC hence BFDP is concyclic.

    So < FDB = < FPB = < GPB so FGE are collinear

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  4. AnonymousJanuary 18, 2017 at 12:22 PM

    Let m(GBC) = m(GCB) = x
    => m(BGC) = 180-2x -----(1)
    From Alternate segment theorem, we have m(BDC) = m(EDC) = m(BAC) = m(BAE) = x
    Join FE and extend it to find a point P such that FCPA and FBPD are concyclic
    => m(FPB) = m(FDB) = x
    Similary m(FPC) = m(FAC) = x
    Therefore m(BPC) = 2x --------(2)
    From (1) and (2) GBPC are collinear and m(GPB) = m(FPB)
    Hence F,G,E are collinear

    ReplyDelete
    Replies
    1. AnonymousJanuary 21, 2017 at 10:56 AM

      need to prove when FCPA is concyclic, FBPD is concyclic also.

      Delete
  5. Peter TranJanuary 18, 2017 at 9:44 PM

    excellent solution
    Peter Tran

    ReplyDelete

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