Tuesday, July 12, 2016

Geometry Problem 1236: Intersecting Circles, Secant, Concyclic Points

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

This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.

Click the figure below to view more details of problem 1236.

Geometry Problem 1236: Intersecting Circles, Secant, Concyclic Points

Posted by Antonio Gutierrez at 3:49 PM
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9 comments:

  1. AnonymousJuly 13, 2016 at 6:55 PM

    Isosceles triangles PEA and QDA are similar => EA/PA = DA/QA => EA.AQ = PA.AD
    Hence E,P,Q and D are concyclic.

    B also lies on the same circle

    ReplyDelete
    Replies
    1. Sumith PeirisJuly 13, 2016 at 9:21 PM

      "B also lies on the same circle"

      Could u pls explain how u arrived at this conclusion?

      Delete
    2. AnonymousJuly 14, 2016 at 2:15 PM

      <PAQ = <PBQ = 135 which are supplimentary. Hence B also lies on the same circle. For some reasons, this did not get printed in my first comment. Faced this trouble couple of times and I stopped posting. ( I have posted <PEA = <PEQ = 45 in the beginning, but the preview is truncating this as well)

      Delete
    3. Sumith PeirisJuly 15, 2016 at 10:39 AM

      < PAQ = < PBQ = 135. How did u arrive at this conclusion?

      Delete
  2. AnonymousJuly 14, 2016 at 9:55 AM

    connect PB, BQ, <PBQ = <PAQ,
    <PAQ + <PAE = 180 = <PAQ + <PEA =< PBQ + <PEA =180,
    SO b ALSO LIES ON THE SAME CIRCLE

    ReplyDelete
  3. AnonymousJuly 16, 2016 at 8:28 PM

    Isosceles triangles PEA and QDA. Since <PBQ+<E=<PAQ+<EAP=180, E,P,B,Q concyclic. Similary, D,Q,B,P concyclic. Therefore, E,P,B,Q,D concyclic.

    ReplyDelete
  4. APOSTOLIS MANOLOUDISJuly 22, 2016 at 1:43 PM

    Problem 1236
    Is <EPA=2.<EBA (PE=PA), <AQD=2.<ABD (AQ=QD).But <EAP=<DAQ=x. Then <EPD=<EPA=
    180-2x=2(90-x)=<AQD=<EQD.But <EBD=<EBA+<ABD=(< EPA)/2+(<AQD)/2=90-x+90-x=180-2x.Therefore points E,P,B,Q and D are concyclic.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete
  5. AnonymousSeptember 16, 2016 at 4:06 AM

    The solution that I propose is the following:
    <PEA=<ADQ=<EAQ=<DAQ=x ; <EPA=AQD=y= 180-2x
    <PAB=<PBA=α ; <APB=K=180-2α
    <QAB=<ABQ=β ;<AQB=R=180-2β ; <ABQ+<ABP = α + β
    To be concyclic the points BPED and Q need that:
    <R+<K+<y=180 or α + β +x= 180 ;
    <R+<K+<y=180-2β+180-2α+180-2x=α + β +x=180
    Therefore point BPED and Q are concyclic
    Reply

    ReplyDelete
  6. Peter TranSeptember 17, 2016 at 10:30 AM

    https://goo.gl/photos/1g61g7m5f8PYXRTeA

    Connect PQ, EB, BD, PB and QB ( see sketch)
    Note that u= ∠ (BEQ)= ∠ (APQ)= ∠ (BPQ)
    So E, P, B and Q are cocyclic.
    Similarly P, B, Q and D are cocyclic

    ReplyDelete

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