Saturday, December 31, 2016

Geometry Problem 1301: Arbelos, Semicircles, Diameters, Circle, Incircle, Incenter, Square, Midpoint, Concurrency

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 1301: Arbelos, Semicircles, Diameters, Circle, Incircle, Incenter, Square, Midpoint, Concurrency.

Posted by Antonio Gutierrez at 12:21 PM
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5 comments:

  1. APOSTOLIS MANOLOUDISJanuary 2, 2017 at 10:00 AM

    Problem 1301
    Let K medium the arc AB and L medium the arc BC, then K is medium M1O1 and L is medium M2O2.Draw IH perpendicular in AC ,
    intersecting the circle (Ι,ΙΤ) in P.So from problem 1298 is IP=PH.
    But IP/PH=M1K/KO1=M2L/LO2 =1 so the points M1,I and C are collinear and points A,I and M2 are collinear.
    APOSTOLIS MANOLOUDIS KORYDALLOS PIRAEUS GREECE

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    Replies
    1. Peter TranJanuary 2, 2017 at 11:17 PM

      To Apostolis

      Referring to last line of your solution " But IP/PH=M1K/KO1=M2L/LO2 =1 so the points M1,I and C are collinear and points A,I and M2 are collinear"
      In my opinion , we need to show that points C, P and K are collinear before conclude that M1, I and C are collinear.

      Delete
    2. APOSTOLIS MANOLOUDISJanuary 3, 2017 at 11:04 AM

      Yes to peter.
      The points C,P and T1 are collinear from problem 1300 , <BT1C+<AT1B+<KT2A=45+90+45=180
      so the points K,T1,P and C are collinear. Thenks

      Delete
  2. c.t.e.oJanuary 4, 2017 at 8:16 AM

    Relation between three inradius (for all three circles inscribed in arbelos)
    Ri=[R1R2(R1+R2)]/(R1²+R1R2+R2²)

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  3. c.t.e.oJanuary 4, 2017 at 2:20 PM

    PN perpendicular to AC, P intersection CM1 and AM2, tr PNA ~ tr M2O2A
    and tr PNC ~ tr M1O1C + Theorem pythagore for tr O1IN and O2IN
    then just to prove PN = 2Ri

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