Saturday, December 24, 2016

Geometry Problem 1298 Arbelos, Semicircles, Diameters, Circle, Incircle, Tangent, Angle Bisector, Perpendicular, Midpoint

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 1298 Arbelos, Semicircles, Diameters, Circle, Incircle, Tangent, Angle Bisector, Perpendicular, Midpoint.

Posted by Antonio Gutierrez at 3:24 AM
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4 comments:

  1. c.t.e.oDecember 24, 2016 at 2:16 PM

    Draw TO. Ang HOI=2A, ang ITP=45-A => ang HIO = 90 - 2A
    => IHO = 90

    ReplyDelete
  2. Peter TranDecember 24, 2016 at 6:08 PM

    https://goo.gl/photos/yYwMDZZwhsdpYQ2w9

    Let D is the midpoint of arc AC with ∠AOD= 90 ( see sketch)
    Since TP is the angle bisector of angle ATC => TP will pass thru D
    Since Triangles TIP and TOD are isoceless and angle ITP= angle OTD=> Angle TOD= angle TIP
    So IP//OD and IP ⊥AC

    ReplyDelete
  3. APOSTOLIS MANOLOUDISDecember 25, 2016 at 9:07 AM

    Problem 1298
    1. Let K medium the arc AB and L medium the arc BC.Is <BTC=<BTA=45 then the points B,P ant T are collinear.But <TAC+<TCA=90 or (45-<KAT)+(<ACT-45)=90 or <KAT=<ACT=<LTC=<KTA so <KTL=90. Is <KTO=<KTA+<ATO=<KAT+<TAO=45=<OTL.So <KTA=<ITP=<IPT=<BIH. But <BIH+<IBH=<KAT+45+<TAB=45+45=90. So IP is perpendicular in AC.
    APOSTOLIS MANOLOUDIS KORYDALLOS PIRAEUS GREECE

    ReplyDelete
  4. Peter TranAugust 5, 2022 at 3:46 PM

    I do not see anybody have solution 0f 2nd part of the question " show that P is the midpoint of IH "

    ReplyDelete

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