Monday, March 4, 2019

Geometry Problem 1421: Right Triangle, Incircle, Excircle, Tangent Lines, Measurement

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 1421: Right Triangle, Incircle, Excircle, Tangent Lines, Measurement, Tutoring.
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7 comments:

  1. Peter TranMarch 4, 2019 at 7:37 PM

    https://photos.app.goo.gl/YvmQY9G5cGXqPdkX9

    We have s= ½(3+4+5)=6
    BD=FC=s-AC= 1=> BI= sqrt(2)
    Since GF//IC=> BG/BI= BF/FC =2/3
    So BG= 2.sqrt(2)/3 and BI= sqrt(2)/2

    ReplyDelete
    Replies
    1. Peter TranMarch 5, 2019 at 9:51 AM

      Minor typo error BI=SQRT(2)/3
      Peter tran

      Delete
    2. UnknownMay 24, 2020 at 8:52 AM

      Sir may I know the book you refer for geometry the way solved the question is outstanding

      Delete
    3. UnknownMay 24, 2020 at 8:55 AM

      I am not getting thought proces similarity

      Delete
    4. UnknownMay 24, 2020 at 8:57 AM

      the similarity in triBGF AND triBIC u applied but how can we say GF is parallel to IC

      Delete
    5. Peter TranMay 25, 2020 at 1:38 PM

      https://photos.app.goo.gl/PxGXK7bKbXqV3Zi6A

      Per the result of problem 1407, we have D, F, M are collinear ( see sketch above)
      and IC ⊥ CE and IC ⊥ FM so IC // GF

      Peter Tran

      Delete
  2. Sumith PeirisMarch 5, 2019 at 12:06 AM

    Let EF & BGI extended meet at X. Let the tangency point of incircle on BC be Y

    From previous problems, BDIY is a square of side b-c = 1 and so FY = 1 and BX = 2.sqrt2

    BD/FX = BG/GX, 1/2 = BG/(2.sqrt2 - BG) since BX = 2.sqrt2
    Hence BG = 2.sqrt2/3 and since BI = sqrt2,

    IG = sqrt2/3

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

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