Monday, February 25, 2019

Geometry Problem 1418: Triangle, Circumcircle, Incenter, Side, Sum of Segments, Tangent, Parallel, Angle

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.
This entry contributed by Ajit Athle.
Details: Click on the figure below.

Geometry Problem 1418: Triangle, Circumcircle, Incenter, Side, Sum of Segments, Tangent, Parallel, Angle

11 comments:

  1. Draw CP=CB (P on AC) Join P to I, P to B, A to I.
    APIB isosceles trapezoid => B/2 = 33°
    => x = 48°

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    Replies
    1. Better replace P by X (say) since P is already given

      How is AXIB an isosceles trapezoid?
      AX = BI but that’s all?

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    2. Ang BAX + AXI = 180°

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    3. Ang of ∆AXI 16,5° + 16,5° +147°
      I was curious to know your solution for 1261

      Delete
    4. To c.t.e.o

      Problem 1261

      Mark X on AB such that BX = BC
      Mark Y on AD such that DC = DY

      Use Pythagoras and show CX = AY and AX = CY that is AXCD is a parallelogram.
      X mid point of AB so C midpoint of BE

      Delete
  2. https://photos.app.goo.gl/kMimcwDGK9WHYCVs9
    Define points D and E per sketch
    BD=BI so CA= CD
    Triangles ACD and IBD are isosceles and CD⊥AD
    Due to symmetry we have ∠ (IDC)= ∠ (IAC)=33/2=16.5
    So ∠ (IBC)=33 and ∠ (ACP)=66
    Note that ACP is isosceles and APCQ is cyclic
    From above we calculate x=48

    ReplyDelete
  3. Mark on CB extended BD = BI

    Tr.s AIC & DIC are congruent SAS
    So < IDC = < B/4 = < CAI 33/2 whence B = 66

    Now < AQP = B = < PAC = < PCA = < PQC since APCQ is concyclic

    So 2B + x = 180 and B = 48

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  4. So we conclude that if in a Tr. ABC, BI + BC = AC where I is the incentre, then B = 2A

    Corollary is also true

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