Wednesday, July 31, 2024

Geometry Problem 1573: Find the Angle at K

Challenging Geometry Puzzle: Problem 1573. Share your solution by posting it in the comment box provided.
Audience: Mathematics Education - K-12 Schools, Honors Geometry, and College Level.

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Illustration of Geometry Problem 1573

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Posted by Antonio Gutierrez at 7:48 AM
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3 comments:

  1. Peter TranAugust 2, 2024 at 11:39 PM

    https://photos.app.goo.gl/PdEUVA2CjPRRJ8sq5

    In triangle GHF we have
    Angle H=56= ∡(CGJ)-∡(CFI)
    56= ½(Arc(CE)+Arc(EJ)-Arc(CI)).....(1)
    In triangle KDE we have
    x= angle(K)= ∡(IDE)-∡(KED)
    x=½(Arc(DC)+Arc(CI)-Arc(EJ)).....(2)
    Add (1) to (2) we have 56+x=½(Arc(CE)+Arc(CD))

    Since AD//BE=> ∡(DAC) supplement to ∡(CBE)=> ½(Arc(CE)+Arc(CD))= 90
    So x= angle(K)= 90-56= 34 degrees

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  2. Sumith PeirisAugust 6, 2024 at 11:53 AM

    Draw the common tangent CM, M on DE
    Now DM = CM = ME, hence M is the centre of right triangle DCE

    Let < CDE = a, < KED = b and < CDI = t = CFI

    Now < KEC = b + (90 - a) = < JGC = 56 + t from which
    t + a = 34 + b ....(1)

    But from Triangle KDE,
    < K + b = < EDI = t + a .....(2)

    Comparing (1) and (2)
    K = 34

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  3. Sumith PeirisAugust 8, 2024 at 5:26 AM

    Much simpler Geometry Solution

    Let < DCF = u

    As before < DCE = 90

    < DIF = u
    < DIH =- 180 - u

    < EJH = 90 + u

    So in quadrilateral KIHJ opposite angles J + I = (180 - u) + (90 + u) = 270

    So the angles K + H = 360 - 270 = 90
    Hence < K = 90 - 56 = 34

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

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