Tuesday, November 6, 2018

Geometry Problem 1398: Isosceles Triangle, Circle, Radius, Perpendicular, 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 1398: Isosceles Triangle, Circle, Radius, Perpendicular, Measurement, Tutoring.
at

5 comments:

  1. Sumith PeirisNovember 6, 2018 at 10:56 PM

    AB = 2^2/1 = 4 = BC

    x = 8^2/BC = 16

    But there appears to be something wrong since in the diagram AB + BC < AC


    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  2. Antonio GutierrezNovember 7, 2018 at 3:42 AM

    Thanks for your comment. Problem has been updated.

    ReplyDelete
  3. Sumith PeirisNovember 7, 2018 at 9:31 AM

    Proof of updated version, same method

    AB = d^2/e and BC = f^2/x
    Since AB = BC the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  4. AnonymousNovember 8, 2018 at 1:45 AM

    f2 =BC*CF = BC*x
    d2 = AE*AB =BC*e
    f2/d2 =x/e
    x = e * (f2)/(d2)
    Florin Popa , Comanesti, Romania

    ReplyDelete
  5. mehmetNovember 8, 2018 at 5:08 AM

    Tangency rule: f^2=x*(x-BF) and d^2=e*(e+EB)
    BAC is isoscele, therefore: d^2=e*(x-BF)
    Dividing side by side:
    f^2/d^2=(x*(x-BF))/(e*(x-BF))
    Result is: x=e*(f^2/d^2)

    ReplyDelete

Share your solution or comment below! Your input is valuable and may be shared with the community.