Sunday, May 12, 2019

Geometry Problem 1433: Tangent Circles, Diameter, Perpendicular, Midpoint, Measurement, Poster

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 1433: Tangent Circles, Diameter, Perpendicular, Midpoint, Measurement, Tutoring.
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3 comments:

  1. Peter TranMay 12, 2019 at 9:27 PM

    https://photos.app.goo.gl/WkovAxHYaaVhSUNF8
    connect QF , TB, TA
    Let QF meet circle Q at H
    Since circle O tangent to circle Q at T so O, Q,T are collinear
    Triangles OBT and QFT are isosceles with ∠ (BOT)= ∠ (FQT)
    So ∠ (OTB)= ∠ (QTF) => B, F, T are collinear
    In the same way we also have A,H,T are collinear
    Triangle BEF and BTA are similar ( case AA)
    So AT=EF. (AB/BF)….(1)
    Triangle BEF similar to HKA ( case AA)
    So AH=BF .(KH/BE)=BF.(EF/BE)…(2)
    Multiply (1) by (2) => AT. AH= AG^2= EF.(AB/BF). (BF/BE).EF
    AG^2= (AB/BE). EF^2= 4 EF^2
    So AG= 2 EF

    ReplyDelete
  2. Sumith PeirisMay 20, 2019 at 11:31 PM

    OCBD being a rhombus, Tr. ACD is equilateral

    However circle Q is not unique.

    Hence Peter's proof above does not/ need not use this fact

    ReplyDelete
  3. Peter TranMay 21, 2019 at 9:49 PM

    the fact that E is the midpoint of OD ( or AB/BE=4) is implicitly indicate that ACD is equilateral.
    In other word , if ACD is not equilateral , AG will not equal 2.EF

    Peter Tran

    ReplyDelete

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