Friday, April 25, 2014

Geometry Problem 1009: Tangent Circles, Common External Tangent, Common Internal Tangent, Arithmetic Mean

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 1009.

Online Math: Geometry Problem 1009: Tangent Circles, Common External Tangent, Common Internal Tangent, Arithmetic Mean

3 comments:

  1. Since EA=ET=EB, FC=FT=FD,
    Also AD//BC (ABCD is trapezium),
    Thus EF=(AD+BC)/2.

    ReplyDelete
  2. Join the points CT and DT.
    Also join the points AT and BT.

    We have

    DF=FT=FC=DC/2

    From the similarity between triangles DFT and DTA we get

    DT/AD=DC/2DT
    (1.)2DT^2=(AD)(DC)

    And from the similarity between triangles CFT and CTB we get

    CT/BC=DC/2CT
    (2.)2CT^2=(BC)(DC)

    Adding equation 1 and 2 we get

    2(DT^2+CT^2)=(AD+BC)DC

    Because DTC is a right triangle we have

    2DC^2=(AD+BC)DC
    2DC=(AD+BC)

    ReplyDelete
    Replies
    1. I forgot the add the final piece:

      DC=EF because AB=DC and therefore DC=2DF=2AE then

      2EF=AD+BC
      EF=(AD+BC)/2

      Delete