Friday, May 4, 2012

Problem 748: Quadrilateral, 90 Degrees, Diagonal, Perpendicular, Metric Relations

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the problem 748 details.

Online Geometry Problem 748: Quadrilateral, 90 Degrees, Diagonal, Perpendicular, Metric Relations.

5 comments:

  1. http://img859.imageshack.us/img859/6765/problem748.png

    let AC cut EF at P ( see sketch)
    ∆DMA similar to ∆DNF so DA/DF=MA/FN=4/7
    ∆BMA similar to ∆BGE so BE/BA=GE/AM= 3/4
    In triangle ∆AEF, applying Ceva’s theorem => DA/DF x PF/PE x BE/BA=1
    Replace DA/DF=4/7 and BE/BA=3/7 we get PF/PE=7/3

    In triangle AEF, applying Von Aubel’s theorem => CF/CB=PF/PE+DF/DA
    Replace PF/PE=7/3 and DF/DA=7/4 we get CF/CB=49/12
    So CB/BF=12/61=x/7
    X=84/61

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  2. By looking my solution as above, I found that angles B and D need not be 90 degrees.
    I am not sure whether I may miss something in my solution or author give us extra information which is not used in the solution. Please clarify.
    Peter Tran

    ReplyDelete
    Replies
    1. To Peter, Problem 748, The extra information is for a short solution. Your solution is great because is more general, see problem 750.

      Delete
  3. Excellent solution by Peter!
    Just wondering if there's any other way to determine x?

    ReplyDelete
    Replies
    1. To Ajit, problem 748, yes, there is another way to determine x.

      Delete