Saturday, May 5, 2012

Problem 750: Complete Quadrilateral, Triangle, Diagonal, Parallel, Similarity, Metric Relations

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

Click the figure below to see the problem 750 details.

Online Geometry Problem 750: Complete Quadrilateral, Triangle, Diagonal, Parallel, Similarity, Metric Relations.

3 comments:

  1. http://img19.imageshack.us/img19/138/problem750.png

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

    In triangle AEF, applying Von Aubel’s theorem => EC/CD=BE/BA+PE/PF
    Replace PE/PF=e/f and BE/BA=e/a we get EC/CD=e(f+a)/af
    CD/DE=CD/(CD+EC)=af/(af+ef+ea)=c/e
    e/c= 1+e/a+e/f
    divide both side by e we get 1/c=1/e+1/a+1/f

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  2. Draw a line through B parallel to CH to meet EC at K and DA extended at K’.
    Let BK = b and BK’ = b’
    Draw a line through D parallel to CH to meet CF at K and BA extended at L’.
    Let DL = d and DL’ = d’
    Apply result of Problem 746 to ∆BCD to obtain
    c = bd /(b + d),
    1/c = 1/b + 1/d ... (i)
    Apply result of Problem 746 to ∆BAD to obtain
    a = b'd' /(b' + d'),
    1/a = 1/b' + 1/d'... (ii)
    From similar ∆s EGB and LDB we get
    e/d' = BG/BD = GD/BD - 1 = (e/b) - 1,
    1/d' = 1/b - 1/e ... (iii)
    Exactly in the same manner, we get
    1/b' = 1/d - 1/f ... (iv)
    Thus by (i) through (iv)
    1/a
    = (1/d - 1/f) + (1/b - 1/e)
    = (1/b + 1/d) - 1/e - 1/f
    = 1/c - 1/e - 1/f
    Hence 1/c = 1/a + 1/e + 1/f as required

    ReplyDelete
  3. Here is a much simpler Proof:
    c/e = DH/DG, c/f = BH/BN
    So BD/c = (BH + DH)/c = BH/c + DH/c = BN/f + DG/e
    Next a/e = BM/BG, a/f = MD/DN
    So BD/a = (BM + MD)/a = BM/a + DM/a = BG/e + DN/f
    Therefore
    BD/c - BD/a = (DG - BG)/e + (BN - DN)/f = BD/f + BD/e
    Follows 1/c - 1/a = 1/f + 1/e
    Hence 1/c = 1/a + 1/e + 1/f

    ReplyDelete