Tuesday, September 23, 2008

Elearn Geometry Problem 183

Right triangle

See complete Problem 183 at:
www.gogeometry.com/problem/p183_right_triangle_hypotenuse.htm

Right Triangle, Hypotenuse Trisection Points, Squares of the Distances. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

Posted by Antonio Gutierrez at 7:15 AM
Labels: , , , , , ,

7 comments:

  1. AnonymousSeptember 23, 2008 at 8:32 PM

    by Cosine Law
    d²+(b/3)²+2d(b/3)Cos[BDC]=AB²
    d²+(2b/3)²-2d(2b/3)Cos[BDC]=BC²
    so
    3d²+2/3b²=2AB²+BC²----(1)
    and similarly
    3e²+2/3b²=2BC²+AB²----(2)
    (1)+(2)
    3(d²+e²)+4/3b²=3(AB²+BC²)=3b²
    d²+e²=5/9b²

    ReplyDelete
  2. c .t . e. oJanuary 8, 2010 at 9:59 AM

    a geometry solution

    Draw DF, EG perpendicular to BC
    mark x = BF = FG = GC ( thales theor )
    mark y = EG => FD = 2y ( middle line )

    tr BDF => d'2 = x'2 + 4y'2 (1)
    tr BEG => e'2 = y'2 + 4x'2 (2)
    from (1) and (2)

    d'2 + e'2 = 5x'2 + 5y'2 = 5( x'2 + y'2 ) (3)

    tr EGC => x'2 + y'2 = 1/9 b'2 (4)
    from (3) and (4)

    d'2 + e'2 = 5/9 b'2

    P.S. d'2 mean d*d ( power 2 )

    ReplyDelete
  3. INDSHASeptember 5, 2010 at 10:22 PM

    Hi,I'm INDSHAMAT from Srilanka.My solution has been denoted as follows
    AB2={2d2+2(b2/9)}-e2 (1)(By Apollonius theorem)
    BC2={2e2+2(b2/9)}-d2 (2)(By Apollonius theorem)
    AB2+BC2=b2 (3)(By pythogorous theorem)
    So {2d2+2(b2/9)}-e2+{2e2+2(b2/9)}-d2=b2
    Therefore d2+e2=5(b2/9)is the Answer

    ReplyDelete
  4. Emil EkkerSeptember 11, 2010 at 1:23 AM

    Let AB=c, BC=a, b/3=f.--->
    0) a^2+c^2=b^2
    1) c^2+e^2=2*(d^2+f^2), and
    2) a^2+d^2=2*(e^2+f^2) --->
    (1)+(2):
    a^2+c^2+d^2+e^2=2*(d^2+e^2+2*f^2)
    b^2+d^2+e^2=2*d^2+2*e^2+4*f^2
    b^2=d^2+e^2+4*(b/3)^2
    --->
    d^2+e^2=b^2*(1-4/9)

    ReplyDelete
  5. Luc GheysensAugust 9, 2011 at 7:09 AM

    Choose a coordinate system so that B(0,0), A(a,0) and C(0,c).
    Then we find that D has coordinates (2a/3, c/3) and that E has coordinates (a/3, 2c/3).

    So by the formula for the distance between two points: d^2 = (a/3)^2 + (2c/3)^2 (1)
    e^2 = (2a/3)^2 + (c/3)^2 (2)

    From (1) + (2) we get: d^2 + e^2 = 5a^2/9 + 5c^2/9 (3) and by Pythagoras' theorem a^2 + c^2 = b^2 (4).

    From (3) and (4) follows that d^2 + e^2 = (5/9)b^2.

    ReplyDelete
  6. PravinAugust 10, 2011 at 9:21 AM

    Let M be the midpoint of AC (DE).
    In triangle BDE:
    BD^2 + BE^2 = 2(BM^2 + ME^2)
    (i.e.)d^2 + e^2 = 2(BM^2 + b^2/36).....(i)
    In triangle ABC:
    b^2 = AC^2 = AB^2 + BC^2 = 2(BM^2 + MC^2)
    (i.e.) b^2 = 2(BM^2 + b^2/4)..... (ii)
    From (i) and (ii);
    d^2 + e^2 - b^2 = 2[(b^2/36) - (b^2/4)]
    d^2 + e^2 = b^2 + (b^2/18) - (b^2/2)= 5b^2/9

    ReplyDelete
  7. PravinAugust 10, 2011 at 9:03 PM

    Post Script:
    Notice that BM = AM = MC = b/2 (since triangle ABC is right angled at B)
    So BD^2 + BE^2 = 2(BM^2 + ME^2)implies
    d^2 + e^2 = 2[(b^2/4) + (b^2/36)]
    =(b^2/2) + (b^2/18) = 5b^2/9

    ReplyDelete

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