Saturday, October 22, 2011

Problem 679: Triangle, Circle, Median, Proportion

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

Click the figure below to see the complete problem 679.

Online Geometry Problem 679: Triangle, Circle, Median, Proportion
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5 comments:

  1. Peter TranOctober 22, 2011 at 8:59 PM

    http://img580.imageshack.us/img580/7519/problem679.png
    Draw RMS // DFE ( see picture)
    Note that Area(ABM)=area(BMC) ( same altitudes and bases )
    Area(ABM)= Area(BMC)=½* c*MN =1/2*a*MP
    So a/c=MN/MP
    Note that triangle RBS is isosceles and triangle MNR similar to tri. MPS
    So MN/MP=MR/MS=FD/FE=x/y
    So a/c=x/y
    Peter Tran

    ReplyDelete
  2. AnonymousOctober 23, 2011 at 5:47 AM

    let be r the radius,B1,B2 the angles at vertex B
    using the sine law
    in DBF:sinB1/x=sinF/r
    in EBF:sinB2/y=sinF/r
    in ABM:sinB1/(0.5b)=sinA/BM
    in CBM:sinB2/(0.5b)=sinC/BM
    in ABC:sinA/a=sinC/c
    hence a/c=x/y

    ReplyDelete
  3. HenkieOctober 24, 2011 at 7:59 AM

    Draw AG//DE and CH//DE, giving K and L the intersections with BM. r = radius circle.
    http://s73.photobucket.com/albums/i207/HenkieR/PuzzleSolutions/GoGeometryProblem679solution_rlg.jpg
    ∆MCL = ∆MAK (opposite angles, Z-angles and AM=CM)
    so LC = AK.
    ∆BEF ~ ∆BCL (aa), so y/r = LC/a => LC = ay/r
    ∆BDF ~ ∆BAK (aa), so x/r = AK/c => AK = cx/r
    Because LC = AK, this gives ay/r = cx/r
    => ay = cx => a/c = x/y
    QED.

    ReplyDelete
  4. Sailendra TSeptember 29, 2017 at 10:32 AM

    Draw a line CPQ parallel to EFD intersecting BM at P and AB at Q
    Since BDE and BQC are similar isosceles triangles, CP = ky and PQ = kx
    Applying Menelaus to triangle ACQ
    => (BQ/BA)*(AM/MC)*(CP/PQ) = 1
    => (a/c)*(1)*(ky/kx) = 1
    => a/c = x/y

    ReplyDelete
  5. MarcoJanuary 4, 2024 at 11:11 PM

    BM is the median
    AM=MC & [ABM]=[CBM]
    (AB)(BM)sin sin<DBF/x=sin<EBF/y
    x/y=sin<DBF/sin<EBF=sin<ABM/sin<CBM=a/c

    ReplyDelete

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