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

3 comments:

  1. 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

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  2. 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

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  3. 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.

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