Thursday, February 28, 2013

Problem 861: Triangle, Incircle, Parallel, Tangent, Metric Relations

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

Click the figure below to see the complete problem 861.

Online Geometry Problem 861: Triangle, Incircle, Parallel, Tangent, Metric Relations

5 comments:

  1. http://img16.imageshack.us/img16/3367/problem861.png
    Let r= radius of incircle and L is tangent point of GH to incircle
    Connect OG,OA,OE,OH,OF,OC
    Note that triangles AOG and HOG are right triangles and OE and OF are altitudes.
    Relation in right triangle AOG give OE^2=r^2= EG.AE= 7
    Relation in right triangle HOC give OF^2=r^2=GH.HC=2.x=7
    So x=7/2

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  2. Let is OE=OF=r
    angle DEM= angle DFM=90 degrees , why MD is diameter of circle
    ED is perpendicular of AO and EM is perpendicular of GO. So, angle GOA=90 degrees.
    Similarly , angle HOC=90 degrees.
    From the right triangle GOA we have, OE^2=AE.EG=7 or, r^2=7 (1)
    From the right triangle COH we have, OF^2=CF.FH=7 o r, r^2=2x (2)
    From (1),(2) ,3x=7 , x=7/2
    http://img268.imageshack.us/img268/5052/74920469.png

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  3. Let BG=y, BH=z.

    Since BG/BA = BH/BC = GH/AC and BE = BF
    y/(y+8) = z/(z+2+x) = 3/(7+x) and y+1 = z+2

    xy+yz+2y = yz+8z
    3x+3z+6 = xz+7z
    xy+7y = 3y+24
    y+1 = z+2

    Solving those equations, we have
    x=7/2, y=16/5, z=11/5

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  4. Proiectam G si H pe AC si daca folosim Teorema lui Pitagora obtinem (x+2)(x+2)-(x-2)(x-2)=8.8-6.6,de unde x=7/2

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  5. Let BG = p so that BH = p-1 since BE = BF

    From similar triangles
    p/(p+8) = (p-1)/(p+1+x) = 3/((7+x)

    Each ratio = 1/(7-x) where I have used the algebraic identity that if a/b = c/d then each ratio = (a-c) /(b-d)

    So 3(7-x) = 7+x and so x = 7/2

    Sumith Peiris
    Moratuwa
    Sri Lanka

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