Thursday, August 25, 2011

Problem 660: Triangle, Circle, Diameter, Perpendicular Lines

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

Click the figure below to see the complete problem 660.

Online Geometry Problem 660: Triangle, Circle, Diameter, Perpendicular Lines.

Posted by Antonio Gutierrez at 9:42 PM
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7 comments:

  1. AjitAugust 28, 2011 at 10:37 PM

    Here’s a proof by brute force. Draw a line // DF from A meeting EF & GD in M & N resply. Now by similar triangles, AM=GE*AF/FG while AN =AG*DF/FG. Now let A be the origin, AC the x-axis and AM, the y-axis so that we can write eqns. for EF & GD as follows:x/AF+y*FG/(GE*AF)= 1 and x/AG+y*FG/(AG*DF)=1 from which we can eliminate y to obtain the x-co-ordinate of H the intersection of EF & GD as x= (GE*AF–AG*DF)/(GE-DF.Now let /_DAC= α=90–C and /_CAE=β=180–A. Hence we can say that:x=[b^2 sinβcosβ(cosα)^2-b^2 sinαcosα(cosβ)^2]/(sinβcosβ-sinαcosα)=2bcosβcosαsin(β-α)/(sin(2β)–sin(2α))
    =bsinC*cosA/sin(B) using standard trigonometric formulae.Further, c/sinC = b/sinB by the sine rule hence, x = c*cosA. In other words, the x-co-ordinate of H is the same as that of B or HB is perpendicular to CA, our x-axis.
    Ajit

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  2. Peter TranAugust 29, 2011 at 4:09 PM

    http://img220.imageshack.us/img220/3329/problem660.png
    Here is the geometry solution.
    Connect EC and AD.
    EG cutt DC at M and DF cut AE at N
    Let H’ and B’ are the projections of H and B over AC
    Note that (ADF)=(ACD) and (GCE)=(FNA) (see picture)
    Triangles AFD similar to triangle GEC and AF/FD=GM/GC ( 1)
    Triangles AFN similar to triangle EGC and AF/FN=GE/GC ( 2)
    Divide expressions (1) to (2) we have FN/FD=GM/GE (3)
    From (3) we also get FD/GE=ND/ME=k
    Triangle HFD similar to triangle HEM with ratio of similarity = k so H’F= k/(1-k) x FG
    Triangle BND similar to triangle BEM with ratio of similarity = k so B’F= k/(1-k) x FG
    H’ coincide with B’ and HB perpendicular to AC
    Peter Tran

    ReplyDelete
  3. PravinSeptember 1, 2011 at 10:14 AM

    Extend DA, CE to meet at X. Join BX.
    Extend CA to meet BX at Y.
    AD ⊥ BC and CE ⊥ AB
    Follows X is the orthocentre of ∆ABC.
    So BYX ⊥ AC
    Let Z be the midpoint of AC.
    Clearly D, Y, E and Z are concyclic (∵they lie on the ninepoint circle of ∆ABC)
    So ∠DYZ = ∠DEZ
    = ∠EDZ (∵DZ = EZ)
    = ∠EYZ
    Same as ∠DYF = ∠EYG
    Follows ∆DYF ∼ ∆EYG
    Hence YF/YG = DF/GE = HD/HG (∵FD ∥ GE)
    Follows HY ∥ DF or HY ⊥ AC
    Already BY ⊥ AC
    Hence HB ⊥ AC

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  4. PravinSeptember 1, 2011 at 9:21 PM

    http://s1092.photobucket.com/albums/i418/vprasad_nalluri/?action=view&current=GoGeom660.jpg

    ReplyDelete
  5. AnonymousSeptember 6, 2011 at 2:33 AM

    Hallo Peter Tran, can you exlain why
    From (3) we also get FD/GE=ND/ME=k. Thank's

    ReplyDelete
  6. Peter TranSeptember 7, 2011 at 6:52 PM

    To Anonymous
    See details below.

    From (3) FN/FD=GM/GE
    we have FN/GM=FD/GE=(FN+FD)/(GM+GE)=ND/ME (Properties of ratio)

    Peter Tran

    ReplyDelete
  7. YOUSHISeptember 10, 2011 at 11:10 PM

    http://hiphotos.baidu.com/gzyoushi/pic/item/18fa9d6d3f5046964216944a.jpg

    CHINESE SORRY

    ReplyDelete

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