Thursday, January 14, 2016

Geometry Problem 1179: Triangle, Centroid, Transversal, Sum of Ratios, Sides

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view more details of problem 1179.

Online Math: Geometry Problem 1179: Triangle, Centroid, Transversal, Sum of Ratios, Sides.

Posted by Antonio Gutierrez at 5:06 PM
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6 comments:

  1. Peter TranJanuary 14, 2016 at 8:46 PM

    http://s9.postimg.org/71ehki32n/pro_1179.png

    Draw median BM and lines AL//DE and CN // DE
    Note that triangles ALM and CNM are congruence… ( case ASA)
    We have a1/a2= GL/GB=(GM-ML)/GB= ½ - ML/ GB …. (1)
    C1/c2= GN/GB= ( GM+MN)/GB = ½ + MN/GB …. (2)
    Add ( 1) to (2) and note that ML=MN we get the result.

    ReplyDelete
  2. APOSTOLIS MANOLOUDISJanuary 15, 2016 at 2:28 AM

    MANOLOUDIS APOSTOLIS 4 HIGH SCHOOL OF KORYDALLOS PIRAEUS GREECE

    BG intersecting AC M is AM=MC and BG=2.MG ,si DE intersecting AC K .
    Apply the theorem Menelaus tri ABM and tri BMC intersecting by KDGE.
    Therefore AK/KM.MG/GB.BD/DA=1 or DA/DB=AK/2.KM and MK/KC.EC/EB.BG/GM=1 or
    EC/EB=KC/2.MK. Therefore DA/DB+EC/EB=AK/2.MK+KC/2.MK=(AK+KC)/2.MK=(MK-AM+KM+MC)/2.MK=2.MK/2.MK=1.

    ReplyDelete
  3. PravinMarch 5, 2018 at 8:33 AM

    Vijaya Prasad Nalluri (Pravin) Rajahmundry, INDIA.

    Let M be the midpoint of AC. Let K, L be the points on the ray BM such that AK//DE//CL.
    Note KL and AC bisect one another at M since AK//LC (ALCK is a parallelogram and so KM = ML)
    Hence a1/a2 + c1/c2 = KG/BG + LG/BG = (KG +LG)/BG
    = (KG + LM +MK + KG) /BG = (2 KG + 2KM)/BG
    = 2GM/BG = BG/BG = 1

    ReplyDelete
  4. AnonymousMarch 6, 2018 at 3:51 PM

    Extend DE to meet AC at P and M be the center of AC
    let PA=x and AP=PC=z
    Apply menelaus to ABM
    =>(PM/AM)(a1/a2)(2/1)=1
    => (x+z/x)=a2/2a1
    => (x+z) = x.a2/2a1 ---------(1)
    and x/z= a2-2a1/2a1-----(2)

    Apply menelaus to ABC
    => (x+2z/x)(a1/a2)(c2/c1)=1
    => (x+2z)(a1/x.a2)=c1/c2
    => (x+2z)(1/x+z)=c1/c2 (from (1))
    => 1+1/(1+x/z)=c1/c2
    => 1+(a2-2a1/a2)=2c1/c2 (from (2))
    => a1/a2+c1/c2=1

    ReplyDelete
  5. PravinMarch 6, 2018 at 6:43 PM

    Does the converse hold good?
    Given D on BA and E on CB such that
    AD = a1, DB = a2, CE = c1, EB = c2
    satisfying a1/a2 + c1/c2 = 1 and
    M is the midpoint of AC,
    is it true that the segments AC and BM
    intersect at the centroid ?

    ReplyDelete
    Replies
    1. Peter TranMarch 6, 2018 at 9:13 PM

      Yes, the converse does hold .
      for example Given D on BA and E on CB such that
      AD = a1, DB = a2, CE = c1, EB = c2
      satisfying a1/a2 + c1/c2 = 1 then DE will pass through the centroid G of triangle ABC.

      see sketch below :
      https://photos.app.goo.gl/60Z3JPEl6dCV325Q2

      Let G is the centroid of triangle ABC.
      DG meet BC at E' . Denote CE'= c1' and BE'=c2'
      per the result of problem 1179 we have a1/a2+c1'/c2'= 1
      compare with original expression we get c1/c2= c1'/c2'
      in segment BC there exist only one point E for each ratio EC/EB so E must coincide to E'.

      Delete

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