Friday, October 28, 2016

Geometry Problem 1279 Parallelogram, Midpoint, Triangle, Area, 3/10

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 1279.

Geometry Problem 1279 Parallelogram, Midpoint, Triangle, Area, 3/10, tutoring.

Posted by Antonio Gutierrez at 10:10 AM
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12 comments:

  1. Peter TranOctober 28, 2016 at 3:34 PM

    https://goo.gl/photos/L9wsdrtPF7pDebWf7
    Let DE extended meet BC at P
    Since EB= ½. CD => PB=BC=AD
    Triangle GAD similar to GFP => GA/GF=AD/PF= 2/3=> GA/AF=2/5 => S(DGF)/S(ADF)=3/5
    S(DGF)=3/5.S(ADF)= 3/5 x ½ S(ABCD)= 3/10. S(ABCD)

    ReplyDelete
  2. Sumith PeirisOctober 29, 2016 at 12:49 AM

    Let M be the mid point of AF.

    EM = AD/4 so AD = 4/5AM

    Now S(AEM) = 1/4 S(ABF) = 1/4(S/4) = S/16
    Hence S(AED) = 4/5 X S/16 = S/20

    S1 = S -S(ABF) - S(ADE) - S(CDF) + S(ADE)

    Hence S1 = S -S/4 - S/4 - S/4 + S/20 = 3S/10

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
    Replies
    1. Sumith PeirisOctober 29, 2016 at 5:33 AM

      In the 2nd line it should read as

      AG = 4/5AM not AD = 4/5AM

      Delete
  3. UnknownOctober 29, 2016 at 7:28 AM

    S(AED) = S/20 ??

    ReplyDelete
    Replies
    1. Sumith PeirisOctober 29, 2016 at 11:21 AM

      S(AEM) = 1/4 S(ABF) = 1/4(S/4) = S/16
      Hence S(AED) = 4/5 X S/16 = S/20

      Delete
    2. Peter TranOctober 29, 2016 at 4:31 PM

      Sumith
      Per observation, S(AED) is not equal S/20.
      Peter Tran

      Delete
    3. UnknownOctober 29, 2016 at 9:43 PM

      Area of triangle AED is 1/4 the area of ABCD, because triangle has same base as ABCD and height 1/2 of height of parallelogram ABCD

      Delete
    4. Sumith PeirisOctober 29, 2016 at 10:56 PM

      Extremely sorry Omid and Peter. My mistake.

      I should have said S(AEG) = S/20 which is easily seen since S(AEM) = S/16 and AG = 4/5AM

      Delete
  4. AnonymousNovember 2, 2016 at 10:45 AM

    Extend AF to meet DC at P
    Since FC is half of AD, from mid-point theorem we have
    CP = CD & FP = AF = l (say) and AG = x (say)
    Also the triangles DGP and EGA are similar
    => DP/EA = PG/AG
    => 4 = 2l-x/x
    => x = AG = (2/5)l and GF = (3/5)l

    Since the area of triangle AFD = 1/2(S) => Area of triangle GFD (S1) = 3/5*1/2*S (S being the are of //gm ABCD)

    Therefore S1 = 3/10*S

    ReplyDelete
  5. c.t.e.oApril 19, 2018 at 12:31 PM

    Draw CG, G midpoint of AD, H on ED. Tr GHD ~ tr AEG => S(AEG)=1/20 (S)
    AGD + AEG + EGFB + FCD + S1 = S => 1/5 + 1/20 + 1/5 + 1/4 + S1 = S
    S1 = 7/10 S

    ReplyDelete
  6. GregNovember 22, 2020 at 2:20 AM

    Another synthetic proof : See graph here
    Let FH // AB intercept ED in I, H on AD
    IH = AE/2 = AB/4 => FI = 3/4.AB = 3/2.AE
    ΔAEG and ΔFIG are similar, ratio 2/3
    AF = AG + GF = 5/2.AG => AG = 2/5.AF
    S(AGE) = 2/5.S(AFE) = 2/5.1/2.S(AFB) = 1/5.1/4.S = 1/20.S
    S(AGD) = S(AED) – S(AGE) = [1/4 - 1/20].S = 1/5.S
    S1 = S(AFD) – S(AGD) = [1/2 - 1/5].S = 3/10.S QED

    ReplyDelete
  7. GregNovember 24, 2020 at 11:12 AM

    I posted here another nice solution, by José Luis da Vila, that I found on Twitter - with a little bit more details than the original.

    ReplyDelete

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