Sunday, October 8, 2017

Geometry Problem 1349: Three Squares, 90 Degrees, Perpendicular Lines, Diagonal. Math Infographic

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

Details: Click on the figure below.

Geometry Problem 1349: Three Squares, 90 Degrees, Perpendicular Lines, Diagonal. Math Infographic.

Posted by Antonio Gutierrez at 1:05 PM
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7 comments:

  1. Sumith PeirisOctober 9, 2017 at 3:21 AM

    Problem 1349
    Let < AHB = < HBG = < AGB = θ.

    AC2 = 2.CE2 = CE.CG, hence < CAE = < AGB = θ.

    Since < CBP = < CAP =θ,
    ABCP is concyclic and so
    < CPE = < ABC = 90

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  2. APOSTOLIS MANOLOUDISOctober 9, 2017 at 4:16 AM

    Problem 1349
    In the extension of AB to B above point K such that AB=BK=>triangleKAD=triangleAFR=E.So
    KD=AE and KD perpendicular to AE.If the AE intersects the KD in Q and the BC in M. Then
    BM=MC and EQ/QA=EM/AD=3/2 =>EQ/EA=3/5 (1).Is EP/PA=BE/AH=2/3 =>EP/EA=2/5(2).
    From (1) and (2)=>EP/EQ=2/3=>EP/PQ=2/1=EC/CM=>CP//MQ=>CP is perpendicular in AE.
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete
  3. Peter TranOctober 9, 2017 at 7:31 AM

    https://photos.app.goo.gl/vydebtzBccTnATBf1

    Let BH meet EF at K
    Let AB=BC=CD=AD= 1
    So EK= 2/3 and AE= sqrt(5)
    Triangle ABP similar to EKP ( case AA)
    PE/PA= EK/AB= 2/3
    So PE=2/sqrt(5) and PA=3/sqrt(5)
    And PE/EB= 1/sqrt(5) and CE/AE= 1/sqrt(5)
    Triangle ECP similar to EBA ( case SAS)
    So ∠ (CPE)= ∠ (EBA)= 90 degrees

    ReplyDelete
  4. AnonymousDecember 2, 2017 at 1:48 AM

    Prove by similar triangles: △CPE similar to △ABE

    Let CE=1. Then BE=2, AH=3
    △BPE similar to △HPA (AAA) so AP:PE=3:2
    AE=SQRT(5)
    PE=2(SQRT(5)) / 5

    Since <AEB=<CEP (common angle)
    AE/BE= SQRT(5)/2
    CE/PE= 1/ 2(SQRT(5)) / 5 = 5 / 2(SQRT(5)) = SQRT(5) / 2
    Hence △CPE similar to △ABE ( ratio of 2 sides, inc.<)
    Since <ABC=90 degrees, <CPE is therefore 90 degrees. (proved)

    ReplyDelete
  5. PravinDecember 21, 2017 at 10:34 PM

    Problem !349 is more amenable for analytic proof:
    Let A = (0,0), D = (1,0), B = (0,1)
    Then C = (1,1), E = (2,1), F = (2,0), H = (3,0)
    Equations: BH: x + 3y = 3, AE: x - 2y = 0,
    their intersection P = (6/5,3/5)
    Slopes: of AE = 1/2, of CP = -2
    Hence their product being -1, they are perpendicular.

    ReplyDelete
  6. Tyler Owen Eli ArianeJanuary 19, 2018 at 8:32 AM

    Great work! you guys really know what you are doing! WOW just WOW!

    ReplyDelete

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