Friday, March 15, 2019

Geometry Problem 1425: Two Squares, Collinear Points, Concurrent Lines

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 1425: Two Squares, Collinear Points, Concurrent Lines, Tutoring.

Posted by Antonio Gutierrez at 9:22 AM
Labels: , , , , , ,

6 comments:

  1. c.t.e.oMarch 15, 2019 at 1:28 PM

    Suppouse : a)AG and BH meet on P b) DF and CE meet on P'. Draw PT, P'T' ꓕ AH
    => From similarity of triangles AB/HG = AT/HT and DC/FE = T'D/T'E
    => AT/HT = T'D/T'E . This happen when P and P' coincide

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  2. Peter TranMarch 15, 2019 at 4:58 PM

    Let u=AB and v= GH
    Triangle ABP simillar to GPH ( case AA)
    So P is a point locate on AG and BH such that PB/PH=PA/PG= u/v
    Let P1 is the intersection of AG and FD
    Triangle ADP1 simillar to GFP1
    So P1 is a point locate on AG such that PA/PG= u/v => P1 coincide to P
    Simillarly to other triangles with P2, P3 .
    P2, and P3 with coincide to P

    ReplyDelete
    Replies
    1. c.t.e.oMarch 16, 2019 at 2:28 AM

      To Peter
      P1A/P1G ≠ PA/PG ?

      Delete
    2. Peter TranMarch 16, 2019 at 8:45 PM

      yes, line 6 of my solution will be:
      "So P1 is a point locate on AG such that P1A/P1G= u/v => P1 coincide to P"
      Peter

      Delete
    3. c.t.e.oMarch 18, 2019 at 1:22 AM

      But A, P1, G are not collinear

      Delete
  3. Sumith PeirisMarch 16, 2019 at 6:22 AM

    Let AG and BH meet at P

    Let AD = a, EF = b, AP = x and PG = y

    From similar triangles ABP & GPH, x/a = y/b

    In Tr.s APD & FPG, x/a = y/b and the included angles < DAP = GPF

    So Tr.s APD & FPG are similar and < APD = < GPF

    So < APF + < APD = < APF + < GPF = 180
    Therefore points DPF are collinear

    Similarly for points CPE

    Sumith Peiris
    Moratuwa
    Sri Lanka

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