Sunday, August 2, 2015

Geometry Problem 1144: Three Equal Squares, Diagonals, Perpendicular, 90 Degrees, Angle Bisector

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

Click the figure below to see the complete problem 1144.

Online Geometry Problem 1144: Three Equal Squares, Diagonals, Perpendicular, 90 Degrees, Angle Bisector

Posted by Antonio Gutierrez at 11:17 AM
Labels: , , , ,

8 comments:

  1. Jacob HA (EMK2000)August 2, 2015 at 1:30 PM

    ∠APB = ∠PEB + ∠PBE = arctan(1/2) + arctan(1/3) = 45°
    thus A B C P D are concyclic.

    Since BD is the diameter, ∠BPD = 90°.
    Thus ∠APB = ∠APD = 45°.

    ReplyDelete
  2. Sumith PeirisAugust 3, 2015 at 12:41 AM

    BD^2 = DF.DH hence BD is tangential to circle BFH and so < DBF = < DHB so < DBH = < BFD = < FAE. Hence ABPD is con cyclic and so < BPD = 90 and < BPA = < DPA = 45

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
    Replies
    1. AjitAugust 4, 2015 at 1:37 AM

      Nice solution, Sumith.

      Delete
    2. Sumith PeirisAugust 5, 2015 at 6:57 AM

      Thanx Ajit. Do u have another method ?

      Delete
    3. AjitAugust 5, 2015 at 3:49 PM

      One could also do this by constructing an additional row of 3 squares underneath the original three.
      (BA extended to BA', GH to GH'). Now H'EA is right isosceles and H'A//HB. The result follows.

      Delete
  3. ion raduAugust 4, 2015 at 12:10 PM

    Notam lungimea seg AB=6a ,a numar real pozitiv.Luam K pe semidreapta (DB astfel incat BK=2a=>HD/HK=12a/20a=3/5 si m(AP/PH=AB/HJ=6a/9a=2/3=>PH/(AP+PH)=3/(3+2)=3/5<=>PH/HA=3/5=HD/HK=>PD||AK<=>m<(DPH)=90

    ReplyDelete
  4. Stan FulgerAugust 5, 2015 at 2:02 AM

    Take Q the symmetrical of B about D; clearly Q is a vertex of the square constructed onto FH, so triangles AFE and BQH are similar (similitude ratio is 1/sqrt{2}) and consequently <PBD=<PAD, so BADP is cyclic, done.

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