Saturday, May 1, 2021

Geometry Problem 1493: Four Squares, Parallelogram, Auxiliary Lines

Geometry Problem 1493. Post your solution in the comment box below.
Level: Mathematics Education, K-12 School, Honors Geometry, College.

Details: Click on the figure below.

Geometry Problem 1493: Four Squares, Parallelogram, Auxiliary Lines

Posted by Antonio Gutierrez at 8:54 AM
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9 comments:

  1. Peter TranMay 1, 2021 at 11:57 AM

    https://photos.app.goo.gl/6SQsFetQuy1gT2zc6

    Let AD=AB=u and DG=GJ=v
    Draw points P and Q per attached sketch
    Right triangle AGF congruence to LPA ( case HA)
    So PA=FG= v and PL=AG=u+v
    Triangle AJB congruence to QFL ( case SAS)
    So LF=BJ and ^(QFL)=^(AJB)
    So BLFJ is a parallelogram

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  2. Sumith PeirisMay 1, 2021 at 5:55 PM

    Solution using Pythagoras and Trigonometry

    Let AD = a, DG = GJ = b, AF = c, BJ = x, FL = y

    Now c^2 = a^2 + 2ab + 2b^2 ....(1) ( Rt.Tr. AFG)
    x^2 = 2a^2 + 4ab + 4b^2........(2) ( Rt.Tr. ABJ)

    So x = sqrt2.c = y ............(3) (Rt. Isosceles Tr. LKF)

    Let < FAJ = p & < FJA = q

    Now tan(<LFE) = tan(45-p) = (1 - tan.p)/(1 + tan.p) = (1 - b/(a+b))/(1 + b/(a+b))
    So tan(<LFE) = a/(a+2b) = tan.q which yields < LFE = q .....(4)

    From (3) & (4), LF and BJ are both equal and parallel
    Hence BLFJ is a parallelogram

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  3. Sumith PeirisMay 1, 2021 at 6:19 PM

    Pure Geometry Solution

    Let FAJ = p as before. So < LAB = p
    Tr.s LAB & FAD are congruent SAS

    Hence < ADF = 135 = < ABL, so L,B,D are collinear
    So LB // FJ and also LB = FD = FJ

    Hence BLFJ is a parallelogram

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  4. Stan FULGERJune 3, 2021 at 2:05 PM

    My solution, if found interesting, can be found at https://stanfulger.blogspot.com/2021/06/problem-1493-gogeometry-my-solution.html

    ReplyDelete
  5. Vijaya Prasad NalluriJune 30, 2021 at 5:22 AM

    Let AL=AF=a,AB=AD=b,DG=GF=GJ=c
    BJ^2=b^2+(b+2c)^2=2b^2+4bc+4c^2 while
    LF^2=2a^2=2[b+c)^2+c^2]=2b^2+4bc+4c^2
    Follows BJ^2=LF^2, BJ=LF
    Next AL=AF,AB=DF and <LAB=<FAD imply triangles LAB and FAD Identical
    and LB=FD=FJ as required.

    ReplyDelete
  6. AnonymousAugust 11, 2021 at 3:30 AM

    Can we expect further problems soon, or has this coloumn ended?

    ReplyDelete
  7. G RíosNovember 20, 2021 at 4:54 PM

    En la R rotación de centro A y ángulo de 90 grados en sentido antihorario
    R(D)=B y R(F)=L.
    Lo anterior implica que BL=DF.
    Como los cuadrados EFGD y FHJG son iguales, DF=FJ.
    En consecuencia BL=FJ.

    Por la rotación mencionada al inicio los ángulos ADF y ABL son iguales, y es claro que ADF mide 135 grados (90+45), esto implica que L, B y D están alineados, y esto implica que LB||FJ.

    Luego el cuadrilátero BLFJ tiene dos lados opuestos paralelos e iguales, por lo que es un paralelogramo.

    ReplyDelete
  8. Arindam BanerjeeJanuary 15, 2022 at 7:08 AM

    Join D,F and B,D. Triangles ABL and ADF are congruent. So, BL=DF. Again, Triangle DJF is isoscales. So DF=FJ. So, BL=FJ. Angle ABL=Angle ADF=135 degree. Therefore, Angle ABL+ Angle ABD=135+45=180 degree. So, L, B and D are collinear. Therefore Angle ADL=Angle GJF=45 which means BL is parallel to FJ. Therefore BLFJ is a parallellogram.

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