Thursday, November 10, 2016

Geometry Problem 1286 Triangle, Quadrilateral, 60, 135, 105 Degree Angles, Congruence, Measurement

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

This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.

Click on the figure below to view more details of problem 1286.

Geometry Problem 1286 Elementts: Triangle, Quadrilateral, 60, 135, 105 Degree Angles, Congruence, Measurement.

Posted by Antonio Gutierrez at 10:57 AM
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2 comments:

  1. APOSTOLIS MANOLOUDISNovember 10, 2016 at 1:24 PM

    Problem 1286
    Let AB=a, BC=b, CD=c,AD=d. The sides AB and CD intersects at point K then triangle KAD is equilateral (<ADC=360-60-135-105=60). Draw BE//AD( the point E lies on the CD).Then ABED is isosceles trapezoid (AB=ED).So AC=BD=AE or triamgle ACE is isosceles .Then <AED=<ACK and <DAE=<KAC.But triangles KAC=DAE and AB=ED=KC, KB=CD. Draw CF
    perpendicular in KB, then KF=KC/2=a/2,CF=√(3 )a/2=BF(<KCF=30,<FCB=<FBC=45).But b^2=2CF^2 or b/a=√(6 )/2.
    c=CD=KB=KF+FB=a/2+√(3 )a/2 =a(√(3 ) +1)/2 or c/a=(√(3 ) +1)/2.
    And d=AD=KA=AB+BK=a+c=a+ a(√(3 ) +1)/2=a (√(3 ) +3)/2 or d/a=(√(3 ) +3)/2 .
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

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  2. Peter TranNovember 10, 2016 at 2:32 PM

    https://goo.gl/photos/YmHy2K7uN38Yp5xY8
    Let AB meet CD at E
    Draw BF//AD , AH⊥CD, CG⊥BE ( see sketch)
    We have AED and BEF are 60-60-60 triangles
    AHD , AHE and CGE are 30-60-90 triangles
    And CCG is 45-45-90 triangle
    Observe that H is the midpoint of CF and DE ;and AF=BD=AC
    Let u= CG
    So BG= u and BC= u.sqrt(2)
    EC=2u/sqrt(3)
    EG= u/sqrt(3)
    EF=BE=u(1+1/sqrt(3))
    CF= EF-EC=u(1-1/sqrt(3))
    AB=FD=CE=2u/sqrt(3)
    CD=EF=u(1+1/sqrt(3))
    AD=2.DH= 2.DF+CF= u(1+sqrt(3))
    Verify that BC/AB=sqrt(6)/2
    CD/AB=(sqrt(3)+1)/2
    And AD/AB=(sqrt(3)+3)/2

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