Thursday, June 11, 2015

Geometry Problem 1120: Isosceles Right Triangle, 120 Degree, Angle, Equilateral, Metric Relations

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

Click the diagram below to enlarge it.

Online Math: Geometry Problem 1120: Isosceles Right Triangle, 120 Degree, Angle, Equilateral, Metric Relations.

Posted by Antonio Gutierrez at 3:03 PM
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4 comments:

  1. Peter TranJune 13, 2015 at 10:31 AM

    http://s1.postimg.org/d51e4r4cf/pro_1120.png

    Let O is circumcentre of triangle BEC
    Let AB=BC= m and AC= m. √ (2)
    Let ∠ (BCE)= θ
    Triangle EDF have 2 60 degrees angles so EDF is equilateral
    1. We have OGC is 30-60-90 triangle so OC= m/√(3)
    And e=EC= 2.m/sqrt(3). cos(θ +30) with chord= m and angle of position= θ
    Similarly with circumcircle of triangle ABF with chord=m and angle of position= θ+30 we have
    F=BE= 2.m/√ (3). cos(θ +60)
    Similarly with circumcircle of triangle ADC with chord= m. sqrt(2) and angle of position= θ +15 we have
    d=AD= 2.m/√ (3). Sqrt(2).cos(θ +45)

    2. e+f= 2.m/√ (3) . [ cos(θ +30)+ cos(θ +60)]
    Replace cos(θ +30)+ cos(θ +60) = 2. Cos(θ +45). Cos(15)
    So d/(e+f)= 1/(√ (2). Cos(15))
    Replace cos(15)= (√(6)+ √(2))/4 and simplify we get
    d/(e+f)= √ (3)-1

    ReplyDelete
  2. smithmathJuly 28, 2015 at 2:05 AM

    Where point G is determine

    ReplyDelete
  3. Peter TranSeptember 11, 2015 at 1:55 AM

    See the sketch

    ReplyDelete

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