## Thursday, June 11, 2015

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

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

Click the diagram below to enlarge it.

1. 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

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

2. Where point G is determine

3. See the sketch

1. Sketch is not open