Geometry Problem. Post your solution in the comments box below.

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

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## Thursday, June 11, 2015

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

Labels:
120,
angle,
equilateral,
isosceles,
metric relations,
right triangle

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ReplyDeleteLet 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

Where point G is determine

ReplyDeleteSee the sketch

ReplyDelete