# Golden Ratio in 3-, 6-fold Geometry

I thought that the Golden Ratio only showed up in Geometry once you introduce the 5-fold symmetry of the pentagon or pentagram (in 2-dimensions) or when you introduce the Icosahedron or the regular Dodecahedron in 3-dimensional geometry.

But a friend of mine e-mailed to me the following link which shows the Golden Ratio occurs when a triangle is surrounded by a circle:
http://www.intent.com/sg/inscribedtrianglephi.html

Here I introduce my own description and Figures to explain this.      To show this is exactly the Golden Ratio, consider the following Figure. The point "O" is the center of the circle and the triangles.

All lengths of the little equilateral triangle (A-C-E) are 1/2 of the corresponding lengths of the big equilateral triangle.

Let the segment AB have a length equal to 2. Then DE = 1.

AB = OB cos(30) = OB sqrt(3)/2

Which gives us AB = 2 = OB sqrt(3)/2.

Which means OB = 4/sqrt(3).

Note that OF is the same length (radius of the circle) as OB.

We also have

OA = OB sin(30) = OB/2 = 2/sqrt(3)

Now, OD = OA/2 = 1/sqrt(3).

From the properties of a right triangle, we have the equation

DF^2 + OD^2 = OF^2
(EF + 1)^2 + (1/sqrt(3))^2 = (4/sqrt(3))^2

We need to solve for the length EF.

(EF^2 + 2EF + 1) + 1/3 = 16/3
EF^2 + 2EF + 1 = 15/3 = 5
EF^2 + 2EF - 4 = 0

Using the quadratic formula, and using the "+" term, we get

EF = (-2 + sqrt(4 - 4(-4)) ) / 2 = (-2 + sqrt(20))/2
EF = (-2 + sqrt(4*5))/2 = (-2 + 2sqrt(5))/2
EF = -1 + sqrt(5)

So the ratio of the lengths are

2ED/EF = 2/(-1 + sqrt(5)) = 2(1 + sqrt(5))/4

2ED/EF = (1 + sqrt(5))/2 = Golden Ratio.

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Last updated: 05-29-2002