Derivation of Golden Ratio by Geometric Shapes

By Golden Triangle

Golden Triangle refers to the isosceles triangle that has its three internal angles in the ratio of 1 : 2 : 2, that is its internal angles have measure of 36°, 72° an 72°, respectively. If measuring the sides of the triangle, the ratio of its hypotenuse to base is equal to the golden ratio.

In the figure is an isosceles triangle with its internal angles equal to 36°, 72° an 72°. Point D is on the side of AC such that BD = BC.

\because \triangle ABC\sim \triangle BCD

\therefore \dfrac{AC}{BC}=\dfrac{BC}{DC}

\because \triangle BCD and \triangle ABD are also isosceles triangles

\therefore BC = BD = AD

Let BC =1 , AC =x

1=x(x-1)

x^2-x-1=0 //now we get golden equation

x = \dfrac{1+\sqrt{5} }{2} //we get golden ratio

Now let's derive from different angle

Let AC = 1, BC = y

which means we are going to derive the ratio of the base to hypotenuse.

Substituting to (1) gives

y^2 = 1\cdotp (1-y)

y^2+y+1 = 0

y = \dfrac{-1+\sqrt{5} }{2}

Coz y is the ratio of the base to hypotenuse, its reciprocal the ratio of the hypotenuse to base is the golden ratio.

\dfrac{1}{y} = \dfrac{1}{\dfrac{-1+\sqrt{5} }{2}} =\dfrac{2}{-1+\sqrt{5} } \cdotp \dfrac{1+\sqrt{5} }{1+\sqrt{5}} = \dfrac{\sqrt{5}+1 }{2 }

By trapezoid

The golden ratio can be derived from the isosceles trapezoid with 72\degree base angles.

In the figure, the length of upper side is equal to the length of legs, the measure of base angles is 72\degree. If the length of upper side is 1, the length of lower side is golden ratio.

\dfrac{\sqrt{5}+1 }{2 }

which means the ratio of the length of the lower side to upper side is equal to golden ratio. Similarly, its reciprocal, the ratio of the length of upper side to lower side is

\dfrac{\sqrt{5}-1 }{2 }

By right triangle

In the figure is a right triangle such that AB = 1, AC = \dfrac{1}{2} . CD bisects \angle ACB and intersects AB at point D. Point E is on the side of AB such that DE = AD. Then AE is the reciprocal of golden ratio.

Since AB = 1, AC = \dfrac{1}{2} , BC = \dfrac{\sqrt{5} }{2}

\sin\angle ACB = \dfrac{AB}{BC} =\dfrac{1}{\dfrac{\sqrt{5} }{2}} =\dfrac{2}{\sqrt{5} }

\cos\angle ACB=\dfrac{AC}{BC} = \dfrac{\dfrac{1}{2} }{ \dfrac{\sqrt{5} }{2} } =\dfrac{1}{\sqrt{5} }

Using half angle identity for tangent function

\tan \dfrac{\angle ACB }{2}= \dfrac{\sin \angle ACB }{1+\cos \angle ACB}

=\dfrac{\sin \angle ACB}{1+\cos \angle ACB}

=\dfrac{\dfrac{2}{\sqrt{5} } }{1+\dfrac{1}{\sqrt{5} } } =\dfrac{2}{\dfrac{1}{\sqrt{5} +1 } }

=\dfrac{2}{\sqrt{5}+1 }

AD = \tan \dfrac{\angle ACB }{2}\cdotp AC= \dfrac{2}{\sqrt{5}+1 }\cdotp \dfrac{1}{2} =\dfrac{1}{\sqrt{5}+1 }

AE = 2AD = \dfrac{2}{\sqrt{5} -1 } =\dfrac{\sqrt{5}-1 }{2}