Aim-To determine the Golden Section and how it

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Transcript Aim-To determine the Golden Section and how it

Aim-To determine the Golden
Section and how it is used in
Geometry
φ
What is the Golden Section

The golden ratio is a ratio that states if
the ratio of the sum of the quantities to the
larger quantity is equal to the ratio of the
larger quantity to the smaller one. It is a
constant irrational number like pi. It is
represented by the greek alphabet phi(φ).
History behind the Ratio

The golden ratio has fascinated
mathematicians since ancient Greece. Early
mathematicians started studying the ratio
because of its constant appearance in
geometry. It is said that it was originally
founded by Pythagoras. It has been used in
ancient art, paintings, architecture, music and
is found in nature.
Constructions using
the Golden Ratio
First, construct a line segment BC, perpendicular to
the original line segment AB, passing through its
endpoint B, and half the length of AB. Draw the
hypotenuse AC.
Draw a circle with center C and radius B. It
intersects the hypotenuse AC at point D.
Draw a circle with center A and radius D. It
intersects the original line segment AB at point S.
This point divides the original segment AB in the
golden ratio.
The Golden Rectangle



Draw a rectangle 1 unit high and √5 = 2·236
long
Draw a square in the center with two triangles
left over.
Each triangle will have the ratio 1 : phi
Hofstetter's 3 Constructions
of Gold Points
On any straight line S, pick two points X
and Y.
With each as centre draw a circle
(green) through the other point
labelling their points of intersection G
(top) and B (bottom) and the points
where they meet line S as P and Q;
With centre X, draw a circle through Q
(blue);
With centre Y, draw a circle through P
(blue); labelling the top point of
intersection of the blue circles A
Used in the Fibonacci sequence
The Fibonacci sequence is:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233,
377, 610, 987,
 Use the following equation when n=223

Example 1

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The diameter AB of a circle is extended to a
point P outside the circle. The tangent
segment PT has length equal to the diameter
AB. Prove that B divides AP in the golden
ratio. The center is O. The point where the
tangent intersects the circle is T.
The radius r is perpendicular to the tangent
at T. This creates a right triangle with the
hypotenuse OP and side lengths r and 2r.
Example 1 Continued

BP is OP - OB