Φ.  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.

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Presentation transcript:

φ

 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).

. 1 1

 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.

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.

 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

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

 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

 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.

 BP is OP - OB