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This paper explains how to obtain the number Φ, using a square with side length equal to a, the right triangle with sides a/2 and a, and a circle with radius equal to the hypotenuse of this right triangle. In particular, from a square whose side length is equal to a, we will show how to obtain a segment b in such a way that the value of a/b is the...
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... is well known that this ratio is also calculated from equating the ratios obtained by dividing a segment of length a + b by a (being a always the largest segment) and a by b, that is, (a + b)/a = a/b. This equality is a consequence of the ratio of proportionality in triangles applying Thales's Theorem, see Fig. 1.1 ...
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... on the stem, in the flowers of artichokes and sunflowers, in the inflorescences of Romanesco broccoli, in the configuration of coniferous conifers, in the reproduction of rabbits and in how DNA encodes the growth of complex organic forms. And allow us to draw a spiral using squares with side length equal to each term of the sequence, as it shows Fig. 1.2. The growth ratio is Φ, that is, the golden ratio. This Fibonacci spiral, it is found in the spiral structure of the shell of some mollusks, such as the nautilus as it shows Fig.1.3, and also in Leonardo da Vinci's masterpieces, as for instance it is shown in Fig. ...
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... growth ratio is Φ, that is, the golden ratio. This Fibonacci spiral, it is found in the spiral structure of the shell of some mollusks, such as the nautilus as it shows Fig.1.3, and also in Leonardo da Vinci's masterpieces, as for instance it is shown in Fig. 1.4. ...
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... using squares with side length equal to each term of the sequence, as it shows Fig. 1.2. The growth ratio is Φ, that is, the golden ratio. This Fibonacci spiral, it is found in the spiral structure of the shell of some mollusks, such as the nautilus as it shows Fig.1.3, and also in Leonardo da Vinci's masterpieces, as for instance it is shown in Fig. ...
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... How to obtain Φ number. First, as it shows the Fig. 2.1 , we draw a square with side length equal to a. Next, we divide this square in two equal rectangles, see Fig. 2.2, and we draw the rectangle diagonal OC (hypotenuse h of the right triangle OCD) as it shows Fig. 2.3. Now, considering the point O as the center and the diagonal OC as the radius, we draw a circle as it shows Fig.2.4, and ...
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... particular, we have shown how from a square whose side measures a, this length being arbitrary, the appropriate segment of length b is the one that allows us to build a circle of radius a 2 + b, centered on the base of the square at point a 2 , and so that the semicircle of this circle circumscribes the upper part of the square, as it is shown in Fig.3.1. ...
