Seccion Aurea

Número de oro pliegues de papel

14 de julio de 2008

contenidos

1

Número de oro

2

regular pentagon

There are many things in mathematics that are beatiful, yet sometimes the beauty is not apparent at first sight.This is not the case with the Golden Section,which ought to be beautiful at first sight, regardless of the form in which it is presented.The Golden Section refers to the proportion in which a line segment is divided by a point. graficamente B − − − − − − − P − − − − − −A

Simply, for the segment AB, the point P partitions (or divides) it into two segment,AP and PB, such that AP PB = PB AB This proportion, apparently already known to the Egyptians and the Greeks,was probably first named the “Golden Section“ or “section aurea“ by Leonardo da Vinci, who drew geometric diagrams for Fra Luca Pacioli’s book,De Divina Proportione(1509), which dealt with this topic.

There are probably endless beauties involving this Golden Section.One of these is the relative ease with which one can construct the ratio by merely folding a strip of paper.

Simply have your student take a strip of paper, say about 1-2 inches widw, and make a knot.Then very carefully flatten the knot as shown in the next figure.Notice the resulting shape appears to be a regular pentagon,that is, a pentagon with all angles congruent and all sides the same length.

regular pentagon

If the student use relatively thin translucent paper and hold it up to a light,they ought to be able to see the pentagon with its diagonals.These diagonals intersect each other in the Golden Section(see below).

regular pentagon

Let’s take a closer look at this pentagon (figure).Point D divides AC in to the Golden Section,since DC AD = AD AC

We can say that the segment of length AD is the mean proportional between the lengths of the shorter segment (DC) and the entire segment (AC).

For some student audiences,it might be useful to show what the value of the Golden Section is.To do this, begin with the isosceles triangle ABC,whose vertex angle has measure 36o .Then consider the bisector BD of ∠ABC(figure)

We find that m∠DBC = 36o .Therefore,4ABC 4BCD.Let AD = x and AB = 1.However,since 4ABC y 4DBC are isosceles,BC = BD = AD = x. From the similiraty above, 1−x x = x 1 This gives us √ 2

x + x − 1 = 0 and x =

5−1 2

(The negative root cannot be used for the length of AD) We recall that

√

5−1 1 = 2 φ

The ratio for 4ABC of side 1 = =φ base x We therefore call this a Golden Triangle.