# The Pentagram & The Golden Ratio

Geometry has two great treasures:
one the Theorem of Pythagoras;
the other, the division of a line into extreme and mean ratio.
The first we may compare to a measure of gold;
the second we may name a precious jewel.
--Johann Kepler (1571-1630)

The 'ratio' has become known as the golden ratio or golden section.
This ratio can be found in many places: in art, architecture, and mathematics.

 Consider the construction of the regular pentagon. If the side AB of a regular pentagon (see figure to the right) has unit length, then any diagonal, such as AC, has length Ø = (1 + 5)/2 = 2cos(/5) = 1.61803... and this is the golden ratio. Indeed, in any similar pentagon/pentagram configuration the ratio of the length of the side of the pentagram (AC) to the length of the side of the pentagon (AB) is the golden ratio . Notice also the diagonals of the pentagon form another regular pentagon in the center of the figure with, of course, the potential for additional diagonals to be drawn, thus generating the golden ratio again as well as another regular pentagon further inside the figure. Presumably this could continue indefinitely. An interesting and important property of Ø is that Ø - 1 is equal to the reciprocal of Ø, that is, the decimal part of Ø, 0.61803, is equal to 1/Ø. You can check this with your calculator, or really show it by solving the equation:              Ø - 1 = 1/Ø

 In this next figure (to the left) is a regular pentagon with an inscribed pentagram. All the line segments found there are equal in length to one of the five line segments described below: The length of the black line segment is 1 unit. The length of the red line seqment, a, is Ø. The length of the yellow line seqment, b, is 1/Ø. The length of the green line seqment, c, is 1, like the black segment. The length of the blue line seqment, d, is (1/Ø)2, or equivalently, 1 - (1/Ø), as can be seen from examining the figure. (Note that b + d = 1). This implies that (1/Ø)2 = 1 - (1/Ø), which you can verify either algebraically or approximately with a calculator. The following ratios of the lengths of the line segments shown in the figure are all equal to the golden ratio, Ø: red to black (also, red to green) black to yellow (also, green to yellow) yellow to blue There are many other interesting relationships to be discovered involving this number, Ø. Here are a few more for you to verify: Ø - 1 = 1/Ø Ø + 1 = Ø2 Ø + 2 = ØÖ(5) Ø3 = Ø2 + Ø Ø3 = 2Ø + 1

 Problem:     Given that the perimeter of the pentagon     shown in the discussion above is 5 units, show that     the perimeter of the pentagram is     10(Ø - 1) units (6.1803... units).        Click here for a hint.

The golden ratio also appears in comparing consecutive elements of certain kinds of sequences, most notably, the Fibonacci sequence, but other sequences also. For instance, take two numbers at random, say 2 and 6. Add them to get 8. Add 6 and 8 to get 14. Add 8 and 14 to get 22. If you keep this going indefinitely, then the ratio of successive numbers approaches the golden ratio as a limit. Hence, for starters:

### ``` 2 + 6 = 8 8/6 = 1.333 6 + 8 = 14 14/8 = 1.750 8 + 14 = 22 22/14 = 1.571 14 + 22 = 36 36/22 = 1.636 22 + 36 = 58 58/36 = 1.611 36 + 58 = 94 94/58 = 1.621 58 + 94 = 152 152/94 = 1.617 . . . . . . ```

You can find many rectangles in an icosahedron whose sides are in the golden ratio.
Here's one.

The Golden Rectangle can be subdivided into squares and additional smaller Golden Rectangles, again a process that seemingly could go on indefinitely. In the figure below the figures 1, 2, 3, 4, and 5 are all squares. In each square a quarter circle can be drawn in such a way that a spiral is created (see figure further below on this page). The spiral is called, surprisingly, the "Golden Spiral"!

What are the dimensions of square #3? And what is the ratio of the side of square #2 to square #3?
(try this calculation using Google Search Calculator for some interesting follow-ups to the Golden Ratio.)

0.61803... : 1.00 :: 1.00 : 1.61803...
or
0.61803... is to 1.00 as 1.00 is to 1.61803...
or
Ø - 1 : 1 :: 1 : Ø    ®    Ø2 - Ø = 1

Pictured here are what might be called a couple of "golden parallelograms", can
you explain why that might be?

 Problem:     Show that in a golden parallelogram the     length of the shorter diagonal is equal to the length     of the longer side of the parallelogram.     That is, show that the three black line     segments in the figure shown are equal.

NOTE: Not every parallelogram whose shorter diagonal equals the length of the
longer side is a "golden parallelogram". The blue arc in the figure below
shows the location of some of the possibilities for setting up these other
parallelograms.

Pictured below are some other "golden figures".
We define them as "golden" because each side can be paired with another
side such that the lengths of the pair of sides are to one another
as the golden ratio.

Golden Triangles (2 sides equal)

Golden Trapezoids (3 sides equal)

Also, we find the golden ratio in the following continued fraction:

 The continued fraction in the white box above is just the original continued fraction, therefore, we have the following calculations shown to the right: Not only does Ø2 = Ø + 1, as shown, but you can show that Ø3 = Ø2 + Ø It's an interesting exploration to find the sequence 1, Ø, Ø2, Ø3, Ø4,... and show that Øn = Øn-1 + Øn-2. Also, 1, Ø, 1 + Ø, 1 + 2Ø, 2 + 3Ø, 3 + 5Ø, ... turns out to be the same sequence! (Notice the Fibonacci numbers emerging as the sequence develops.)

 Problem:     Golden Right Triangle I     Given that the lengths of the sides     of the triangle shown are Ö(Ø), 1 and Ø,     show that the triangle is a right triangle.     (Don't make this too hard!     And most certainly do not use a calculator.     And no, you don't know yet that the hypotenuse is     the diameter of the circle.) Problem:     Golden Right Triangle II     Given triangle ABC is inscribed in     a circle whose diameter is AB, segment CD     is perpendicular to diameter AB, and the     lengths of segments AD, BD, and CD are as shown,     find the length of segments AC and BC. Problem:     Golden Right Triangle III     Show that the triangle ABC in Problem II above is similar to the blue triangle     shown in Problem I.

 Want to draw the Pentagram? Want to find the area of the Pentagram?

 Problem:     Given that the length of the side of the     regular pentagon shown in the discussions     above is 1 unit, express the area of the inscribed     pentagram (star) in terms of Ø.        This is a good algebra problem and     will challenge some, and there are many different     expressions that one can derive.     Here is one you can compare your result with: ½(3 - Ø)3/2     If you don't get that exact expression, use a     calculator to compare your result to this. Problem:     Given that the length of the side of the     larger regular pentagon shown in the figure to the right     is 1 unit, show that the ratio of the area of larger     pentagon to the area of the smaller (white) pentagon is Ø4 to 1. Problem:     And show that the ratio of the area of the     larger regular pentagon shown in the figure to the area     of the pentagram (star - red & white combined) is Ø3 to 2.

For more about The Golden Ratio, see P. Davis & R. Hersh, The Mathematical Experience, Birkhauser Pub., Chap. 4, "Inner Issues: The Aesthetic Component".

OnLine you can follow the links below:
Pentagonal Geometry and the Golden Ratio
David Eppstein, Computer Science Department University of California, Irvine
http://www.ics.uci.edu/~eppstein/junkyard/pent.html

Also OnLine
Museum of Harmony and Golden Section
Authored by Alexey Stakhov and Anna Sluchenkova
http://www.goldenmuseum.com/

Also OnLine
Fibonacci Numbers and the Golden Section
This is the Home page for Ron Knott's Surrey University multimedia web site on the Fibonacci numbers, the Golden section and the Golden string.
Hosted by the Department of Mathematics of Surrey University in Guildford, UK
http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fib.html

StarDimensions tg

A Venn diagram made from five congruent ellipses.
From F. Ruskey's Combinatorial Object Server, University of Victoria.
http://www.theory.csc.uvic.ca/~cos/venn/VennSymmEJC.html

send comments to Thomas M. Green c/o CCC Math Dept. hwalters@contracosta.cc.ca.us