STAR GEOMETRY
Pythagoras, Fibonacci and Beard
R.S. Beard
The diagonal of the pentagon is made the unit of length in the
upper left c i r c l e of the accompanying d r a w i n g s .
K is the r a t i o of the
side of the pentagon to its diagonal.
Line 4. 3 is the s h o r t side of t r i a n g l e 1-4-3 and the long side of
triangle 3-4-6.
Line 4. 6 is the s h o r t side of t r i a n g l e 3-4-6 and the long side of
triangle 4-6-7.
T h e r e f o r e the sides of t h e s e t h r e e s i m i l a r i s o s e e l e s t r i a n g l e s
2
2
3
a r e r e s p e c t i v e l y 1 and K, K and K , K and K . Lines 1-6 and
3-6 have the s a m e length, K, as they a r e the equal sides of the i s o s c e l e s t r i a n g l e 1-3-6. Diagonal 1-4 of unit length is thus divided into
2
2
s e g m e n t s K and K , that is K + K = 1.
It follows from this equation that
K = \\HT-
i = 0.618034
and that K n = K n + 1 + K n + 2 .
Since each power of K is the sum of its next two higher p o w e r s ,
t h e s e powers form a Fibonacci s e r i e s when a r r a n g e d in t h e i r d e s c e n d ing o r d e r .
The r a d i u s is m a d e the unit of length in the upper right c i r c l e .
2
S i m i l a r t r i a n g l e s divide r a d i u s O - l into s e g m e n t s K and K . This
c o n s t r u c t i o n d e m o n s t r a t e s that each side of a r e g u l a r decagon has the
golden section r a t i o to the r a d i u s of its c i r c u m s c r i b i n g c i r c l e .
The right t r i a n g l e in the lower c i r c l e has sides of 1, 1/2 and
1 / 2 V 5 7 The d i m e n s i o n K with the value of 1 / 2 V 5 - 1 / 2 is the length
of the side of the i n s c r i b e d decagon.
The leaf shaped figure shows one simple way to c o n s t r u c t a five
pointed s t a r of a given width.
The Fibonacci s t a r of s t a r s is proportioned in the ten s u c c e s s i v e
0
9
powers of the golden section from K to K .
70
1966
STAR GEOMETRY.
PYTHAGORAS,
71
72
FIBONACCI AND BEARD
Feb.
AB is m a d e the unit of length or K .
AC, the side
The side
of the bounding pentagon h a s the length of K.
of each r a y of the m a s t e r s t a r is K
long.
3
The bounding pentagon of the c e n t r a l s t a r h a s sides of K length.
Since one diagonal of each of the s m a l l e r s t a r s is a side of the
bounding pentagon of the next l a r g e r s t a r all of the c o r r e s p o n d i n g d i m e n s i o n s of these s u c c e s s i v e s t a r s a r e in golden section r a t i o .
The bounding pentagons of the s u c c e s s i v e s t a r s have s i d e s of
K , K , K
and K
respectively.
8
The r a y s of the s m a l l e s t s t a r s have K edges and base widths
of K 9 .
This figure can be used to d e m o n s t r a t e that any power of the
golden section K , is the sum of a i l of its higher powers from
K
to K°°.
The lines connecting the tips of the c e n t r a l s t a r and the c e n t e r s
of the five next s m a l l e r s t a r s form a decagon.
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All s u b s c r i p t i o n c o r r e s p o n d e n c e should be a d d r e s s e d to B r o t h e r U.
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