Download Zomes and Geometry

Summary of Zome/phi Discussions
In the first session the purpose was to build regular polygons, hexagonal prisms, right
polygonal pyramids, pyramidal solids (5, Platonic Solids) to illustrate the mathematics
involved with determining area of regular polygons. This skill gives us the ability to
determine surface area of regular polyhedral. Prerequisite to this was verifying that
the Zome struts determine the Golden Ratio.
We developed a formula for area of square (atypical method), area of equilateral
triangle, and area of hexagon. These use 300-600-right, or 45-isosceles right triangles. (*)
Central angles equal the intercepted arc measure, so angle BOC =
1200
A
B
Radii are equal so each central triangle is isosceles. Angles in
O
triangle BOC have sum of 1800, so base angles OCB, OBC each are
D
300. Triangle BOD is 300-600-900. Assume side of equilateral
s
C
triangle ABC is s. Then OD =
, and DB = (s/2)
2 3
s2
s2 3
s
Area of triangle BOC = (1/2)(
)(s) =
=
12
4 3
2 3
s2 3
However, there are three of these central triangles so the area of ABC =
4
The formula permits us to find the area of a hexagon (6 equilateral triangles) as well as
the surface area of an icosahedron.
The concept of using central angles for regular polygons inscribed in a circle provides a
determination of the central angle and thus the measure of the polygonal angle. If we
have an n-gon, each central angle is 1800/n and each n-gon angle is the supplement.
The four Platonic Solids: tetrahedron, cube, octahedron, and icosahedron can be
constructed using these two special right triangles (*).
The dodecahedron is another story. It involves a more special irrational number than
3 or 2 . It requires the Golden Ratio.
So, let’s first look at the Golden Ratio and some properties before we apply it like the
Greeks did to the pentagon so they could construct the dodecahedron.
The division of a line segment defines the Golden Ratio so that the whole segment is to
a longer part as the longer part is to the short segment. This ratio is called f.
(AB + BC)/AB = AB/BC
Lodholz summary of Zome discussions
page 1
A
B
a
C
b
Let’s use this notation for convenience in evaluating, let AB = a, and BC = b.
ab a
a
 = f by definition. So, to evaluate
Then we have:
= f or a = bf
a
b
b
b  b b
 1 
Substituting for a in the defined ratio gives us
or


b
b

1
2
2
So f + 1 = f
we wrote this in quadratic equation form as f - f - 1
Using the quadratic formula: f =
b  b 2  4ac
2a
because we want a positive value. So,
= 0
Note we take only the +
1 5
(1)  (1)2  4(1)(1)
=
which is
2
2(1)
1.618034 …
Go back to equation above with f + 1 = f2, this says that the square of this
marvelous number is determined by adding 1. Manipulating the symbols by dividing
by f we also see that, subtracting 1 finds the reciprocal from the number.
In the second Zome session we showed a connection of the Golden Ratio to the
Fibonacci numbers.
f2 = f + 1
f3 = f2 + f = (f + 1) + f = 2f2 + 1
f4 = f(2f2 + 1) = 2f2 + f = 2(f+1) + f
= 3f + 2
f5 = f(3f + 2) = 3f2 + 2f = 3(f+1) + 2f = 5f + 3
f6 = f(5f + 3) = 5f2 + 3f = 5(f+1) + 3f = 8f + 5
f7 = f(8f + 5) = 8f2 + 5f = 8(f+1) + 5f =13f + 8
.
The pattern is really created with the Fibonacci rule of adding to the previous term.
The Fibonacci series
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, … shows that the
fourth number 3 relates to the 4th power of f. And, that the 6th number (8) relates to f6.
In general to represent this pattern symbolically, if we let F represent a Fibonacci
number, we have fn = (Fn)f + (F(n-1))
We also reversed the scheme to show that dividing two consecutive Fibonacci numbers
closely approximated f
Fn/F(n-1) - f
With some understanding of f we can appreciate the Zomes much more than just fun
toys. They are powerful tools for geometry because they are produced in Golden
ratios.
node
b3
b2
b1
r
s
Lodholz
summary of Zome discussions
t
3
Qu i ck Ti me ™ an d a
TIFF (LZW) d ec om pres so r
are n ee de d to s ee th is pi ctu re .
r
u
t
s
r2
r1
y3
y2
y1
page 2
This we verified with two procedures. First we showed that b1 + b2 = b3. Actually,
with similar triangles we verified that x1 + x2 = x3, for each color: b, r, y
The second task was to show they determine the Golden Ratio. We did this in two
ways. I will review the one with two similar triangles, one triangle contained in the
other.
ABC and BCD are each isosceles and share a common base
angle. This means the two triangles are similar and thus
AB/BC = BC/CD.
A
D
We built this “Golden Triangle” using one b3, 3 b2’s, and a b1
strut. See the illustration below.
With the stated similar triangles we have
b3 b2
C
b2

but from our first procedure we
b
b
2
1
b3
showed that b1 + b2 = b3 so we can substitute for b3
b1
B
b2
Then we obtain two segments in the defined ratio of the Golden
(b1  b2 ) b2

Ratio.
=f
b2
b1
b2
To set up the formula for the area of this Golden Triangle, we need to determine the
angle measures. We are now getting serious, so as Professor Otto
we take off our jackets. We will even employ Greek letters to give
a
the illusion of higher math. Consider the illustration below, same
triangle with angle measures labeled with variables.
d
In the larger triangle the sum of the measures of the angles is 1800
b
1800
So 2a + b + a =
, but b = 2a , so we really have 5a
0
= 180 ,
which means that a = 360 Then 2b = 1800 – a or b = 720 .
a
a
b
That means that d is equal to 1080 This is really interesting since the angle of an
inscribed regular polygon is also 1080
A
E
B
Find the Golden Triangle in the pentagon (made from
two b3’s and one b2 blue struts) and the isosceles obtuse
triangle made from b2’s and a b3.
LodholzDsummary of Zome discussions
C
page 3
AC is a diagonal of the pentagon and we have shown AC/AB = f . We have then
illustrated that the diagonals of a pentagon are in Golden Ratio with the sides using the
Zomes.
Take the reciprocal, to find: AB/AC = 1/f
The patterns are endless once we put in all diagonals and begin to expand the
structure.
The AGF is a Golden Triangle,
also.
A
b3
b2
F
E
b2
G
b1
J
B
b2
FBCD is a rhombus and of course the
polygon FGHIJ is a regular pentagon.
The figure AGBHCIDJEF is called a
pentagram. The vertices are A, B, C, D, & E.
H
I
D
ADC is
C
We showed that if the perimeter of the
pentagon is 5 then the perimeter of its
pentagram is 10f – 10.
A
How for the area of the Golden Triangle.
Suppose the length of side of pentagon is
represented by variable s.
E
B
D
Lodholz summary of Zome discussions
C
page 4
DC = s which means AC/s = f , or AC = sf
A
Then using the Pythagorean Theorem on the
right triangle ACT we have
D
T
(AC)2 = (AT)2 + (TC)2 or (AT)2 = (sf)2 - (s/2)2
s2
= s 2 2 
4
2 2
4s   s 2
=
4
C
s
So the altitude needed (AT) =
s 2 (4 2  1)
4
s 4(  1)  1 s 4(2.618)  1
4 2  1

=
2
2
2
Now using a calculator this is about 1.538 s
Then AT =
s
Then the area of the Golden Triangle ACD = 1/2 (AT)(DC) = 1/2 (1.539 s) s = .769 s2
You should check my calculations. I am tired from typing all these silly subscripts and
exponents. Easily, could be an error.
I leave the area of ABC to you. But see AB = s , and AC = fs, so the altitude from B to
AC could be found by using a right triangle in which
(altitude)2 = (AB)2 – {(1/2)(AC)}2
I got about .4755 s2 for the area of ABC. Since there are two of these triangles, I
added .9511 s2 to the Golden Triangle area of .7694 s2. This gave me an area of for the
pentagon of approximately 1.72 s2 .
Of course we could use the central triangles and find length from center to side
(apothem). But we would have to assume a value for the radius, or use some right
triangular trigonometry.
Once we have the area of the polygon with side length s formulated we can find the
surface area of the dodecahedron.
I hope this reads well enough and is of some help. Again, please check the
calculations.
Lodholz summary of Zome discussions
page 5
Some websites I have found very informative relating to this topic are listed
below.

http://www.contracosta.edu/math/pentagrm.htm
Kepler quote, fun stuff, perimeter of pentagram)

http://www.goldenmuseum.com/index_engl.html
(Huge collection, connections and history)

http://mathworld.wolfram.com/GoldenTriangle.html
(Develop golden triangle, vertex 36 degrees, and decagon)

http://en.wikipedia.org/wiki/Golden_ratio
(Algebraic, history,)

http://www.math.rutgers.edu/~erowland/polyhedra.html
(Regular polyhedral, volume,)
Lodholz summary of Zome discussions
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