zome system

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Zome system in rational trigonometry and rational projectivetrigonometry

M M AlmanjumiSchool of Maths UNSW

Sydney 2052 Australia

AbstractThe work is devoted to the study of The Zome System in rational trigonometry and projective rational trigonom-

etry. By using The Zome System no more the fteen triangles and nineteen tetrahedrons have been found. Majorresults also include six new theorems which have been developed using rational trigonometry.

1 Introduction

Geometry continues to fascinate us over time with the patterns of sizes, angles and shapes using the universal

language of mathematics to reveal those patterns. The Zome system is an exceptional ball-and-strut system,designed for children but also appeals to professionals in both eld of science and art. The system consists ofgeometric holes in sphere-shaped connectors and each ball is linked to another by means of sticks with dierentshapes and colours to make the construction simple.

A new theory of trigonometry, called rational trigonometry, was developed in 2005 by N. J. wildberger (UNSW)in Divine Proportions: Rational Trigonometry to Universal Geometry, Wild Egg Books, Sydney, 2005. Rationaltrigonometry is a new framework that replaces distance and angle with quadratic concepts called quadrance andspread. The projective plane inherits a rich metrical structure which extends to higher dimensions and arbitraryelds, a fact which has major implications for algebraic geometry, and possibly also for dierential geometry.Thales theorem and Pythagoras theorem are particularly important here with the wide variety of classical sphericalformulas being replaced by simpler, polynomial relations.

In this project we are going to rstly look at the background of the Zome system. Secondly we will then give basicdenitions and facts of rational trigonometry covering four theorems proposed by N. J. Wildberger. We used these

to establish 6 new theorems. This is done by constructing fteen possible triangles using the Zome system derivedfrom a regular pentagon. Finally we give main denitions and laws of projective trigonometry covering twelvetheorems given by N. J. Wildberger. We then construct nineteen possible tetrahedrons using the fteen trianglesthat we found by using the Zome system. After determining the projective quadrances for each tetrahedron, wethen used the Projective Cross Law Theorem in order to determine the projective spreads for each tetrahedron.

2 The Zome System

In structural design or construction, a Zome is a large family of geometric shapes that can be built and givesself-supporting volumes without the need for internal support and in which one can live and enjoy all the space,cosmetics lines, the roundness of the dome, polygonal and no right angles. The structure is like a skeleton, and the

organization of the balanced facets is of an amazing strength. As it is easy to construct, one can direct what hewants: from the niche of a dog, a bird aviary, a workshop, shelter with tools or materials, greenhouse as extra roomand of course as a full habitat for the largest.

The mathematics set of the Zome system is a plastic construction composed of balls and sticks called respectivelynodes and struts. When assembled together they form an amazing mathematically and creatively fascinatingstructures.

The Zome tool struts colours are usually red, yellow, blue, and green. They have respectively pentagon, tri-angular, rectangle and rhombus end shapes. They come in small, medium and large sizes. However, we will notuse the green ones in this project. Green struts are those generally necessary for building regular tetrahedrons,octahedrons and are a little harder to work with.

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Figure 1: White node and strutsh t t p : / / w w w . g e o r g e h a r t . c o m / v i r t u a l - p o l y h e d r a / z o m e t o o l . h t m l

2.1 Possible constructions

Below are some possible constructions that be made from using the Zome system.

2.1.1 Tetrahedron

This has four faces, four vertexes and six edges.

Figure 2: Tetrahedron

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2.1.2 Cube

This has six faces, eight vertexes and twelve edges.

Figure 3: Cube h t t p : / / w w w . g e o r g e h a r t . c o m / v i r t u a l -p o l y h e d r a / z o m e t o o l . h t m l

2.1.3 Octahedron

This has eight faces, six vertexes and twelve edges.

Figure 4: Terahedron and Octahedronh t t p : / / w w w . g e o r g e h a r t . c o m / v i r t u a l - p o l y h e d r a / z o m e t o o l . h t m l

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2.1.4 Dodecahedron

This has twelve faces, twenty vertexes and thirty edges.

Figure 5: Dodecahedronh t t p : / / w w w .g e o r g e h a r t . c o m / v i r t u a l -

p o l y h e d r a / z o m e t o o l . h t m l

2.1.5 Building Icosahedron

This has twenty faces, twelve vertexes and thirty edges.This can be done by rst connecting twelve red (pentagon) struts to a white node and connecting twelve white

nodes to each of the ends of those struts. This gives the foundation for the main structure, where each of twelve

external white nodes are connected to the neighboring ve white nodes using blue (rectangle) struts. After removingthe internal white node and connected twelve struts it gives the Icosahedron.

Figure 6: Icosahedronh t t p : / / w w w . g e o r g e h a r t . c o m / v i r t u a l - p o l y h e d r a / z o m e t o o l . h t m l

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3 Rational trigonometry

It is a new framework for planar trigonometry and been proposed [N J Wildberger]). Rational trigonometry replacesdistance and angle with quadratic concepts called quadrance and spread. The usual laws are replaced by purelyalgebraic analogs, with the consequence that they hold in much wider generality, allow more accurate calculations,and are much easier to learn. The usual menagerie of transcendental circular functions and their inverses play no

role.

3.1 Basic denitions and facts of rational trigonometry

Denition 1 The quadrance Q(A1; A2) between the points A1 [x1; y1] and A2 [x2; y2] is the number

Q (A1; A2) (x2 x1)2 + (y2 y1)2 :

Denition 2 The spread s (l1; l2) ,which is a number between 0 and 1, is the notion of angle between to lines l1and l2

Theorem 3 The Triple quad formula suppose that A1; A2 and A3 are points with Q1 Q (A1; A2) ; Q2 Q (A1; A3) and Q3 Q (A2; A3) : Then

(Q1 + Q2 + Q3)2 = 2

Q21 + Q

22 + Q

23

:

precisely when A1; A2 and A3 are collinear.

Theorem 4 Pythagoras theorem

One of the fundamental theorems, in rational trigonometry becomes more general, extending to an arbitraryeld, not of characteristic two.(Pythagoras theorem) Suppose that the triangle A1A2A3 has qudrances Q1 Q (A2; A3) ; Q2 Q (A1; A3) and

Q3 Q (A1; A2) : Then

Q1 + Q2 = Q3

precisely when A1A3 and A2A3 are perpendicular.

Theorem 5 The spread law

Suppose three points A1; A2 and A3 form non-zero quadrances Q1 Q (A2; A3) ; Q2 Q (A1; A2) and Q3 Q (A1; A3) : Dene the spreads s1 s (A1A2; A1A3) ; s2 s (A2A1; A2A3) and s3 s (A3A1; A3A2) : Then

s1

Q1=

s2

Q2=

s3

Q3:

Theorem 6 The cross law

Suppose three points A1; A2 and A3 form quadrances Q1 Q (A2; A3) ; Q2 Q (A1; A3) and Q3 Q (A1; A2) ;and dene the cross c3 c (A3A1; A3A2) : Then

(Q1 + Q2 Q3)2 = 4Q1Q2c3:

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3.2 Triangles, consisting of zome system

Using the zome system we can construct many triangle, each formed by 3 3 = 9 dierent sticks: 3 blue calledB1; B2; B3 ( respectively small, medium and large), also 3 red called R1; R2; R3 ( respectively small, medium andlarge), also 3 yellow called Y1; Y2; Y3 ( respectively small, medium and large).

Figure 7: Struts

These can be used to create triangles, and we are going to study some of these triangles to determine thequadrances of each of the sticks above, and also the spreads formed by any two sticks intersecting from a vertex (white ball ).

Consider rst the following pentagon.

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Figure 8: Pentagon

We will draw a line from A to C which will be B2:

We have a triangle ABC and we are going to study this triangle as a rst triangle.

3.2.1 The First Triangle:

Figure 10: The rst triangle

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As you can see this triangle ABC, we take this triangle from the pentagon above.So, we have in this triangle

s1 = s3 = =5 p5

8

s2 = =5 + p5

8

We assume that

Q(A; B) = Q(B; C) = B1 = 1

Now, we want to nd B2; we can apply the spread law to nd B2; the spread law is

s1

Q1=

s2

Q2=

s3

Q3

By substituting the values that we have in the spread law

s1

B1=

s2

B2=

s3

B1

Hence,

1=

B2

Now, we have that

B2 =

=

5+p5

8

5p5

8

=1

2

p5 +

3

2=

we can also infer that

B3 = 2

3.2.2 The Second Triangle:

We can construct the second triangle from three sticks of B1; and we are also going to study this triangle.

Figure 11: The second triangle

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we know in this triangle the three quadrances

Q(A; B) = Q(A; C) = Q(B; C) = B1 = 1:

So, we can use the cross law to nd one of the three spreads.The cross law is

(Q1 + Q2 Q3)2 = 4Q1Q2 (1 s3)By substituting the values in the cross law

(B1 + B1 B1)2 = 4B1B1 (1 s3)Hence,

(1 + 1 1)2 = 4 (1) (1) (1 s3)We will have that

s3 =3

4

Because the triangle is equilateral.Hence,

s2 = s1 = s3 =3

4

3.2.3 The third triangle:

we can construct the third triangle from two sticks of B2 and one stick of B1:

Figure 12: The third triangle

We know in this triangle that

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Q(A; B) = B1 = 1

Q(A; C) = Q(B; C) = B2 = =1

2

p5 +

3

2

We know the three quadrances. So, we can use the cross law to nd one of the three spreads.

(Q1 + Q2 Q3)2 = 4Q1Q2 (1 s3)By substituting the values in the cross law

(B1 + B2 B2)2 = 4B1B2 (1 s3)Hence,

(1 + )2 = 4(1) () (1 s3)We have that

s3 =1

4(4 1)

Thus,

s3 = 143+

p5

2

43 + p52

! 1! = 18p5 + 5

8

Because the triangle is isoscelesHence,

s3 =5 +

p5

8= = s1

Now, we can apply the spread law to nd s3:

s3

Bs=

s1

Bm

Hence,

s3

1=

=

5+p5

8

3+p5

2

=5 p5

8=

3.2.4 The fourth triangle:

We can construct the fourth triangle from two sticks of R1 and one stick of B1:

Figure 13: The fourth triangle

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we have that

Q(A; C) = Q(B; C) = R1 = =5 +

p5

8Q(A; B) = B1 = 1


(Q1 + Q2 + Q3)2 = 4Q1Q2 (1 s3)

By substituting the values in the cross law

(Rs + Rs Bs)2 = 4RsRs (1 s3)Hence,

(+ 1)2 = 4 () () (1 s3)We have that

s3 = 142

(4 1) = 145+

p5

8

2 45 + p58 ! 1! = 45Now, we can apply the spread law to nd s1.

s1

Rs=

s3

Bs

By substituting the values.

s1

=

4

5

1

Thus,

s1 =4

5

Because the triangle is isosceles.Hence,

s1 = s2 =4

5

3.2.5 The fth triangle:

We can construct the fth triangle from one stick of B1; one stick of R2 and one stick of Y1:

Figure 14: The fth triangle

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we have that

B1 = 1; R2 =

s3 =4

5

In this case we have two quadrances and one spread, so, by using the cross law we will nd the third quadrancewhich is Y1:

(B1 + R2 Y1)2 = 4B1R2 (1 s3)Hence

(1 + Y1)2 = 4

+ 1

We have that

(1 + )2 + Y21

2 (1 + ) Y1 =

2

Now, we have a quadratic equation.

Y21 2 (1 + ) Y1 + (1 + )2 2 = 0

By solving the quadratic equation, we have that

Y1 =3

4

From this result we infer that

Y2 = 34

Y3 =3

42

Now, we can apply the spread law to nd s1:

s1

B1=

s3

Y1

By substituting the values.

s11 =

4

53

4

Hence,

s1 =16

15

We can also apply the spread law to nd s2:

s2

R2=

s3

Y1

By substituting the value.

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s2

=

45

3

4

Hence,

s2 =1

6

p5 +

1

2=

3

3.2.6 The sixth triangle:

We can construct the sixth triangle from one stick of R1; one stick of R2 and one stick of Y2:

Figure 15: The sixth triangle

we know in this triangle (by comparing this triangle with the previous triangles which we previously solvedthem)

Q(A; C) = Y2 =3

4 ; Q(B; C) = R2 = ; Q(A; B) = R1 =

s2 =4

5

Now, we can apply the spread law to nd s1.

s1

R2=

s2

Y2=

s3

R1

By substituting the values in the spread law.s1

Rm=

s2

Ym

Hence,s1

=

4

5

3

4

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We have that

s1 =16

15We can also apply the spread law to nd s3:

s3

Rs=

s2

Ym

By substituting the values in the spread law.

s3

=

4

5

3

4

Hence,

s3 =16

15=

165+

p5

8

153+

p5

2

= 23 2

15

p5 =

16

15

3.2.7 The seventh triangle:

we can construct the seventh triangle from one sticks of medium blue (B2), one stick of medium red (R2) and onestick of small yellow (Y1).

Figure 16: The seventh triangle

In this triangle we know the spreads s2 and the three quadrances B2; R2 and Y1 by comparing this triangle withthe previous triangles.

Q(B; C) = R2 = ; Q(A; C) = B2 = ; Q(A; B) = Y1 =3

4

s2 =16

15

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Now, we want to nd s2 by applying the spread law we will have:

s1

R2=

s2

B2

Hence,

s1

=1615

we have that

s1 =162

15

Notice that

=5 +

p5

8; =

3 +p

5

2

Thus,

s1 =165+

p5

8

215

=1

6

p5 +

1

2=

3 +p

5

6=

3

We can also apply the spread law to nd s3:

s3

Ys=

s2

Bm

Hence,

s33

4

=1615

We have that

s3 =4860

=

4

5

Notice that

=

Thus,

s3 =4

5

3.2.8 The eighth triangle:

We can construct the eighth triangle from two sticks of Y2 and one stick of B1:

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Figure 17: The eighth triangle

In this triangle we have the two spreads (s1 , s2) and the three quadrances by comparing this triangle with theprevious triangle.

Q(A; C) = Q(B; C) = Y2 =3

4 ; Q(A; B) = B1 = 1

s1 =162

15=

1

6

p5 +

1

2=

3= s2

Now, by using the spread law, we can nd s3

s3

B1=

s1

Y2

Hence,

s3 =162

15

3

4

We have that

s3 =642

45

Notice that =

5 + p58

; =3 + p5

2

we can substitute for and :Thus,

s3 =645+

p5

8

2453+

p5

2

= 49

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3.2.9 The ninth triangle:

We can construct the ninth triangle from two stick of Y1 and one stick of B1:

Figure 18: The ninth triangle


Q(A; C) = Q(B; C) = Y1 =3

4; Q(A; B) = B1 = 1


(B1 + Y1 + Y1)2 = 4B1Y1 (1 s1)

Hence,

1 + 34

3

42

= 4(1)34 (1

s1)

We have that

s1 =2

3

Because the triangle is isosceles.Thus,

s1 = s2 =2

3

Now, we can use the spread law to nd s3

s3

B1=

s2

Y1

Hence;s3

1=

2

3

3

4

We have that,

s3 =8

9

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3.2.10 The tenth triangle:

Figure 19: The tenth triangle

We know in this triangle the following,

Q(A; B) = Y2 =3

4 ; Q(B; C) = Y1 =

3

4; Q(A; C) = R1 = =

5 +p

5

8

s1 =4

3 + 3=

16

15

s2 =642

45=

4

9

In this case we know two spreads and the the three quadrances. So, we can use the spread law to nd the thirdspread s3:

s3

Y2= s

1

Y1

Thus;

s33

4

=4

3+3

3

4

We have that,

s3 =4

3 + 3

we can substitute for

=3 +

p5

2.

Hence;

s3 =43+

p5

2

33+

p5

2

+ 3

=2

15

p5 +

2

3=

165+

p5

8

15

Notice that

=5 +

p5

8

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Thus,

s3 =16

15

3.2.11 The eleventh triangle:

Figure 20: The eleventh triangle


Q(A; C) = Q(B; C) = Y1 =3

4

Q(A; B) = B2 = =3 +

p5

2

For this triangle we know the three quadrances. So, we can use the cross law to nd s1:B2 +

3

4 3

4

2= 4B2Y1 (1 s1)

Thus, +

3

4 3

4

2= 4 ()

3

4

(1 s1)

We have that

s1 =3

3=

3 3+

p5

2

3

=3

p5

2

3

We will call3 p5

2=

Thus,

s1 = 3

Because the triangle is isoscelesHence,

s1 = s2 =

3

Now we know the three quadrances and two spreads. So, We can use the spread law to nd the third spread s3:

s3

Bm=

s1

Ys

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Hence;s3

=

33

3

4

Notice that

=3 +

p5

2

Thus,

s3 =4

9:

3.2.12 The twelfth triangle:

Figure 21: The twelfth triangle

We know in this triangle the following

R1 =

B2 =

s1 = s2 =1

+ 1=

4

5

We know three quadrances and two spreads in this case we can apply the spread law to nd the third spread s3:

s3

B2=

s1

R1

Hence;s3

=

45

We have that

s3 =4

5

= 4

5

Notice that

=

1

Thus,

s3 =4

5

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3.2.13 The thirteenth triangle:

We can construct the thirteenth triangle from one stick of B3; one stick of R1 and one stick of Y3:

Figure 22: The thirteenth triangle

when we compare this triangle with the previous triangles we would have the all three spreads which appear inthe previous triangles and the three quadrances as well.

Q(A; C) = Y3 =3

42

Q(A:b) = B3 = 2

Q(B; C) = R1 =

s1 =3

3=

3 3+

p5

2 3

=1

2 1

6

p5 =

3s2 =

4

5

s3 =16

15

3.2.14 The fourteenth triangle:

We can construct the fourteenth triangle from one stick of B2; one stick of R1 and one stick of Y3:

Figure 23: The fourteenth triangle

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We know in this triangle the following:

Q(A; C) = B2 =

Q(A:b) = Y3 =3

42

Q(B; C) = R1 =

s1 =3

3=

3 3+

p5

2

3

=1

2 1

6

p5 =

3

s2 =4

3 + 3=

4

33+

p5

2

+ 3

=2

3 2

15

p5 =

165

p5

8

15

=16

15

Now, we know three quadrances and two spreads. So, we can use the spread law to nd the third spread s3:

s3

Yb=

s1

Rs

Hence,

s33

42

=1

2 1

6

p5

we have that

s3 = 18

p52 32

Notice that

=5 +

p5

8; =

3 +p

5

2

Thus,

s3 = 185+

p5

8

0@p5

3 +

p5

2

!2 3

3 +

p5

2

!21A = 110

p5 +

1

2=

45+

p5

8

5

We have that

s3 =4

5

3.2.15 The fteenth triangle:

We can construct the fteenth triangle from one stick of B1; two stick of R1 and one stick of B2:

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Figure 24: The fteenth triangle

when we compare this triangle with the previous triangles we would have the all three quadrances which appearin the previous triangles and the three spreads as well instead of s2 which is equal to 1.

The three quadrances are:

Q(A; B) = B2 =

Q(A:C) = 2

Q(B; C) = B1 = 1

The three spreads are:

s1 =4

5s2 = 1

s3 =4

5

3.3 Some theorems that we can get from the triangles above.

Theorem 7 The possible spreads between two blue sticks is

;; 1 and 34

:

Theorem 8 The possible spreads between two red sticks is

4

5:

Theorem 9 The possible spreads between two yellow sticks is

4

9and

8

9

:

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Theorem 10 The possible spreads between blue stick and red stick is

4

5and

4

5:

Theorem 11 The possible spreads between blue stick and yellow stick is

3

;

3

and2

3

:

Theorem 12 The possible spreads between red stick and yellow stick is

16

15and

16

15:

In brief, this a catalog of ZOME triangles which is made by using ZOME system.

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Figure 25: A catalog of ZOME triangles

So far we have found no more than fteen triangles by using The Zome System.

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4 Rational projective trigonometry

As rotation of earth aects our view of sky and the objects in it. Though the longitudinal angle becomes animportant and unavoidable concept but for many spherical geometrical applications, there is no uniform motionaround a xed axis that plays such a distinguished role. For this kind of stationary spherical geometry there is arational version of the classical theory which again is simpler, more elegant and accurate. This theory is developed

in the more natural setting of projective trigonometry. The projective plane inherits a rich metrical structure whichextends to higher dimensions and arbitrary elds.The sphere has equation x2 + y2 + z2 = 1 and center O = [0; 0; 0] : Any two non-antipodal points A and B lying

on it determine a unique spherical line, or great circle arc, which is the intersection of the sphere with the planeOBA. Any two such spherical lines intersect at a pair of antipodal points.

In nineteen century, an alternative of nding the antipodal points on the sphere is consider the associated linethrough the origin O passing through the antipodal points. Such a line will be called a projective point. Similarlya plane through O will be called a projective line.

Figure below shows a spherical triangle formed by three spherical points A, B and C and three great circle arcs,and on the right the corresponding projective triangle, consisting of three projective points a, b and c, and the threeprojective lines that they form.

Figure 25: Spherical and projective triangle h t t p : / / w i l d e g g . c o m / p a p e r s / P r o j e c t i v e T r i g . p d f

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4.1 Main denitions and laws of Projective trigonometry

Denition 13 A projective triangle a1a2a3 is a set of three non-collinear projective points.

Figure 27: Three views of a projective triangle http://wildegg.com/papers/ProjectiveTrig.pdf

Denition 14 The projective quadrance q(a1; a2) between two projective points a1 and a2 is dened to be thespread between them.

Denition 15 Theprojective spread S(L1; L2) between two projective lines L1 andL2 is dened to be the spreadbetween them.

4.1.1 Theorem 1 (projective Thales theorem)

suppose L1 and L2 are distance projective lines intersection at the projective point a and with a projective spreadof S. Choose a projective point b

6= a on one of the lines, say L1; and let c be the projective point which is the foot

of the (or a) perpendicular projective line N from b to L2 as in gure . If q(b; c) = q and q(a; b) = r then

S =q

r

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Figure 28: Projective Thales theoremspherical and projective views h t t p : / / w i l d e g g . c o m / p a p e r s / P r o j e c t i v e T r i g . p d f

4.1.2 Theorem 2 (Projective triple quad formula)

If the three projective points a1; a2 and a3 are collinear, then

(q1 + q2 + q3)

2

= 2

q

2

1 + q2

2 + q2

3

+ 4q1q2q3:

4.1.3 Theorem 3 (Dual projective triple quad formula)

If the three projective lines L1; L2 and L3 are concurrent, then

(S1 + S2 + S3)2 = 2

S21 + S

22 + S

23

+ 4S1S2S3:

4.1.4 Theorem 4 (Projective Pythagoras theorem)

Suppose that a1a2a3 is a projective triangle with projective quadrances q1; q2 and q3; and projective spreads S1; S2and S3: If S3 = 1 then

q3 = q1 + q2 q1q2:

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4.1.5 Theorem 4 (Dual projective Pythagoras theorem)

Suppose that a1a2a3 is a projective triangle with projective quadrances q1; q2 and q3; and projective spreads S1; S2and S3: If q3 = 1 then

S3 = S1 + S2 S1S2:

Figure 29: Pythagoras theoremspherical viewh t t p : / / w i l d e g g . c o m / p a p e r s / P r o j e c t i v e T r i g . p d f

4.1.6 Theorem 6 (Projective spread law)

Suppose that a1a2a3 is a projective triangle with projective quadrances q1; q2 and q3; and projective spreads S1; S3and S3:Then

S1

q1=

S2

q2=

S3

q3:

4.1.7 Theorem 7 (Projective cross law)

Suppose that a1a2a3 is a projective triangle with projective quadrances q1; q2 and q3; and projective spreads S1; S2and S3: Then

(S3q1q2 q1 q2 q3 + 2)2 = 4(1 q1) (1 q2) (1 q3) :

4.1.8 Theorem 8 (Dual projective cross law)

Suppose that a1a2a3 is a projective triangle with projective quadrances q1; q2 and q3; and projective spreads S1; S2and S3: Then

(S1S2q3 S1 S2 S3 + 2)2

= 4(1 S1) (1 S2) (1 S3) :

4.1.9 Theorem 9

The projective quadrea A = A (a1; a2; a3) of the projective points a1 = [x1 : y1 : z1] ; a2 = [x2 : y2 : z2] and a3 =[x3 : y3 : z3] i

A =(x1y2z3 x1y3z2 x2y1z3 + x2z1y3 + y1x3z2 x3y2z1)2

(x21 + y21 + z

21) (x

22 + y

22 + z

22) (x

23 + y

23 + z

23)

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4.1.10 Theorem 10 (Right projective triangle)

Suppose a right projective triangle has projective quadrances q1; q2 and q3; and projective spreads S1; S2 and S3 = 1:Then any two of the ve quantities fq1; q2; q3; S1; S2g determine the other three, solely through the three basicequation

q3 = q1 + q2

q1q2 S1 =q1

q2S2 =

q2

q3

Figure 30: Right projective triangleh t t p : / / w i l d e g g . c o m / p a p e r s / P r o j e c t i v e T r i g . p d f

4.1.11 Theorem 11 (Projective isosceles triangle)

Suppose a projective isosceles triangle has projective quadrances q1 = q2 = q and q3; and projective spreadsS1 = S2 = S and S3:Then

q3 =4q(1 S) (1 q)

(1 Sq)2and S3 =

4S(1 S) (1 q)(1 Sq)2

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Figure 31: Projective isosceles triangleh t t p : / / w i l d e g g . c o m / p a p e r s / P r o j e c t i v e T r i g . p d f

4.1.12 Theorem 12 (Equilateral projective triangles)

Suppose that a projective triangle is equilateral with common projective quadrance q1 = q2 = q3 = q; and withcommon projective spread S1 = S2 = S3 = S: Then

(1 Sq)2 = 4(1 S) (1 q) :

4.2 Tetrahedrons, consisting of ZOME triangles.Using ZOME triangles we can construct many tetrahedrons based on the triangles that we have previously studiedin subsection 3.2. This is made possible to achieve by projecting a line from each angle (ball) of the trianglespreviously built to a single point connected by another ball forming a tetrahedron and we are going to study someof these tetrahedrons to determine the projective spread and the projective quadrnace.

In the following subsection 4 random tetrahedrons have been studied to determine the projective spread andthe projective quadrance.

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4.2.1 The rst tetrahedron

Figure 32: The rst tetrahedron

Figure 33: The rst tetrahedron

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Figure 34: The rst tetrahedron consisting of four zome triangles

4.2.2 The main results of the rst tetrahedron.

Projective spreads S1 =5

6S2 =

5

6S3 =

5

6S4 =

15

16S5 =

15

16S6 =

15

16

Vertex Projective quadrancesA1 q11 =

8

9; q12 =

8

9and q13 =

8

9

A2 q21 =2

3; q22 =

2

3and q23 =

3

4

A3 q31 =3

4; q32 =

2

3and q33 =

2

3

A4 q41 =2

3; q42 =

2

3and q43 =

3

4

There are 4 vertexes and each vertex has a projective triangle.

33


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a) Projective triangle at vertex A1:

Figure 35: Projective triangle at vertex A1

Now, we know q11; q12 and q13:

q11 = q12 = q13 =8

9:

So, we can use the projective cross law to nd the three face spreads (Projective spreads).

Firstly, we want to compute the projective spread S4 by using the projective cross law.This will lead to

(S4q12q13 q11 q12 q13 + 2)2

= 4(1 q11) (1 q12) (1 q13)Therefore, susbtituting the values in the law we get:

S4 8

9 8

9 8

9 8

9 8

9+ 2

2= 4

1 8

9

1 8

9

1 8

9

4

6561(32S4 27)2 = 4

729

The solution is:

S4 =15

16or

3

4

Secondly, we want to compute the projective spread S5 by using the projective cross law.This will lead to

(S5q11q13 q11 q12 q13 + 2)2 = 4(1 q11) (1 q12) (1 q13)Therefore, susbtituting the values in the law we get:

S5 89 8

9 8

9 8

9 8

9+ 2

2= 4

1 8

9

1 8

9

1 8

9

4

6561(32S5 27)2 = 4

729

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The solution is:

S5 =15

16or

3

4

Thirdly, we want to compute the projective spread S6 by using the projective cross law.

This will lead to (S6q11q12 q11 q12 q13 + 2)2 = 4(1 q11) (1 q12) (1 q13)Therefore, susbtituting the values in the law we get:

S6 8

9 8

9 8

9 8

9 8

9+ 2

2= 4

1 8

9

1 8

9

1 8

9

4

6561(32S6 27)2 = 4

729

The solution is:

S6 =15

16or

3

4

b) Projective triangle at vertex A2:



q21 =2

3

q22 =2

3

q23 =3

4

So, we can use the projective cross law to nd the three face spreads (Projective spreads)

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S5 2

3 2

3 2

3 2

3 3

4+ 2

2= 4

1 2

3

1 2

3

1 3

4

1

1296(16S5 3)2 = 1

9

The solution is:

S5 =15

16or 9

16

) S5 =15

16



S3 2

3 3

4 2

3 2

3 3

4+ 2

2= 4

1 2

3

1 2

3

1 3

4

1

144(6S3 1)2 = 1

9

The solution is:

S3 =5

6or 1

2

Thirdly, we want to compute the projective spread S1 by using the projective cross law.This will lead to


S1 2

3 3

4 2

3 2

3 3

4+ 2

2= 4

1 2

3

1 2

3

1 3

4

1

144(6S1 1)2 = 1

9

The solution is:

S1 =5

6or 1

2

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c) Projective triangle at vertex A3:



q31 =3

4

q32 =2

3

q33 =2

3




S6 2

3 2

3 3

4 2

3 2

3+ 2

2= 4

1 3

4

1 2

3

1 2

3

11296

(16S6 3)2 = 19

The solution is:

S6 =15

16or 9

16

) S6 =15

16

Secondly, we want to compute the projective spread S3 by using the projective cross law.

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This will lead to


S3 3

4 2

3 3

4 2

3 2

3+ 2

2

= 41 3

41 2

31 2

31

144(6S3 1)2 = 1

9

The solution is:

S3 =5

6or 1

2

) S3 =5

6

Thirdly, the projective spread S2 by using the projective cross law.This will lead to


S2 34 2

3 3

4 2

3 2

3+ 2

2= 4

1 3

4

1 2

3

1 2

3

1

144(6S2 1)2 = 1

9

The solution is:

S2 =5

6or 1

2

d) Projective triangle at vertex A4:


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q41 =2

3

q42 =2

3

q43 =34




S4 2

3

2

3

2

3

2

3

3

4

+ 22

= 41 2

31

2

31

3

4

1

1296(16S4 3)2 = 1

9

The solution is:

S4 =15

16or 9

16

) S4 =15

16



S1 2

3 3

4 2

3 2

3 3

4+ 2

2= 4

1 2

3

1 2

3

1 3

4

1

144(6S1 1)2 = 1

9

The solution is: S1 =56

or 12

) S1 =5

6


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S2 2

3 3

4 2

3 2

3 3

4+ 2

2= 4

1 2

3

1 2

3

1 3

4

1144

(6S2 1)2 = 19

The solution is:

S2 =5

6or 1

2

) S2 =5

6

4.2.3 The second tetrahedron

Figure 39: The second tetrahedron

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Figure 40: The second tetrahedron

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Figure 41: The second tetrahedron consisting of four zome triangle

4.2.4 The main results of the second tetrahedron.


6

p5 + 1

2S2 =

1

6

p5 + 1

2S3 =

1

6

p5 + 1

2S4 = S5 = S6 =

Vertex Projective quadrancesA1 q11 =

4

5; q12 =

4

5and q13 =

4

5

A2 q21 =45

; q22 =45

and q23 =3

4

A3 q31 =3

4; q32 =

45

and q33 =45

A4 q41 =45

; q42 =45

and q43 =3

4


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q11 = q12 = q13 =4

5:




S4 45 4

5 4

5 4

5 4

5+ 2

2= 4

1 4

5

1 4

5

1 4

5

4

625(8S4 5)2 = 4

125

The solution is:

S4 =1

8

p5 +

5

8= or

5

8 1

8

p5 =



S5 45 4

5 4

5 4

5 4

5+ 2

2= 4

1 4

5

1 4

5

1 4

5

4

625(8S5 5)2 = 4

125

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The solution is:

S5 =1

8

p5 +

5

8= or

5

8 1

8

p5 =


This will lead to (S6q11q12 q11 q12 q13 + 2)2 = 4(1 q11) (1 q12) (1 q13)Therefore, susbtituting the values in the law we get:

S6 4

5 4

5 4

5 4

5 4

5+ 2

2= 4

1 4

5

1 4

5

1 4

5

4

625(8S6 5)2 = 4

125

The solution is:

S6 =1

8

p5 +

5

8= or

5

8 1

8

p5 =




q21 =4

5

q22 =4

5

q23 =3

4


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S5 4

5 4

5 4

5 4

5 3

4 + 22

= 4

1 4

5

1 4

5

1 3

4

Notice that

=1

8

p5 +

5

8

Hence, 3

10S5 +

1

10

p5S5 1

5

p5 +

1

4

2=

3

10 1

10

p5

The solution is:

S5 =33

8

p5 75

8or

1

8

p5 +

5

8=

) S5 =



S3 45 3

4 4

5 4

5 3

4+ 2

2= 4

1 4

5

1 4

5

1 3

4

Notice that

=

1

8p5 +5

8

Hence,

3

8S3 +

3

40

p5S3 1

5

p5 +

1

4

2=

3

10 1

10

p5

The solution is:

S3 =3

2

p5 7

2or

1

6

p5 +

1

2


This will lead to


S1 45 3

4 4

5 4

5 3

4+ 2

2= 4

1 4

5

1 4

5

1 3

4

Notice that

=1

8

p5 +

5

8

45


46/71

Hence,

3

8S1 +

3

40

p5S1 1

5

p5 +

1

4

2=

3

10 1

10

p5

The solution is:

S1 =3

2

p5 7

2or

1

6

p5 +

1

2




q31 =3

4

q32 =4

5

q33 =4

5




S6 4

5 4

5 3

4 4

5 4

5+ 2

2= 4

1 3

4

1 4

5

1 4

5

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47/71

Notice that

=1

8

p5 +

5

8

Hence,

3

10S6 +

1

10

p5S6 1

5

p5 +

1

42

=3

10 1

10

p5

The solution is:

S6 =33

8

p5 75

8or

1

8

p5 +

5

8=

) S6 =



S3 34 4

5 3

4 4

5 4

5+ 2

2

= 4

1 34

1 4

5

1 4

5

Notice that

=1

8

p5 +

5

8Hence,

3

8S3 +

3

40

p5S3 1

5

p5 +

1

4

2=

3

10 1

10

p5

The solution is:

S3 =3

2

p5 7

2or

1

6

p5 +

1

2

) S3 =16p5 + 1

2



S2 34 4

5 3

4 4

5 4

5+ 2

2= 4

1 3

4

1 4

5

1 4

5

Notice that

= 18p5 + 5

8Hence,

3

8S2 +

3

40

p5S2 1

5

p5 +

1

4

2=

3

10 1

10

p5

The solution is:

S2 =3

2

p5 7

2or

1

6

p5 +

1

2

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Figure 45: Projective triangle at vetrex A4


q41 =4

5

q42 =4

5

q43 =3

4So, we can use the projective cross law to nd the three face spreads (Projective spreads)



S4 34 4

5 3

4 4

5 4

5+ 2

2= 4

1 3

4

1 4

5

1 4

5

Notice that

= 18p5 + 5

8

Hence, 3

10S4 +

1

10

p5S4 1

5

p5 +

1

4

2=

3

10 1

10

p5

The solution is:

S4 =33

8

p5 75

8or

1

8

p5 +

5

8=

) S4 =

48


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S1 3

4 4

5 3

4 4

5 4

5+ 2

2= 4

1 3

4

1 4

5

1 4

5

Notice that

=1

8

p5 +

5

8

Hence, 3

8S1 +

3

40

p5S1 1

5

p5 +

1

4

2=

3

10 1

10

p5

The solution is:

S1 =3

2

p5 7

2or

1

6

p5 +

1

2

) S1 =1

6

p5 +

1

2



S2 3

4 4

5 3

4 4

5 4

5+ 2

2= 4

1 3

4

1 4

5

1 4

5

Notice that

=1

8

p5 +

5

8

Hence,

3

8S2 +

3

40

p5S2 1

5

p5 +

1

4

2=

3

10 1

10

p5

The solution is:

S2 = 32p5 7

2or 1

6p5 + 1

2

) S2 =1

6

p5 +

1

2

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4.2.5 The third tetrahedron

Figure 46: The third tetrahedron

Figure 47: The third tetrahedron

50


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Figure 48: The third tetrahedron consisting of four zome triangles

4.2.6 The main results of the second tetrahedron.


6

p5 + 1

2S2 =

1

6

p5 + 1

2S3 =

1

6

p5 + 1

2S4 = S5 = S6 =

Vertex Projective quadrances

A1 q11 =1615

; q12 =8

9and q13 =

1615

A2 q21 =45

; q22 =45

and q23 = A3 q31 = ; q32 =

2

3and q33 =

3

A4 q41 =3

; q42 =2

3and q43 =


51


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q11 =16

15

q12 =8

9

q13 = 1615




S4 8

9 16

15 16

15 8

9 16

15+ 2

2

= 41 16

15 1 8

91 16

15 Notice that

=1

8

p5 +

5

8

Hence, 16

27S4 +

16

135

p5S4 4

15

p5 2

9

2=

4

45 16

405

p5

The solution is:

S4 =3

4

p5 3

4or

3

16

p5 +

9

16

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S5

16

15 16

15 16

15 8

9 16

15 + 22

= 4

1 16

15

1 8

9

1 16

15

Notice that

=1

8

p5 +

5

8

Hence, 8

15S5 +

8

45

p5S5 4

15

p5 2

9

2=

4

45 16

405

p5

The solution is:

S5 =1

8

p5 +

5

8= or

3

2

p5 5

2



S6 16

15 8

9 16

15 8

9 16

15+ 2

2= 4

1 16

15

1 8

9

1 16

15

Notice that

=1

8

p5 +

5

8

Hence, 16

27S6 +

16

135

p5S6 4

15

p5 2

9

2=

4

45 16

405

p5

The solution is:

S6 =3

4

p5 3

4or

3

16

p5 +

9

16


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q21 =4

5

q22 =4

5q23 =




S5 4

5 4

5 4

5 4

5 + 2

2= 4

1 4

5

1 4

5

(1 )

Notice that

=5 p5

8

Hence, 3

10S5 1

10p5S5 + 13

40p5 + 3

8

2

= 310p5 + 7

10

The solution is:

S5 = 558

p5 115

8or

1

8

p5 +

5

8=

) S5 =

Secondly, we want to compute the projective spread S3 by using the projective cross law.

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This will lead to


S3

4

5 4

5 4

5 + 22

= 4

1 4

5

1 4

5

(1 )Notice that

=5 p5

8

Hence,

3

8S3 1

8

p5S3 +

13

40

p5 +

3

8

2=

3

10

p5 +

7

10

The solution is:

S3 =

11

2

p5

23

2

or1

10

p5 +

1

2



S1 4

5 4

5 4

5 + 2

2= 4

1 4

5

1 4

5

(1 )

Notice that

= 5 p58

Hence,

3

8S1 1

8

p5S1 +

13

40

p5 +

3

8

2=

3

10

p5 +

7

10

The solution is:

S1 =1

10

p5 +

1

2or 11

2

p5 23

2

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q31 =

q32 =2

3

q33 =

3




S6 2

3

3 2

3

3+ 2

2= 4(1 )

1 2

3

1

3

Notice that =

1

8

p5 +

5

8

Hence, 1

3S6 +

1

9

p5S6 7

24

p5 +

5

24

2=

7

18 1

6

p5

The solution is:

S6 =75

16

p5 159

16or

3

16

p5 +

9

16

56


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) S6 =3

16

p5 +

9

16


(S3q31q33 q31 q32 q33 + 2)2

= 4(1 q31) (1 q32) (1 q33)Therefore, susbtituting the values in the law we get:

S3

3 2

3

3+ 2

2= 4(1 )

1 2

3

1

3

Notice that

=1

8

p5 +

5

8

Hence,

5

12S3 +

1

6

p5S3 7

24

p5 +

5

242

=7

18 1

6

p5

The solution is:

S3 =89

10

p5 39

2or

1

10

p5 +

1

2

) S3 =1

10

p5 +

1

2


(S2q31q32 q31 q32 q33 + 2)2 = 4(1 q31) (1 q32) (1 q33)

Therefore, susbtituting the values in the law we get:S2 2

3 2

3

3+ 2

2= 4(1 )

1 2

3

1

3

Notice that

=1

8

p5 +

5

8

Hence,

5

12S2 +

1

12

p5S2 7

24

p5 +

5

24

2=

7

18 1

6

p5

The solution is:

S2 =9

5

p5 7

2or

1

5

p5 +

1

2

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q41 =

3

q42 =2

3q43 =




S4 3 2

3

3 2

3 + 2

2= 4

1

3

1 2

3

(1 )

Notice that

= 18p5 + 5

8and = 3 + p5

2

Hence, 1

3S4 +

1

9

p5S4 7

24

p5 +

5

24

2=

7

18 1

6

p5

The solution is:

S4 =75

16

p5 159

16or

3

16

p5 +

9

16

) S4 =

58


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S1

3

3 2

3 + 2

2= 4

1

3

1 2

3

(1 )

Notice that

=1

8

p5 +

5

8and =

3 +p

5

2

Hence, 5

12S1 +

1

6

p5S1 7

24

p5 +

5

24

2=

7

18 1

6

p5

The solution is:

S1 =1

10

p5 +

1

2

or89

10

p5

39

2

) S1 =1

6

p5 +

1

2

Thirdly, we want to compute the projective spread S2 by using the projective cross law.This will lead law

(S2q42q43 q41 q42 q43 + 2)2 = 4(1 q41) (1 q42) (1 q43)

Therefore, susbtituting the values in the law we get:

S2 2

3

3 2

3 + 2

2= 4

1

3

1 2

3

(1 )

Notice that

=1

8

p5 +

5

8and =

3 +p

5

2

Hence,

5

12S2 +

1

12

p5S2 7

24

p5 +

5

24

2=

7

18 1

6

p5

The solution is:

S2 =1

5

p5 +

1

2or

9

5

p5 7

2

) S2 =1

5

p5 +

1

2

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4.2.7 The fourth tetrahedron

Figure 53: The fourth tetrahedron

Figure 54: The fourth tetrahedron consisting of four zome triangles

4.2.8 The main results of the fourth tetrahedron.

Projective spreads S1 =3p5

16+ 9

16S2 = S3 = S4 =

3p5

16+ 9

16S5 =

1

2S6 =

60


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Vertex Projective quadrances

A1 q11 =45

; q12 =1615

and q13 =2

3

A2 q21 =45

; q22 =2

3and q23 =

1615

A3 q31 =4

5; q32 =

4

5and q33 =

4

5

A4 q41 =8

9; q42 =

1615

and q43 =1615

There are 4 vertexes and each vertex has projective triangle.




q11 =4

5

q12 =16

15

q13 =2

3

So, we can use the projective cross law to nd the three face spreads (Projective spread).

Firstly, the projective spread S4 by using the projective cross law.This will lead to


S4 16

15 2

3 4

5 16

15 2

3+ 2

2= 4

1 4

5

1 16

15

1 2

3

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Notice that

=5 +

p5

8and =

5 p58

Hence,

49 S4

4

45p5S4 +1

30p5 +1

62

=

2

45p5 +2

15

The solution is:

S4 = 916

p5 27

16or

3

16

p5 +

9

16

Secondly, the projective spread S5 by using the projective cross law.This will lead to


S5 45 2

3 4

5 16

15 2

3+ 2

2

= 4

1 4

5

1 16

15

1 2

3

Notice that

=5 +

p5

8and =

5 p58

Hence,1

900

p5 + 5

2(2S5 + 1)

2 =2

45

p5 +

2

15

The solution is:

S5 =1

2or

3

2



S6 45 16

15 4

5 16

15 2

3+ 2

2= 4

1 4

5

1 16

15

1 2

3

Notice that

=5 +

p5

8 and =5p

5

8Hence,

4

15S6 +

1

30

p5 +

1

6

2=

2

45

p5 +

2

15

The solution is:

S6 = 38

p5 15

8or

1

8

p5 +

5

8=

62


63/71




q21 =4

5

q22 =2

3

q23 =

16

15

So, we can use projective cross law to nd the three face spreads (Projective spread)



S5 45 2

3 4

5 16

15 2

3+ 2

2= 4

1 4

5

1 16

15

1 2

3

Notice that

=5 +

p5

8and =

5 p58

Hence,1

900

p5 + 5

2(2S5 + 1)

2 =2

45

p5 +

2

15

The solution is:

S5 =1

2or 3

2

63


64/71



S3 4

5 16

15 4

5 2

3 16

15+ 2

2= 4

1 4

5

1 2

3

1 16

15

Notice that

=5 +

p5

8and =

5 p58

Hence,

4

15S3 +

1

30

p5 +

1

6

2=

2

45

p5 +

2

15

The solution is:

S3 = 38

p5 15

8or

1

8

p5 +

5

8=



S1 2

3 16

15 4

5 2

3 16

15+ 2

2

= 4

1 4

5

1 2

3

1 16

15

Notice that

=5 +

p5

8and =

5 p58

Hence,

4

9

S1

4

45

p5S1 +

1

30

p5 +

1

62

=2

45

p5 +

2

15

The solution is:

S1 = 916

p5 27

16or

3

16

p5 +

9

16

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q31 =4

5

q32 =4

5

q33 = 45




S6 4

5 4

5 4

5 4

5 4

5+ 2

2

= 41 4

51

4

51

4

5

4

625(8S6 5)2 = 4

125

The solution is

S6 =1

8

p5 +

5

8= or

5

8 1

8

p5 =

) S6 =

Secondly, the projective spread S3 by using the projective the cross law.

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This will lead to

(S3q31q33 q31 q32 q33 + 2)2 = 4(1 q31) (1 q32) (1 q33)Therefore, substituting the values in the law we get:

S3 4

5 4

5 4

5 4

5 4

5+ 2

2

= 41 4

51 4

51 4

54

625(8S3 5)2 = 4

125

The solution is:

S3 =1

8

p5 +

5

8= or

5

8 1

8

p5 =

) S3 =



S2 45 4

5 4

5 4

5 4

5+ 2

2= 4

1 4

5

1 4

5

1 4

5

4

625(8S2 5)2 = 4

125

The solution is:

S2 =1

8

p5 +

5

8= or

5

8 1

8

p5 =



66


67/71


q41 =8

9

q42 =16

15

q43 =1615




S4 8

9 16

15 8

9 16

15 16

15+ 2

2

= 41 8

91 16

15 1 16

15 Notice that

=5 +

p5

8

Hence, 16

27S4 +

16

135

p5S4 4

15

p5 2

9

2=

4

45 16

405

p5

The solution is:

S4 =3

4

p5

3

4

or3

16

p5 +

9

16

) S4 =3

16

p5 +

9

16



S1 8

9 16

15 8

9 16

15 16

15+ 2

2

= 41 8

91

16

151

16

15

Notice that

=5 +

p5

8

Hence, 16

27S1 +

16

135

p5S1 4

15

p5 2

9

2=

4

45 16

405

p5

The solution is:

67


68/71

S1 =3

4

p5 3

4or

3

16

p5 +

9

16

) S1 =3

16

p5 +

9

16



S2 16

15 16

15 8

9 16

15 16

15+ 2

2= 4

1 8

9

1 16

15

1 16

15

Notice that =

5 + p58

Thus,

8

15S2 +

8

45

p5S2 4

15

p5 2

9

2=

4

45 16

405

p5

The solution is:

S2 =3

2

p5 5

2or

1

8

p5 +

5

8=

) S2 =

68


69/71

The same process leads us to nd no more than 19 tetrahedrons as shown in the following catalog.

69


70/71

Figure 59: A catalog of ninteen tetrahedrons

5 Conclusion

The Zome system remains an outstanding and sophisticated tool to express mathematical thoughts into tangiblestructures. We have used this system to nd six new theorems by rational trigonometry which has proved to be veryuseful for locating and determining any spreads and quadrances in triangles. A good idea for this research will be the

use of these results in the future to investigate the case of determining the solid spread in the nineteen tetrahedronsdeveloped in this completed work. It would be recommended for further research in rational trigonometry to explorethe possibility of still more theorems. The Egyptians and Greeks would have loved this tool!

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6 Bibliography

[1] D.Booth. The new Zome Primer in Five Fold Symmetry. Hargittal , World Scientic Publishing Company, 1992.

[2] G.Hart, H.Picciotto, Zome Geometry: Hands-on with Zome Models, Key Curriculum Press, 2001.

[3 ] G. Hart, Zometool Polyhedra. [online] available from [1 June 2011]< http://www.georgehart.com/virtual-polyhedra/zometool.html >

[ 4] P .Hildebrandt.. Zome-inspired Sculpture. Colorado: Zometool .2006 Inc. online] available from [1 June 2011]

[5 ] J. Kappra, (2001). Connections: The Geometric Bridge Between Art and Science. Massachusetts: WorldScientic Publishing. [1 June 2011]

[ 6] D.A. Richter. Two results concerning the Zome model of the 600-cell. [Online]. Available from [1 June2011]

[7 ] N.J Wildberger. Projective and spherical trigonometry, School of Mathematics UNSW Sydney. (2007).

[8 ] N.J Wildberger, Divine Proportion: Rational trigonometry to Universal Geometry, Wild Egg Books, Sydney,2005.

[ 9] N.J Wildberger, Greek Geometry, Rational Trigonometry, and the Snellius - Pothenot Surveying ProblemChamchuri Journal of Mathematics : 2(2010).

[10 ] N.J Wildberger .The ancient Greeks present: Rational Trigonometry. School of Mathematics and StatisticsUNSW Sydney. 2008 http://web.maths.unsw.edu/~norman/.

[ 11] Zometool Inc. Educators. [online] available from [1 June 2011].

zome system

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