φ6R Extreme and Mean Isosceles Triangle Spiral

This spiral study, suggested by my friend Alexis, was meant to be a sandbox of ideas on different types of spirals. Here are the results of this sandbox study on spirals using Euclid's Extreme and Mean Ratio, the Golden Ratio, Bach's music theory on the 6th root, and drawings by Rafael Araujo. 

Euclid's Elements Book IV Proposition 10

Alexis Doval's drawing of the Isosceles triangles cut into extreme and mean ratios of the Golden Ratio.

If the lines AF and CF were a single straight line, they could be cut into extremes and mean and equal to the Golden Ratio.

If the lines CF and BF were a single straight line, they could be cut into extremes and mean and equal to the Golden Ratio.

After the Isosceles triangles are developed and cut into the Extreme and Mean ratios, the radius of the cone is taken and used to calculate the length of each construction ray in plan view by the 6th root. 

Then, you bring the lengths of the construction rays that define the spiral up to the profile view. Where these lengths intersect, the cone slant line will determine the arcs for the cone surface. Then, use the intersection of the arcs and construction rays on the cone surface to develop the spiral on the cone surface. 

The sixth root of a number is the number that would have to be multiplied by itself 6 times to get the original number.

Golden Ratio Spiral based on Rafael Araujo drawing. 

The vertex of each isosceles triangle is the center point of each arc.

Note:  φ6R Extreme and Mean Isosceles Triangle Spiral is Logarithmic

The Golden Ration Spiral is not a Logarithmic or an Archimedean Spiral.

Archimedean spiral increases in constant distance in plan view but not on the cone's surface.

Logarithmic spiral increases in constant geometric progression in plan view and on the cone's surface.

Fibonacci Icosahedron based on the drawing from Rafael Araujo

Note: The Fibonacci Spiral has nothing to do with the Icosahedron; it just looks neat. 

The Golden Ratio by Rafael Araujo

Golden Shell with tapered torus based on the Golden Ratio

            The golden ratio and Fibonacci numbers in Bach’s music

 Euclid's Elements Book XIII Proposition 18

In a polygonal pyramid, the angle between a base edge and a lateral edge (edge connecting the base to the apex) is called the base-edge angle or edge angle. It's one of several angles you can consider in a pyramid, including angles between faces (dihedral angles). 

Relating the polygonal pyramid edge angle to a roof-framing hip rafter is the easiest way to prove Euclid's Book XIII Proposition 18 is correct. As well, as my drawing of Proposition 18.  Euclid states that the line HC is 5 times the square of lines HK and KC. So, how did Euclid know this was true? How do we prove it's true.  How do we know the lines representing the edge length of the Platonic Solids are the edge lengths of the Platonic Solids when the Platonic Solid is circumscribed in a sphere with a radius of one?

For some reason, Euclid didn't use Euclid's Elements Book II Proposition 11 Extreme and Mean Ratio for The Golden Ratio as the base drawing for Proposition 18.  As, seen on the next couple of pages the Extreme and Mean drawing provides all the proof he needed for comparing the edge lengths of the Platonic Solids when inscribed in a sphere with a radius of one. 

Radius EK = 1

AN side of Cube = 1.15470

AK side of Octahedron=1.41421356

AM side of Icosahedron = 1.051462

CN side of Tetrahedron = 1.632993

AQ side of Dodecahedron = 0.713644

Icosahedron Development

Line AM is the edge length of the Icosahedron inscribed in a sphere with a radius of 1. Point E is the center of the sphere. Points A-M are points on the sphere. Line SM is a plane perpendicular to the line AE. Line SM is the dimension to the center of the deck-plan view drawing of the pentagon with a sphere that has a radius of 1. 

Line AH is the extreme ratio of the Golden Ratio. Angle ACH is the Golden Slope Triangle. Line EB is the square root of 5. Both lines CH and EB intersect at point M. Therefore, we can say the Edge of the Icosahedron is developed from the Golden Ratio and the square root of 5. 

Dodecahedron Development

Line AQ is the edge length of the Dodecahedron inscribed in a sphere with a radius of 1. Point E is the center of the sphere. Points A-Q are points on the sphere. Line YQ is a plane perpendicular to the line AE. Line YQ is the dimension to the center of the deck-plan view drawing of the Equilateral Triangle with a sphere that has a radius of 1. 

Line AH is the extreme ratio of the Golden Ratio. Angle GCF is developed from the extreme ratio of the Golden Ratio. Line XB is developed from the Golden Ratio of 1.61803. Both lines GC and XB intersect at point Q. Therefore, we can say the Edge of the Dodecahedron is developed from the Golden Ratio and the extreme ratio. 

To prove the lines forming the triangles containing the edge length in the semi-circle of the sphere are correct, you can develop the deck-plan view for each of the Platonic Solids as a roof framing exercise. Then, develop the Edge angle triangle(hip rafter slope triangle) and use a bevel square to compare the angles in the triangle to the angles in the semi-circle. 

The deck plan view drawing is developed from a vertex point and the face that intersects the vertex point. 

Tetrahedron

is defined by four equal equilateral triangles and four Vertices

with three edges meeting at each vertex

Cube

is defined by six equal squares and eight Vertices

with three edges meeting at each vertex

Octahedron

is defined by eight equal equilateral triangles and six Vertices

with four edges meeting at each vertex

Icosahedron

is defined by twenty equal equilateral triangles and twelve Vertices

with five edges meeting at each vertex

Dodecahedron

is defined by twelve equal equilateral triangles and twenty Vertices

with three edges meeting at each vertex

To complete the proof of my drawing for proposition 18, you can draw out the hip rafter slope triangles using any radius for the semi-circle. The edge length of the Platonic Solid in the semi-circle is equal to 2 times the radius of the semi-circle times the "rise" leg of the hip rafter slope triangle when the hypotenuse of the hip rafter slope triangle is equal to the radius of the sphere. This is supported by using congruent angles for the edge length of the Platonic Solid. 

A couple of extra drawings of the Icosahedron with the √5 triangles that develop point M. 

2 x Radius of the sphere  x gd = Edge Length in the sphere

Icosahedron developed from √5 Triangle.

Icosahedron Developed from Five Squared

Euclid states that HK is quadruple of the square at KC. Then he says that the square on HK is five times the square of HK and KC. This is the same for line aE that develops the Golden Ratio of 1.618. 

As, seen in this drawing the line AF ,Golden Ratio, is also the straight line of a Hexagon & Decagon. Then we can use the line AF, Golden Ratio, to develop the deck-plan view of the pentagon of the Icosahedron. Line HK is the radius of the deck-plan view of the pentagon and line AH is the length of each side in the pentagon, as well as each side of an Icosahedron.

Only an Alien would have known that cutting the line AL into extreme and mean ratio would have developed the Golden Ratio of AX that is part of the straight line containing the Hexagon and Decagon inscribed in a circle with the radius equal to one.

The vertical lines EC,FD, ML, and QH are the radius's used for the development of the deck-plan view drawings. These radius's develop the Platonic Solid with a sphere circumradius of 1.

 Euclid's Elements Book XIII Proposition 17

I have always been fascinated by “who knew what and when they knew it. “ As well as how they knew it. So, yesterday I was searching for a download link on the book De quinque corporibus regularibus by Pierodella Francesca. That has the Platonic Solids and the Archimedean solids. I had already downloaded the book Divina proportione by Luca Pacioli. That supposedly plagiarized Piero della Francesca book. I saw in Luca Pacioli book that he had drawn Euclid's Elements Book XIII Proposition 17, but not Euclid's Elements Book XIII Proposition 18. So, I was wondering if Piero della Francesca had a drawing of Euclid's Elements Book XIII Proposition 18. I never did find a link to download Piero della Francesca's book. Even though I read on a page at the JLC forum, I thought I should study Stereotomy by reading the book De quinque corporibus regularibus by Piero della Francesca 15 years ago.

 

Over the last two months, I’ve been studying Euclid. Stereotomy is taking a point in 3D space and laying down that 3D point on a 2D drawing. So you can scribe the wood. After studying Euclid for the last two months, I realized I needed to take the next step in my journey in geometry and mathematics. Where you take a point in 3D space and transfer that point to another point in 3D space.

 

Hippasus, who was part of the Pythagorean Brotherhood, a secret society of mathematicians, could draw the Platonic Solids, the Icosahedron or the Dodecahedron. I don't know which one it was. But he was stoned to death for having too much knowledge in 500 BC.

 

My goal in my journey in geometry and mathematics is to learn Critical Thinking in Geometry so I can lay down the 3D points in space like Hippasus and Euclid. Euclid's Elements Book XIII Proposition 17 is an excellent example of Critical Thinking in Geometry. 

Representing a three-dimensional 3D point in space, the objects that compose it are studied analytically, in their shapes and position in relationship to the place that contains them. 

My study on nesting the Platonic Solids is based on Euclid's proposition 18. Here I'm able to draw out the edge lengths of the 5 Platonic Solids in a sphere using 1.618 and √2. 

Tetrahedron, Cube, Octahedron, Icosahedron, and Dodecahedron

Sphere Radius = EC = AC = 1

FB side of Cube = 1.15470

EB side of Octahedron = 1.41421356

MB side of Icosahedron = 1.051462

AF side of Tetrahedron = 1.632993

QB = NB side of Dodecahedron = 0.713644 

(√5 + 1)÷2= 1.618033

2÷√3 = 1.1570

√2 = 1.41421

√(2- ((2√5 )÷5))= 1.051462

2÷√3 *√2 = 1.632993

(2÷√3 ) ÷ ((√5 + 1)÷2)= 0.713644

Euclid's Elements Book II Proposition 11 

Extreme and Mean Ratio

 The Golden Ratio 

To cut a given straight line so that the rectangle contained by the whole and one of the segments equals the square on the remaining segment.

 

Triangle ABE is the √5 triangle, and triangle GFC is the Dodecahedron Hip Rafter Slope.

Euclid's Elements Book XIII Proposition 18

To set out the sides of the five figures and compare them with one another.

Set out AB the diameter of the given sphere, and cut it at C so that AC equals CB, and at D so that AD is

double DB. Describe the semicircle AEB on AB, draw CE and DF from C and D at right angles to AB,

and join AF, FB, and EB. Then, since AD is double DB, therefore AB is triple BD. In conversion, therefore, BA is one and a half times AD. But BA is to AD as the square on BA is to the square on AF, for

the triangle AFB is equiangular with the triangle AFD. Therefore the square on BA is one and a half times the square on AF. But the square on the diameter of the sphere is also one and a half times the square on the side of the pyramid. And AB is the diameter of the sphere, therefore AF equals the side of the pyramid. cube.

Euclid's Elements Book XIII Proposition 17

To construct a dodecahedron and comprehend it in a sphere, like the aforesaid figures; and to

prove that the square on the side of the dodecahedron is the irrational straight line called

apotome.

BCEF is the top of the Cube. H,N,M,O are midpoints of the square on top of the cube. Draw lines HM, NO, GK, and HL. Mark off point J, the midpoint of HP, and draw line OJ. Then draw line JY the length of OJ.  Develop the square PYIS. Point S will divide the line PO, apotome, into the Golden Ratio. PS is the extreme, and SO is the mean. Mark off points UXV perpendicular to the plane of the cube, using the length PS. Mark off point T, using the length SO. Mark off point W perpendicular to the plane on the side of the cube, using length PS. 

The "Tint Roof" is also the first bastard hip roof ever drawn.(230 BC)

Triangle WXW' is the dihedral triangle for the Dodecahedron. 

The angle between two planes in the Dodecahedron.

The Dodecahedron dihedral triangle angle is 116.5650°

Luca Pacioli's portrait with him drawing the edge length for the Octahedron on the slate board.

 Luca Pacioli 

 Piero della Francesca 

Icosahedron inscribed in a cube, from De quinque corporibus regularibus, and a modern illustration of the same construction

The edge length of the Icosahedron inscribed inside a cube is:

 cube length ÷ 1.618 = Icosahedron edge length

Then, connect the ends of the Icosahedron edges to the 20 triangular faces of the Icosahedron.