The Golden Ladder

The Golden Ladder

By

Weixin Wu

Submitted to the Board of Mathematics/Computer Science

School of Natural and Social Sciences

In partial fulfillment of the requirements

For the degree of Bachelor of Arts

Purchase College

State University of New York

May 2015

Sponsor: Irina Shablinsky

Second Reader: Knarik Tunyan

Table of ContentsAbstract.......................................................................................................................................................3

1. Introduction to the Golden Ladder..................................................................................4

2. Recursive Properties of the Golden Ladder.................................................................8

2.1 Theorem 1.................................................................................................................................8

2.2 Theorem 2...............................................................................................................................10

2.3 Theorem 3...............................................................................................................................12

3. More about the Golden Ladder.........................................................................................14

3.1 Spiral in the Golden Ladder.........................................................................................14

3.2 Find the End Point of the Golden Ladder.............................................................15

3.3 The Golden Ladder in Three Dimension................................................................16

4. A “Processing” Simulation of the Golden Ladder...................................................17

4.1 How the Program Works?..............................................................................................17

4.2 Program Output..................................................................................................................19

4.3 How to Use this Program...............................................................................................21

4.4 Why Writing this Program?..........................................................................................21

5. Results.............................................................................................................................................21

6. References.....................................................................................................................................22

7. Appendix.........................................................................................................................................23

2

Abstract

In this work, a new Euclidean structure the Golden Ladder is introduced. The Golden

Ladder was first created in visual software by applying the Golden Ratio to the diameters of

tangent circles. For this work three recursive properties of the circumference and area have

been proven, and a 2D simulation of the Golden Ladder was written in processing. The Golden

Ladder has been expanded to the third dimension, and it is possible to expand the definition to

the nth dimension. The volume will be the property of interest in the third dimension. Additional

interesting properties have been found for the Golden Ladder and are currently under

investigation.

3

1. Introduction to the Golden Ladder

I am going to introduce some facts, properties and applications of the Golden Ratio

before I get into the Golden Ladder. The Golden Ratio is one of the most famous irrational

numbers. Also known as the divine proportion, golden mean, or golden section. The term

“golden section” was first used in the textbook Die Reine Elementar-Mathematik in 1835 by

Martin Ohm. [9] Two quantities a and b are in the Golden Ratio if their ratio is same as the ratio

between the sum and the larger quantity. As shown in the equation below and figure 1 for a >

b.

a+ba

=ab=φ

The lower case Greek letter phi (φ) is used to represent the Golden Ratio. Phi (φ) is one

of the roots of the quadratic equation x2 – x – 1. The exact value of the Golden Ratio is (√5 + 1) /

2 or ca 1.6180339887… expressed as an infinite decimal number. The upper case Greek letter

phi (φ) is used to represent the reciprocal of the Golden Ratio; the exact value is (√5 -1) / 2 or ca

0.6180339887… expressed as an infinite decimal fraction.

4

Figure 1 The Golden Ratio

There is an interesting relation between the Golden Ratio and the Fibonacci sequence.

[1][4] Fibonacci sequence is an integer sequence that start with 1 and 1, and each subsequent

number is the sum of the previous two. It’s known that the limit of the ratio of two consecutive

Fibonacci numbers is φ as shown in the equation below. φ can also be expressed as exact

trigonometric formulas including2 cos( π5 ), 12 sec (2 π5

) and 12csc ( π

10).

limn→∞

Fn+ 1

Fn=φ

φ is the “worst” real number for rational approximation because its continued fraction

representation as shown in the equation below.

φ=[ 1,1,1,1,… ]

φ=1+ 1

1+ 1

1+ 11+…

φ can also be represented in terms of a nested radical as shown below.

φ=√1+√1+√1+…

The Golden Ratio is given by the series

φ=138

+∑n=0

∞ (−1 )n+1(2n+1) !(n+2) !n !42n+3

5

The Golden Ratio is widely used from Renaissance artists of the 1500’s to modern

graphic designers as exemplified by the painting of Mona Lisa by Da Vinci to the logo design of

both Apple iPhone and Toyota. Now I know why the apple logo looks so nice and became

famous quickly. The Golden Ratio has a great influence on our everyday life without noticing. As

a conclusion, use the Golden Ratio to design your logo if you want to start a new business.

The Golden Rectangle shown in Fig 2 above was constructed by laying out Golden Ratio

rectangles, and then drawing the quarter-circle arcs in the square part of each rectangle.

Similarly a Golden Triangle may be drawn and appropriately divided and a spiral may also be

drawn as shown in Fig 3. The Golden Ladder is similar to the Golden Rectangle and the Golden

Triangle. I am going to introduce this new Euclidean geometric structure by applying the Golden

Ratio to the diameter of each adjacent circles. The structure is shown below in Fig 4, and I

called it the Golden Ladder.

6

Figure 2 Golden Rectangle Figure 3 Golden Triangle

I am going to walk you through the process of construction of a Golden Ladder. Draw two

circles, C1 with the diameter d1= φ and C2 with the diameter d2= φφ tangent to each other. Add another

circle C3 with the diameter d3= φφ2 tangent to both C1 and C2. Note that d1 > d2 > d3. The circles C1, C2, and

C3 are mutually tangent. Connect the centers of the three circles to create a triangle. This triangle is the

first step of the Golden Ladder. The fourth circle C4 with the diameter d4= φφ3 must be placed tangent to

both C2 and C3. Again connect the centers of the circles C2, C3, and C4 to get the second triangle. Keep

going. Eventually you will get a sequence of circles C1 , C2,...,Cn,... with the diameters d1, d2, ... , dn, ...,

where dn= φφn-1, and each of the three circles Cn, Cn+1, Cn+2 are mutually tangent. You will also get a

sequence of triangles T1, T2,..., Tn,... . This sequence is called the Golden Ladder.

7

Figure 4 The Golden Ladder

2. Recursive Properties of the Golden Ladder

I want to introduce some recursive properties of the Golden Ladder. The first recursive

property is that all triangles in the Golden Ladder are similar triangle, the ratio between

corresponding sides of any triangle in the Golden Ladder and its next triangle is the Golden

Ratio (φ).

2.1 Theorem 1

1) All triangles in the Golden Ladder are similar.

2) The ratio of corresponding sides of triangles Tn and Tn+1 is the Golden Ratio φ.

Proof. Let the two triangles in figure 5 and 6 be Tn and Tn+1, which are taken from the

Golden Ladder. Triangle Tn was surrounded by three circles with diameters φn, φn+1 and φn+2,

triangle Tn+1 was surrounded by three circles with diameters φn+1, φn+2 and φn+3.

8

Figure 6 n+1th Triangle (Tn+1) in Golden LadderFigure 5 nth Triangle (Tn) in Golden Ladder

Sn

Sn+1

Sn+2 Sn+2

Sn+3

Sn+4

Tn Tn+1

The three sides of Tn are Sn, Sn+1 and Sn+2, and the three sides of Tn+1 are Sn+2, Sn+3 and Sn+4. Both

triangles share the same side Sn+2 as we see in Fig 4 & 5. In order to prove this recursive

property, we need to prove the following:

Table 1 Length of each side

S nS n+2

= S n+1S n+3

= S n+2S n+4

= φ

Side Length

Sn (φn + φn+1 )/2

Sn+1 (φn + φn+2 )/2

Sn+2 (φn+1 + φn+2 )/2

Sn+3 (φn+1 + φn+3 )/2

Sn+4 (φn+2 + φn+3 )/2

The table above shows the length of each side, calculated using the diameter of the

circle. Now we can calculate Sn / Sn+2, Sn+1 / Sn+3 and Sn+2 / Sn+4.

S nS n+2

= (ϕn+ϕn+1)/2(ϕn+1+ϕn+2)/2

=(1+ϕ)/2(ϕ+ϕ2)/2

= 1+ϕϕ+ϕ2=

1+ϕϕ∗(1+ϕ)

=1ϕ=φ

S n+1S n+3

= (ϕn+ϕn+2)/2(ϕn+1+ϕn+3)/2

=(1+ϕ2) /2(ϕ+ϕ3)/2

=1+ϕ2

ϕ+ϕ3 =1+ϕ2

ϕ∗(1+ϕ2)=1ϕ=φ

S n+2S n+4

= (ϕn+1+ϕn+2)/2

(ϕn+2+ϕn+3)/2=

(1+ϕ) /2(ϕ+ϕ2) /2

= 1+ϕϕ+ϕ2 =

1+ϕϕ∗(1+ϕ )

= 1ϕ=φ

S nS n+2

= φ (1)

9

Thus, we proved the first recursive property.

The second recursive property is related to the area of the triangles. The ratio between

the area of any triangle in the Golden Ladder and its next triangle is the Golden Ratio square

(φ2).

2.2 Theorem 2

1) The ratio between the area of Tn and Tn+1 is φ2.

Proof. I will use figure 5 and figure 6 again for this proof. The area of Tn is An and the

area of Tn+1 is An+1. Let’s find the area of Tn and Tn+1 using Heron’s Formula, Pn represents the

perimeter.

P n = (Sn + S n+1+S n+2 ¿ /2

A n=√P n∗(P n−Sn )∗(P n−S n+1 )∗( P n−S n+2 )

P n+1 = (Sn+2 + S n+3+S n+4 ¿ /2

A n+1=√P n+1∗(P n+1−S n+2 )∗(P n+1−S n+3 )∗(P n+1−S n+4 )

Next, we will divide An by An+1 to find the ratio:

A nA n+1

= √P n∗( P n−S n )∗(P n−S n+1 )∗(P n−S n+2 )

√P n+1∗(P n+1−S n+2 )∗(P n+1−Sn+3 )∗(P n+1−S n+4 )

10

A nA n+1

=√P nP n+1

∗(P n−S n )

(P n+1−S n+2 )∗(P n−S n+1 )

(P n+1−S n+3 )∗(P n−S n+2 )

(P n+1−S n+4 )

Next, we will be using equation (1) above for substitution to find the value of each term

above, including PnPn+1 , (Pn−Sn )

(Pn+1−Sn+2 ), (Pn−Sn+1 )

(Pn+1−Sn+3 ), and

(Pn−Sn+2 )(Pn+1−Sn+4 )

.

P nP n+1

=(S n + S n+1+S n+2 ¿ /2 ¿(S n+2 + S n+3+S n+4 ¿

/2¿=φ∗Sn+2 +φ∗S n+3+φ∗S n+4 S n+2 + S n+3+S n+4

=φ∗(S n+2 + S n+3+S n+4)S n+2 + S n+3+S n+4

=φ

(P n−S n )(P n+1−S n+2 )

=(S n + S n+1+S n+2 −S n ¿/2 ¿(S n+2 + S n+3+S n+4 −Sn+2 ¿

/2¿= S n+1+S n+2S n+3+S n+4

=φ∗Sn+3+φ∗S n+4 S n+3+Sn+4

=φ∗(S n+3+S n+4)S n+3+S n+4

=φ

(Pn−Sn+1 )( Pn+1−Sn+3 )

∧(Pn−Sn+2 )

(Pn+1−Sn+4 )are evaluated usinga similar procedure ¿ (Pn−Sn )

( Pn+1−Sn+2 ) above

which also equal to φ.

Now, we can substitution back to calculate the ratio between An and An+1,

A nA n+1

=√φ∗φ∗φ∗φ

A nA n+1

=φ2 (2)

11

The third recursive property is also involves areas, but in this case the global area, the

sum of all triangle areas not the local area, will be used. The ratio between the area of any

triangle in the Golden Ladder and the total area of all triangles after this triangle is equal to the

Golden Ratio.

2.3 Theorem 3

1) The ratio between area of Tn and total area of Tn+1, Tn+2,…, Tm is the Golden Ratio φ as

m goes to infinite.

Proof. Our goal is to find the ratio below, where An is the area starting triangle and Am is

the area ending triangle as m goes to infinite for m > n:

AnAn+1+An+2+…+Am−1+Am

Let’s starting by calculating the total area from Tn to Tm as m increases without limit.

A n+A n+1+A n+2+…+A m-1+A m

Next, let’s substitute all area using equation 2:

A n+A n /φ2+A n /φ4+…+A n /φ2 (m−1)+A n /φ2m

Let r = 1 / ϕ2, then the expression becomes:

A n+A n∗r+A n∗r2+…+A n∗rm−1+A n∗rm

This is a geometric series and r < 1, so the total area becomes:

12

An∗11−r

Next let’s subtract An from the above expression, because we don’t want to include the

area of the first triangle:An∗11−r

−An

Then, we divide An by the expression above, the result is equivalent to the ratio we

wanted:An

An∗11−r

−An

=An

An+1+An+2+…+Am−1+Am

Let’s simplify the expression:

AnAn∗11−r

−An

=An

An∗11−r

−An∗1−r

1−r

=AnAn∗r1−r

=1−rr

Finally, we substitute the definition of r into the result:

1−rr

=1− 1φ2

1φ2

=φ2−1=φ

An

∑i=n+1

m

Ai

=φ(3)

3. More about the Golden Ladder

3.1 Spiral in the Golden Ladder

13

There is another version of the Golden Ladder. From the Golden Ladder in figure 4, flip

all triangles into its pervious triangle starting at the second triangle. The result Golden Ladder is

shown in figure 7. In this version, it’s possible to draw a spiral that connect the center of all

tangent circles which is similar to the Golden Rectangle.

14

Figure 7 embedded Golden Ladder

3.2 Find the End Point of the Golden Ladder

In figure 8, three line can be drawn in the Golden Ladder to find the end point. The first

line would be drawing a straight line that passes through the intersections of every two

adjacent circles. The second line is also a straight line, which is tangent to all circles that on the

top portion of the Golden Ladder. And the third line is similar to the second line, but in this

case, it will be tangent to all circles on the bottom portion of the Golden Ladder. The

intersection of these three lines would be the end point, which is where the circles are

approaching to in the Golden Ladder.

15

Figure 8 End point of the Golden Ladder

3.3 The Golden Ladder in Three Dimension

The Golden Ladder may be presented in three dimension by applying the Golden Ratio

to the diameters of tangent spheres. I am calling it the Golden Tower (figure 9). This idea of

dimensional increase suggests that it is possible to extend the concept to the nth dimension

after completing some research on the three dimensional case. Instead of finding the ratio

between the areas of triangles, the ratio between the volumes of the tetrahedron would be our

interest. A tetrahedron can be constructed by connecting the centers of any four consecutive

spheres.

16

Figure 9 The Golden Tower

4. A “Processing” Simulation of the Golden Ladder

4.1 How the Program Works?

Processing is a java based virtual design programming language, which is a free software

that provides a great environment to simulate simple 2D and 3D objects. Drawing tangent

circles is very easy in real life, but to calculate the center of the next tangent circle with precise

diameter would not be as easy as you think. The program was written based on the math

showing below.

17

Figure 10 processing sketch

In the processing sketch, start with the two tangent circles. In order to find the center of

the next circle, we will need to setup two distances. Let the center of the first circle be X1, Y1,

and the distance from the first circle to the third circle be D1. For the second circle, the center

be X2, Y2, distance be D2. The unknown would be the center of the third center X3, Y3. We will

setup and solve equations below before writing the program.

√(X3−X 1)2+(Y 3−Y 1)2=D1 (4)

√ (X 3−X 2 )2+ (Y 3−Y 2 )2=D 2 (5)

18

Two unknowns and two equations, very simply, right? You can try to solve it or think

about it before I show you my approach. My procedure would surprise you and involves solving

a quadratic equation. This makes sense because quadratic equations can have two solutions

and there are two possible circles that can be tangent to the two pervious circles.

1. Square both sides of equation 4 and 5, and find the difference between them.

(X 3−X 2 )2+ (Y 3−Y 2 )2−(X 3−X1 )2−(Y 3−Y 1)2=D 22−D12

2. By a little simplification, we obtain the equation below,

2 (X 1−X 2 ) X 3+2 (Y 1−Y 2 )Y 3=D 22−D 12−X 22−Y 22−X 12−Y 12

3. Denote by temp = D22−D12−X22−Y 22−X 12−Y 12

2,simplify, and the expression

becomes:

(X 1−X2 )X 3+(Y 1−Y 2 )Y 3=temp

Solving for X3 gives us,

X 3=temp−(Y 1−Y 2 )Y 3(X 1−X 2 )

Now substitute X3 into equation 4,

√( temp−(Y 1−Y 2 )Y 3(X1−X 2 )

−X 1)2

+ (Y 3−Y 1 )2=D1

4. Square the equation above and transform it to a standard quadratic equation a(Y3)2 +

b(Y3) + c = 0. Solve the quadratic equation and substitute each of the two values of Y3 into

19

equations (4) or (5). That would give us two corresponding X3 values. Below is snippet of code

showing the calculation of the quadratic coefficient a, b and c.

float a = pow( (y1 - y2), 2) / pow( ( x1 - x2), 2) + 1;

float b = -2 * (y1 - y2) * temp / pow ( ( x1 - x2), 2) - 2 * y1 + 2 * x1 * (y1 - y2) / (x1 - x2);

float c = pow( temp, 2) / pow ( (x1 - x2), 2) + pow(x1, 2) + pow(y1, 2) - pow(distance1, 2) - 2 * x1

* temp / (x1 - x2);

The above process would repeat for each tangent circle. The program runs very well,

that mean the math approach is one correct way to solve the problem, I am sure there is a way

to solve this problem much easier.

4.2 Program Output

There are two modes you can select after you start the program. Mode 1 is for the

Golden Ladder and Mode 2 is for embedded Golden Ladder.

20

Figure 11 Mode 1 of the Golden Ladder simulation

4.3 How to Use this Program

You can drag the Golden Ladder with your mouse, and rotate it by pressing the

rotation icon. Press space to add circle to the Golden Ladder, and press ‘b’ on your

keyboard to remove circle from the Golden Ladder. Avoid to delete circle while there

are two circles left. Press 1 or 2 to switch mode. Press the ‘up arrow’ key to zoom in and

the ‘down arrow’ key to zoom out.

4.4 Why Writing this Program?

From this program, we can find out the exact location or the center of each

circle. These new data of the Golden Ladder would give us more understanding of the

structure and its properties. Also these information would be helpful for solving future

problem.

5. Results

In this work, a new Euclidean structure the Golden Ladder is introduced. Three recursive

properties have been proven, which relative to the area and circumference of triangles within

the Golden Ladder. And this structure has been expanded to the three dimension. More

properties have been found and will be proven in the near future.

21

6. References[1] Alberto, Fiorenza. Giovanni, Vincenzi. From Fibonacci Sequence to the Golden Ratio. Journal

of Mathematics Hindawi Publishing Corporation(2013).

[2] Pamela, M. Pallett. Stephen, Link. Kang, Lee. New “golden” ratios for facial beauty. Vision

Research 50 (2010) 149-154.

[3]. “Debunking The Myth Of Apple’s ‘Golden Ratio’ ”. fastcodesign.com. Mark, Wilson. June 4,

2013.

[4] “Golden Ratio”. mathworld.wolfram.com. Eric, Weisstein. May 4, 2015.

[5] "Golden Ratio in

Geometry." http://www.cut-the-knot.org/do_you_know/GoldenRatio.shtml. Bogomolny, A. n.d.

[6] Boyer, C. B. History of Mathematics. New York: Wiley, p. 56, 1968.

[7] Brown, D. The Da Vinci Code. New York: Doubleday, 2003.

[8] Coxeter, H. S. M. “The Golden Section, Phyllotais, and Whythoff’s Game.” Scripta

Mathematica 19, 135-143, 1953.

[9] Martin Ohm. Die Reine Elementar-Mathematik. 1835.

22

7. AppendixThe program consists of two classes Circle and Triangle other than the main code.

Circle class

class Circle

{

float x, y, d, r, g, b;

Circle(float x, float y, float d, color r, color g, color b)

{

this.x = x;

this.y = y;

this.d = d;

this.r = r;

this.g = g;

this.b = b;

}

void display()

{

fill(r, g, b);

noStroke();

ellipse(x, y, d, d);

}

}

Triangle class

23

class Triangle

{

float x1, y1, x2, y2, x3, y3;

Triangle(float x1, float y1, float x2, float y2, float x3, float y3)

{

this.x1 = x1;

this.y1 = y1;

this.x2 = x2;

this.y2 = y2;

this.x3 = x3;

this.y3 = y3;

}

void display()

{

stroke(0);

line(x1, y1, x2, y2);

line(x2, y2, x3, y3);

line(x1, y1, x3, y3);

}

}

Main code

ArrayList<Circle> Circles = new ArrayList<Circle>();

24

ArrayList<Triangle> Triangles = new ArrayList<Triangle>();

ArrayList<Circle> Circles2 = new ArrayList<Circle>();

ArrayList<Triangle> Triangles2 = new ArrayList<Triangle>();

float phi = 1.618034;

float phi2 = phi - 1;

float tempPhi = phi * phi2;

float tempPhi2 = tempPhi;

float ratio;

int biggerCircle = 0, smallerCircle = 1;

int biggerCircle2 = 0, smallerCircle2 = 1;

boolean press = false;

float px, py, tranX, tranY, rotateAngle = 0;

PImage rotate1, rotate2;

float rotate2X, rotate2Y, rotate1X, rotate1Y;

float zoom = 1;

int mode = 1;

float mode1X, mode1Y, mode2X, mode2Y;

float count1 = 0, count2 = 0;

void setup()

{

size(displayWidth - 50, displayHeight - 100);

smooth();

25

textSize(30);

textAlign(CENTER, CENTER);

rotate1 = loadImage("rotate1.png");

rotate2 = loadImage("rotate2.png");

rotate2X = width/20;

rotate2Y = height*9/10;

rotate1X = width/7;

rotate1Y = height*9/10;

mode1X = width*8.8/10;

mode1Y = height/10;

mode2X = width*9.5/10;

mode2Y = height/10;

ratio = 150;

Circles.add(new Circle(phi * ratio, phi * ratio, phi * ratio * 2, (int)random(255), (int)random(255), (int)random(255)));

Circles.add(new Circle(phi * ratio + ratio, (tempPhi + phi * 2) * ratio - ratio/5, tempPhi * ratio * 2, (int)random(255), (int)random(255), (int)random(255)));

Circles2.add(new Circle(phi * ratio, phi * ratio, phi * ratio * 2, (int)random(255), (int)random(255), (int)random(255)));

Circles2.add(new Circle(phi * ratio + ratio, (tempPhi + phi * 2) * ratio - ratio/5, tempPhi * ratio * 2, (int)random(255), (int)random(255), (int)random(255)));

}

26

void draw()

{

background(255);

image(rotate2, rotate2X, rotate2Y);

image(rotate1, rotate1X, rotate1Y);

imageMode(CENTER);

stroke(0);

strokeWeight(2);

fill(0);

text("Mode:", width*4/5, height/10);

if(mode == 1)

{

fill(65, 247, 78);

}

else

{

fill(255);

}

ellipse(width*8.8/10, height/10, width/20, width/20);

fill(0);

text("1", width*8.8/10, height/10);

if(mode == 2)

{

27

fill(65, 247, 78);

}

else

{

fill(255);

}

ellipse(width*9.5/10, height/10, width/20, width/20);

fill(0);

text("2", width*9.5/10, height/10);

scale(zoom);

if(press)

{

tranX = mouseX - px;

tranY = mouseY - py;

}

translate(tranX, tranY);

if(mousePressed)

{

if(dist(mouseX, mouseY, rotate2X, rotate2Y) < 50)

{

rotateAngle -= 0.01;

}

28

if(dist(mouseX, mouseY, rotate1X, rotate1Y) < 50)

{

rotateAngle += 0.01;

}

}

rotate(rotateAngle);

translate(-phi * ratio, -phi * ratio);

if(mode == 1)

{

for(int i = 0; i < Circles.size(); i++)

{

Circles.get(i).display();

}

for(int i = 0; i < Triangles.size(); i++)

{

Triangles.get(i).display();

}

}

else

{

for(int i = 0; i < Circles2.size(); i++)

{

Circles2.get(i).display();

29

}

for(int i = 0; i < Triangles2.size(); i++)

{

Triangles2.get(i).display();

}

}

}

void keyPressed()

{

if(key == ' ')

{

if(mode == 1)

{

count1++;

float d1 = Circles.get(biggerCircle).d;

float d2 = Circles.get(smallerCircle).d;

float d3 = d2 * phi2;

float distance1 = (d1 + d3)/2;


float x1 = Circles.get(biggerCircle).x;

float x2 = Circles.get(smallerCircle).x;

float y1 = Circles.get(biggerCircle).y;

float y2 = Circles.get(smallerCircle).y;

float temp = pow(distance2, 2) - pow(distance1, 2);

30

temp = temp + pow(x1, 2) - pow(x2, 2) + pow(y1, 2) - pow(y2, 2);

temp = temp / 2;


float b = -2 * (y1 - y2) * temp / pow( ( x1 - x2), 2) - 2 * y1 + 2 * x1 * (y1 - y2) / (x1 - x2);

float c = pow( temp, 2) / pow ( (x1 - x2), 2) + pow(x1, 2) + pow(y1, 2) - pow(distance1, 2) - 2 * x1 * temp / (x1 - x2);

float yRoot1 = ( -b + sqrt( pow(b, 2) - 4 * a * c) ) / (2 * a);

float yRoot2 = ( -b - sqrt( pow(b, 2) - 4 * a * c) ) / (2 * a);

float xRoot1, xRoot2;

if(count1%2 == 1){

xRoot1 = -sqrt( pow(distance2, 2) - pow( (yRoot1 - y2), 2)) + x2;

xRoot2 = sqrt( pow(distance2, 2) - pow( (yRoot2 - y2), 2)) + x2;

}

else

{



}

float realRootY, realRootX;

31

if( (yRoot1 + xRoot1) > (yRoot2 + xRoot2))

{

realRootY = yRoot1;

realRootX = xRoot1;

}

else

{

realRootY = yRoot2;

realRootX = xRoot2;

}

tempPhi = tempPhi * phi2;

Circles.add(new Circle(realRootX, realRootY, tempPhi * ratio * 2, (int)random(255), (int)random(255), (int)random(255)));

Triangles.add(new Triangle(x1, y1, x2, y2, realRootX, realRootY));

println("root1: " + xRoot1 + ", " + yRoot1);


biggerCircle += 1;

32

smallerCircle += 1;

}

}

if(key == ' ')

{

if(mode == 2)

{

count2++;

float d1 = Circles2.get(biggerCircle2).d;

float d2 = Circles2.get(smallerCircle2).d;

float d3 = d2 * phi2;



float x1 = Circles2.get(biggerCircle2).x;

float x2 = Circles2.get(smallerCircle2).x;

float y1 = Circles2.get(biggerCircle2).y;

float y2 = Circles2.get(smallerCircle2).y;

float temp = pow(distance2, 2) - pow(distance1, 2);

temp = temp + pow(x1, 2) - pow(x2, 2) + pow(y1, 2) - pow(y2, 2);

temp = temp / 2;


float b = -2 * (y1 - y2) * temp / pow( ( x1 - x2), 2) - 2 * y1 + 2 * x1 * (y1 - y2) / (x1 - x2);

33

float c = pow( temp, 2) / pow ( (x1 - x2), 2) + pow(x1, 2) + pow(y1, 2) - pow(distance1, 2) - 2 * x1 * temp / (x1 - x2);

float yRoot1 = ( -b + sqrt( pow(b, 2) - 4 * a * c) ) / (2 * a);

float yRoot2 = ( -b - sqrt( pow(b, 2) - 4 * a * c) ) / (2 * a);

float xRoot1, xRoot2;

if(count2%3 == 1){



}

else

{



}

tempPhi2 = tempPhi2 * phi2;

if(count2%3 == 0)

{

Circles2.add(new Circle(xRoot1, yRoot1, tempPhi2 * ratio * 2, (int)random(255), (int)random(255), (int)random(255)));

Triangles2.add(new Triangle(x1, y1, x2, y2, xRoot1, yRoot1));

34

}

else

{

Circles2.add(new Circle(xRoot2, yRoot2, tempPhi2 * ratio * 2, (int)random(255), (int)random(255), (int)random(255)));

Triangles2.add(new Triangle(x1, y1, x2, y2, xRoot2, yRoot2));

}



biggerCircle2 += 1;

smallerCircle2 += 1;

}

}

if(key == 'b')

{

if(mode == 1)

{

count1--;

biggerCircle -= 1;

smallerCircle -= 1;

35

tempPhi = tempPhi / phi2;

Circles.remove(Circles.get(Circles.size() - 1));

Triangles.remove(Triangles.get(Triangles.size() - 1));

}

else

{

count2--;

biggerCircle2 -= 1;

smallerCircle2 -= 1;

tempPhi2 = tempPhi2 / phi2;

Circles2.remove(Circles2.get(Circles2.size() - 1));

Triangles2.remove(Triangles2.get(Triangles2.size() - 1));

}

}

if(key == CODED)

{

if(keyCode == UP)

{

zoom += 0.2;

}

else if(keyCode == DOWN)

{

zoom -=0.2;

}

36

}

}

void mousePressed()

{

press = true;

px = mouseX - tranX;

py = mouseY - tranY;

println(mouseX + " " + mouseY);

if(dist(mouseX, mouseY, mode1X, mode1Y) < width/20)

{

mode = 1;

}

else if(dist(mouseX, mouseY, mode2X, mode2Y) < width/20)

{

mode = 2;

}

}

void mouseReleased()

{

press = false;

}

37

Documents

The Golden Ladder