mscroggs.co.uk
mscroggs.co.uk

subscribe

Blog

 2019-04-09 
In the latest issue of Chalkdust, I wrote an article with Edmund Harriss about the Harriss spiral that appears on the cover of the magazine. To draw a Harriss spiral, start with a rectangle whose side lengths are in the plastic ratio; that is the ratio \(1:\rho\) where \(\rho\) is the real solution of the equation \(x^3=x+1\), approximately 1.3247179.
A plastic rectangle
This rectangle can be split into a square and two rectangles similar to the original rectangle. These smaller rectangles can then be split up in the same manner.
Splitting a plastic rectangle into a square and two plastic rectangles.
Drawing two curves in each square gives the Harriss spiral.
A Harriss spiral
This spiral was inspired by the golden spiral, which is drawn in a rectangle whose side lengths are in the golden ratio of \(1:\phi\), where \(\phi\) is the positive solution of the equation \(x^2=x+1\) (approximately 1.6180339). This rectangle can be split into a square and one similar rectangle. Drawing one arc in each square gives a golden spiral.
A golden spiral

Continuing the pattern

The golden and Harriss spirals are both drawn in rectangles that can be split into a square and one or two similar rectangles.
The rectangles in which golden and Harriss spirals can be drawn.
Continuing the pattern of these arrangements suggests the following rectangle, split into a square and three similar rectangles:
Let the side of the square be 1 unit, and let each rectangle have sides in the ratio \(1:x\). We can then calculate that the lengths of the sides of each rectangle are as shown in the following diagram.
The side lengths of the large rectangle are \(\frac{1}{x^3}+\frac{1}{x^2}+\frac2x+1\) and \(\frac1{x^2}+\frac1x+1\). We want these to also be in the ratio \(1:x\). Therefore the following equation must hold:
$$\frac{1}{x^3}+\frac{1}{x^2}+\frac2x+1=x\left(\frac1{x^2}+\frac1x+1\right)$$
Rearranging this gives:
$$x^4-x^2-x-1=0$$ $$(x+1)(x^3-x^2-1)=0$$
This has one positive real solution:
$$x=\frac13\left( 1 +\sqrt[3]{\tfrac12(29-3\sqrt{93})} +\sqrt[3]{\tfrac12(29+3\sqrt{93})} \right).$$
This is equal to 1.4655712... Drawing three arcs in each square allows us to make a spiral from a rectangle with sides in this ratio:
A spiral which may or may not have a name yet.

Continuing the pattern

Adding a fourth rectangle leads to the following rectangle.
The side lengths of the largest rectangle are \(1+\frac2x+\frac3{x^2}+\frac1{x^3}+\frac1{x^4}\) and \(1+\frac2x+\frac1{x^2}+\frac1{x^3}\). Looking for the largest rectangle to also be in the ratio \(1:x\) leads to the equation:
$$1+\frac2x+\frac3{x^2}+\frac1{x^3}+\frac1{x^4} = x\left(1+\frac2x+\frac1{x^2}+\frac1{x^3}\right)$$ $$x^5+x^4-x^3-2x^2-x-1 = 0$$
This has one real solution, 1.3910491... Although for this rectangle, it's not obvious which arcs to draw to make a spiral (or maybe not possible to do it at all). But at least you get a pretty fractal:

Continuing the pattern

We could, of course, continue the pattern by repeatedly adding more rectangles. If we do this, we get the following polynomials and solutions:
Number of rectanglesPolynomialSolution
1\(x^2 - x - 1=0\)1.618033988749895
2\(x^3 - x - 1=0\)1.324717957244746
3\(x^4 - x^2 - x - 1=0\)1.465571231876768
4\(x^5 + x^4 - x^3 - 2x^2 - x - 1=0\)1.391049107172349
5\(x^6 + x^5 - 2x^3 - 3x^2 - x - 1=0\)1.426608021669601
6\(x^7 + 2x^6 - 2x^4 - 3x^3 - 4x^2 - x - 1=0\)1.4082770325090774
7\(x^8 + 2x^7 + 2x^6 - 2x^5 - 5x^4 - 4x^3 - 5x^2 - x - 1=0\)1.4172584399350432
8\(x^9 + 3x^8 + 2x^7 - 5x^5 - 9x^4 - 5x^3 - 6x^2 - x - 1=0\)1.412713760332943
9\(x^{10} + 3x^9 + 5x^8 - 5x^6 - 9x^5 - 14x^4 - 6x^3 - 7x^2 - x - 1=0\)1.414969877544769
The numbers in this table appear to be heading towards around 1.414, or \(\sqrt2\). This shouldn't come as too much of a surprise because \(1:\sqrt2\) is the ratio of the sides of A\(n\) paper (for \(n=0,1,2,...\)). A0 paper can be split up like this:
Splitting up a piece of A0 paper
This is a way of splitting up a \(1:\sqrt{2}\) rectangle into an infinite number of similar rectangles, arranged following the pattern, so it makes sense that the ratios converge to this.

Other patterns

In this post, we've only looked at splitting up rectangles into squares and similar rectangles following a particular pattern. Thinking about other arrangements leads to the following question:
Given two real numbers \(a\) and \(b\), when is it possible to split an \(a:b\) rectangle into squares and \(a:b\) rectangles?
If I get anywhere with this question, I'll post it here. Feel free to post your ideas in the comments below.
                        
(Click on one of these icons to react to this blog post)

You might also enjoy...

Comments

Comments in green were written by me. Comments in blue were not written by me.
@g0mrb: CORRECTION: There seems to be no way to correct the glaring error in that comment. A senior moment enabled me to reverse the nomenclature for paper sizes. Please read the suffixes as (n+1), (n+2), etc.
(anonymous)
                 Reply
I shall remain happy in the knowledge that you have shown graphically how an A(n) sheet, which is 2 x A(n-1) rectangles, is also equal to the infinite series : A(n-1) + A(n-2) + A(n-3) + A(n-4) + ... Thank-you, and best wishes for your search for the answer to your question.
g0mrb
                 Reply
 Add a Comment 


I will only use your email address to reply to your comment (if a reply is needed).

Allowed HTML tags: <br> <a> <small> <b> <i> <s> <sup> <sub> <u> <spoiler> <ul> <ol> <li> <logo>
To prove you are not a spam bot, please type "p" then "r" then "i" then "m" then "e" in the box below (case sensitive):

Archive

Show me a random blog post
 2024 

Feb 2024

Zines, pt. 2

Jan 2024

Christmas (2023) is over
 2023 
▼ show ▼
 2022 
▼ show ▼
 2021 
▼ show ▼
 2020 
▼ show ▼
 2019 
▼ show ▼
 2018 
▼ show ▼
 2017 
▼ show ▼
 2016 
▼ show ▼
 2015 
▼ show ▼
 2014 
▼ show ▼
 2013 
▼ show ▼
 2012 
▼ show ▼

Tags

talking maths in public game show probability oeis people maths numbers hyperbolic surfaces raspberry pi game of life databet christmas card mean programming ucl trigonometry wave scattering interpolation games folding paper data logs pythagoras sobolev spaces squares plastic ratio stickers gather town puzzles recursion final fantasy matrices triangles convergence data visualisation golden ratio javascript rugby royal institution standard deviation correlation weak imposition mathsjam dates datasaurus dozen live stream sound exponential growth light hats signorini conditions national lottery wool tennis harriss spiral noughts and crosses graphs chess guest posts machine learning inline code countdown tmip chalkdust magazine dinosaurs errors statistics arithmetic manchester science festival edinburgh coins determinants frobel pi approximation day football finite element method folding tube maps crossnumber big internet math-off a gamut of games fractals logic craft probability manchester runge's phenomenon binary youtube news turtles pac-man geogebra christmas braiding error bars computational complexity crochet european cup fence posts map projections accuracy captain scarlet graph theory preconditioning cross stitch matrix of minors hannah fry bodmas london approximation matt parker nine men's morris radio 4 propositional calculus numerical analysis matrix multiplication advent calendar flexagons the aperiodical palindromes mathslogicbot realhats pizza cutting gaussian elimination reddit fonts rhombicuboctahedron polynomials weather station logo video games platonic solids dragon curves curvature finite group quadrilaterals geometry sorting mathsteroids php simultaneous equations inverse matrices speed london underground ternary royal baby bempp stirling numbers misleading statistics chebyshev zines martin gardner dataset go 24 hour maths world cup pi draughts python asteroids gerry anderson estimation pascal's triangle sport books golden spiral electromagnetic field latex bubble bobble menace matrix of cofactors cambridge anscombe's quartet reuleaux polygons boundary element methods newcastle phd hexapawn

Archive

Show me a random blog post
▼ show ▼
© Matthew Scroggs 2012–2024